Aptitude & reasoning

 

Pipes and Cistern

 

1. Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

 

Outlet:

A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

 

2. If a pipe can fill a tank in xhours, then:

part filled in 1 hour =	1	.
	x

3. If a pipe can empty a tank in yhours, then:

part emptied in 1 hour =	1	.
	y

4. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then

the net part filled in 1 hour =		1	–	1		.
		x		y

5. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then

the net part emptied in 1 hour =		1	–	1		.
		y		x

 

 

Questions:

 

Level-I:

 

1.	Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
	A. 5 11 B. 6 11 C. 7 11 D. 8 11

 

2.	Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
	A. 1 13 hours 17 B. 2 8 hours 11 C. 3 9 hours 17 D. 4 1 hours 2

 

3.	A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
	A. 4 1 hours 3 B. 7 hours C. 8 hours D. 14 hours
4.	Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
	A. 5 min. B. 9 min. C. 10 min. D. 15 min.

 

5.	A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
	A. 6 hours B. 10 hours C. 15 hours D. 30 hours
6.	Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
	A. 60 gallons B. 100 gallons C. 120 gallons D. 180 gallons

 

7.	A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
	A. 20 hours B. 25 hours C. 35 hours D. Cannot be determined E. None of these

 

8.	Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
	A. 1 hour B. 2 hours C. 6 hours D. 8 hours

 

9.	Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?
	A. 12 min B. 15 min C. 25 min D. 50 min

 

10.	Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
	A. 10 min. 20 sec. B. 11 min. 45 sec. C. 12 min. 30 sec. D. 14 min. 40 sec.
11.	Level-II:   One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
	A. 81 min. B. 108 min. C. 144 min. D. 192 min.

 

12.	A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
	A. 15 min B. 20 min C. 27.5 min D. 30 min

 

13.	A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
	A. 3 hrs 15 min B. 3 hrs 45 min C. 4 hrs D. 4 hrs 15 min

 

14.	Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
	A. 6 hours B. 6 2 hours 3 C. 7 hours D. 7 1 hours 2

 

15.	Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
	A. 10 B. 12 C. 14 D. 16
16.	How much time will the leak take to empty the full cistern? I. The cistern is normally filled in 9 hours.  II. It takes one hour more than the usual time to fill the cistern because of la leak in the bottom.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

17.	How long will it take to empty the tank if both the inlet pipe A and the outlet pipe B are opened simultaneously? I. A can fill the tank in 16 minutes.  II. B can empty the full tank in 8 minutes.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

18.	If both the pipes are opened, how many hours will be taken to fill the tank? I. The capacity of the tank is 400 litres. II. The pipe A fills the tank in 4 hours.  III. The pipe B fills the tank in 6 hours.
	A. Only I and II B. Only II and III C. All I, II and III D. Any two of the three E. Even with all the three statements, answer cannot be given.

 

 

Answers:

 

Level-I:

 

Answer:1 Option B

 

Explanation:

Part filled by (A + B + C) in 3 minutes = 3		1	+	1	+	1		=		3 x	11		=	11	.
		30		20		10					60			20

 

Part filled by C in 3 minutes =	3	.
	10

 

Required ratio =		3	x	20		=	6	.
		10		11			11

 

Answer:2 Option C

 

Explanation:

Net part filled in 1 hour		1	+	1	–	1		=	17	.
		5		6		12			60

 

The tank will be full in	60	hours i.e., 3	9	hours.
	17		17

 

 

Answer:3 Option D

 

Explanation:

Work done by the leak in 1 hour =		1	–	3		=	1	.
		2		7			14

Leak will empty the tank in 14 hrs.

 

 

Answer:4 Option B

 

Explanation:

Let B be turned off after x minutes. Then,

Part filled by (A + B) in x min. + Part filled by A in (30 –x) min. = 1.

x		2	+	1		+ (30 – x).	2	= 1
		75		45			75

 

	11x	+	(60 -2x)	= 1
	225		75

11x + 180 – 6x = 225.

x = 9.

 

 

Answer:5 Option C

 

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x – 9) hours respectively to fill the tank.

