RELATIVE SPEEED AND TRAIN QUESTIONS

 

Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.

Important Formulas – Problems on Trains

  1. x km/hr = (x×5)/18 m/s

 

  1. y m/s = (y×18)/5 km/hr

 

  1. Speed = distance/time, that is, s = d/t

 

  1. velocity = displacement/time, that is, v = d/t

 

  1. Time taken by a train x meters long to pass a pole or standing man or a post
    = Time taken by the train to travel x meters.

 

  1. Time taken by a train x meters long to pass an object of length y meters

= Time taken by the train to travel (x + y) metres.

 

  1. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 – v2) m/s

 

  1. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

 

  1. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

 

  1. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds

 

  1. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,

A’s speed: B’s speed = √q: √p

 

 

Solved Examples

Level 1

1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?
A. 190 metres B. 160 metres
C. 200 metres

Answer : Option C

D. 120 metres

 

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s

Time taken to cross, t = 18 s

Distance Covered, d = vt = (400/36)× 18 = 200 m

Distance covered is equal to the length of the train = 200 m

2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?
A. 120 sec B. 99 s
C. 89 s D. 80 s

 

Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s

t = (240+650)/10 = 89 seconds

3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is
A. 10.8 s B. 12 s
C. 9.8 s D. 8 s

 

Answer : Option A

Explanation :

Distance = 140+160 = 300 m

Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s

Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s

4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
A. 79.2 km/hr B. 69 km/hr
C. 74 km/hr D. 61 km/hr

 

Answer : Option A

Explanation :

Let x is the length of the train and v is the speed

Time taken to move the post = 8 s

=> x/v = 8

=> x = 8v — (1)

Time taken to cross the platform 264 m long = 20 s

(x+264)/v = 20

=> x + 264 = 20v —(2)

Substituting equation 1 in equation 2, we get

8v +264 = 20v

=> v = 264/12 = 22 m/s

= 22×36/10 km/hr = 79.2 km/hr

5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is
A. 2 : 3 B. 2 :1
C. 4 : 3 D. 3 : 2

 

Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train

= √16: √ 9

= 4:3

 6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A. 320 m B. 190 m
C. 210 m D. 230 m

 

Answer : Option D

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s

time = 9s

Total distance covered = 270 + x where x is the length of other train

(270+x)/9 = 500/9

=> 270+x = 500

=> x = 500-270 = 230 meter

7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A. 10.30 a.m B. 10 a.m.
C. 9.10 a.m. D. 11 a.m.

 

Answer : Option B

Explanation :

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
A. 42 B. 36
C. 28 D. 20

 

Answer : Option B

Explanation :

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v+v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10×36/10 km/hr = 36 km/hr

 

Level 2

1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is
A. 270 m B. 245 m
C. 235 m D. 220 m

 

Answer : Option B

Explanation :

Assume the length of the bridge = x meter

Total distance covered = 130+x meter

total time taken = 30s

speed = Total distance covered /total time taken = (130+x)/30 m/s

=> 45 × (10/36) = (130+x)/30

=> 45 × 10 × 30 /36 = 130+x

=> 45 × 10 × 10 / 12 = 130+x

=> 15 × 10 × 10 / 4 = 130+x

=> 15 × 25 = 130+x = 375

=> x = 375-130 =245

2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.
A. 182 km/hr B. 180 km/hr
C. 152 km/hr D. 169 km/hr

 

Answer : Option A

Explanation :

Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr

3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds?
A. Insufficient data B. 3 : 1
C. 1 : 3 D. 3 : 2

 

Answer : Option D

Explanation :

Let the speed of the trains be x and y respectively

length of train1 = 27x

length of train2 = 17y

Relative speed= x+ y

Time taken to cross each other = 23 s

=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)

=> 4x = 6y => x/y = 6/4 = 3/2

4.A jogger is running at 9 kmph alongside a RAILWAY track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger?
A. 46 B. 36
C. 18 D. 22

 

Answer : Option B

Explanation :

Distance to be covered = 240+ 120 = 360 m

Relative speed = 36 km/hr = 36×10/36 = 10 m/s

Time = distance/speed = 360/10 = 36 seconds

5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is
A. None of these B. 280 meter
C. 240 meter D. 200 meter

 

Answer : Option C

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s

Length of the train = speed × time taken to cross the man = 15×20 = 300 m

Let the length of the platform = L

Time taken to cross the platform = (300+L)/15

=> (300+L)/15 = 36

=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter

6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?
A. 62 m B. 54 m
C. 50 m D. 55 m

 

Answer : Option C

Explanation :

Let x is the length of the train in meter and v is its speed in kmph

x/9 = (v-2) (10/36) — (1)

x/10 = (v-4) (10/36) — (2)

Dividing equation 1 with equation 2

10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22

Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
A. 500 m B. 360 m
C. 480 m D. 400 m

 

Answer : Option D

Explanation :

Speed of train1 = 48 kmph

Let the length of train1 = 2x meter

Speed of train2 = 42 kmph

Length of train 2 = x meter (because it is half of train1’s length)

Distance = 2x + x = 3x

Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s

Time = 12 s

Distance/time = speed => 3x/12 = 25

=> x = 25×12/3 = 100 meter

Length of the first train = 2x = 200 meter

Time taken to cross the platform= 45 s

Speed of train1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y where y is the length of the platform

=> 200 + y = 45×40/3 = 600

=> y = 400 meter

8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?
A. 440 m B. 500 m
C. 260 m D. 430 m

 

Answer : Option B

Explanation :

Distance = 800+x meter where x is the length of the tunnel

Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s

Distance/time = speed

(800+x)/60 = 65/3 => 800+x = 20×65 = 1300

=> x = 1300 – 800 = 500 meter

9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is :
A. 19 m B. 2779 m
C. 1329 m D. 33 m

 

Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s

Time = 5 s

Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train

Important Formulas – Percentage

 

  • Percentage

    Percent means for every 100

    So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value.

    percent is denoted by the symbol %. For example, x percent is denoted by x%

  • x%=x/100

    Example : 25%=25/100=1/4

  • To express x/y as a percent,we have x/y=(x/y×100)%

    Example : 1/4=(1/4×100)%=25%

  • If the price of a commodity increases by R%, the reduction in consumptionso as not to increase the expenditure = [R/(100+R)×100]%
  • If the price of a commodity decreases by R%, the increase in consumptionso as not to decrease the expenditure = [R/(100−R)×100]%
  • If the Population of a town = P and it increases at the rate of R% per annum, thenPopulation after n years = P((1+R)/100))n
  • If the population of a town = P and it increases at the rate of R% per annum, thenPopulation before n years = P((1+R)/100))n
  • If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine after n years = P((1-R)/100))n

  • If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine before n years = P((1-R)/100))n

 

Solved Examples

Level 1

1.    If A = x% of y and B = y% of x, then which of the following is true?
A. None of these B. A is smaller than B.
C. Relationship between A and B cannot be determined. D. If x is smaller than y, then A is greater than B.
E. A is greater than B.

 

   

Answer : Option A

Explanation :

A = xy/100 ………….(Equation 1)

B = yx/100……………..(Equation 2)

From these equations, it is clear that A = B

 

 

2.If 20% of a = b, then b% of 20 is the same as:
A. None of these B. 10% of a
C. 4% of a D. 20% of a

 

Answer :Option C

Explanation :

20% of a = b

=> b = 20a/100

b% of 20 = 20b/100=(20a/100) × 20/100

=(20×20×a)/(100×100)=4a/100 = 4% of a

 

3.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 1 B. 1 : 2
C. 1 : 1 D. 4 : 3

 

 

Answer :Option D

Explanation :

5% of A + 4% of B = 2/3(6% of A + 8% of B)

5A/100+4B/100=2/3(6A/100+8B/100)

⇒5A+4B=2/3(6A+8B)

⇒15A+12B=12A+16B

⇒3A=4B

⇒AB=43⇒A:B=4:3

4.The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the Average percent increase of population per year?
A. 4% B. 6%
C. 5% D. 50%

 

Answer :Option C

Explanation :

Increase in the population in 10 years = 2,62,500 – 1,75,000 = 87500

% increase in the population in 10 years = (87500/175000)×100=8750/175=50%

Average % increase of population per year = 50%/10=5%

 

5.Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
A. 57% B. 50%
C. 52% D. 60%

 

Answer :Option A

Explanation :

Votes received by the winning candidate = 11628

Total votes = 1136 + 7636 + 11628 = 20400

Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%

 

6.A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally?
A. 420 B. 700
C. 220 D. 400

 

Answer :Option B

Explanation :

He sells 40% of oranges and still there are 420 oranges remaining

=> 60% of oranges = 420

⇒(60×Total Oranges)/100=420

⇒Total Oranges/100=7

⇒ Total Oranges = 7×100=700

7.A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 499/11 % B. 45 %
C. 500/11 % D. 489/11 %

 

Answer :Option C

Explanation :

Total runs scored = 110

Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60

Total runs scored by running between the wickets = 110 – 60 = 50

Required % = (50/110)×100=500/11%

 

8.What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 2023% B. 20%
C. 21% D. 2223%

 

 

Answer :Option B

Explanation :

Total numbers = 70

Total numbers in 1 to 70 which has 1 in the unit digit = 7

Total numbers in 1 to 70 which has 9 in the unit digit = 7

Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14

Required percentage = (14/70)×100=140/7=20%

 

Level 2

 

1.In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
A. 2800 B. 2700
C. 2100 D. 2500

 

Answer :Option B

Explanation :

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = (7500×80)/100

1st candidate got 55% of the total valid votes.

Hence the 2nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = (total valid votes ×45)/100

=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700

 

2.In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?
A. 8200 B. 7500
C. 7000 D. 8000

 

Answer :Option D

Explanation :

State A and State B had an equal number of candidates appeared.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

 

3.In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A. 100 B. 102
C. 110 D. 90

 

 

Answer :Option A

Explanation :

Let the total number of students = x

Given that 20% of students are below 8 years of age

then The number of students above or equal to 8 years of age = 80% of x —–(Equation 1)

Given that number of students of 8 years of age = 48 —–(Equation 2)

Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age

=>number of students above 8 years of age = (2/3)×48=32—–(Equation 3)

From Equation 1,Equation 2 and Equation 3,
80% of x = 48 + 32 = 80

⇒80x/100=80

⇒x100=1⇒x=100

4.In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination?
A. 28000 B. 30000
C. 32000 D. 33000

 

Answer :Option B

Explanation :

Let the number of candidates applied for the examination = x

Given that 5% of the applicants were found ineligible.

It means that 95% of the applicants were eligible (∴ 100% – 5% = 95%)

Hence total eligible candidates = 95x/100

Given that 85% of the eligible candidates belonged to the general category

It means 15% of the eligible candidates belonged to other categories(∴ 100% – 85% = 15%)
Hence Total eligible candidates belonged to other categories

=(total eligible candidates×15)/100=(95x/100)×(15/100)

=(95x×15)/(100×100)

Given that Total eligible candidates belonged to other categories = 4275

⇒(95x×15)/(100×100)=4275

⇒(19x×15)/(100×100)=855

⇒(19x×3)/(100×100)=171

⇒(x×3)/(100×100)=9

⇒x/(100×100)=3

⇒x=3×100×100=30000

 

5.A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation?
A. 64% B. 32%
C. 34% D. 42%

 

Answer :Option A

Explanation :

Let the number = 1

Then, ideally he should have multiplied 1 by 5/3.

Hence the correct result was 1 x (5/3) = (5/3)

By mistake, he multiplied 1 by 3/5.

Hence the result with the error = 1 x (3/5) = (3/5)

Error = 5/3−3/5=(25−9)/15=16/15

percentage error = (Error/True Value)×100={(16/15)/(5/3)}×100

=(16×3×100)/(15×5)=(16×100)/(5×5)=16×4=64%

 

6.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the Insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?
A. Rs. 76,375 B. Rs. 34,000
C. Rs. 82,150 D. Rs. 70,000

 

Answer :Option A

Explanation :

Price of the car = Rs.3,25,000

Car insured to 85% of its price

=>Insured price=(325000×85)/100

Insurance company paid 90% of the insurance
⇒Amount paid by Insurance company =(Insured price×90)/100

=325000×(85/100)×(90/100)=325×85×9=Rs.248625

Difference between the price of the car and the amount received

= Rs.325000 – Rs.248625 = Rs.76375

 

7.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?

A. 8% B. 7%
C. 10% D. 6%

 

 

Answer :Option A

Explanation :

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = (100×(100+25))/100=Rs.125

Since Benson spends additional 15% on petrol,

amount spent by Benson = (100×(100+15))/100=Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced = (1−115/125) Litre

Percentage Quantity of reduction = ((1−115/125))/1×100=(10/125)/×100=(10/5)×4=2×4=8%

8.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 60% B. 70%
C. 80% D. 90%

 

Answer :Option C

Explanation :

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

=> 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒Number of men above the age of 50 who play football = (20×20)/100=4

Number of men who play football = 20 (Since 20% of all men play football)
Percentage of men who play football above the age of 50 = (4/20)×100=20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

Important Formulas – Mixtures and Alligations

 

  1. Alligation

    It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.

