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# Ratio and Proportion Questions for IBPS PO Prelims

Question 1:Â A tank contains 120 litres of milk and 80 litres of water. If __ litres of mixture is taken out from the tank and __ litres of milk is added to the tank then the final quantity of milk is 57.14% more than the final quantity of water.
Which of the following values can fill in the blanks appropriately?
(i) 60, 4
(ii) 95, 3
(iii) 140, 2

a)Â Only (ii) and (iii)

b)Â Only (i) and (ii)

c)Â All (i), (ii) and (iii)

d)Â Only (iii)

e)Â Only (i) and (iii)

Solution:

The final quantity of milk is 57.14% more than the final quantity of water.
Final quantity of milk = Final quantity of water + $\frac{4}{7}\times$Final quantity of water
Final quantity of milk = $\frac{11}{7}\times$Final quantity of water
Ratio of the final quantity of milk and water is 11:7 respectively.

(i) 60, 4
60 litres of mixture is taken and 4 litres of milk is added from the tank.
Ratio of initial quantity of milk and water = 120 : 80
= 3 : 2
Final quantity of milk = 120 – $\frac{3}{3+2}\times$60 + 4 = 88 litres
Final quantity of water = 80 – $\frac{2}{3+2}\times$60 = 56
Ratio of final quantity of milk and water = 88 : 56
= 11:7
Given values satisfy the conditions of the question.

(ii) 95, 3
95 litres of mixture is taken and 3 litres of milk is added from the tank.
Ratio of initial quantity of milk and water = 120 : 80
= 3 : 2
Final quantity of milk = 120 – $\frac{3}{3+2}\times$95 + 3 = 66 litres
Final quantity of water = 80 – $\frac{2}{3+2}\times$95 = 42
Ratio of final quantity of milk and water = 66 : 42
= 11:7
Given values satisfy the conditions of the question.

(iii) 140, 2
140 litres of mixture is taken and 2 litres of milk is added from the tank.
Ratio of initial quantity of milk and water = 120 : 80
= 3 : 2
Final quantity of milk = 120 – $\frac{3}{3+2}\times$140 + 2 = 38 litres
Final quantity of water = 80 – $\frac{2}{3+2}\times$140 = 24
Ratio of final quantity of milk and water = 38 : 24
= 19:12
Given values do not satisfy the conditions of the question.
Hence, the correct answer is Option B

Question 2:Â A certain number of toffees are divided among four people such that the ratio between the number of toffees obtained by B to C is 9:8 respectively. The number of toffees obtained by A is 20% less than the number of toffees obtained by D. The ratio between the number of toffees obtained by A and B is 4:3 respectively. If the number of toffees obtained by C is 40, then find out the sum of the number of toffees obtained by A, B and D.

a)Â 145

b)Â 175

c)Â 180

d)Â 160

e)Â None of the above

Solution:

The ratio between the number of toffees obtained by B to C is 9:8 respectively.
If the number of toffees obtained by C is 40.
number of toffees obtained by B = 40 of (9/8) = 45
The ratio between the number of toffees obtained by A and B is 4:3 respectively.
number of toffees obtained by A = 45 of (4/3) = 60
The number of toffees obtained by A is 20% less than the number of toffees obtained by D.
60 = 80% of the number of toffees obtained by D
the number of toffees obtained by D = 75
Sum of the number of toffees obtained by A, B and D = 60+45+75
= 180
Hence, option c is the correct answer.

Question 3:Â An Organisation distributed chocolates to its employees on an occasion. The number of chocolates received by employee Y is 77.77% of the number of chocolates received by employee X. Ratio of the number of chocolates received by employees Y and Z is 9:8. If the average of the number of chocolates received by X, Y and Z is 66$\frac{2}{3}$, then what is the number of chocolates received by Z?

a)Â 81

b)Â 72

c)Â 63

d)Â 56

e)Â None of the above

Solution:

The number of chocolates received by employee Y is 77.77% of the number of chocolates received by employee X.
Let the number of chocolates received by employee X = 9p
Number of chocolates received by employee Y = $\frac{7}{9}\times$9p = 7p

Ratio of the number of chocolates received by employees Y and Z is 9:8.
Number of chocolates received by employee Z = $\frac{8}{9}\times$7p = $\frac{56}{9}$p

Average of the number of chocolates received by X, Y and Z is 66$\frac{2}{3}$.
$\Rightarrow$ $\frac{9p+7p+\frac{56}{9}p}{3}$ = 66$\frac{2}{3}$
$\Rightarrow$ $\frac{81p+63p+56p}{9\times3}$ = $\frac{200}{3}$
$\Rightarrow$ 200p = 200 x 9
$\Rightarrow$ p = 9
Number of chocolates received by employee Z = $\frac{56}{9}$p
= $\frac{56}{9}\times$9
= 56
Hence, the correct answer is Option D