	1	+	1	=	1
	x		(x – 5)		(x – 9)

 

	x – 5 + x	=	1
	x(x – 5)		(x – 9)

(2x – 5)(x – 9) = x(x – 5)

x² – 18x + 45 = 0

(x – 15)(x – 3) = 0

x = 15.    [neglecting x = 3]

 

 

Answer:6 Option C

 

Explanation:

Work done by the waste pipe in 1 minute =	1	–		1	+	1
	15			20		24

 

=		1	–	11
		15		120

 

= –	1	.    [-ve sign means emptying]
	40

 

Volume of	1	part = 3 gallons.
	40

Volume of whole = (3 x 40) gallons = 120 gallon

 

 

Answer:7 Option C

 

Explanation:

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take	x	and	x	hours respectively to fill the tank.
	2		4

 

	1	+	2	+	4	=	1
	x		x		x		5

 

	7	=	1
	x		5

x = 35 hrs.

 

Answer:8 Option C

 

Explanation:

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

	1	+	1	=	1
	x		(x + 6)		4

 

	x + 6 + x	=	1
	x(x + 6)		4

x² – 2x – 24 = 0

(x -6)(x + 4) = 0

x = 6.     [neglecting the negative value of x]

 

 

Answer:9 Option A

 

Explanation:

Part filled by A in 1 min =	1	.
	20

 

Part filled by B in 1 min =	1	.
	30

 

Part filled by (A + B) in 1 min =		1	+	1		=	1	.
		20		30			12

Both pipes can fill the tank in 12 minutes.

 

 

Answer:10 Option D

 

Explanation:

Part filled in 4 minutes = 4		1	+	1		=	7	.
		15		20			15

 

Remaining part =		1 –	7		=	8	.
			15			15

 

Part filled by B in 1 minute =	1
	20

 

	1	:	8	:: 1 : x
	20		15

 

x =		8	x 1 x 20		= 10	2	min = 10 min. 40 sec.
		15				3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

 

Level-II:

 

Answer:11 Option C

 

Explanation:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in	x	minutes.
	3

 

	1	+	3	=	1
	x		x		36

 

	4	=	1
	x		36

x = 144 min.

 

 

 

Answer:12 Option D

 

Explanation:

Part filled by (A + B) in 1 minute =		1	+	1		=	1	.
		60		40			24

Suppose the tank is filled in x minutes.

Then,	x		1	+	1		= 1
	2		24		40

 

	x	x	1	= 1
	2		15

x = 30 min.

 

Answer:13 Option B

 

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =		4 x	1		=	2	.
			6			3

 

Remaining part =		1 –	1		=	1	.
			2			2

 

	2	:	1	:: 1 : x
	3		2

 

x =		1	x 1 x	3		=	3	hours i.e., 45 mins.
		2		2			4

So, total time taken = 3 hrs. 45 mins.

 

Answer:14 Option C

 

Explanation:

(A + B)’s 1 hour’s work =		1	+	1		=	9	=	3	.
		12		15			60		20

 

(A + C)’s hour’s work =		1	+	1		=	8	=	2	.
		12		20			60		15

 

Part filled in 2 hrs =		3	+	2		=	17	.
		20		15			60

 

Part filled in 6 hrs =		3 x	17		=	17	.
			60			20

 

Remaining part =		1 –	17		=	3	.
			20			20

 

Now, it is the turn of A and B and	3	part is filled by A and B in 1 hour.
	20

Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

 

Answer:15 Option C

 

Explanation:

Part filled in 2 hours =	2	=	1
	6		3

 

Remaining part =		1 –	1		=	2	.
			3			3

 

(A + B)’s 7 hour’s work =	2
	3

 

(A + B)’s 1 hour’s work =	2
	21

C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }

=		1	–	2		=	1
		6		21			14

C alone can fill the tank in 14 hours.

 

Answer:16 Option E

 

Explanation:

1. Time taken to fill the cistern without leak = 9 hours.

Part of cistern filled without leak in 1 hour =	1
	9

1. Time taken to fill the cistern in presence of leak = 10 hours.

Net filling in 1 hour =	1
	10

 

Work done by leak in 1 hour =		1	–	1		=	1
		9		10			90

Leak will empty the full cistern in 90 hours.

Clearly, both I and II are necessary to answer the question.

Correct answer is (E).

 

 

 

 

Answer:17 Option E

 

Explanation:

I. A’s 1 minute’s filling work =	1
	16

 

II. B’s 1 minute’s filling work =	1
	8

 

(A + B)’s 1 minute’s emptying work =		1	–	1		=	1
		8		16			16

Tank will be emptied in 16 minutes.

Thus, both I and II are necessary to answer the question.

Correct answer is (E).

 

Answer:18 Option B

 

Explanation:

II. Part of the tank filled by A in 1 hour =	1
	4

 

III. Part of the tank filled by B in 1 hour =	1
	6

 

(A + B)’s 1 hour’s work =		1	+	1		=	5
		4		6			12

 

A and B will fill the tank in	12	hrs = 2 hrs 24 min.
	5

So, II and III are needed.