  2. Mean Price

    Mean price is the cost price of a unit quantity of the mixture

 

  1. Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid = [x(1−y/x)n]

 

  1. Rule of Alligation

    If two ingredients are mixed, then

    (Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)

Cost Price(CP) of a unit quantity
of cheaper (c)
Cost Price(CP) of a unit quantity
of dearer (d)
   
Mean Price
(m)
(d – m) (m – c)
  1. => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

 

Solved Examples

Level 1

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26 litres B. 29.16 litres
C. 28 litres D. 28.2 litres

 

 

Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced

by water. After n operations, the quantity of pure liquid

=[x(1−y/x)n]

Hence milk now contained by the container = 40(1−4/40)3=40(1−1/10)3

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

2.A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the Percentage of alcohol was found to be 26%. The quantity of whisky replaced is
A. 43 B. 34
C. 32 D. 23

 

 

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in 1st Jar Concentration of alcohol in 2nd Jar
40% 19%
Mean
26%
7 14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

 

3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ?
A. 7 : 8 B. 8 : 7
C. 6 : 7 D. 7 ; 6

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

 

Cost of 1 kg rice of 1st kind Cost of 1 kg rice of 2nd kind
9.3 10.80
Mean Price
10
10.8-10 = .8 10 – 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

 

 

 

 4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3 B. 2 : 2
C. 1 : 2 D. 3 : 1

 

 

Answer : Option C

 

Explanation :

By the rule of alligation, we have

Cost Price of 1 litre of water Cost Price of 1 litre of milk
0 12
Mean Price
8
12-8=4 8-0=8

Required Ratio = 4 : 8 = 1 : 2

 

 

 

5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%
A. 1.5 Kg B. 2 Kg
C. .5 Kg D. 1 Kg

 

 

 

 

Answer : Option D

 

Explanation :

By the rule of alligation, we have

Percentage concentration of
manganese in the mixture : 20
Percentage concentration of
manganese in pure iron : 0
Percentage concentration of manganese in the final mixture
10
10 – 0 = 10 20 – 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

 

6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%.
A. 1200 Kg B. 1400 Kg
C. 1600 Kg D. 800 Kg

 

 

 

Answer : Option A

 

Explanation :

By the rule of alligation, we have

 

% Profit by selling part1 % Profit by selling part2
8 12
Net % Profit
11
12 – 11 = 1 11 – 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200

 

 

7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it
A. 4 litre B. 2 litre
C. 1 litre D. 3 litre

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

% Concentration of water
in pure water : 100
% Concentration of water
in the given mixture : 10
Mean % concentration
20
20 – 10 = 10 100 – 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

 

8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300 B. 400
C. 600 D. 500

 

 

Answer : Option C

 

Explanation :

By the rule of alligation,we have

Profit% by selling 1st part Profit% by selling 2nd part
8 18
Net % profit
14
18-14=4 14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg

 

Level 2

 

1.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?
A. Rs.182.50 B. Rs.170.5
C. Rs.175.50 D. Rs.180

 

 

Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their Average price = (126+135)2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e.,

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of tea Cost of 1 kg of 2nd kind of tea
130.50 x
Mean Price
153
(x – 153) 22.50

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

 

2.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 5litres, 7 litres B. 7litres, 4 litres
C. 6litres, 6 litres D. 4litres, 8 litres

 

 

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

Milk in 1 litre mix in 1st can = 3/4 litre

Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4

Milk in 1 litre mix in 2nd can = 1/2 litre

Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2

Milk in 1 litre of the final mix=5/8

Cost Price(CP) of 1 litre final mix = Rs.5/8

=> Mean price = 5/8
By the rule of alligation, we can write as

CP of 1 litre mix in 2nd can CP of 1 litre mix in 1st can
1/2 3/4
Mean Price
5/8
3/4 – 5/8 = 1/8 5/8 – 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1

ie, from each can, 12×12=6 litre should be taken

 

3.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?
A. 3: 4 B. 4 : 3
C. 9 : 7 D. 7 : 9

 

 

Answer : Option D

 

Explanation :

 

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

 

CP of 1 litre mixture from vessel A CP of 1 litre mixture from vessel B
5/7 7/13
Mean Price
8/13
8/13 – 7/13 = 1/13 5/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

 

4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
A. 60 Kg B. 63 kg
C. 58 Kg D. 56 Kg

 

 

Answer : Option B

Explanation :

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind CP of 1 kg sugar of 2nd kind
Rs. 9 Rs. 7
Mean Price
Rs.8.4
8.4 – 7 = 1.4 9 – 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

ratio 1.4 : .6 = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23 B. 21
C. 19 D. 17

 

 

Answer : Option B

Explanation :

Let’s initial quantity of P in the container be 7x

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4

Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54

HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4

Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4

Since the container is filled with Q after 9 litres of mixture is drawn off,

Quantity of Q in the mixtures=5x−15/4+9=5x+21/4

Given that the ratio of P and Q becomes 7 : 9

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21

 

6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25% B. 20%
C. 22% D. 24%

 

 

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1

Given than Gain = 25%

Hence CP of 1 litre mixture = 100/(100+Gain%)×SP

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

CP of 1 litre milk CP of 1 litre water
1 0
CP of 1 litre mixture
4/5
4/5 – 0 = 4/5 1- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1/5×100=20%

MENSURATION

 

MENSURATION is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebraic equations and geometric calculations to provide measurement data regarding the width, depth and volume of a given object or group of objects

  • Pythagorean Theorem (Pythagoras’ theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c2 = a2 + b2 where c is the length of the hypotenuse and a and b are the lengths of the other two sides

  • Pi is a mathematical constant which is the ratio of a circle’s circumference to its diameter. It is denoted by π

π≈3.14≈227

  • Geometric Shapes and solids and Important Formulas
Geometric Shapes Description Formulas
Rectangle

l = Length

b = Breadth

d= Length of diagonal

Area = lb

Perimeter = 2(l + b)

d = √l2+b2

Square

a = Length of a side

d= Length of diagonal

Area= a*a=1/2*d*d

Perimeter = 4a

d = 2√a

Parallelogram

b and c are sides

b = base

h = height

                                 Area = bh

Perimeter = 2(b + c)

Rhombus

a = length of each side

b = base

h = height

d1, d2 are the diagonal

Area = bh(Formula 1)

Area = ½*d1*d2 (Formula 2 )

Perimeter = 4a

Triangle

a , b and c are sides

b = base

h = height

Area = ½*b*h (Formula 1) Area(Formula 2)                         = √S(Sa)(Sb)(Sc              where S is the semiperimeter
S  =(a+b+c)/2 (Formula 2 for area          -Heron’s formula) Perimeter = a + b + c

Radius of incircle of a triangle of area A =AS
where S is the semiperimeter
=(a+b+c)/2

Equilateral Triangle

a = side

Area = (√3/4)*a*a               Perimeter = 3a

Radius of incircle of an equilateral                                                                  triangle of side a = a/2*√3

Radius of circumcircle of an equilateral triangle
of side a = a/√3

 

Base a is parallel to base b Trapezium(Trapezoid in American English)

h = height

Area = 12(a+b)h

 

Circle

r = radius

d = diameter

d = 2r

Area = πr2 = 14πd2

Circumference = 2πr = πd

 

Sector of Circle

r = radius

θ = central angle

Area  = (θ/360) *π*r*r
Arc Length, s = (θ/180)* π*r

In the radian system for angular measurement,
2π radians = 360°
=> 1 radian = 180°π
=> 1° = π180 radians
Hence,
Angle in Degrees
= Angle in Radians × 180°π
Angle in Radians
= Angle in Degrees × π180°

 

Ellipse

Major axis length = 2a

Minor axis length = 2b

Area = πab

Perimeter ≈

Rectangular Solid

l = length

w = width

h = height

Total Surface Area
= 2lw + 2wh + 2hl
= 2(lw + wh + hl)

Volume = lwh

Cube

s = edge

Total Surface Area = 6s2

Volume = s3

Right Circular Cylinder

h = height

r = radius of base

Lateral Surface Area
= (2 π r)h

Total Surface Area
= (2 π r)h + 2 (π r2)

Volume = (π r2)h

Pyramid

h = height

B = area of the base

Total Surface Area = B +                 Sum of  the areas of the triangular sides

Volume = 1/3*B*h

Right Circular Cone

h = height

r = radius of base

Lateral Surface Area=πrs
where s is the slant height =√r*r+h*h
Total Surface Area
                                =πrs+πr2
Sphere

r = radius

d = diameter

 d = 2r

Surface Area =4πr*r=πd*d

Volume =4/3πr*r*r=16πd*d*d

     
 
 
 

 

 
 
 

 

 

 

 
 
 
 
 

 

  • Important properties of Geometric Shapes
    1. Properties of Triangle
      1. Sum of the angles of a triangle = 180°
      2. Sum of any two sides of a triangle is greater than the third side.
  • The line joining the midpoint of a side of a triangle to the positive vertex is called the Median
  1. The median of a triangle divides the triangle into two triangles with equal areas
  2. Centroid is the point where the three medians of a triangle meet.
  3. Centroid divides each median into segments with a 2:1 ratio
  • Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.
  • An equilateral triangle is a triangle in which all three sides are equal
  1. In an equilateral triangle, all three internal angles are congruent to each other
  2. In an equilateral triangle, all three internal angles are each 60°
  3. An isosceles triangle is a triangle with (at least) two equal sides
  • In isosceles triangle, altitude from vertex bisects the base.

 

  1. Properties of Quadrilaterals
  2. Rectangle
    1. The diagonals of a rectangle are equal and bisect each other
    2. opposite sides of a rectangle are parallel
  • opposite sides of a rectangle are congruent
  1. opposite angles of a rectangle are congruent
  2. All four angles of a rectangle are right angles
  3. The diagonals of a rectangle are congruent
  4. Square
  • All four sides of a square are congruent
  • Opposite sides of a square are parallel
  1. The diagonals of a square are equal
  2. The diagonals of a square bisect each other at right angles
  3. All angles of a square are 90 degrees.
  • A square is a special kind of rectangle where all the sides have equal length
  1. Parallelogram
  • The opposite sides of a parallelogram are equal in length.
  • The opposite angles of a parallelogram are congruent (equal measure).
  1. The diagonals of a parallelogram bisect each other.
  • Each diagonal of a parallelogram divides it into two triangles of the same area
  1. Rhombus
  • All the sides of a rhombus are congruent
  • Opposite sides of a rhombus are parallel.
  • The diagonals of a rhombus bisect each other at right angles
  1. Opposite internal angles of a rhombus are congruent (equal in size)
  • Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)
  • If each angle of a rhombus is 90°, it is a square

Other properties of quadrilaterals

  • The sum of the interior angles of a quadrilateral is 360 degrees
  • If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)
  • A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
  • Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
  • Each diagonal of a parallelogram divides it into two triangles of the same area
  • A square is a rhombus and a rectangle.
  1. Sum of Interior Angles of a polygon
    1. The sum of the interior angles of a polygon = 180(n – 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 – 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 – 2) = 180 × 2 = 360.

 

 

Solved Examples

Level 1

  1. An error 2% in excess is made while measuring the side of a square. What is the Percentage of error in the calculated area of the square?
  2. 4.04 %
  3. 2.02 %
  4. 4 %
  5. 2 %

Answer : Option A

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = (100×(100+2))/100=102 (∵ error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000
Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 – 10000 = 404
Percentage Error = (Error/Actual Value)×100=(404/10000)×100=4.04%

 

  1. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?
  2. 30 %
  3. 28 %
  4. 32 %
  5. 26 %

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Lost 20% of length
=> New length =( Original length × (100−20))/100
=(100×80)/100=80

Lost 10% of breadth
=> New breadth= (Original breadth × (100−10))/100
=(100×90)/100=90

New area = 80 × 90 = 7200

Decrease in area
= Original Area – New Area
= 10000 – 7200 = 2800

Percentage of decrease in area
=(Decrease in Area/Original Area)×100=(2800/10000)×100=28%

  1. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
  2. 25 % Increase
  3. 25 % Decrease
  4. 50 % Decrease
  5. 50 % Increase

Answer : Option D

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Length of the rectangle is halved
=> New length = (Original length)/2=100/2=50

breadth is tripled
=> New breadth= Original breadth × 3 = 100 × 3 = 300

New area = 50 × 300 = 15000

Increase in area = New Area – Original Area = 15000 – 10000= 5000
Percentage of Increase in area =( Increase in Area/OriginalArea)×100=(5000/10000)×100=50%

  1. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?
  2. 14 metres
  3. 20 metres
  4. 18 metres
  5. 12 metres

Answer : Option B

Explanation:

lb = 460 m2 ——(Equation 1)

Let the breadth = b
Then length, l =( b×(100+15))/100=115b/100——(Equation 2)

From Equation 1 and Equation 2,
115b/100×b=460b2=46000/115=400⇒b=√400=20 m

 

  1. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?
  2. equal to ½
  3. equal to ¾
  4. greater than 1
  5. equal to 1

Answer : Option C

Explanation :

If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)

 

Hence greater than 1 is the more suitable choice from the given list

================================================================
Note : Proof

Consider a square and rhombus standing on the same base ‘a’. All the sides of a square are of equal length. Similarly all the sides of a rhombus are also of equal length. Since both the square and rhombus stands on the same base ‘a’,

Length of each side of the square = a
Length of each side of the rhombus = a

Area of the sqaure = a2 …(1)

From the diagram, sin θ = h/a
=> h = a sin θ

Area of the rhombus = ah = a × a sin θ = a2 sin θ …(2)

From (1) and (2)

Area of the square/Area of the rhombus= a2 /a2sinθ=1/sinθ

Since 0° < θ < 90°, 0 < sin θ < 1. Therefore, area of the square is greater than that of rhombus, provided both stands on same base.

