Aptitude & reasoning

 

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 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.

 

1. Alligation

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

2. Mean Price

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

 

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

 

4. Rule of Alligation

   If two ingredients are mixed, then

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

Cost Price(CP) of a unit quantity of cheaper (c)	Cost Price(CP) of a unit quantity of dearer (d)
Mean Price
(m)
(d – m)	(m – c)

5. => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

 

Solved Examples

Level 1

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

 

 

Answer : Option B

Explanation :

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

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

=[x(1−y/x)n]

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

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

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

 

 

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

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

By the rule of alligation,

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

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

 

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

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

 

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

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

 

 

 

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

 

 

Answer : Option C

 

Explanation :

By the rule of alligation, we have

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

Required Ratio = 4 : 8 = 1 : 2

 

 

 

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

 

 

 

 

Answer : Option D

 

Explanation :

By the rule of alligation, we have

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

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

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

 

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

 

 

 

Answer : Option A

 

Explanation :

By the rule of alligation, we have

 

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

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

Given that total quantity = 1600 Kg

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

 

 

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

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

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

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

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

 

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

 

 

Answer : Option C

 

Explanation :

By the rule of alligation,we have

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

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

Total quantity is given as 1000Kg

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

 

Level 2

 

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

 

 

Answer : Option C

Explanation :

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

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

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

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

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

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

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

 

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

 

 

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

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

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

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

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

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

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

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

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

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

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

 

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

 

 

Answer : Option D

 

Explanation :

 

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

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

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

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

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

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

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

By the rule of alligation, we can write as

 

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

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

 

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

 

 

Answer : Option B

Explanation :

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

Profit = 10%

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

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

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

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

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

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

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

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

 

 

Answer : Option B

Explanation :

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

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

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

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

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

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

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

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

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

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

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

 

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

 

 

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

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

Given than Gain = 25%

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

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

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

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

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

Age Problems

 

Important Formulas on “Problems on Ages”:

 

1. If the current age is x, then ntimes the age is nx.
2. If the current age is x, then age nyears later/hence = x+ n.
3. If the current age is x, then age nyears ago = x– n.
4. The ages in a ratio a: bwill be ax and bx.

5. If the current age is x, then	1	of the age is	x	.
	n		n

Example:

A problem with one variable: How old is Al?

Many single-variable algebra word problems have to do with the relations between different people’s ages. For example:

Al’s father is 45. He is 15 years older than twice Al’s age. How old is Al?

We can begin by assigning a variable to what we’re asked to find. Here this is Al’s age, so let Al’s age = x.

We also know from the information given in the problem that 45 is 15 more than twice Al’s age. How can we translate this from words into mathematical symbols? What is twice Al’s age?

Well, Al’s age is x, so twice Al’s age is 2x, and 15 more than twice Al’s age is 15 + 2x.That equals 45, right? Now we have an equation in terms of one variable that we can solve for x: 45 = 15 + 2x.

original statement of the problem:	45 = 15 + 2x
subtract 15 from each side:	30 = 2x
divide both sides by 2:	15 = x

Since x is Al’s age and x = 15, this means that Al is 15 years old.

It’s always a good idea to check our answer:

twice Al’s age is 2 x 15:	30
15 more than 30 is 15 + 30:	45

This should be the age of Al’s father, and it is.

 

 

Questions:

Level-I:

 

1.	Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?
	A. 2 times B. 2 1 times 2 C. 2 3 times 4 D. 3 times

 

2.	The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
	A. 4 years B. 8 years C. 10 years D. None of these

 

3.	A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:
	A. 14 years B. 19 years C. 33 years D. 38 years

 

4.	A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
	A. 7 B. 8 C. 9 D. 10 E. 11

 

5.	Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
	A. 24 B. 27 C. 40 D. Cannot be determined E. None of these

 

6.	A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
	A. 14 years B. 18 years C. 20 years D. 22 years

 

7.	Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?
	A. 16 years B. 18 years C. 20 years D. Cannot be determined E. None of these

 

8.	The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:
	A. 12 years B. 14 years C. 18 years D. 20 years

 

9.	At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?
	A. 12 years B. 15 years C. 19 and half D. 21 years

 