Correct answer is (B).

 

• 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

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

Hence the 2^nd 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%

 

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

 

2. Basic Trigonometric Values

 

θ in degrees	θ in radians	sinθ	cosθ	tanθ
0°	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

 

3. 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

 

4. 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

5. 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

6. 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)

7. 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 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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

9. Find the wrong number in the series

441, 484, 529, 566, 625

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

 

10. 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.

 

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

 

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

 

4. B. 81*3=243

64*3=192

25*3=75

But     16*4=64

 

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

 

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

 

7. 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

 

8. C. A+2=C    &   C+2=E

F+2=H     &   H+2=J

But      K+1=L     &   L+1=M

 

9. D. 21^2=441

22^2=484

23^2=529

25^2=625

But   (23.79)^2=566

 

10. D. 232+111=343

343+111=454

454+111=565 (but given 564)

 

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*(a−b) 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)(x−y))/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* x –y* 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* x –y* 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 11⁄2 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 1¹⁄₂ hour.

ie, distance travelled downstream in 1 hour = distance travelled upstream in 1¹⁄₂ 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 111⁄4 minutes and down the stream in 71⁄2minutes. 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 = 11¹⁄₄ 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 = 7¹⁄₂minutes = 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

 

 

 

Introduction:

There are four main directions – East, West, North and South as shown below:

 

 

 

 

There are four cardinal directions – North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:

 

 

 

Key points

 

1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.
2. At the time of sunset the shadow of an object is always in the east.
3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow

 

 

 

 

 

 

 

 

 

Practice Questions

Type 1:

Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?

Solution:

From third position it is clear he is 4 km from his house and is in North direction.

 

 

 

 

 

 

Type 2:

Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?

Solution:

 

Type 3:

One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli’s shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?

 

Solution: In the morning sunrises in the east.

So in morning the shadow falls towards the west.

Now Lalli’s shadow falls to the right of the Juhi. Hence Juhi is facing South.

 

 

 

 

Type 4: Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?

Solution:

From II it is clear that the road which goes to IT-Park is left to Hema.

 

 

 

 

 

 

 

 

Questions

 

Level-1

 

1.	One morning Udai and Vishal were talking to each other face to face at a crossing. If Vishal’s shadow was exactly to the left of Udai, which direction was Udai facing?
	A. East B. West C. North D. South

2.	Y is in the East of X which is in the North of Z. If P is in the South of Z, then in which direction of Y, is P?
	A. North B. South C. South-East D. None of these

3.	If South-East becomes North, North-East becomes West and so on. What will West become?
	A. North-East B. North-West C. South-East D. South-West

4.	A man walks 5 km toward south and then turns to the right. After walking 3 km he turns to the left and walks 5 km. Now in which direction is he from the starting place?
	A. West B. South C. North-East D. South-West

5.	Rahul put his timepiece on the table in such a way that at 6 P.M. hour hand points to North. In which direction the minute hand will point at 9.15 P.M. ?
	A. South-East   B. South   C. North D. West
6.	Rasik walked 20 m towards north. Then he turned right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Finally he turns left and walks 15 m. In which direction and how many metres is he from the starting position?
	A. 15 m West B. 30 m East C. 30 m West D. 45 m East

7.

Two cars start from the opposite places of a main road, 150 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?

A. 65 km B. 75 km C. 80 km D. 85 km

8.	Starting from the point X, Jayant walked 15 m towards west. He turned left and walked 20 m. He then turned left and walked 15 m. After this he turned to his right and walked 12 m. How far and in which directions is now Jayant from X?
	A. 32 m, South B. 47 m, East C. 42 m, North D. 27 m, South

 

9.	One evening before sunset Rekha and Hema were talking to each other face to face. If Hema’s shadow was exactly to the right of Hema, which direction was Rekha facing?
	A. North B. South C. East D. Data is inadequate

10.	A boy rode his bicycle Northward, then turned left and rode 1 km and again turned left and rode 2 km. He found himself 1 km west of his starting point. How far did he ride northward initially?
	A. 1 km B. 2 km C. 3 km D. 5 km

 

 

Answers:

1Answer: Option C

Explanation:

 

2Answer: Option D

Explanation:

P is in South-West of Y.

 

3Answer: Option C

Explanation:

It is clear from the diagrams that new name of West will become South-East.

 

4Answer: Option D

Explanation:

Hence required direction is South-West.

 

5Answer: Option D

Explanation:

At 9.15 P.M., the minute hand will point towards west.