(Note that, when each angle of the rhombus is 90°, rhombus is also a square (can be considered as special case) and in that case, areas will be equal.

 

  1. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.
  2. 37500 m2
  3. 30500 m2
  4. 32500 m2
  5. 40000 m2

Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length
b=(60/100)* l =(3/5)* l

perimeter of the field = 800 m
=> 2 (l + b) = 800
⇒2(l+(3/5)* l)=800⇒l+(3/5)* l =400⇒(8/5)* l =400⇒l/5=50⇒l=5×50=250 m

b = (3/5)* l =(3×250)/5=3×50=150 m

Area = lb = 250×150=37500 m2

 

  1. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
  2. 45%
  3. 44%
  4. 40%
  5. 42%

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Increase in 20% of length.
=> New length = (Original length ×(100+20))/100=(100×120)/100=120

Increase in 20% of breadth
=> New breadth= (Original breadth × (100+20))/100=(100×120)/100=120

New area = 120 × 120 = 14400

Increase in area = New Area – Original Area = 14400 – 10000 = 4400
Percentage increase in area =( Increase in Area /OriginalArea)×100=(4400/10000)×100=44%

  1. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
  2. 814
  3. 802
  4. 836
  5. 900

Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cm
b = 9 m 2 cm = 902 cm
Area = 1517 × 902 cm2

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.
Let’s find out the HCF using long division method for quicker results

902)  1517  (1

 

-902

—————–

615)  902  (1

 

  • 615

————–

 

287)  615 (2

 

-574

—————–

 

41)  287  (7

 

-287

————

0
————

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm
Area of each square tile = 41 × 41 cm2

Number of tiles required = (1517×902)/(41×41)=37×22=407×2=814

 

Level 2

  1. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?
  2. 126 sq. ft.
  3. 64 sq. ft.
  4. 100 sq. ft.
  5. 102 sq. ft.

Answer : Option A

Explanation :

Let l = 9 ft.

Then l + 2b = 37
=> 2b = 37 – l = 37 – 9 = 28
=> b = 282 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

 

  1. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the Average of the two areas. What is the area of the smaller part in hectares?
  2. 400
  3. 365
  4. 385
  5. 315

Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 – x) hectares.

Difference of the areas of the two parts = x – (700 – x) = 2x – 700

one-fifth of the average of the two areas = 15[x+(700−x)]2
=15×7002=3505=70

Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x – 700 = 70
=> 2x = 770
x=7702=385

Hence, area of smaller part = (700 – x) = (700 – 385) = 315 hectares.

 

  1. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?
  2. 18 cm
  3. 16 cm
  4. 40 cm
  5. 20 cm

Answer : Option C

Explanation :

Let breadth = x cm
Then length = 2x cm
Area = lb = x × 2x = 2x2

New length = (2x – 5)
New breadth = (x + 5)
New Area = lb = (2x – 5)(x + 5)

But given that new area = initial area + 75 sq.cm.
=> (2x – 5)(x + 5) = 2x2 + 75
=> 2x2 + 10x – 5x – 25 = 2x2 + 75
=> 5x – 25 = 75
=> 5x = 75 + 25 = 100
=> x = 1005 = 20 cm

Length = 2x = 2 × 20 = 40cm

 

  1. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)?
  2. 142000
  3. 112800
  4. 142500
  5. 153600

Answer : Option D

Explanation :

l : b = 3 : 2 —-(Equation 1)

Perimeter of the rectangular park
= Distance travelled by the man at the speed of 12 km/hr in 8 minutes
= speed × time = 12×860     (∵ 8 minute = 860 hour)
= 85 km = 85 × 1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600
=> l + b = 16002 = 800 m —-(Equation 2)

From (Equation 1) and (Equation 2)
l = 800 × 35 = 480 m
b = 800 × 25 = 320 m (Or b = 800 – 480 = 320m)

Area = lb = 480 × 320 = 153600 m2

 

  1. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction?
  2. Rs.3500
  3. Rs. 4200
  4. Insufficient Data
  5. Rs. 4400

Answer : Option C

Explanation :
Let length and width of the rectangular plot be l and b respectively
Total area of the rectangular plot = 96 sq.m.
=> lb = 96

Width of the pathway = 2 m
Length of the remaining area in the plot = (l – 4)
breadth of the remaining area in the plot = (b – 4)
Area of the remaining area in the plot = (l – 4)(b – 4)

Area of the pathway
= Total area of the rectangular plot – remaining area in the plot
= 96 – [(l – 4)(b – 4)] = 96 – [lb – 4l – 4b + 16] = 96 – [96 – 4l – 4b + 16] = 96 – 96 + 4l + 4b – 16
= 4l + 4b – 16
= 4(l + b) – 16

We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of the construction.

 

  1. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?
  2. 144√3−48π cm2
  3. 121√3−36π cm2
  4. 144√3−36π cm2
  5. 121√3−48π cm2

Answer : Option A

Explanation :
Area of an equilateral triangle = (3/√4)*a *a where a is length of one side of the equilateral triangle
Area of the equilateral Δ ABC = (3/√4)*a *a = (3/√4)*24*24=144√3 cm2⋯ (1)

Area of a triangle = 12bhwhere b is the base and h is the height of the triangle
Let r = radius of the inscribed circle. Then
Area of Δ ABC
= Area of Δ OBC + Area of Δ OCA + area of Δ OAB
= (½ × r × BC) + (½ × r × CA) + (½ × r × AB)
= ½ × r × (BC + CA + AB)
= ½ x r x (24 + 24 + 24)
= ½ x r x 72 = 36r cm2 —-(2)

From (1) and (2),
144√3=36rr=144√3/36=4√3−−−−(3)

Area of a circle = πr2 where = radius of the circle
From (3), the area of the inscribed circle = πr2=π(4√3)* (4√3)=48π⋯(4)

Hence, area of the remaining portion of the triangle
= Area of Δ ABC – Area of inscribed circle
144√3−48π cm2

 

  1. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
  2. √11600 cm
  3. √14400 cm
  4. √10000 cm
  5. √12040 cm

Answer : Option A

Explanation :
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal).
Hence the length of the longest rod = AG

Initially let’s find out AC. Consider the right angled triangle ABC

AC2 = AB2 + BC2 = 402 + 802 = 1600 + 6400 = 8000
⇒AC = √8000 cm

Consider the right angled triangle ACG

AG2 = AC2 + CG2
(√8000) 2+602=8000+3600=11600
=> AG = √11600 cm
=> Length of the longest rod = √11600cm

 

  1. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?
  2. 30
  3. 44
  4. 56
  5. 60

Answer : Option C

Explanation :

Perimeter of a rectangle = 2(l + b)
where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.
Hence number of poles required = 280/5 = 56

Height and Distance

 

This topic has many practical application in day to day life. In engineering stage it is used in surveying. The basic purpose is to find the unknown variables by observing the angle of the line of sight. This is done by using some the fact that in a right angled triangle the ratio of any two sides is a function of the angle between them. From exam point of view this is one of the more tough sections and tedious to some extent. So an aspirant must thoroughly solve all the questions given here.

Important Formulas

  1. Trigonometric Basics

sinθ=oppositeside/hypotenuse=y/r

cosθ=adjacentside/hypotenuse=x/r

tanθ=oppositeside/adjacentside=y/x

cosecθ=hypotenuse/oppositeside=r/y

secθ=hypotenuse/adjacentside=r/x

cotθ=adjacentside/oppositeside=x/y

From Pythagorean theorem, x2+y2=r2 for the right angled triangle mentioned above

 

  1. Basic Trigonometric Values

 

θ
in degrees
θ
in radians
sinθ cosθ tanθ
0 0 1 0
30° π/6 1/2 3/√2 1/√3
45° π/4 1/√2 1/√2 1
60° π/3 3/√2 1/2 √3
90° π/2 1 0 Not defined

 

  1. Trigonometric Formulas

Degrees to Radians and vice versa

360°=2π radian

 

Trigonometry – Quotient Formulas

tanθ=sinθ/cosθ

cotθ=cosθ/sinθ

 

Trigonometry – Reciprocal Formulas

cosecθ=1/sinθ

secθ=1/cosθ

cotθ=1/tanθ

 

Trigonometry – Pythagorean Formulas

sin2θ+cos2θ=1

sec2θ−tan2θ=1

cosec2θ−cot2θ=1

 

  1. Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight).

i.e., angle of elevation =  AOP

  1. Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer’s line of sight

i.e., angle of depression =  AOP

  1. Angle Bisector Theorem

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then BD/DC=AB/AC

(Note that an angle bisector divides the angle into two angles with equal measures.
i.e., BAD = CAD in the above diagram)

  1. Few Important Values to memorize

√2=1.414, √3=1.732, √5=2.236

 

Solved Examples

Level 1

1.The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
A. 14.8 m B. 6.2 m
C. 12.4 m D. 24.8 m

 

 

Answer : Option D

Explanation :

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60° = PQ/PR

1/2=12.4/PR

PR=2×12.4=24.8 m

2.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
A. 346 m B. 400 m
C. 312 m D. 298 m

 

 

Answer : Option A

Explanation :

tan 30°=RQ/PQ

1/√3=200/PQ

PQ=200√3=200×1.73=346 m

3.The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
A. None of these B. 60°
C. 45° D. 30°

 

 

Answer : Option C

Explanation :

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let QPR = θ

tan θ = QR/PQ=1 (since QR = PQ)

=> θ = 45°

i.e., required angle of elevation = 45°

4.An observer 2 m tall is 103√ m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. None of these B. 12 m
C. 14 m D. 10 m

 

 

Answer : Option B

Explanation :

SR = PQ = 2 m

PS = QR = 10√3m

tan 30°=TS/PS

1/3=TS/10√3

TS=10√3/√3=10 m

TR = TS + SR = 10 + 2 = 12 m

5.From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
A. 40 m B. 138.4 m
C. 46.24 m D. 160 m

 

 

Answer : Option B

Explanation :

Let AC be the tower and B be the position of the bus.

Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30°

ABC = DAB = 30° (Because DA || BC)

tan 30°=AC/BC=>tan 30°=80/BC=>BC = 80/tan 30°=80/(1/√3)=80×1.73=138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m

6.Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long?
A. 30° B. 60°
C. 45° D. None of these

 

 

Answer : Option B

Explanation :

Let RQ be the pole and PQ be the shadow

Given that RQ = 18 m and PQ = 6√3 m

Let the angle of elevation, RPQ = θ

From the right  PQR,

tanθ=RQ/PQ=18/6√3=3/√3=(3×√3)/( √3×√3)=3√3/3=√3

θ=tan−1(3√)=60°

 

Level 2

1.A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
A. 8 min 17 second B. 10 min 57 second
C. 14 min 34 second D. 12 min 23 second

 

 

Answer : Option B

Explanation :

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car

Then, ADC = 30° , ACB = 45°

Let AB = h, BC = x, CD = y

tan 45°=AB/BC=h/x

=>1=h/x=>h=x——(1)

tan 30°=AB/BD=AB/(BC + CD)=h/(x+y)

=>1/√3=h/(x+y)

=>x + y = √3h

=>y = √3h – x

=>y = √3h−h(∵ Substituted the value of x from equation 1 )

=>y = h(√3−1)

Given that distance y is covered in 8 minutes
i.e, distance h(√3−1) is covered in 8 minutes

Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes

since distance is proportional to the time when the speed is constant, we have

h(√3−1)∝8—(A)

h∝t—(B)

(A)/(B)=>h(√3−1)/h=8/t

=>(√3−1)=8/t

=>t=8/(√3−1)=8/(1.73−1)=8/.73=800/73minutes ≈10 minutes 57 seconds

2.The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 5 metres B. 8 metres
C. 10 metres D. 12 metres

 

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the tower and DE represents the pole

Given that AC = 15 m , ADB = 30°, AEC = 60°

Let DE = h

Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE

tan 60°=AC/CE=>√3=15/CE=>CE = 15√3— (1)

tan 30°=AB/BD=>1/√3=(15−h)/BD

=>1/√3=(15−h)/(15/√3)(∵ BD = CE and Substituted the value of CE from equation 1)

=>(15−h)=(1/√3)×(15/√3)=15/3=5

=>h=15−5=10 m

i.e., height of the electric pole = 10 m

 

3.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 300 m B. 173 m
C. 273 m D. 200 m

 

 

Answer : Option C

Explanation :

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,  BAD = 30° ,  BCD = 45°

tan 30° = BD/BA⇒1/√3=100/BA

⇒BA=100√3

tan 45° = BD/BC

⇒1=100/BC

⇒BC=100

Distance between the two ships = AC = BA + BC
=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273 m

4.From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
A. 52 m B. 50 m
C. 66.67 m D. 33.33 m

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the hill and DE represents the pole

Given that AC = 100 m

XAD = ADB = 30° (∵ AX || BD )
XAE = AEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE

=>√3=100/CE=>CE = 100/√3— (1)

tan 30°=AB/BD=>1/√3=(100−h)/BD

=>1/√3=(100−h)/(100/√3)(∵ BD = CE and Substituted the value of CE from equation 1 )

=>(100−h)=1/√3×100/√3=100/3=33.33=>h=100−33.33=66.67 m

i.e., the height of the pole = 66.67 m

5.A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?
A. 9 m B. 10.40 m
C. 15.57 m D. 12 m

 

 

Answer : Option A

Explanation :

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.