10.	Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
	A. 16 years B. 18 years C. 28 years D. 24.5 years E. None of these

 

 

11.	Level-II:         The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
	A. 8, 20, 28 B. 16, 28, 36 C. 20, 35, 45 D. None of these

 

12.	Ayesha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
	A. 2 years B. 4 years C. 6 years D. 8 years

 

13.	A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?
	A. 32 years B. 36 years C. 40 years D. 48 years

 

14.	Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?
	A. 1 year B. 2 years C. 25 years D. Data inadequate E. None of these

 

15.	The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. The ratio of their present ages is:
	A. 5 : 2 B. 7 : 3 C. 9 : 2 D. 13 : 4

 

16.	What is Sonia’s present age? I. Sonia’s present age is five times Deepak’s present age.  II. Five years ago her age was twenty-five times Deepak’s age at that time.
	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.	Average age of employees working in a department is 30 years. In the next year, ten workers will retire. What will be the average age in the next year? I. Retirement age is 60 years.  II. There are 50 employees in the department.
	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.	Divya is twice as old as Shruti. What is the difference in their ages? I. Five years hence, the ratio of their ages would be 9 : 5.  II. Ten years back, the ratio of their ages was 3 : 1.
	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

 

 

 

Answers:

Level-I:

 

Answer:1 Option A

 

Explanation:

Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.

	(4x + 8) =	5	(x + 8)
		2

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio =	(4x + 16)	=	48	= 2.
	(x + 16)		24

 

 

Answer:2 Option A

 

Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

 

 

 

Answer:3 Option A

 

Explanation:

Let the son’s present age be x years. Then, (38 – x) = x

2x = 38.

x = 19.

Son’s age 5 years back (19 – 5) = 14 years.

 

Answer:4 Option D

 

Explanation:

Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.

(2x + 2) + 2x + x = 27

5x = 25

x = 5.

Hence, B’s age = 2x = 10 years.

 

Answer:5 Option A

 

Explanation:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then,	5x + 3	=	11
	4x + 3		9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

 

Answer:6 Option D

 

Explanation:

Let the son’s present age be x years. Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

 

Answer:7 Option A

 

Explanation:

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then,	(6x + 6) + 4	=	11
	(5x + 6) + 4		10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

Sagar’s present age = (5x + 6) = 16 years.

 

Answer:8 Option D

 

Explanation:

Let the present ages of son and father be x and (60 –x) years respectively.

Then, (60 – x) – 6 = 5(x – 6)

54 – x = 5x – 30

6x = 84

x = 14.

Son’s age after 6 years = (x+ 6) = 20 years..

 

Answer:9 Option B

 

Explanation:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26        4x = 20

x = 5.

Deepak’s age = 3x = 15 years.

 

Answer:10 Option D

 

Explanation:

Let Rahul’s age be x years.

Then, Sachin’s age = (x – 7) years.

	x – 7	=	7
	x		9

9x – 63 = 7x

2x = 63

x = 31.5

Hence, Sachin’s age =(x – 7) = 24.5 years.

 

Answer:11 Option B

 

Explanation:

Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x – 8) + (7x – 8) + (9x – 8) = 56

20x = 80

x = 4.

Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

 

Answer:12 Option C

 

Explanation:

Mother’s age when Ayesha’s brother was born = 36 years.

Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.

Required difference = (42 – 36) years = 6 years.

 

Answer:13 Option C

 

Explanation:

Let the mother’s present age be x years.

Then, the person’s present age =		2	x		years.
		5

 

		2	x + 8		=	1	(x + 8)
		5				2

2(2x + 40) = 5(x + 8)

x = 40.

 

Answer:14 Option D

 

Explanation:

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R – Q = ?.

Explanation:

R – Q = Q – T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R – Q) = ?

Here we know the value(age) of Q (25), but we don’t know the age of R.

Therefore, (R-Q) cannot be determined.

 

Answer:15 Option B

 

Explanation:

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

 

 

Answer:16  Option E

 

Explanation:

I. S = 5D     D =	S	….(i)
	5

1. S – 5 = 25 (D – 5)    S = 25D – 120 ….(ii)

Using (i) in (ii), we get S =		25 x	S		– 120
			5

4S = 120.