Question 4:Â Certain amount of money was divided among four people such that the ratio between the amount obtained by B and C is 13:14 respectively. The difference between the amount obtained by A and D is Rs. 300. If the average of the amount obtained by C and D is Rs. 4900 and the amount obtained by A is 25% less than the amount obtained by B, then find out the average of the amount obtained by all the four people.

a)Â 4195

b)Â 4365

c)Â 4725

d)Â 4575

e)Â 4945

Solution:

The ratio between the amount obtained by B and C is 13:14 respectively.
Letâ€™s assume the amount obtained by B and C is 13y and 14y respectively.
the average of the amount obtained by C and D is Rs. 4900.
$\frac{14y+\text{amount obtained by D}}{2} = 4900$
14y+amount obtained by D = 9800
amount obtained by D = (9800-14y)
the amount obtained by A is 25% less than the amount obtained by B
the amount obtained by A = 75% of 13y
= 9.75y
The difference between the amount obtained by A and D is Rs. 300.
9.75y-(9800-14y) = 300
9.75y-9800+14y = 300
23.75y = 9800+300 = 10100
y = 425.263158 Eq.(i)
Or (9800-14y)-9.75y = 300
9800-14y-9.75y = 300
9800-300 = 23.75y
23.75y = 9500
y = 400 Eq.(ii)
Average of the amount obtained by all the four people = $\frac{9.75y+13y+14y+(9800-14y)}{4}$
= $\frac{22.75y+9800}{4}$
= (5.6875y+2450) Eq.(iii)
Put the value of â€˜yâ€™ from Eq.(i) to the above given equation.
= (5.6875\times425.263158+2450)
After solving this, we will get a fractional value which is not available in any of the options. So the value of â€˜yâ€™ which is given in Eq.(i) is not possible.
Put the value of â€˜yâ€™ from Eq.(ii) to the equation Eq.(iii).
= $(5.6875\times400+2450)$
= 2275+2450
= 4725
Hence, option c is the correct answer.

Question 5:Â Certain amount is divided among P, Q and R such that the amount obtained by Q is 11.11% less than the amount obtained by P. The amount obtained by P is 14.28% more than the amount obtained by R. If the total amount divided among them is â‚¹1194, then what is the amount obtained by R?

a)Â â‚¹400

b)Â â‚¹432

c)Â â‚¹384

d)Â â‚¹378

e)Â None of the above

Solution:

Let the amount obtained by P = 72p
The amount obtained by Q is 11.11% less than the amount obtained by P.
Amount obtained by Q = $\frac{8}{9}$ of the amount obtained by P
= $\frac{8}{9}\times$72p
= 64p
The amount obtained by P is 14.28% more than the amount obtained by R.
Amount obtained by P = $\frac{8}{7}$ of the amount obtained by R
72p = $\frac{8}{7}\times$Amount obtained by R
Amount obtained by R = 63p
Total amount divided among them is â‚¹1194
72p + 64p + 63p 1194
199p = 1194
p = 6
Amount obtained by R = 63p
= 63(6)
= â‚¹378
Hence, the correct answer is Option D

Question 6:Â There are two tanks A and B which contain a mixture of milk and water. Tank A contains milk and water in the ratio of 4 : 1 and tank B contains milk and water in the ratio of 3 : 1. The total quantity of mixture is 60 litres when $\frac{1}{5}$ of mixture from tank A and $\frac{1}{4}$ of mixture from tank B are mixed. The initial quantity of milk in tank A is 40 litres more than the initialÂ quantity of water in tank B. What is the initial quantity of water in tank A?

a)Â 20 litres

b)Â 24 litres

c)Â 16 litres

d)Â 12 litres

e)Â None of the above

Solution:

Ratio of milk and water in tank A is 4 : 1
Let the quantity of milk and water in tank A is 4p and p respectively.
Ratio of milk and water in tank B is 3 : 1
Let the quantity of milk and water in tank B is 3q and q respectively.
The total quantity of mixture is 60 litres when $\frac{1}{5}$ of mixture from tank A and $\frac{1}{4}$ of mixture from tank B are mixed.
$\Rightarrow$ $\frac{1}{5}$(4p+p) + $\frac{1}{4}$(3q+q) = 60
$\Rightarrow$ p + q = 60â€¦â€¦â€¦.(1)
The quantity of milk in tank A is 40 litres more than quantity of water in tank B.
$\Rightarrow$ 4p = q + 40â€¦….(2)
5p + q = q + 100
p = 20
$\therefore$ Initial quantity of water in tank A = p = 20 litres
Hence, the correct answer is Option A