Given that AD = 18 m, ABC = 60°, DBC = 30°

Let DC be h.

tan 30°=DC/BC

1/√3=h/BC

h=BC√3—— (1)

tan 60°=AC/BC

√3=(18+h)/BC

18+h=BC×√3—— (2)

(1)/(2)=>h/(18+h)=(BC/√3)/(BC×√3)=1/3

=>3h=18+h=>2h=18=>h=9 m

i.e., the height of the tower = 9 m

6.A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?
A. 0.63 meter/sec B. 2.16 meter/sec
C. 3.87 meter/sec D. 0.72 meter/sec

 

 

Answer : Option B

Explanation :

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.

Given that CA = 150 m, BCA = 60°

tan 60°=BA/CA

√3=BA/150

BA=150√3

i.e, the distance travelled by the balloon = 150√3meters

time taken = 2 min = 2 × 60 = 120 seconds

Speed = Distance/Time=150√3/120=1.25√3=1.25×1.73=2.16 meter/second

7. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
A. 22 m B. 44 m
C. 33 m D. None of these

 

 

Answer : Option B

Explanation :

Let DC be the wall, AB be the tree.

Given that DBC = 30°, DAE = 60°, DC = 11 m

tan 30°=DC/BC

1/√3=11/BC

BC = 11√3 m

AE = BC =11√3 m—— (1)

tan 60°=ED/AE

√3=ED/11√3[∵ Substituted the value of AE from (1)]

ED =11√3×√3=11×3=33

Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m

 

8. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.
A. 141 m and 282 m B. 70.5 m and 141 m
C. 65 m and 130 m D. 130 m and 260 m

 

 

Answer : Option B

Explanation :

Let AB and CD be the poles with heights h and 2h respectively

Given that distance between the poles, BD = 200 m

Let E be the middle point of BD.

Let AEB = θ and CED = (90-θ) (∵ given that angular elevations are complementary)

Since E is the middle point of BD, we have BE = ED = 100 m

From the right  ABE,
tanθ=AB/BE and tanθ=h/100

h = 100tanθ—— (1)

From the right  EDC,

tan(90−θ)=CD/ED

cotθ=2h/100[∵tan(90−θ)=cotθ]

2h =100cotθ—— (2)

(1) × (2) => 2h2=1002[∵tanθ×cotθ=tanθ×1/tanθ=1]

=>√2h=100

=>h=100/√2=(100×√2)/( √2×√2)=50√2=50×1.41=70.5

2h=2×70.5=141

i.e., the height of the poles are 70.5 m and 141 m.

9. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?
A. 8.65 m B. 2 m
C. 2.5 m D. 3.65 m

 

 

Answer : Option D

Explanation :

Let AB be the man and CD be the window

Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
DAF = 45° , CAF = 60°

From the diagram, AF = BE = 5 m

From the right  AFD, tan45°=DF/AF

1=DF/5

DF = 5—— (1)From the right  AFC, tan60°=CF/AF

√3=CF/5

CF=5√3—— (2)

Length of the window = CD = (CF – DF)

=5√3−5[∵ Substitued the value of CF and DF from (1) and (2)]=5(√3−1)=5(1.73−1)=5×0.73=3.65 m

10.The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. What is the approximate height of the mountain?
A. 1.2 km B. 0.6 km
C. 1.4 km D. 2.7 km

 

 

Answer : Option D

Explanation :

Let A be the foot and C be the summit of a mountain.

Given that CAB = 45°

From the diagram, CB is the height of the mountain. Let CB = x

Let D be the point after ascending 2 km towards the mountain such that
AD = 2 km and given that DAY = 30°

It is also given that from the point D, the elevation is 60°

i.e., CDE = 60°

From the right  ABC,

tan45°=CB/AB

=>1=x/AB[∵ CB = x (the height of the mountain)]

=>AB = x—— (eq:1)

From the right  AYD,

sin30°=DY/AD

=>1/2=DY/2(∵ Given that AD = 2)

=> DY=1—— (eq:2)

cos30°=AY/AD=>√3/2=AY/2(∵ Given that AD = 2)=> AY=√3—— (eq:3)

From the right  CED, tan60°=CE/DE=>tan60°=(CB – EB)/YB∵ [CE = (CB – EB) and DE = YB)]

=>tan60°=(CB – DY)/(AB – AY)[ ∵ EB = DY and YB = (AB – AY)]

=>tan60°=(x – 1)/(x -√3)∵ [CB = x, DY = 1(eq:2), AB=x (eq:1) and AY = 3√(eq:3)]

=>√3=(x – 1)/(x -√3)=>x√3−3=x−1=>x(√3−1)=2=>0.73x=2=>x=2/0.73=2.7

i.e., the height of the mountain = 2.7 km

CLASSIFICATION LEVEL 1

 

Classification involves putting things into a class or group according to particular characteristics so it’s easier to make sense of them, whether you’re organizing your shoes, your stock portfolio, or a group of invertebrates.  From all competitive examination classification is one of the most important topics, this pattern come with lot of questions minimum they asking the 4 to 5 question from the classification topic. In the SSC CGL or SSC constable GD examination having the same topics from the reasoning section but the standard of the topic will be different, so most of the candidates preference for this topic to get the best score in the written examination.

 

 

Directions: Find the odd one out

 

  1. A. Square B. Circle                     C. Rectangle             D. Triangle

 

  1. A. https://exam.pscnotes.com/cotton”>Cotton B. Terene                  C. Silk                         D. Wool

 

  1. A. Light B. Wave                    C. Heat                      D. Sound

 

  1. A. 81 : 243 B. 16 :64                   C. 64 : 192                D. 25 : 75

 

  1. A. 64 : 8 B. 80 : 9                     C. 7 : 49                     D. 36 : 6

 

  1. A. 26 : 62 B. 36 : 63                  C. 46 : 64                  D. 56 : 18

 

  1. A. ABZY B. BCYX                      C. CDVW                   D. DEVU

 

  1. A. ACE B. FHJ                         C. KLM                       D. SUW

 

  1. Find the wrong number in the series

441, 484, 529, 566, 625

  1. 484 B. 529                                    C. 625                                    D. 566

 

  1. Find the wrong number in the series

232, 343, 454, 564, 676

  1. 676 B. 454                                    C. 343                                    D. 564

 

 

SOLUTION TO CLASSIFICATION LEVEL 1

 

 

  1. B. Except circle, all others are geometrical figures consisting straight lines.

 

  1. B. Except terene, all others are natural fibres.

 

  1. B. Except wave, all others are different form of energy.

 

  1. B. 81*3=243

64*3=192

25*3=75

But     16*4=64

 

  1. D. Except D, in each pair one number is square root of the other.

 

  1. D. Except D, in each pair the position of digits has been interchanged.

 

  1. C. A+1=B   &   Z-1=Y

B+1=C   &   Y-1=X

D+1=E   &   V-1=U

But   C+1=D   &   V+1=W

 

  1. C. A+2=C    &   C+2=E

F+2=H     &   H+2=J

But      K+1=L     &   L+1=M

 

  1. D. 21^2=441

22^2=484

23^2=529

25^2=625

But   (23.79)^2=566

 

  1. D. 232+111=343

343+111=454

454+111=565 (but given 564)

CHAIN RULE

 

This module will teach you the basics of direct and indirect proportions. These concepts will further help you in time and work questions.

Important Formulas – CHAIN RULE

  • Direct Proportion

    Two quantities are said to be directly proportional, if on the increase or decrease of the one, the other increases or decreases the same extent.
    Examples

    1. Cost of the goods is directly proportional to the number of goods. (More goods, More cost)
    2. Amount of work done is directly proportional to the number of persons who did the work. (More persons, More Work)
  • Indirect Proportion (inverse proportion)

    Two quantities are said to be indirectly proportional (inversely proportional) if on the increase of the one, the other decreases to the same extent and vice-versa.

Examples

    1. Number of days needed to complete a work is indirectly proportional (inversely proportional) with the number of persons who does the work (More Persons, Less Days needed)
    2. The time taken to travel a distance is indirectly proportional (inversely proportional) with the speed in which one is travelling (More Speed, Less Time)

 

Solved Examples

Level 1

1. If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate?
A. Rs. (xd/y) B. Rs. x/d
C. Rs. (yd/x) D. Rs. y/d

 

Answer : Option C

Explanation :

cost of x metres of wire = Rs. d

cost of 1 metre of wire = Rs.(d/x)

cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)

2. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?
A. 50 B. 30
C. 40 D. 10

 

Answer : Option B

Explanation :

Meal for 200 children = Meal for 120 men

Meal for 1 child = Meal for 120/200 men

Meal for 150 children = Meal for (120×150)/200 men=Meal for 90 men

Total mean available = Meal for 120 men

Renaming meal = Meal for 120 men – Meal for 90 men = Meal for 30 men

 

3. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work?
A. 26 B. 22
C. 12 D. 24

 

Answer : Option D

Explanation :
Let the required number of days be x

More men, less days (indirect proportion)

Hence we can write as

Men36:27}::x:18 ⇒36×18=27×x ⇒12×18=9×x

⇒12×2=x

⇒x=24

4. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel?
A. 15 B. 12
C. 21 D. 9

 

Answer : Option D

Explanation :

Let the number of revolutions made by the larger wheel be x

More cogs, less revolutions (Indirect proportion)

Hence we can write as

Cogs 6:14}: x: 21⇒6×21=14×x ⇒6×3=2×x ⇒3×3=x ⇒x=9

5. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day?
A. 10 B. 12
C. 8 D. 15

 

Answer : Option B

Explanation :

Let the required hours needed be x

More pumps, less hours (Indirect proportion)
More Days, less hours (Indirect proportion)

Hence we can write as

Pumps  3:4

::x:8

Days                      2:1

⇒3×2×8=4×1×x

⇒3×2×2=x

⇒x=12

6. 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?
A. 9 B. 12
C. 10 D. 13

 

Answer : Option D

Explanation :
Let the required number of days be x

More persons, less days (indirect proportion)
More hours, less days (indirect proportion)

Hence we can write as

Persons                39:30

::x:12

Hours    5:6
⇒39×5×12=30×6×x ⇒39×5×2=30×x ⇒39=3×x ⇒x=13

7. A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth?
A. 205 B. 200
C. 180 D. 195

 

Answer : Option D

Explanation :

Let the required number of seconds be x

More cloth, More time, (direct proportion)

Hence we can write as

Cloth         0.128:25} :: 1:x

⇒0.128x=25 ⇒x=25/0.128 ⇒25000/128=3125/16≈195

 

8. 21 goats eat as much as 15 cows. How many goats each as much as 35 cows?
A. 49 B. 32
C. 36 D. 41

 

Answer : Option A

Explanation :

15 cows ≡ 21 goats

1 cow ≡21/15 goats

35 cows ≡ (21×35)/15 goats≡(21×7)/3 goats≡7×7 goats ≡ 49 goats

 

Level 2

 

1. In a https://exam.pscnotes.com/dairy”>Dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?
A. 1 B. 40
C. 20 D. 26

 

Answer : Option B

Explanation :

Assume that in x days, one cow will eat one bag of husk.