S = 30.

Thus, I and II both together give the answer. So, correct answer is (E).

 

Answer:17 Option E

 

Explanation:

1. Retirement age is 60 years.
2. There are 50 employees in the department.

Average age of 50 employees = 30 years.

Total age of 50 employees = (50 x 30) years = 1500 years.

Number of employees next year = 40.

Total age of 40 employees next year (1500 + 40 – 60 x 10) = 940.

Average age next year =	940	years = 23	1	years.
	40		2

Thus, I and II together give the answer. So, correct answer is (E).

 

Answer:18   Option C

 

Explanation:

Let Divya’s present age be D years and Shruti’s present age b S years

Then, D = 2 x S        D – 2S = 0 ….(i)

I.	D + 5	=	9	….(ii)
	S + 5		5

 

II.	D – 10	=	3	….(iii)
	S – 10		1

From (ii), we get : 5D + 25 = 9S + 45        5D – 9S = 20 ….(iv)

From (iii), we get : D – 10 = 3S – 30        D – 3S = -20 ….(v)

Thus, from (i) and (ii), we get the answer.

Also, from (i) and (iii), we get the answer.

I alone as well as II alone give the answer. Hence, the correct answer is (C).

Simplification

Simplification is one of the most important part of Quantitative Aptitude section of any competitive exam. Today I am sharing all the techniques to solve Simplification questions quickly.

Rules of Simplification

V → Vinculum

B → Remove Brackets – in the order ( ) , { }, [ ]

O → Of
D → Division

M → Multiplication

A → Addition

S → Subtraction

 

Classification

Types	Description
Natural Numbers:	all counting numbers ( 1,2,3,4,5….∞)
Whole Numbers:	natural number + zero( 0,1,2,3,4,5…∞)
Integers:	All whole numbers including Negative number + Positive number(∞……-4,-3,-2,-1,0,1,2,3,4,5….∞)
Even & Odd Numbers :	All whole number divisible by 2 is Even (0,2,4,6,8,10,12…..∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19….∞)
Prime Numbers:	It can be positive or negative except 1, if the number is not divisible by any number except the number itself.(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61….∞)
Composite Numbers:	Natural numbers which are not prime
Co-Prime:	Two natural number a and b are said to be co-prime if their HCF is 1.

 

Divisibility

Numbers	IF A Number	Examples
Divisible by 2	End with 0,2,4,6,8 are divisible by 2	254,326,3546,4718 all are divisible by 2
Divisible by 3	Sum of its digits  is divisible by 3	375,4251,78123 all are divisible by 3.  [549=5+4+9][5+4+9=18]18 is divisible by 3  hence 549 is divisible by 3.
Divisible by 4	Last two digit divisible by 4	5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4.
Divisible by 5	Ends with 0 or 5	225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5.
Divisible by 6	Divides by Both 2 & 3	4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6.
Divisible by 8	Last 3 digit divide by 8	746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8.
Divisible by 10	End with 0	220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10.
Divisible by 11	[Sum of its digit in odd places-Sum of its digits in even places]= 0 or multiple of 11	Consider the number 39798847 (Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3) (23-12) 23-12=11, which is divisible by 11. So 39798847 is divisible by 11.

Division & Remainder Rules

Suppose we divide 45 by 6

 

hence ,represent it as:

dividend = ( divisor✘quotient ) + remainder

or

divisior= [(dividend)-(remainder] / quotient

could be write it as

x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)

 

 

Rules

1. Modulus of a Real Number:

Modulus of a real number a is defined as

|a| =		a, if a > 0
		–a, if a < 0

Thus, |5| = 5 and |-5| = -(-5) = 5.

2. Virnaculum (or Bar):

When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.

 

 

Example:

On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?

Number = 342k + 47

( 18 ✘19k ) + ( 18 ✘2 ) + 11

18 ✘( 19k + 2 ) +11.