 

6Answer: Option D

Explanation:

 

7Answer: Option A

Explanation:

 

 

 

 

8Answer: Option A

Explanation:

 

9Answer: Option B

Explanation:

In the evening sun sets in West. Hence then any shadow falls in the East. Since Hema’s shadow was to the right of Hema. Hence Rekha was facing towards South.

 

10Answer: Option B

Explanation:

The boy rode 2 km. Northward

 

 

Level – 2

 

Dev, Kumar, Nilesh, Ankur and Pintu are standing facing to the North in a playground such as given below:

1. Kumar is at 40 m to the right of Ankur.
2. Dev is are 60 m in the south of Kumar.
3. Nilesh is at a distance of 25 m in the west of Ankur.
4. Pintu is at a distance of 90 m in the North of Dev

 

 

1.	Which one is in the North-East of the person who is to the left of Kumar?
	A. Dev B. Nilesh C. Ankur D. Pintu

2.	If a boy starting from Nilesh, met to Ankur and then to Kumar and after this he to Dev and then to Pintu and whole the time he walked in a straight line, then how much total distance did he cover?
	A. 215 m B. 155 m C. 245 m D.   185 m

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U.
2. Q gets a North facing flat and is not next to S.
3. S and U get diagonally opposite flats.
4. R next to U, gets a south facing flat and T gets North facing flat.

 

 

3.	If the flats of P and T are interchanged then whose flat will be next to that of U?
	A. P B. Q C. R D. T

4.	Which of the following combination get south facing flats?
	A. QTS B. UPT C. URP D. Data is inadequate

5.	The flats of which of the other pair than SU, is diagonally opposite to each other?
	A. QP B. QR C. PT D. TS
6.	Whose flat is between Q and S?
	A. T B. U C. R D. P

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. 8-trees → mango, guava, papaya, pomegranate, lemon, banana, raspberry and apple are in two rows 4 in each facing North and South.
2. Lemon is between mango and apple but just opposite to guava.
3. Banana is at one end of a line and is just next in the right of guava or either banana tree is just after guava tree.
4. Raspberry tree which at one end of a line, is just diagonally opposite to mango tree.

 

 

	7 .Which of the following statements is definitely true?
	A. Papaya tree is just near to apple tree. B. Apple tree is just next to lemon tree. C. Raspberry tree is either left to Pomegranate or after. D. Pomegranate tree is diagonally opposite to banana tree.

8	Which tree is just opposite to raspberry tree?
	A. Papaya B. Pomegranate C. Papaya or Pomegranate D. Data is inadequate
9	Which tree is just opposite to banana tree?
	A. Mango B. Pomegranate C. Papaya D. Data is inadequate

 

 

Answer: 1 Option D

Explanation:

Ankur is in the left of Kumar. Hence Pintu is in North-East of Ankur

 

 

Answer: 2 Option A

Explanation:

Required distance = 25 m + 40 m + 60 m + 90 m

Required distance = 215 m

 

 

Answer:3 Option C

Explanation:

Hence flat R will be next to U.

 

 

Answer:4 Option C

 

Explanation:

Hence URP flat combination get south facing flats.

 

Answer:5 Option A

 

Explanation:

Hence QP is diagonally opposite to each other.

 

 

 

 

 

 

Answer:6 Option A

 

Explanation:

Hence flat T is between Q and S.

 

Answer: 7 Option B

 

Explanation:

 

 

Answer:8 Option C

 

Explanation:

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

 

Ratio

Introduction:

Ratio is the relation which one quantity bears to another of the same kind. The ratio of two quantities a and b is the fraction a/b and we write it as a: b.

In the ratio a: b, we call a as the first term or antecedent and b, the second term or consequent.

 

Note: The multiplication or division of each term of a ratio by the same non- zero number does not affect the ratio.

 

Compound Ratio: – It is obtained by multiplying together the numerators for new numerator and denominators for new denominator.

 

 

Example 1. If the ratios are 4:3, 15:20, 2:6 and 3:5 find the compound ratio?

 

 

 

 

 

 

 

 

Example2. If we divide 4185 into two parts such that they are in ratio 7:2, then find the values of both the parts?

Sol 2. Let the actual variable be 7x and 2x.

So, the 1^st part = 7 ×465=3255

The 2^nd part = 2 ×465=930

 

 

Note:

The ratio of first , second and third quantities is given by

ac : bc : bd

 

If the ratio between first and second quantity is a:b and third and fourth is c:d .