More cows, less days (Indirect proportion)
More bags, more days (direct proportion)
Hence we can write as

Cows    40:1         ::x:40

Bags     1:40

⇒40×1×40=1×40×x ⇒x=40

2. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost?
A. 65 paise B. 70 paise
C. 52 paise D. 48 paise

 

Answer : Option D

Explanation :
Let 200 gm potato costs x paise

Cost of ¼ Kg potato = 60 Paise
=> Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = 1000/4 gm = 250 gm)

More quantity, More Paise (direct proportion)

Hence we can write as

Quantity  200:250} :: x:60

⇒200×60=250×x ⇒4×60=5×x ⇒4×12=x ⇒x=48

3. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, 2/5 of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time, each person’s now working 9 hours a day.
A. 160 B. 150
C. 24 D. 56

 

Answer : Option D

Explanation :

Persons worked = 104
Number of hours each person worked per day = 8
Number of days they worked = 30
Work completed = 2/5

Remaining days = 56 – 30 = 26
Remaining Work to be completed = 1 – 2/5 = 3/5
Let the total number of persons who do the remaining work = x
Number of hours each person needs to be work per day = 9

More days, less persons(indirect proportion) More hours, less persons(indirect proportion)
More work, more persons(direct proportion)

Hence we can write as

Days     30:26

Hours    8:9                                   ::x:104

Work     35:25
⇒30×8×3/5×104=26×9×2/5×x

⇒x=(30×8×3/5×104)/(26×9×2/5)=(30×8×3×104)/(26×9×2)

=(30×8×104)/(26×3×2)=(30×8×4)/(3×2)=5×8×4=160

Number of additional persons required = 160 – 104 = 56

 

4. x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days?
A. x2/y2 units B. y3/x2 units
C. x3/y2 units D. y2/x2 units

 

Answer : Option B

Explanation :
Let amount of work completed by y men working y hours per in y days = w units

More men, more work(direct proportion)
More hours, more work(direct proportion)
More days, more work(direct proportion)

Hence we can write as

Men                      x:y

Hours    x:y          ::x:w

Days                      x:y
⇒x3w=y3x ⇒w=y3x/x3=y3/x2

5. A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions?
A. 12.5 m B. 10.5 m
C. 14 D. 12

 

Answer : Option A

Explanation :
Let the required height of the building be x meter

More shadow length, More height (direct proportion)

Hence we can write as

Shadow length 40.25:28.75}:: 17.5:x

⇒40.25×x=28.75×17.5 ⇒x=(28.75×17.5)/40.25=(2875×175)/40250

= (2875×7)/1610=2875/230=575/46=12.5

 

6. If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of such apples?
A. Rs. 2500 B. Rs. 2300
C. Rs. 2200 D. Rs. 1400

 

Answer : Option A

Explanation :

Let the required price be x

More apples, More price (direct proportion)

Hence we can write as

Apples 357:(49×12)} :: 1517.25:x

⇒357x = (49×12)×1517.25⇒x = (49×12×1517.25)/357=(7×12×1517.25)/51

= (7×4×1517.25)/17

=7×4×89.25≈2500

7. 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type?
A. 20 metric tonnes B. 22 metric tonnes
C. 24 metric tonnes D. 26 metric tonnes

 

Answer : Option D

Explanation :

Let required amount of coal be x metric tonnes

More engines, more amount of coal (direct proportion)

If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption
More rate of consumption, more amount of coal (direct proportion)

More hours, more amount of coal(direct proportion)

Hence we can write as

Engines                                                                9:8

rate of consumption                       13:14                     ::24:x

hours                                                                    8:13
⇒9×1/3×8×x=8×1/4×13×24 ⇒3×8×x=8×6×13 ⇒3xX=6×13⇒x=2×13=26

8. in a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came?
A. 1900 B. 1800
C. 1940 D. 2000

 

Answer : Option A

Explanation :

Given that food was sufficient for 2000 people for 54 days
Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 – 15 =39)
Let x number of people came after 15 days.
Then, total number of people after 15 days = (2000 + x)
Then, the remaining food was sufficient for (2000 + x) people for 20 days

More men, Less days (Indirect Proportion)⇒Men        2000:(2000+x)}  ::  20:39

⇒2000×39=(2000+x)20⇒100×39=(2000+x)⇒3900=2000+x⇒x=3900−2000=1900

BOAT AND STREAM

 

BOAT AND STREAM problems is a sub-set of time, speed and distance type questions where in relative speed takes the foremost role. We always find several questions related to the above concept in SSC common graduate level exam as well as in bank PO exam. Upon listing the brief theory of the issue below we move to the various kinds of problems asked in the competitive examination.

Important Formulas – Boats and Streams

  • Downstream
    In running/moving water, the direction along the stream is called downstream.
  • Upstream
    In running/moving water, the direction against the stream is called upstream.

 

  • Let the speed of a boat in still water be u km/hr and the speed of the stream be v km/hr, then

    Speed downstream = (u+v) km/hr
    Speed upstream = (u – v) km/hr

 

  • Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

    Speed in still water =1/2*(a+b)km/hr
    Rate of stream = 1/2*(ab) km/hr

Some more short-cut methods

  • Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his Average speed throughout the journey

    = (Speed downstream × Speed upstream)/Speed in still water=((x+y)(xy))/xkm/hr

 

  • Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance

    =((x* xy* y)*t)/2ykm

 

  • A man rows a certain distance downstream in t1 hours and returns the same distance upstream in t2 If the speed of the stream is y km/hr, then the speed of the man in still water

    =y((t2+t1) / (t2−t1)) km/hr

 

  • A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places

    =t((x* xy* y))/2xkm

 

  • A man takes n times as long to row upstream as to row downstream the river. If the speed of the man is x km/hr and the speed of the stream is y km/hr, then

    x=y*((n+1)/(n−1))

 

 

Solved Examples

 

Level 1

 

1. A man’s speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man’s speed against the current is:
A. 8.5 km/hr B. 10 km/hr.
C. 12.5 km/hr D. 9 km/hr

 

Answer : Option B

Explanation :

Man’s speed with the current = 15 km/hr

=>speed of the man + speed of the current = 15 km/hr

speed of the current is 2.5 km/hr

Hence, speed of the man = 15 – 2.5 = 12.5 km/hr

man’s speed against the current = speed of the man – speed of the current

= 12.5 – 2.5 = 10 km/hr

2. In one hour, a boat goes 14 km/hr along the stream and 8 km/hr against the stream. The speed of the boat in still water (in km/hr) is:
A. 12 km/hr B. 11 km/hr
C. 10 km/hr D. 8 km/hr

 

Answer : Option B

Explanation :

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2(a+b) km/hr and Rate of stream =1/2(a−b) km/hr
Speed in still water = 1/2(14+8) kmph = 11 kmph.

3. A boatman goes 2 km against the current of the stream in 2 hour and goes 1 km along the current in 20 minutes. How long will it take to go 5 km in stationary water?
A. 2 hr 30 min B. 2 hr
C. 4 hr D. 1 hr 15 min

 

Answer : Option A

Explanation :

Speed upstream = 2/2=1 km/hr

Speed downstream = 1/(20/60)=3 km/hr

Speed in still water = 1/2(3+1)=2 km/hr

Time taken to travel 5 km in still water = 5/2= 2 hour 30 minutes

4. Speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph. A man rows to a place at a distance of 4864 km and comes back to the starting point. The total time taken by him is:
A. 700 hours B. 350 hours
C. 1400 hours D. 1010 hours

 

Answer : Option A

Explanation :

Speed downstream = (14 + 1.2) = 15.2 kmph

Speed upstream = (14 – 1.2) = 12.8 kmph

Total time taken = 4864/15.2+4864/12.8 = 320 + 380 = 700 hours

 

 

5. The speed of a boat in still water in 22 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 24 minutes is:
A. 9.4 km B. 10.2 km
C. 10.4 km D. 9.2 km

 

   

Answer : Option C

Explanation :

Speed downstream = (22 + 4) = 26 kmph

Time = 24 minutes = 24/60 hour = 2/5 hour

distance travelled = Time × speed = (2/5)×26 = 10.4 km

6. A boat covers a certain distance downstream in 1 hour, while it comes back in 112 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?
A. 14 kmph B. 15 kmph
C. 13 kmph D. 12 kmph

 

Answer : Option B

Explanation :

Let the speed of the boat in still water = x kmph

Given that speed of the stream = 3 kmph

Speed downstream = (x+3) kmph

Speed upstream = (x-3) kmph

He travels a certain distance downstream in 1 hour and come back in 112 hour.

ie, distance travelled downstream in 1 hour = distance travelled upstream in 112 hour

since distance = speed × time, we have

(x+3)×1=(x−3)*3/2

=> 2(x + 3) = 3(x-3)

=> 2x + 6 = 3x – 9

=> x = 6+9 = 15 kmph

7. A boat can travel with a speed of 22 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 54 km downstream
A. 5 hours B. 4 hours
C. 3 hours D. 2 hours

 

Answer : Option D

Explanation :

Speed of the boat in still water = 22 km/hr

speed of the stream = 5 km/hr

Speed downstream = (22+5) = 27 km/hr

Distance travelled downstream = 54 km

Time taken = distance/speed=54/27 = 2 hours

 

8. A boat running downstream covers a distance of 22 km in 4 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water?
A. 5 kmph B. 4.95 kmph
C. 4.75 kmph D. 4.65

 

Answer : Option B

Explanation :

Speed downstream = 22/4 = 5.5 kmph

Speed upstream = 22/5 = 4.4 kmph

Speed of the boat in still water = (½) x (5.5+4.42) = 4.95 kmph

9. A man takes twice as long to row a distance against the stream as to row the same distance in favor of the stream. The ratio of the speed of the boat (in still water) and the stream is:
A. 3 : 1 B. 1 : 3
C. 1 : 2 D. 2 : 1

 

Answer : Option A

Explanation :

Let speed upstream = x

Then, speed downstream = 2x

Speed in still water = (2x+x)2=3x/2

Speed of the stream = (2x−x)2=x/2

Speed of boat in still water: Speed of the stream = 3x/2:x/2 = 3 : 1

 

Level  2

1. A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
A. 10 B. 6
C. 5 D. 4

 

   

Answer : Option C

Explanation :

Speed of the motor boat = 15 km/hr

Let speed of the stream = v

Speed downstream = (15+v) km/hr

Speed upstream = (15-v) km/hr

Time taken downstream = 30/(15+v)

Time taken upstream = 30/(15−v)

total time = 30/(15+v)+30/(15−v)

It is given that total time is 4 hours 30 minutes = 4.5 hour = 9/2 hour

i.e., 30/(15+v)+30/(15−v)=9/2

⇒1(15+v)+1(15−v)=(9/2)×30=3/20

⇒(15−v+15+v)/(15+v)(15−v)=3/20

⇒30/(15*15−v*v)=3/20

⇒30/(225−v*v)=3/20

⇒10/(225−v* v)=1/20

⇒225−v* v =200

⇒v* v =225−200=25

⇒v=5 km/hr

2. A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:
A. 1 km/hr. B. 2 km/hr.
C. 1.5 km/hr. D. 2.5 km/hr.

 

Answer : Option A

Explanation :

Assume that he moves 4 km downstream in x hours

Then, speed downstream = distance/time=4/x km/hr

Given that he can row 4 km with the stream in the same time as 3 km against the stream

i.e., speed upstream = 3/4of speed downstream=> speed upstream = 3/x km/hr

He rows to a place 48 km distant and come back in 14 hours

=>48/(4/x)+48/(3/x)=14

==>12x+16x=14

=>6x+8x=7

=>14x=7

=>x=1/2

Hence, speed downstream = 4/x=4/(1/2) = 8 km/hr

speed upstream = 3/x=3/(1/2) = 6 km/hr

Now we can use the below formula to find the rate of the stream

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2*(a+b) km/hr

Rate of stream =12*(a−b) km/hr
Hence, rate of the stream = ½*(8−6)=1 km/hr

 

3. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

A. 5 : 6 B. 6 : 5
C. 8 : 3 D. 3 : 8

 

Answer : Option C

Explanation :

Let the rate upstream of the boat = x kmph

and the rate downstream of the boat = y kmph

Distance travelled upstream in 8 hrs 48 min = Distance travelled downstream in 4 hrs.

Since distance = speed × time, we have

x×(8*4/5)=y×4

x×(44/5)=y×4

x×(11/5)=y— (equation 1)

Now consider the formula given below

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2(a+b) km/hr

Rate of stream =1/2(a−b) km/hr
Hence, speed of the boat = (y+x)/2

speed of the water = (y−x)/2

Required Ratio = (y+x)/2:(y−x)/2=(y+x):(y−x)=(11x/5+x):(11x/5−x)

(Substituted the value of y from equation 1)

=(11x+5x):(11x−5x)=16x:6x=8:3

 

4. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?
A. 3.2 km B. 3 km
C. 2.4 km D. 3.6 km

 

Answer : Option C

Explanation :

Speed in still water = 5 kmph
Speed of the current = 1 kmph

Speed downstream = (5+1) = 6 kmph
Speed upstream = (5-1) = 4 kmph

Let the requited distance be x km

Total time taken = 1 hour

=>x/6+x/4=1

=> 2x + 3x = 12

=> 5x = 12

=> x = 2.4 km

5. A man can row three-quarters of a kilometer against the stream in 1114 minutes and down the stream in 712minutes. The speed (in km/hr) of the man in still water is:
A. 4 kmph B. 5 kmph
C. 6 kmph D. 8 kmph

 

Answer : Option B

Explanation :

Distance = 3/4 km

Time taken to travel upstream = 1114 minutes

= 45/4 minutes = 45/(4×60) hours = 3/16 hours

Speed upstream = Distance/Time= (3/4)/ (3/16) = 4 km/hr

Time taken to travel downstream = 712minutes = 15/2 minutes = 15/2×60 hours = 1/8 hours

Speed downstream = Distance/Time= (3/4)/ (1/8) = 6 km/hr

Rate in still water = (6+4)/2=10/2=5 kmph

6. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is:
A. 4 mph B. 2.5 mph
C. 3 mph D. 2 mph

 

   

Answer : Option D

Explanation :

Speed of the boat in still water = 10 mph

Let speed of the stream be x mph

Then, speed downstream = (10+x) mph

speed upstream = (10-x) mph

Time taken to travel 36 miles upstream – Time taken to travel 36 miles downstream= 90/60 hours