Instructions

Ravi and kavitha take x ml, y ml of drink which is a mixture of Tonic and Water. The ratio of Tonic to water in Ravi is 3:1 and that of in kavitha’s drink is 9:2. Both pay a total amount of Rs. 1200and the price of each ml is Rs. 20. If Ravi mixes 10ml moreÂ water in his drink then quantiy of his drink becomes equals toÂ  kavitha’s drink

Question 7:Â What is the total amount costed forÂ Tonic in both Ravi and Kavitha’s drink ?

a)Â 845/-

b)Â 756.4/-

c)Â 936.6/-

d)Â 786.8/-

e)Â 990.8/-

Solution:

The ratio of Tonic and water in Ravi’s drink is 3:1

Let this be 3x:x

in Kavitha’s drink is 9:2

let this be 9y :2y

The amount paid by both = 1200
Price of each ml = 20

1200/20 = 60

3x+x+9y+2y = 60

4x+11y = 60 ——-(1)

Given that If Ravi mixes 10ml of water in his drink then qty of drink becomes half of kavitha

4x+10 =1 1y——(2)

solving both equations,

x = 6.25ml y = 3.12ml

so the amount of Tonic and water in Ravi’s drink is 3(6.25) = 18.75

and 6.25mlml

amount of tonic = 18.75ml and amount of water = 6.25ml

The ratio of Tonic and water

in Kavitha’s drink is 9:2

9y :2y

9(3.12) = 28.08

2(3.12) = 6.24

Tonic = 28.08ml Water = 6.24ml

So, the total amount of tonic is 18.75+28.08 =Â  = 46.83

Given that price per ml is Rs. 20

936.6 rupees

Question 8:Â What is the amount of Tonic and water in Kavitha’s drink ?

a)Â 28.08ml , 6.24ml

b)Â 8.4ml, 37.8ml

c)Â 5.5ml, 4ml

d)Â 4.8ml, 15ml

e)Â 15ml, 4.8ml

Solution:

The ratio of Tonic and water in Ravi’s drink is 3:1

Let this be 3x:x

in Kavitha’s drink is 9:2

let this be 9y :2y

The amount paid by both = 1200
Price of each ml = 20

1200/20 = 60

3x+x+9y+2y = 60

4x+11y = 60 ——-(1)

Given that If Ravi mixes 10ml of water in his drink then qty of drink becomes half of kavitha

4x+10 =1 1y——(2)

solving both equations,

x = 6.25ml y = 3.12ml

The ratio of Tonic and water

in Kavitha’s drink is 9:2

9y :2y

Tonic = 28.08ml Water = 6.24ml

Question 9:Â What is the amount of Tonic and Water in Ravi’s drink ?

a)Â 18.75 ml, 6.25ml

b)Â 14.2ml, 9.9ml

c)Â 15.5ml, 6.4ml

d)Â 16.4ml, 5.5ml

e)Â 18.5ml, 4.5ml

Solution:

The ratio of Tonic and water in Ravi’s drink is 3:1

Let this be 3x:x

in Kavitha’s drink is 9:2

let this be 9y :2y

The amount paid by both = 1200
Price of each ml = 20

1200/20 = 60

3x+x+9y+2y = 60

4x+11y = 60 ——-(1)

Given thatÂ Â If Ravi mixes 10ml more water in his drink then qty of drink becomes equal toÂ  kavitha’s drink

4x+10 =1 1y——(2)

solving both equations,

x = 6.25ml y = 3.12ml

so the amount ofÂ Â Tonic and water in Ravi’s drink is 3(6.25) = 18.75ml

and 6.25ml

amount of tonic = 18.75ml and amount of water = 6.25ml

Question 10:Â In a drum A, the ratio between the quantity of milk and water is 23:13 respectively. In drum B, the quantity of milk is 28.57% more than the quantity of water. If both the mixture of both of the drums are mixed together with â€˜yâ€™ litres of water, then the ratio of milk and water in the new mixture is 5:4 respectively. Find out the value of â€˜yâ€™. It is assumed that the sum of the quantities of milk from both the drums together is 800 litres.

a)Â 180

b)Â 240

c)Â 160

d)Â 140

e)Â Cannot be determined

Solution:

It is assumed that the sum of the quantities of milk from both the drums together is 800 litres.
If both the mixture of both of the drums are mixed together with â€˜yâ€™ litres of water, then the ratio of milk and water in the new mixture is 5:4 respectively.
5c = 800
c = 160
4c = 640
Total mixture = 9c = 1440
In a drum A, the ratio between the quantity of milk and water is 23:13 respectively.
Letâ€™s assume the quantity of milk and water in drum A is 23a and 13a respectively.
In drum B, the quantity of milk is 28.57% more than the quantity of water.
Letâ€™s assume the quantity of water in drum B is 7b.
quantity of milk in drum B = 7b of (9/7) = 9b
13a+7b+y = 640
23a+9b = 800
Now further, we cannot solve it. Because one equation has three variables and another one has two variables. So we cannot determine the answer from the given information.
Hence, option e is the correct answer.