Similarly, the ratio of first, second, third and fourth quantities is given by
ace : bce : bde : bdf
If the ratio between first and second quantity is a: b and third and fourth is c:d.

 

                                                 Proportion

 
Introduction:-
Four quantities are said to be proportional if the two ratios are equal i.e.  the A, B, C and D are proportion. It is denoted by “::” it is written as A : B : C : D where A and D are extremes and B and C are called means .
                             Product of the extreme = Product of the means

 

 

Direct proportion: – The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also increases (or decreases).

Example 1. If 5 pens cost Rs 10 then 15 pen cost?

Sol 1. It is seen that if number of pens increases then cost also increases. So,

5 pens: 15 pens:: Rs 10 : required cost

 

 

Inverse proportion: – The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also decreases (or increases).

Example 2.If 10 men can do a work in 20 days then in how many days 20 men can do that work?

Sol 2. Here if men increase then days should decrease, so this is a case of inverse proportion, so

10 men: 20 men :: required days : 20 days

 

 

Rule of three: It Is the method of finding 4th term of a proportion if all the other three are given, if ratio is a:b :: c:d then ,

 

 

 

                                             ALLIGATION

Introduction:-

The word allegation means linking. It is used to find:

1. The proportion in which the ingredients of given price are mixed to produce a new mixture at a given price.
2. The mean or Average value of mixture when the price of the two or more ingredients and the proportion in which they are mixed are given.

Mathematical Formula:

 

For two ingredient:-

 

 

Example 1: If the rice at Rs 3.20 per kg and the rice at Rs 3.50 per kg be mixed then what should be their proportion so that the new mixture be worth Rs 3.35 per kg ?

Sol 1: CP of 1 kg of cheaper rice                          CP of 1 kg of dearer rice

Hence they must be mixed in equal proportion i.e. 1:1

 

 

Example 2: Find out the ratio of new mixture so that it will cost Rs 1.40 per kg from the given three kinds of rice costing Rs 1.20, Rs 1.45 and Rs 1.74?

 

Sol 2: 1^st rice cost = 120, 2^nd rice cost = 145 and 3^rd rice cost = 174 paisa.

From the above rule: we have,

Therefore, three rice must be mixed in 39: 20: 20 ratios to have a new mixture of rice.

 

 

Questions

Level-I

..	1.   A  and B together have Rs. 1210. If  of A’s amount is equal to  of B’s amount, how much amount does B have?
	A. Rs. 460 B. Rs. 484 C. Rs. 550 D. Rs. 664

 

2.	Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
	A. 2 : 5 B. 3 : 5 C. 4 : 5 D. 6 : 7

 

 

3.	A sum of https://exam.pscnotes.com/money”>Money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?
	A. Rs. 500 B. Rs. 1500 C. Rs. 2000 D. None of these

 

 

 

4.	Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
	A. 2 : 3 : 4 B. 6 : 7 : 8 C. 6 : 8 : 9 D. None of these

 

 

5.	In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:
	A. 20 litres B. 30 litres C. 40 litres D. 60 litres
6.	The ratio of the number of boys and girls in a college is 7 : 8. If the Percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
	A. 8 : 9 B. 17 : 18 C. 21 : 22 D. Cannot be determined

 

7.	Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit’s salary?
	A. Rs. 17,000 B. Rs. 20,000 C. Rs. 25,500 D. Rs. 38,000

 

8.	If 0.75 : x :: 5 : 8, then x is equal to:
	A. 1.12 B. 1.2 C. 1.25 D. 1.30

 

 

9.	The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:
	A. 20 B. 30 C. 48 D. 58

 

 

	10 .If Rs. 782 be divided into three parts, proportional to  :  : , then the first part is:
	A. Rs. 182 B. Rs. 190 C. Rs. 196 D. Rs. 204

 

 

 

 

Answers

1. Answer:Option B

 

Explanation:

	4	A	=	2	B
	15			5

 

A =		2	x	15	B
		5		4

 

A =	3	B
	2

 

	A	=	3
	B		2

A : B = 3 : 2.

B’s share = Rs.		1210 x	2		= Rs. 484.
			5

 

 

 

 

2 .Answer: Option C

 

Explanation:

Let the third number be x.

Then, first number = 120% of x =	120x	=	6x
	100		5

 

Second number = 150% of x =	150x	=	3x
	100		2

 

Ratio of first two numbers =		6x	:	3x		= 12x : 15x = 4 : 5.

 

 

3 .Answer: Option C

Explanation:

Let the Shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x – 3x = 1000

x = 1000.