Remainder = 11

 

Sum Rules

(1+2+3+………+n) = ¹/₂ n(n+1)

(1²+2²+3²+………+n²) = ¹/₆ n (n+1) (2n+1)

(1³+2³+3³+………+n³) = ¹/4 n² (n+1)²

 

Questions:

Level-I:

 

1.	A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
	A. 45 B. 60 C. 75 D. 90

 

2.	There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
	A. 20 B. 80 C. 100 D.   200

 

3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:

A. Rs. 3500 B. Rs. 3750 C. Rs. 3840 D. Rs. 3900

 

4. If a – b = 3 and a2 + b2 = 29, find the value of ab.

A. 10 B. 12 C. 15 D. 18

 

5.	The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
	A. Rs. 1200 B. Rs. 2400 C. Rs. 4800 D. Cannot be determined E. None of these

 

6.	A sum of Rs. 1360 has been divided among A, B and C such that A gets  of what B gets and B gets  of what C gets. B’s share is:
	A. Rs. 120 B. Rs. 160 C. Rs. 240 D. Rs. 300

 

7.	One-third of Rahul’s savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?
	A. Rs. 30,000 B. Rs. 50,000 C. Rs. 60,000 D. Rs. 90,000

 

8.	A fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
	A. 30 birds B. 60 birds C. 72 birds D. 90 birds

 

9.	Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:
	A. 1 7 B. 1 8 C. 1 9 D. 7 8

 

10.	To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?
	A. 10 B. 35 C. 62.5 D. Cannot be determined E. None of these

 

Level-II:

1.	In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?
	A. 160 B. 175 C. 180 D. 195

 

12.	Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?
	A. 256 B. 432 C. 512 D. 640 E. None of these

 

13.	A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:
	A. 22 B. 23 C. 24 D. 26

 

14.	(469 + 174)2 – (469 – 174)2 = ? (469 x 174)
	A. 2 B. 4 C. 295 D. 643

 

15.	David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
	A. 19 B. 28 C. 30 D. 37

 

16. Find the value of 1/(3+1/(3+1/(3-1/3)))

A.) 3/10	B.) 10/3
C.) 27/89	D.) 89/27

17. 
    Find the value of

A.) 3½ 99;	B.) 34/99
C.) 2.131313	D.) 3.141414

 

18.Find the value of

((0.1)³ + (0.6)³ + (0.7)³ − (0.3)(0.6)(0.7))/((0.1)² + (0.6)² + (0.7)² − 0.006 − 0.42 − 0.07)

 

A.) 14/10	B.) 1.35
C.) 13/10	D.) 0

 

 

 

 

19. Solve(0.76 × 0.76 × 0.76 − 0.008)/(0.76 × 0.76 + 0.76 × 0.2 + 0.04)

A.) 0.56	B.) 0.65
C.) 0.54	D.) 0.45

20. Find the value of

A.) 1.5	B.) -1.5
C.) 1	D.) 0

 

Answers:

Level-I

 

Answer:1 Option D

 

Explanation:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

 

 

Answer:2 Option C

 

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10      x – y = 20 …. (i)

and x + 20 = 2(y – 20)      x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

 

 

Answer:3 Option D

 

Explanation:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

Then, 10x = 4y   or   y =	5	x.
	2

15x + 2y = 4000

15x + 2 x	5	x = 4000
	2

20x = 4000

x = 200.

So, y =		5	x 200		= 500.
		2

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

= Rs. 3900.

 

 

 

 

 

Answer:4 Option A

 

Explanation:

2ab = (a² + b²) – (a – b)²

= 29 – 9 = 20

ab = 10.

 

 

 

Answer:5 Option B

 

Explanation:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 …. (i)

and x + 6y = 1600 …. (ii)

 

Divide equation (i) by 2, we get the below equation.

 

=> x +  2y =  800. — (iii)

 

Now subtract (iii) from (ii)

 

x +  6y = 1600  (-)

x +  2y =  800

—————-

4y =  800

—————-

 

Therefore, y = 200.

 

Now apply value of y in (iii)

 

=>  x + 2 x 200 = 800

 

=>  x + 400 = 800

 

Therefore x = 400

 

Solving (i) and (ii) we get x = 400, y = 200.

Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

 

 

Answer:6 Option C

 

Explanation:

Let C’s share = Rs. x

Then, B’s share = Rs.	x	,   A’s share = Rs.		2	x	x		= Rs.	x
	4			3		4			6

 

	x	+	x	+ x = 1360
	6		4

 

	17x	= 1360
	12

 

x =	1360 x 12	= Rs. 960
	17

 

Hence, B’s share = Rs.		960		= Rs. 240.

 

 

Answer:7 Option C

 

Explanation:

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 – x) respectively. Then,

1	x =	1	(150000 – x)
3		2

 

	x	+	x	= 75000
	3		2

 

	5x	= 75000
	6

 

x =	75000 x 6	= 90000
	5

Savings in Public Provident Fund = Rs. (150000 – 90000) = Rs. 60000

 

 

Answer:8 Option A

 

Explanation:

Let the total number of shots be x. Then,

Shots fired by A =	5	x
	8

 

Shots fired by B =	3	x
	8

 

Killing shots by A =	1	of	5	x	=	5	x
	3		8			24

 

Shots missed by B =	1	of	3	x	=	3	x
	2		8			16

 

	3x	= 27 or x =		27 x 16		= 144.
	16			3

 

Birds killed by A =	5x	=		5	x 144		= 30.
	24			24

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

Original share of 1 person =	1
	8

 

New share of 1 person =	1
	7

 

Increase =		1	–	1		=	1
		7		8			56

 

Required fraction =	(1/56)	=		1	x	8		=	1
	(1/8)			56		1			7

 

 

Answer:10 Option C

 

Explanation:

Let the capacity of 1 bucket = x.

Then, the capacity of tank = 25x.

New capacity of bucket =	2	x
	5

 

Required number of buckets =	25x
	(2x/5)

 

=	25x	x	5
			2x

 

=	125
	2

= 62.5

 

Level-II:

Answer:11 Option B

 

Explanation:

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

160 x 2.40 + x x 3.20 = 432

3.20x = 432 – 384 = 48

x = 15.

Hence, total hours of work = (160 + 15) = 175.

 

Answer:12 Option C

 

Explanation:

Let total number of children be x.

Then, x x	1	x =	x	x 16     x = 64.
	8		2

 

Number of notebooks =	1	x2 =		1	x 64 x 64		= 512

 

Answer:13 Option D

 

Explanation:

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 …. (i)

and 2x + 4y = 140      x + 2y = 70 …. (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

 

Answer:14 Option B

 

Explanation:

Given exp. =	(a + b)2 – (a – b)2
	ab

 

=	4ab
	ab

= 4 (where a = 469, b = 174.)

 

Answer:15 Option C

 

Explanation:

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 – 63x        120x = 40

x =	1
	3

 

Number of floors covered by David in (1/3) min. =		1	x 57		= 19.
		3

So, their paths cross at (11 +19) i.e., 30^th floor.

 

Answer:16 Option ‘C’

Explanation:

1/[3 + (1/(3+1/(3 – 1/3)))]

=> 1/[3 + 1/(3 + 1/(8/3))]

=> 1/[3 + 1/(3 + 3/8)]

=> 1/[3 + 8/27]

=> 1/(89/27)

=> 27/89

 

Answer:17 Option ‘D’

Explanation:

6/9 + 7/9 + 9/9 + 69/99

2/3 + 7/9 + 1 + 69/99

(66 + 77 + 99 + 69)/99

311/99 => 3.141414

 

Answer:18 Option ‘A’

Explanation:

((0.1)³ + (0.6)³ + (0.7)³ − (0.3)(0.6)(0.7))/((0.1)² + (0.6)² + (0.7)² − 0.006 − 0.42 − 0.07)

=> (0.1 + 0.6 + 0.7)³/(0.1 + 0.6 + 0.7)²

=> 0.1 + 0.6 + 0.7 => 1.4 = 14/10

 

Answer:19 Option ‘A’

 

Answer:20 Option ‘D’

11/30 − [1/6 + 1/5 + [7/12 − 7/12]]

11/30 − [1/6 + 1/5 + [0]]

11/30 − [(5 + 6)/30]

11/30 − 11/30 = 0.

 

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

 

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

 

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

 

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).