=>36/(10−x)−36/(10+x)=3/2=>12/(10−x)−12/(10+x)=1/2=>24(10+x)−24(10−x)=(10+x)(10−x)

=>240+24x−240+24x=(100−x* x)=>48x=100− (x* x)=> x* x +48x−100=0

=>(x+50)(x−2)=0=>x = -50 or 2; Since x cannot be negative, x = 2 mph

7. At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 24-mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
A. 2*1/3 mph B. 1*1/3 mph
C. 1*2/3 mph D. 2*2/3 mph

 

Answer : Option D

Explanation :

Let the speed of Rahul in still water be x mph
and the speed of the current be y mph

Then, Speed upstream = (x – y) mph
Speed downstream = (x + y) mph

Distance = 12 miles

Time taken to travel upstream – Time taken to travel downstream = 6 hours

⇒12/(x−y)−12/(x+y)=6

⇒12(x+y)−12(x−y)=6(x*x−y*y)

⇒24y=6(x*x−y*y)

⇒4y= x*x−y*y

⇒x * x =(y* y +4y)⋯(Equation 1)

Now he doubles his speed. i.e., his new speed = 2x

Now, Speed upstream = (2x – y) mph

Speed downstream = (2x + y) mph

In this case, Time taken to travel upstream – Time taken to travel downstream = 1 hour

⇒12/(2x−y)−12/(2x+y)=1

⇒12(2x+y)−12(2x−y)=4*x* x –y* y

⇒24y=4*x* x –y* y

⇒4*x* x = y* y +24y⋯(Equation 2)

(Equation 1 × 4)⇒4x* x =4(y* y +4y)⋯(Equation 3)

(From Equation 2 and 3, we have)

y* y +24y=4(y* y +4y)⇒y* y +24y=4y* y +16y⇒3y* y =8y⇒3y=8

y=8/3 mphi.e., speed of the current = 8/3 mph=2*2/3 mph

ANALOGY LEVEL 1

 

When you draw an analogy between two things we compare them for the purpose of explanation. If a scientist says that earth’s forest functions as human lungs then we instantly draw an explanation that both lungs and trees intake important Elements from air. As far as SSC exam is concerned this is one of the trickiest section. We intend to comprehend it by solving as many different types of questions as is asked in papers. Presently in the level 1 exam a good aspirant must be able to solve at least 7 out of 10 questions given.

 

In each of the following questions, select the related word/number from the given alternative:

 

  1. Flow : River :: Stagnant : ?                                                                                                                  A. Rain              B. Stream             C. Pool                       D. Canal

 

 

  1. Ornithologist : Bird :: Archaeologist : ?                                                                                          A.Islands             B. Mediators                        C. Archaeology       D.Aquatic

 

 

  1. Peacock : India :: Bear : ?                                                                                                         A.Australia          B. America             C. Russia        D. England

 

 

  1. Given set: (3,7,15)                                                                                                                                A. 2,6,10     B. 4,8,18       C. 5,9,17       D. 7,12,19

 

 

  1. Given set: (63,49,35)                                                                                                                            A. 81,63,45                  B. 64,40,28              C. 72,40,24              D. 72,48,24

 

 

  1. 3 : 243 :: 5 : ?                                                                                                                                            A. 405               B. 465                   C. 3125                     D. 546

 

 

  1. 5 : 36 :: 6 : ?                                                                                                                                                A. 48               B. 50                 C. 49              D. 56

 

 

  1. TALE : LATE :: ? : CAFE                                                                                                                          A. FACE                 B. CAEF                  C. CAFA                     D. FEAC

 

 

  1. AZBY : DWEV :: HSIR : ?                                                                                                                   A. JQKO                B. KPOL                  C. KPLO                     D. KOLP

 

 

  1. DE : 45 :: BC : ?                                                                                                                                           A. 34              B. 23              C. 56              D. 43

 

 

 

 

SOLUTION TO ANALOGY LEVEL 1

 

 

 

  1. Answer: Option C

Explanation: As Water of a River flows similarly water of Pool is Stagnant. Answer & Explanation

 

 

  1. Answer: Option C

Explanation: As Ornithologist is a specialist of Birds similarly Archaeologist is a specialist of Archaeology.

 

 

  1. Answer: Option C

Explanation: As Peacock is the national bird of India, similarly Bear is the national animal of Russia.

 

 

  1. Ans. Option C

Explanation:  1st number+4 = 2nd number

2nd number+8= 3rd number

 

 

  1. Ans. Option A

Explanation:  63= 7*9

49= 7*7

35= 7*5

 

 

  1. Ans. Option C

Explanation: 3^5=243

5^5=3125

 

 

  1. Ans. Option A

Explanation: 13^2+13=182

18^2+18=342

19^2+19=380

 

 

  1. Ans. Option A

Explanation: The first and the third letters has been interchanged

 

 

  1. Ans. Option C

Explanation: Pairs of opposite letter

A&Z,  B&Y,  similarly H&S,  I&R,  K&P,  L&O

 

 

  1. Ans. Option B

Explanation: D=4,  E=5,  similarly  B=2,  C=3

Inflation & Control Mechanism

https://exam.pscnotes.com/inflation”>Inflation is a sustained increase in the general price level of goods and Services in an economy over a period of time. When the price level rises, each unit of currency buys fewer goods and services.It is the Percentage change in the value of the Wholesale Price Index (WPI) on a year-on year basis. It effectively measures the change in the prices of a basket of goods and services in a year. In India, inflation is calculated by taking the WPI as base.

Formula for calculating Inflation=

(WPI in month of current year-WPI in same month of previous year)
————————————————————————————– X 100
WPI in same month of previous year

Inflation occurs due to an imbalance between demand and supply of Money, changes in production and distribution cost or increase in taxes on products. When economy experiences inflation, i.e. when the price level of goods and services rises, the value of currency reduces. This means now each unit of currency buys fewer goods and services.

It has its worst impact on consumers. High prices of day-to-day goods make it difficult for consumers to afford even the basic commodities in life. This leaves them with no choice but to ask for higher incomes. Hence the government tries to keep inflation under control.

Contrary to its negative effects, a moderate level of inflation characterizes a good economy. An inflation rate of 2 or 3% is beneficial for an economy as it encourages people to buy more and borrow more, because during times of lower inflation, the level of interest rate also remains low. Hence the government as well as the central bank always strive to achieve a limited level of inflation.

Various measures of Inflation are:-

There are following types on Inflation based on their causes:-

  • Demand pull inflation
  • cost push inflation
  • structural inflation
  • speculation
  • cartelization
  • hoarding

Various control measures to curb rising inflation are:-

  • Fiscal measures like reduction in indirect taxes
  • Dual pricing
  • Monetary measures
  • Supply side measures like importing the shortage goods to meet the demand
  • Administrative measures to curb hoarding, Cratelization.

 

 

 

 

 

 

 

Tax Reforms in India

Sience 1990 ie the Liberalization of Indian economy saw the beginning of Taxation reforms in the nation. The taxation system in the nation has been subjected to consistent and comprehensive reform. Following factors arise the need for TAX REFORMS IN INDIA:-

  • Tax Resources must be maximized for increased social sector Investment in the economy.
  • International competitiveness must be imparted to Indian economy in the globalized world.
  • Transaction costs are high which must be reduced.
  • Investment flow should be maximized.
  • Equity should be improved
  • The high cost nature of Indian economy should be changed.
  • Compliance should be increased.

Direct & Indirect Tax Reforms

Direct Tax reforms undertaken by the government are as follows:-

  • Reduction and rationalization of tax rates, India now has three rates of Income tax with the highest being at 30%.
  • SIMPLIFICATION of process, through e-filling and simplifying the tax return forms.
  • Strengthening of administration to check the leakage and increasing the tax base.
  • Widening of tax base to include more tax payers in the tax net.
  • Withdrawal of tax exceptions gradually.
  • Minimum Alternate Tax (MAT) was introduced for the ‘Zero Tax’ companies.
  • The direct tax code of 2010 replace the outdated tax code of 1961.

Indirect tax reforms undertaken by the government are as follows:-

  • Reduction in the peak tariff rates.
  • reduction in the number of slabs
  • Progressive change from specific duty to ad valor-em tax.
  • VAT is introduced.
  • GST has been planned to be introduced.
  • Negative list of services since 2012.

Subsidies- Cash Transfer of Subsidy Issue.

A subsidy is a benefit given by the government to groups or individuals usually in the form of a cash payment or tax reduction. The subsidy is usually given to remove some type of burden and is often considered to be in the interest of the public.

Direct Cash Transfer Scheme is a POVERTY reduction measure in which government subsidies and other benefits are given directly to the poor in cash rather than in the form of subsidies.

It can help the government reach out to identified beneficiaries and can plug leakages. Currently, ration shop owners divert subsidised PDS grains or kerosene to open market and make fast buck. Such Leakages could stop. The scheme will also enhance efficiency of welfare schemes.

The money is directly transferred into bank accounts of beneficiaries. LPG and kerosene subsidies, pension payments, scholarships and EMPLOYMENT guarantee scheme payments as well as benefits under other government welfare programmes will be made directly to beneficiaries. The money can then be used to buy services from the market. For eg. if subsidy on LPG or kerosene is abolished and the government still wants to give the subsidy to the poor, the subsidy portion will be transferred as cash into the banks of the intended beneficiaries.

It is feared that the money may not be used for the intended purpose and men may squander it.

Electronic Benefit Transfer (EBT) has already begun on a pilot basis in Andhra Pradesh, Chhattisgarh, Punjab, Rajasthan, Tamil Nadu, West Bengal, Karnataka, Pondicherry and Sikkim. The government claims the results are encouraging.

Only Aadhar card holders will get cash transfer. As of today, only 21 crore of the 120 crore people have Aadhar cards. Two other drawbacks are that most BPL families don’t have bank accounts and several villages don’t have any bank branches. These factors can limit the reach of cash transfer.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Recent Trends in Indian Economy: Role of Foreign Capital, FDI

FDI refers to capital inflows from abroad that are invested in or to enhance the production capacity of the economy. Despite Globalization, the essential role of Foreign Direct Investment (FDI) in Economic Development has not changed.

Foreign Direct Investment (FDI) plays an important role in global business. It can provide a firm with new Marketing channels, cheaper production facilities, access to technology transfer, product, skills and financing. With the advent of globalization and strong governmental support, foreign investment has helped the Indian economy grow tremendously. India has continuously sought to attract investment from the world’s major investors. In 1998 and 1999, the Government of India announced a number of reforms designed to encourage and promote a favorable business Environment for investors. Foreign investments in the country can take in the form of investments in listed companies i.e., Foreign Institutional Investors’(FIIs) investments, investments in listed/unlisted companies other than through Stock Exchanges i.e., through the foreign direct investment or Private Equity/foreign Venture Capital investment route, investments through American Depository Receipts (ADR), Global Depository Receipts (GDR), or investments by Non-Resident Indians (NRIs) and Persons of Indian Origin (PIOs) in various forms.

ROUTES OF FOREIGN INVESTMENT INFLOW
DIRECT INVESTMENT
I) Equity
(a) Government (SIA/FIPB)
(b) RBI
(c) NRI
(d) Acquisition of Shares
(e) Equity capital of
(f) Unincorporated bodies
II) Re-invest Dearing
III) Other capital
INDRECT INVESTMENT
(I) GDRs/ADRs
(II) FIIs
(III) off-shore funds and others

Main advantages of FDI are:-
1. Inflow of Foreign Capital. Capital base of domestic country increases.
2. Increase in tax revenue.
3. Boost economy by GDP Growth.
4. Increase competition, productivity and efficiency.
5. Large employment opportunities -FDI in retail will create lakhs of jobs.
6. Inflow of technology, expertise and know how.
7. Infrastructure facilities improve and it will bring growth and prosperity.
8. Reduce cost of production. Prices of products will come down. This will tame inflationary pressure in the economy.
9. Increase in international trade.
10. High quality products that will help them develop local businesses and industries.
11. Decrease in food wastage: Today a major chunk of the food that is almost 30%, 40% of the produce is wasted in transportation. A lot of grains are also wasted in the government storage and go-downs. The government has made it compulsory to invest 50% of the investment in the development of infrastructure in Logistics. Thus it will become critical to save a lot in storage and logistics. More investments in the end to end supply chain and world class cold storage facilities.
12. Benefits to the farmers: Farmers were long been left behind and squeezed between the price raise. Worldwide the big retail giants buy the produce directly from the farmers eliminating the middle men and offering them at least 15% – 20% higher prices then they get.
13. Increase in Forex reserves: As per Government’s proposal in increasing the FDI in retail the each retail giant is supposed to invest a minimum of 100 million dollars. Each retail giant is expected to open atleast 15 stores across India and to open each
14. Better consumer choice: Since most of the retail giants work on a large scale, they have large number product varieties which generally the kirana stores in your neighbourhood are not able to store. Better Options and offers to the consumer.
15. Reduction in food inflation: The increase in FDI will create stronger competition among the retailers and will eliminate the middle man, which will eventually help in reducing food prices and the stocks will help in reducing the supply constraint.
16. Increase in economic growth by dealing in various international products.
17. Billion dollars will be invested in Indian retail market.
18. FDI in DEFENCE sector will reduce imports; improve country’s capacity to produce defence equipment locally and save foreign money. Definitely, it will create employment opportunities. It will give them a hope that Indian defence equipment will become globally competitive. High technology and expertise will flow to the country.