B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

 

 

4 .Answer: Option A

 

Explanation:

 

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

 

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

 

		140	x 5x		,		150	x 7x		and		175	x 8x
		100					100					100

 

7x,	21x	and 14x.
	2

 

The required ratio = 7x :	21x	: 14x
	2

 

14x : 21x : 28x

 

2 : 3 : 4.

 

 

 

 

 

 

 

 

 

 

 

5 .Answer: Option D

 

Explanation:

Quantity of milk =		60 x	2	litres = 40 litres.
			3

Quantity of water in it = (60- 40) litres = 20 litres.

New ratio = 1 : 2

Let quantity of water to be added further be x litres.

 

 

Then, milk : water =		40		.
		20 + x

 

Now,		40		=	1
		20 + x			2

 

20 + x = 80

 

x = 60.

Quantity of water to be added = 60 litres.

 

6 .Answer: Option C

 

Explanation:

 

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

 

Their increased number is (120% of 7x) and (110% of 8x).

 

		120	x 7x		and		110	x 8x
		100					100

 

	42x	and	44x
	5		5

 

The required ratio =		42x	:	44x		= 21 : 22

 

7 .Answer: Option D

 

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

 

Then,	2x + 4000	=	40
	3x + 4000		57

57(2x + 4000) = 40(3x + 4000)

 

6x = 68,000

 

3x = 34,000

 

Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

 

 

 

8 .Answer: Option B

 

Explanation:

(x x 5) = (0.75 x 8)    x =		6		= 1.20
		5

 

 

 

 

 

 

 

 

9 .Answer: Option B

 

Explanation:

Let the three parts be A, B, C. Then,

 

A : B = 2 : 3 and B : C = 5 : 8 =		5 x	3		:		8 x	3		= 3 :	24
			5					5			5

 

A : B : C = 2 : 3 :	24	= 10 : 15 : 24
	5

 

B =		98 x	15		= 30.
			49

 

 

 

10 .Answer: Option D

 

 

Explanation:

 

Given ratio =  :  :  = 6 : 8 : 9.

 

1st part = Rs.		782 x	6		= Rs. 204

 

 

 

 

Level-II

11.	The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
	A. 3 : 3 : 10 B. 10 : 11 : 20 C. 23 : 33 : 60 D. Cannot be determined     Answer: Option C   Explanation: Let A = 2k, B = 3k and C = 5k. A’s new salary = 115 of 2k = 115 x 2k = 23k 100 100 10   B’s new salary = 110 of 3k = 110 x 3k = 33k 100 100 10   C’s new salary = 120 of 5k = 120 x 5k = 6k 100 100    New ratio 23k : 33k : 6k = 23 : 33 : 60 10 10
12.	If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
	A. 2 : 5 B. 3 : 7 C. 5 : 3 D. 7 : 3 Answer: Option C   Explanation: Let 40% of A = 2 B 3   Then, 40A = 2B 100 3   2A = 2B 5 3   A = 2 x 5 = 5 B 3 2 3 A : B = 5 : 3.

 

13.	The fourth proportional to 5, 8, 15 is:
	A. 18 B. 24 C. 19 D. 20     Answer: Option B   Explanation: Let the fourth proportional to 5, 8, 15 be x. Then, 5 : 8 : 15 : x 5x = (8 x 15)   x = (8 x 15) = 24. 5

 

14.	Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:
	A. 27 B. 33 C. 49 D. 55 Answer: Option B   Explanation: Let the numbers be 3x and 5x. Then, 3x – 9 = 12 5x – 9 23 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.

 

15.	In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
	A. 50 B. 100 C. 150 D. 200 Answer: Option C   Explanation: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively. Then, sum of their values = Rs. 25x + 10 x 2x + 5 x 3x = Rs. 60x 100 100 100 100   60x = 30     x = 30 x 100 = 50. 100 60 Hence, the number of 5 p coins = (3 x 50) = 150.

 

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(S−a)(S−b)(S−c              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
   3. 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%

 

2. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?
3. 30 %
4. 28 %
5. 32 %
6. 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%

4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
5. 25 % Increase
6. 25 % Decrease
7. 50 % Decrease
8. 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%

5. 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?
6. 14 metres
7. 20 metres
8. 18 metres
9. 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

 

6. 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?
7. equal to ½
8. equal to ¾
9. greater than 1
10. 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.

 

7. 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.
8. 37500 m²
9. 30500 m²
10. 32500 m²
11. 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

 

8. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
9. 45%
10. 44%
11. 40%
12. 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%

9. 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?
10. 814
11. 802
12. 836
13. 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.