 

Multinational Corporations

MNC may be defined as a company, which operates in number of countries and has production and service facilities outside the country of its origin. They are also called Trans National Company (TNC) Their activities have both good and bad impacts on the economy. They take decisions on a global context or basis. Their maximum profit objectives take no account of the reactions produced in the countries felling in their orbit. They operate in different institutional forms Some are: Subsidiaries companies wholly owned by MNC in other countries Subsidiary company enter into joint venture with a company another company Agreement among companies of different countries regarding production and discussion of market.

Role of MNC’s

1. Promotion of Foreign Investment:

MNCs can bridge the gap between the requirements of foreign capital for increasing foreign investment in India.The liberalized foreign investment pursued since 1991, allows MNCs to make investment in India subject to different ceilings fixed for different industries or projects.

 

2. Non-Debt Creating Capital inflows:

The direct foreign investment by multinational corporations represents non-debt creating capital inflows we can avoid the liability of debt-servicing payments. Moreover, the advantage of investment by MNCs lies in the fact that servicing of non-debt capital begins only when the MNC firm reaches the stage of making profits to repatriate Thus, MNCs can play an important role in reducing Stress strains and on India’s Balance of Payments (BOP).

3. Technology Transfer:

 

Transfer high sophisticated technology to developing countries which are essential for raising productivity of working class and enable us to start new productive ventures requiring high technology is possible due to mnc’s. Whenever, multinational firms set up their subsidiary production units or joint-venture units, they not only import new equipment and machinery embodying new technology but also skills and technical know-how to use the new equipment and machinery.

4. Promotion of Exports:

With extensive links all over the world and producing products efficiently and therefore with lower costs multinationals can play a significant role in promoting exports of a country in which they invest.

5. Investment in Infrastructure:

With a large command over financial resources and their superior ability to raise resources both globally and inside India it is said that multinational corporations could invest in infrastructure such as power projects, modernisation of Airports and posts, Telecommunication.

The investment in infrastructure will give a boost to industrial growth and help in creating income and employment in the India economy. The external economies generated by investment in infrastructure by MNCs will therefore crowd in investment by the indigenous private sector and will therefore stimulate economic growth.

 

 

 

 

 

 

 

Food Security & Public Distribution System(PDS)

WHO Defines Food security to exists when all people, at all times, have physical, social and economic access to sufficient, safe and nutritious food which meets their dietary needs and food preferences for an active and healthy life.
Food security has three interlinked contents such as :-

  1. Availability of food,
  2. Access to food and
  3. absorption of food.

Food security is a multidimensional concept covering even the  micro level household food security,energy intakes and indicators of Malnutrition.

 

Major components of food security are:-

  1. Production and Procurement
  2. Storage
  3. Distribution

Agriculture-notes-for-state-psc-exams”>Agriculture“>Indian Agriculture is rightly called as a gamble with Monsoon, variability in food production and rising Population creates food insecurity in the nation and worst effected are the downtrodden section of the Society.

While India has seen impressive economic growth in recent years, the country still struggles with widespread poverty and hunger. India’s poor population amounts to more than 300 million people, with almost 30 percent of India’s rural population living in poverty. The good news is, poverty has been on the decline in recent years. According to official government of India estimates, poverty declined from 37.2% in 2004-05 to 29.8% in 2009-10.

Need for Self-Sufficiency:

India suffered two very severe droughts in 1965 and 1966. Food Aid to India was restricted to a monthly basis by USA under the P.L. 480 programme.  The Green Revolution made a significant change in the scene. India achieved self-sufficiency in food grains by the year 1976 through the implementation of the seed- water-fertilizer policy adopted by the Government of India.

Food grain production increased four-fold during 1950-51 and 2001-2002 from 51 million tons to 212 million tones. The country is no longer exposed to real famines. But the regional variation in the success of Green Revolution which was chiefly limited to northern- Western states has lead to the divide in the nation. Evergreen revoloution and Bringing green revolution to eastern India is the need of the hour.

Green revolution was focused on wheat and rice and thus the production of pulses was stagnant.

National Food Security Mission comprising rice, wheat and pulses to increase the production of rice by 10 million tons, wheat by 8 million tons and pulses by 2 million tons by the end of the Eleventh Plan (2011-12). The Mission is being continued during 12th Five Year Plan with new targets of additional production of food grains of 25 million tons of food grains comprising of 10 million tons rice, 8 million tons of wheat, 4 million tons of pulses and 3 million tons of coarse Cereals by the end of 12th Five Year Plan.
The National Food Security Mission (NFSM) during the 12th Five Year Plan will have five components

(i) NFSM- Rice;

(ii) NFSM-Wheat;

(iii) NFSM-Pulses,

(iv) NFSM-Coarse cereals and

(v) NFSM-Commercial Crops.

Government through Public Distribution System has tried to counter the problem of food insecurity by providing the food grains through fair price shops.

The central Government through Food Corporation of India has assumed the responsibilities of  procurement,storage,transfer and bulk allocation of food grains to state governments.

The public distribution system (PDS) has played an important role in attaining higher levels of the household food security and completely eliminating the threats of famines from the face of the country, it will be in the fitness of things that its evolution, working and efficacy are examined in some details.

PDS was initiated as a deliberate social policy of the government with the objectives of:

  1. i) Providing foodgrains and other essential items to vulnerable sections of the society at resonable (subsidised) prices;
  2. ii) to have a moderating influence on the open market prices of cereals, the distribution of which constitutes a fairly big share of the total marketable surplus; and

iii) to attempt socialisation in the matter of distribution of essential commodities.

 

The focus of the Targeted Public Distribution System (TPDS) is on “poor in all areas” and TPDS involves issue of     35 Kg of food grains per family per month for the population Below Poverty Line (BPL) at specially subsidized prices. The TPDS requires the states to Formulate and implement :-

  1. foolproof arrangements for identification of poor,
  2. Effective delivery of food grains to Fair Price Shops (FPSs)
  3. Its distribution in a transparent and accountable manner at the FPS level.

 

Export Import (EXIM) Policy  of India

 

Export Import Policy or  Exim Policy or Foreign Trade Policy is a set of guidelines and instructions related to the import and export of goods.

Various Objectives of Exim Policy are :-

  • To facilitate sustained growth in exports from India and import in India.
  • To stimulate sustained economic growth by providing access to essential raw materials, intermediates, components, consumables and Capital Goods scheme required for augmenting production and providing services.
  • To enhance the technological strength and efficiency of Industry Agriculture industry and services, thereby improving their competitive strength while generating new employment opportunities, and to encourage the attainment of internationally accepted standards of quality.
  • To provide clients with high-quality goods and services at globally competitive rates. Canalization is an important feature of Exim Policy under which certain goods can be imported only by designated agencies. For an example, an item like gold, in bulk, can be imported only by specified banks like SBI and some Foreign Banks or designated agencies.

The new five year Foreign Trade Policy, 2015-2020 provides a framework for increasing exports of goods and services as well as generation of employment and increasing value addition in the country, in keeping with the “Make in India” vision of our Hon’blc Prime Minister. The focus of the government is to support both the manufacturing and services sectors, with a special emphasis on improving the ‘Ease of Doing Business‘.

Merchandise Exports from India Scheme (MEIS):-To offset infrastructural inefficiencies and the associated costs of exporting products produced in India giving special emphasis on those which are of India’s export interest and have the capability to generate employment and enhance India’s competitiveness in the world market.With the aim in making India’s products more competitive in the global markets, the scheme provides incentive in the form of duty credit scrip to the exporter to compensate for his loss on payment of duties.

Service Exports from India Scheme (SEIS) :-Service Provider of eligible services shall be entitled to Duty Credit Scrips at notified rates.

Export Promotion Capital Goods (EPCG) scheme allows import of capital goods including spares for pre production, production and post production at zero duty.

Other Specific steps taken for the developement of international trade are:-

 

  • Trade Facilitation & Ease Of Doing Business
  • DGFT as a facilitator of exports/imports
  • Niryat Bandhu – Hand Holding Scheme for new export / import entrepreneurs
  • Online Complaint Registration and Citizen’s Charter
  • Monitoring System
  • Issue of e-IEC (Electronic-Importer Exporter Code)
  • e-BRC
  • MoU with State Governments for sharing of e-BRC data
  • Exporter Importer Profile
  • Reduction in mandatory documents required for Export and Import
  • Online Inter-ministerial consultation
  • Facility of online filing of applications
  • Facility to upload documents by Chartered Accountant / Company Secretary / Cost Accountant
  • Electronic Data Interchange (EDI)
  • Message Exchange with Community partners
    (a) Message Exchange with Customs
    (b) Message Exchange with eBiz
    (c) Message Exchange with Banks
    (d) Message Exchange with EPCs
  • Encouraging development of Third Party API
  • Forthcoming e-Governance Initiatives
  • Free passage of Export consignment
  •  No seizure of export related Stock
  • 24 X 7 Customs clearance
  • Single Window in Customs
  • Self-Assessment of Customs Duty
  • Authorised Economic Operator (AEO) Programme
  • Prior filing facility for Shipping Bills
  • Cutting down delay in filing of Export General Manifest (EGM) for duty drawback
  • Facility of Common Bond / LUT against authorizations issued under different EP Schemes
  • Exemption from Service Tax on Services received abroad
  • Export of perishable agricultural Products
  • Time Release Study (TRS)
  • Towns of Export Excellence (TEE)

 

 

12 Finance Commission

The Twelfth Finance Commission  was appointed under the chairmanship of C. Rangarajan on November 1, 2002 to make recommendations regarding the distribution between the Union and the States of net proceeds of shareable taxes, the principles which should govern the grants- in-aid of the revenues of States from the Consolidated Fund of India and the measures needed to augment the Consolidated Fund of a State to supplement the resources of local bodies in the State on the basis of the recommendations made by the Finance Commission of the State.

 

Recommendations of the Twelfth Finance Commission

Restructuring public finances

  • Centre and States to improve the combined tax-GDP ratio to 17.6 per cent by 2009-10.
  • Combined debt-GDP ratio, with External Debt measured at historical exchange rates, to be brought down to 75 percent by 2009-10.
  • Fiscal Deficit to GDP targets for the Centre and States to be fixed at 3 per cent.
  • Revenue Deficit of the Centre and States to be brought down to zero by 2008-09.
  • Interest payments relative to Revenue Receipts to be brought down to 28 per cent and 15 per cent in the case of the Centre and States, respectively.
  • States to follow a recruitment policy in a manner so that the total salary bill, relative to Revenue Expenditure, net of interest payments, does not exceed 35 per cent.
  • Each State to enact a fiscal responsibility legislation providing for elimination of revenue deficit by 2008-09 and reducing fiscal deficit to 3 per cent of State Domestic Product.
  • The system of on-lending to be brought to an end over time. The long term goal should be to bring down debt-GDP ratio to 28 per cent each for the Centre and the States.

Sharing of Union tax revenues

  •  The share of States in the net proceeds of shareable Central taxes fixed at 30.5 per cent, treating additional excise duties in lieu of sales tax as part of the general pool of Central taxes. Share of States to come down to 29.5 per , when States are allowed to levy sales tax on sugar, textiles and tobacco.
  • In case of any legislation enacted in respect of service tax, after the notification of the eighty eighth amendment to the Constitution, revenue accruing to a State should not be less than the share that would accrue to it, had the entire service tax proceeds been part of the shareable pool.
  • The indicative amount of overall transfers to States to be fixed at 38 per cent of the Centre’s gross revenue receipts.

Local bodies

  • A grant of Rs.20,000 crore for the Panchayati Raj institutions and Rs.5,000 crore for Urban Local Bodies to be given to States for the period 2005-10.
  • Priority to be given to expenditure on operation and maintenance (O&M) costs of Water supply and sanitation, while utilizing the grants for the Panchayats. At least 50 per cent of the grants recommended for urban local bodies to be earmarked for the scheme of solid Waste Management through public-private PARTNERSHIP.

Calamity relief

  •  The scheme of Calamity Relief Fund (CRF) to continue in its present form with contributions from the Centre and States in the ratio of 75:25. The size of the Fund worked out at Rs.21,333 crore for the period 2005-10.
    The outgo from the Fund to be replenished by way of collection of National Calamity Contingent Duty and levy of special surcharges.
  • The definition of natural calamity to include landslides, avalanches, cloud burst and pest attacks.
    Provision for disaster preparedness and mitigation to be part of State Plans and not calamity relief.