 

2. 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?
3. 400
4. 365
5. 385
6. 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.

 

3. 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?
4. 18 cm
5. 16 cm
6. 40 cm
7. 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

 

4. 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)?
5. 142000
6. 112800
7. 142500
8. 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

 

5. 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?
6. Rs.3500
7. Rs. 4200
8. Insufficient Data
9. 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.

 

6. 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?
7. 144√3−48π cm²
8. 121√3−36π cm²
9. 144√3−36π cm²
10. 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=36r⇒r=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

 

7. 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?
8. √11600 cm
9. √14400 cm
10. √10000 cm
11. √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

 

8. 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?
9. 30
10. 44
11. 56
12. 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

Profit and loss

 

IMPORTANT FACTS

Cost Price:

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

 

Selling Price:

The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

 

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

 

Loss:

If S.P. is less than C.P., the seller is said to have incurred a loss.

 

IMPORTANT FORMULAE

1. Gain = (S.P.) – (C.P.)
2. Loss = (C.P.) – (S.P.)
3. Loss or gain is always reckoned on C.P.
4. Gain Percentage: (Gain %)

Gain % =		Gain x 100
		C.P.

5. Loss Percentage: (Loss %)

Loss % =		Loss x 100
		C.P.

6. Selling Price: (S.P.)

SP =		(100 + Gain %)	x C.P
		100

7. Selling Price: (S.P.)

SP =		(100 – Loss %)	x C.P.
		100

8. Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 + Gain %)

9. Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 – Loss %)

10. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.
11. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.
12. When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by:

Loss % =		Common Loss and Gain %		2	=		x		2	.
		10					10

13. If a trader professes to sell his goods at cost price, but uses false weights, then

Gain % =		Error	x 100	%.
		(True Value) – (Error)

 

Questions:

Level-I:

 

 

1.	Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:
	A. 4 4 % 7 B. 5 5 % 11 C. 10% D. 12%

 

2.	The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of xis:
	A. 15 B. 16 C. 18 D. 25

 

3.	If selling price is doubled, the profit triples. Find the profit percent.
	A. 66 2 3 B. 100 C. 105 1 3 D. 120

 

4.	In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
	A. 30% B. 70% C. 100% D. 250%

 

 

5.	A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
	A. 3 B. 4 C. 5 D. 6

 

6.	The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
	A. Rs. 2000 B. Rs. 2200 C. Rs. 2400 D. Data inadequate

 

7.	A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?
	A. Rs. 18.20 B. Rs. 70 C. Rs. 72 D. Rs. 88.25

 

8.	A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?
	A. Rs. 1090 B. Rs. 1160 C. Rs. 1190 D. Rs. 1202

 

9.	Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
	A. 3.5 B. 4.5 C. 5.6 D. 6.5

 

10.	Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:
	A. 30% B. 33 1 % 3 C. 35% D. 44%
11.	Level-II:     On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
	A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60

 

 

12.	When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?
	A. Rs. 21,000 B. Rs. 22,500 C. Rs. 25,300 D. Rs. 25,800

 

13.	100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:
	A. 14 2 % gain 7 B. 15% gain C. 14 2 % loss 7 D. 15 % loss

 

14.	A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:
	A. 5 15 % loss 17 B. 5 15 % gain 17 C. 6 2 % gain 3 D. None of these

 

 

15.	A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
	A. No profit, no loss B. 5% C. 8% D. 10% E. None of these

 

16. A man buys an article for Rs. 27.50 and sells it for Rs 28.60. Find his gain percent
17. 1%
18. 2%
19. 3%
20. 4%

 

 

17. A TV is purchased at Rs. 5000 and sold at Rs. 4000, find the lost percent.
18. 10%
19. 20%
20. 25%
21. 28%

 

 

18. In terms of percentage profit, which among following the best transaction.
    1. P. 36, Profit 17
    2. P. 50, Profit 24
    3. P. 40, Profit 19
    4. P. 60, Profit 29

 

 

 

 

Answer:1 Option B

 

Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.

Selling Price (S.P.) = Rs. 5800.

Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.

Gain % =		300	x 100	%	= 5	5	%
		5500				11

 

Answer:2 Option B

 

Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 – x).

		20 – x	x 100 = 25
		x

2000 – 100x = 25x

125x = 2000

x = 16.

 

 

Answer:3 Option B

 

Explanation:

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y – x) = (2y – x)    y = 2x.

Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.

Profit % =		x	x 100	% = 100%

 

 

Answer:4 Option B

 

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.