Grants-in-aid to States

  •  The present system of Central assistance for State Plans, comprising grant and loan components, to be done away with, and the Centre should confine itself to extending plan grants and leaving it to States to decide their borrowings.
  • Non-plan revenue deficit grant of Rs.56,856 crore recommended to 15 States for the period 2005-10. Grants amounting to Rs.10,172 crore recommended for the Education-sector”>Education sector to eight States. Grants amounting to Rs.5,887 crore recommended for the Health-sector”>Health sector for seven States. Grants to education and health sectors are additionalities over and above the normal expenditure to be incurred by States.
  • A grant of Rs.15,000 crore recommended for roads and bridges, which is in addition to the normal expenditure of States.
  • Grants recommended for maintenance of public buildings, forests, heritage conservation and specific needs of States are Rs. 500 crore, Rs.1,000 crore, Rs.625 crore, and Rs.7,100 crore, respectively.

Fiscal reform facility

  •  With the recommended scheme of debt relief in place, fiscal reform facility not to continue over the period 2005-10.

Debt relief and corrective measures

  •  Central loans to States contracted till March,2004 and outstanding on March 31, 2005 amounting to Rs.1,28,795 crore to be consolidated and rescheduled for a fresh term of 20 years, and an interest rate of 7.5 per cent to be charged on them. This is subject to enactment of fiscal responsibility legislation by a State.
  • A debt write-off scheme linked to reduction of revenue deficit of States to be introduced. Under this scheme,
    repayments due from 2005-06 to 2009-10 on Central loans contracted up to March 31,2004 will be eligible for write- off.
  • Central Government not to act as an intermediary for future lending to States, except in the case of weak States,
    which are unable to raise funds from the market.
  • External assistance to be transferred to States on the same terms and conditions as attached to such assistance by external funding agencies.
  • All the States to set up sinking funds for amortization of all loans.
  • States to set up guarantee redemption funds through earmarked guarantee fees.

Others

  •  The Centre should share ‘profit petroleum’ from New Exploration and Licensing Policy (NELP) areas in the ratio of 50:50 with States where mineral oil and natural gas are produced. No sharing of profits in respect of nomination fields and non-NELP blocks.
  • Every State to set up a high level committee to monitor the utilization of grants recommended by the TFC.
    Centre to gradually move towards accrual basis of accounting.

Source:Ministry of Finance

 

 

 

Poverty Alleviation Schemes

  • Poverty alleviation programmes can be in form of employment generation programmes or social assistance programmes so that different dimensions of poverty are addressed.
  • At present there are three centrally sponsored employment programmes in operation
    • MNREGS: Rural, wage employment
    • SGSY: Rural, self-employment
    • SJSRY: Urban, self and wage employment
  • MNREGS
    • 2006
    • Launched in 200 most backward districts in the first phase. At present 619 districts are covered under the NREGS
    • During 2008-09, 4.51 crore households were provided employment under the scheme
  • Swarnajayanti Gram Swarozgar Yojana
    • 1999 after restructuring the Integrated Rural Development Programme (IRDP) and allied programmes, viz., Development of Women and Children in Rural Areas (DWCRA), Training of Rural Youth for Self-Employment (TRYSEM), Supply of improved tool-kits to rural artisans (SITRA), Ganga Kalyan Yojana (GKY) and Million Wells Scheme (MWS)
    • Self-employment programme for rural poor
    • Objective is to bring the assisted swarozgaris above the poverty line by providing them income generating assets through bank credit and government subsidy
    • Centre: State – 75:25; 90:10 for NE states
  • Swarna Jayanti Shahari Rozgar Yojana (SJSRY)
    • It is a unified centrally sponsored scheme launched a fresh in lieu of the erstwhile Urban Poverty alleviation programmes, viz, Nehru Rozgar Yojana (NRY), PM’s Integrated Urban Poverty Eradication Programme (PMIUPEP), and Urban Basic Services for the Poor (UBSP)
    • Revamped in 2009
    • Self-employment + Wage employment

 

  • The revamped SJSRY has 5 components
    • Urban Self-Employment Programme (USEP)
    • Urban Woman Self-help Programme (UWSP)
    • Skill Training for Employment Promotion amongst urban poor (STEP-UP)
    • Urban Wage Employment Programme (UWEP)
  • Pradhan Mantri Jan Dhan Yojana (PMJDY):-National Mission for Financial Inclusion to ensure access to financial services, namely, Banking/ Savings & Deposit Accounts, Remittance, Credit, Insurance, Pension in an affordable manner.
  • Pradhan Mantri Sukanya Samriddhi Yojana (PMSSY)
  • Pradhan Mantri MUDRA Yojana (PMMY):-To create an inclusive, sustainable and value based entrepreneurial culture, in collaboration with our partner institutions in achieving economic success and financial security.
  • Pradhan Mantri Jeevan Jyoti Bima Yojana (PMJJBY)
  • Pradhan Mantri Suraksha Bima Yojana (PMSBY)
  • Atal Pension Yojana (APY)
  • Pradhan Mantri Fasal Bima Yojana (PMFBY)
  • Pradhan Mantri Gram Sinchai Yojana (PMGSY)
  • Pradhan Mantri Garib Kalyan Yojanaye (PMGKY)
  • Pradhan Mantri Jan Aushadhi Yojana (PMJAY)
  • Deen Dayal Upadhyaya Grameen Kaushalya Yojana (DDUGKY)
  • Pradhan Mantri Ujjwala Yojana
  • Rajasthan Mission on Skill and Livelihoods
  • End to end computerization of PDS
  • Bhamashah Yojana
  • Primary Health Centre (PHC) Scheme

 

 

Role of World Bank, IMF WTO & other Important International Organisations in world Economy

World Bank

The International Bank for Reconstruction and Development (IBRD), commonly referred to as the World Bank, is an international financial institution whose purposes include assisting the development of its member nation’s territories, promoting and supplementing private foreign investment and promoting long-range balance growth in international trade.

The World Bank was established in December 1945 at the United Nations Monetary and Financial Conference in Bretton Woods, New Hampshire. It opened for business in June 1946 and helped in the reconstruction of nations devastated by World War II. Since 1960s the World Bank has shifted its focus from the advanced industrialized nations to developing third-world countries.

Organization and Structure:

The organization of the bank consists of the Board of Governors, the Board of Executive Directors and the Advisory Committee, the Loan Committee and the president and other staff members. All the powers of the bank are vested in the Board of Governors which is the supreme policy making body of the bank.

Capital Resources of World Bank:

The initial authorized capital of the World Bank was $ 10,000 million, which was divided in 1 lakh shares of $ 1 lakh each. The authorized capital of the Bank has been increased from time to time with the approval of member countries.Member countries repay the share amount to the World Bank in the following ways:

  1. 2% of allotted share are repaid in gold, US dollar or Special Drawing Rights (SDR).
  2. Every member country is free to repay 18% of its capital share in its own currency.
  3. The remaining 80% share deposited by the member country only on demand by the World Bank.

Objectives:

The following objectives are assigned by the World Bank:

 

  1. To provide long-run capital to member countries for economic reconstruction and development.

 

  1. To induce long-run capital investment for assuring Balance of Payments (BoP) equilibrium and balanced development of international trade.

 

  1. To provide guarantee for loans granted to small and large units and other projects of member countries.

 

  1. To ensure the implementation of development projects so as to bring about a smooth transference from a war-time to peace economy.

 

  1. To promote capital investment in member countries by the following ways;

 

(a) To provide guarantee on private loans or capital investment.

 

(b) If private capital is not available even after providing guarantee, then IBRD provides loans for productive activities on considerate conditions.

 

Functions:

 

World Bank is playing main role of providing loans for development works to member countries, especially to underdeveloped countries. The World Bank provides long-term loans for various development projects of 5 to 20 years duration.

 

The main functions can be explained with the help of the following points:

 

  1. World Bank provides various technical services to the member countries. For this purpose, the Bank has established “The Economic Development Institute” and a Staff College in Washington.

 

  1. Bank can grant loans to a member country up to 20% of its share in the paid-up capital.

 

  1. The quantities of loans, interest rate and terms and conditions are determined by the Bank itself.

 

  1. Generally, Bank grants loans for a particular project duly submitted to the Bank by the member country.

 

  1. The debtor nation has to repay either in reserve currencies or in the currency in which the loan was sanctioned.

 

  1. Bank also provides loan to private investors belonging to member countries on its own guarantee, but for this loan private investors have to seek prior permission from those counties where this amount will be collected.

The major roles of the International Monetary Fund are as follows:

  1. To promote international monetary cooperation through a permanent institution which provides the machinery for consultation and collaboration on international monetary problems.
  2. To facilitate the expansion and balanced growth of international trade, and to contribute thereby to the promotion and maintenance of high levels of employment and real income and to the development of the productive resources of all members as primary objectives of economic policy.
  3. To promote exchange stability, to maintain orderly exchange arrangements among members, and to avoid competitive exchange depreciation.
  4. To assist in the establishment of a multilateral system of payments in respect of current transactions between members and in the elimination of Foreign Exchange restrictions which hamper the growth of world trade.
  5. To give confidence to members by making the general resources of the Fund temporarily available to them under adequate safeguards, thus providing them with opportunity to correct maladjustments in their balance of payments without resorting to measures destructive of national or international prosperity.
  6. In accordance with the above, to shorten the duration and lessen the degree of disequilibrium in the international balances of payments of members.“Articles of Agreement: Article I—Purposes,” International Monetary Fund

World Trade Organization(WTO

The important Objectives of WTO are:

  1. To improve the standard of living of people in the member countries.
  2. To ensure full employment and broad increase in effective demand.
  3. To enlarge production and trade of goods.
  4. To increase the trade of services.
  5. To ensure optimum utilization of world resources.
  6. To protect the environment.
  7. To accept the concept of Sustainable Development.

Functions:

The main functions of WTO are discussed below:

  1. To implement rules and provisions related to trade policy review mechanism.
  2. To provide a platform to member countries to decide future strategies related to trade and tariff.
  3. To provide facilities for implementation, administration and operation of multilateral and bilateral agreements of the world trade.
  4. To administer the rules and processes related to dispute settlement.
  5. To ensure the optimum use of world resources.
  6. To assist international organizations such as, IMF and IBRD for establishing coherence in Universal Economic Policy determination.

 

 

 

Concept of Developing, Emerging and Developed countries.

In 1978, the World Bank, for the first time, constructed an analytical country Classification system. The occasion was the launch of the World Development Report. Annexed to the report was a set of World Development Indicators (WDI), which provided the statistical underpinning for the analysis. The first economic classification in the 1978 WDI divided countries into three categories: (1) developing countries, (2) industrialized countries, and (3) capital-surplus oil-exporting countries. Developing countries were categorized as low- income (with GNI/n of US$250 or less) and middle-income (with GNI/n above US$250).

Major Characteristics of Developing Countries are:-

  1.  Lower per-capita income
  2.  Low levels of Human Capital
  3. High levels of poverty and under-Nutrition
  4. Higher Population Growth rates
  5. Predominance of agriculture and low levels of industrialization
  6. Low level of Urbanization but rapid rural-to-urban Migration
  7. Dominance of informal sector
  8. Underdeveloped labor, financial, and other markets.

Major Characteristics of Emerging Countries are:-

  1. the small size of the economy,
  2. GNP/Capita much lower than in developed countries,
  3. a reduced opening for accepting foreign investors,
  4. a high volatility of the exchange rate which implies greater risk in trading.

Major Characteristics of Developed Countries are:-

  1. Average income per capita of the population is generally high.
  2. Education level of high average population.
  3. Life expectancy of the population average height.
  4.  Population growth rate per year is relatively small.
  5. The death rate per year is relatively small population.
  6. Life-style market economy.
  7. His wide and varied field.
  8. Economic activity in most industry sectors, as well as export commodities.
  9. The majority of the population lives in cities.
  10. Relatively high level of population health.

 

 

 

 

Indian Economy in global Scenario

The global macroeconomic landscape is currently chartering a rough and uncertain terrain characterized by weak growth of world output. The situation has been exacerbated by;
(i) declining prices of a number of commodities, with reduction in crude oil prices being the most visible of them,
(ii) turbulent fnancial markets (more so equity markets), and
(iii) volatile exchange rates.

These conditions refect extreme risk-aversion behaviour of global investors, thus putting many, and in particular, commodities exporting economies under considerable stress.

Even in these trying and uncertain circumstances, India’s growth story has largely remained positive on the strength of domestic absorption, and the country has registered a robust and steady pace of economic growth in 2015-16 as it did in 2014-15. Additionally, its other macroeconomic parameters like infation, fscal defcit and Current Account balance have exhibited distinct signs of improvement. Wholesale price infation has been in negative territory for more than a year and the all-important consumer prices infation has declined to nearly half of what it was a few years ago.

However, weak growth in advanced and emerging economies has taken its toll on India’s exports. As imports have also declined, principally on account of reduced prices of crude oil for which the country is heavily dependent on imports, trade and current account defcits continue to be moderate. Growth in agriculture has slackened due to two successive years of less-than-normal monsoon rains. Saving and investment rates are showing hardly any signs of revival. The rupee has depreciated vis-à-vis the US dollar, like most other currencies in the world, although less so in magnitude. At the same time, it has appreciated against a number of other major currencies. Given the fact that the government is committed to carrying the reform process forward, aided by the prevailing macroeconomic stability, it appears that conditions do exist for raising the economy’s growth momentum and achieving growth rates of 8 per cent or higher in the next couple of years.