Required percentage =		295	x 100	%	=	1475	% = 70% (approximately).
		420				21

 

 

Answer:5 Option C

 

Explanation:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs.	6
	5

 

For Rs.	6	, toffees sold = 6.
	5

 

For Re. 1, toffees sold =		6 x	5		= 5.
			6

 

 

Answer:6 Option A   Explanation: Let C.P. be Rs. x. Then, 1920 – x x 100 = x – 1280 x 100 x x 1920 – x = x – 1280 2x = 3200 x = 1600  Required S.P. = 125% of Rs. 1600 = Rs. 125 x 1600 = Rs 2000. 100

 

 

Answer:7 Option C

 

Explanation:

C.P. = Rs.		100	x 392		= Rs.		1000	x 392		= Rs. 320
		122.5					1225

Profit = Rs. (392 – 320) = Rs. 72.

 

Answer:8 Option C

 

Explanation:

S.P. = 85% of Rs. 1400 = Rs.		85	x 1400		= Rs. 1190
		100

 

 

 

Answer:9 Option C

 

Explanation:

Cost Price of 1 toy = Rs.		375		= Rs. 31.25
		12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 – 31.25) = Rs. 1.75

Profit % =		1.75	x 100	%	=	28	% = 5.6%
		31.25				5

 

 

 

Answer:10 Option D

 

Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs.		5	x 30		= Rs. 25.
		6

 

S.P. of 30 articles = Rs.		6	x 30		= Rs. 36.
		5

 

Gain % =		11	x 100	% = 44%.
		25

 

 

Answer:11 Option D

 

Explanation:

(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)

C.P. of 12 balls = S.P. of 17 balls = Rs.720.

C.P. of 1 ball = Rs.		720		= Rs. 60.
		12

 

 

Answer:12 Option C

 

Explanation:

85 : 18700 = 115 : x

x =		18700 x 115		= 25300.
		85

Hence, S.P. = Rs. 25,300.

 

Answer:13 Option A

 

Explanation:

C.P. of 1 orange = Rs.		350		= Rs. 3.50
		100

 

S.P. of 1 orange = Rs.		48		= Rs. 4
		12

 

Gain% =		0.50	x 100	%	=	100	% = 14	2	%
		3.50				7		7

 

 

 

Answer:14 Option B

 

Explanation:

C.P. of 1st transistor = Rs.		100	x 840		= Rs. 700.
		120

 

C.P. of 2nd transistor = Rs.		100	x 960		= Rs. 1000
		96

So, total C.P. = Rs. (700 + 1000) = Rs. 1700.

Total S.P. = Rs. (840 + 960) = Rs. 1800.

Gain % =		100	x 100	%	= 5	15	%
		1700				17

 

 

 

Answer:15 Option B

 

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =		80	x 100	% = 5%.
		1600

 

Answer:16 Option D

 

Explanation:

So we have C.P. = 27.50
S.P. = 28.60

Gain = 28.60 – 27.50 = Rs. 1.10

Gain%=(Gain/Cost∗100)%=(1.10/27.50∗100)%=4%

 

 

 

 

Answer:17 Option B

 

Explanation:

We know, C.P. = 5000
S.P. = 4000
Loss = 5000 – 4000 = 1000
Loss%=(Loss/Cost∗100)%=(1000/5000∗100)%=20%

 

 

Answer:18 Option D

 

Explanation:

Hint: Calculate profit percent as

Profit% = (profit/cost) * 100

 

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

 

 

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

 

 

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

 

 

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

 

 

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

 

 

6. 3 : 243 :: 5 : ?                                                                                                                                            A. 405               B. 465                   C. 3125                     D. 546

 

 

7. 5 : 36 :: 6 : ?                                                                                                                                                A. 48               B. 50                 C. 49              D. 56

 

 

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

 

 

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

 

 

10. 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

 

 

2. Answer: Option C

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

 

 

3. Answer: Option C

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

 

 

4. Ans. Option C

Explanation:  1st number+4 = 2nd number

2nd number+8= 3rd number

 

 

5. Ans. Option A

Explanation:  63= 7*9

49= 7*7

35= 7*5

 

 

6. Ans. Option C

Explanation: 3^5=243

5^5=3125

 

 

7. Ans. Option A

Explanation: 13^2+13=182

18^2+18=342

19^2+19=380

 

 

8. Ans. Option A

Explanation: The first and the third letters has been interchanged

 

 

9. Ans. Option C

Explanation: Pairs of opposite letter

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

 

 

10. Ans. Option B

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