Question 11:Â An amount was divided among three people such as the ratio between the amount obtained by Anu and Chotu is 3:2 respectively. The amount obtained by Bindu is 83.33% more than the amount obtained by Chotu. The average of the amount obtained by Anu and Bindu is 12000. If the amount will be divided between two people Anu and Bindu such as that Anu got Rs. 2600 less than Bindu, then find out the amount obtained by Anu is how much more than the amount obtained by her earlier.

a)Â Rs. 3900

b)Â Rs. 3500

c)Â Rs. 3000

d)Â Rs. 3200

e)Â None of the above

Solution:

When divided among three people â‡’
An amount was divided among three people such as the ratio between the amount obtained by Anu and Chotu is 3:2 respectively.
Anu : Chotu â‡’ 3:2 Eq.(i)
The amount obtained by Bindu is 83.33% more than the amount obtained by Chotu.
amount obtained by Bindu = (11/6) amount obtained by Chotu
Bindu : Chotu â‡’ 11:6 Eq.(ii)
From Eq.(i) and Eq.(ii).
Anu : Bindu : Chotu â‡’ 18:22:12
â‡’ 9:11:6
The average of the amount obtained by Anu and Bindu is 12000.
9y+11y = $12000\times2$ = 24000
20y = 24000
y = 1200
Total amount = 9y+11y+6y = 26y = $26\times1200$
= 31200
When divided between two people â‡’
Bindu + Anu = 31200 Eq.(iii)
If the amount will be divided between two people Anu and Bindu such as that Anu got Rs. 2600 less than Bindu.
Anu = Bindu – 2600
Bindu – Anu = 2600 Eq.(iv)
2Bindu = 31200+2600 = 33800
Bindu = 16900
Put the above value in Eq.(iv).
16900 – Anu = 2600
Anu = 16900-2600
= 14300
Amount obtained by Anu is more than the amount obtained by her earlier = 14300-$1200\times9$
= 14300-10800
= Rs. 3500
Hence, option b is the correct answer.

Question 12:Â In a 50 litres mixture of milk and water, percentage of water is 20%. The milkman sold 10 litres of this mixture and added 32 litres of milk and 24 litres of water in the remaining mixture. What is the percentage of water in the new mixture?

a)Â 33$\frac{1}{3}$%

b)Â 31$\frac{1}{3}$%

c)Â 31$\frac{2}{3}$%

d)Â 33$\frac{2}{3}$%

e)Â None of the above

Solution:

Total initial mixture = 50 litres
Milkman sold 10 litres of this mixture
Remaining mixture = 40 litres
Percentage of water = 20%
Quantity of water in the remaining mixture = $\dfrac{20}{100}\times$40 = 8 litres
Quantity of milk in the remaining mixture = 40 – 8 = 32 litres
After adding 32 litres of milk and 24 litres of water in the remaining mixture,
Quantity of milk in the new mixture = 32 + 32 = 64 litres
Quantity of water in the new mixture = 8 + 24 = 32 litres
$\therefore$ Percentage of water in the new mixture = $\dfrac{32}{64+32}\times$100
= $\dfrac{32}{96}\times$100
= 33$\frac{1}{3}$%
Hence, the correct answer is Option A

Question 13:Â An amount of Rs. 4410 was divided among three people Aman, Bhanu and Chandu such as the ratio between the amount obtained by Aman and Chandu is 1:2 respectively. The amount obtained by Bhanu is 25% less than the amount obtained by Aman. Find out the sum of the amount obtained by Bhanu and Chandu.

a)Â Rs. 3520

b)Â Rs. 3234

c)Â Rs. 3300

d)Â Rs. 3410

e)Â None of the above

Solution:

The ratio between the amount obtained by Aman and Chandu is 1:2 respectively.
Letâ€™s assume the amount obtained by Aman and Chandu is y and 2y respectively.
The amount obtained by Bhanu is 25% less than the amount obtained by Aman.
amount obtained by Bhanu = Â¾ of y = 0.75y
An amount of Rs. 4410 was divided among three people Aman, Bhanu and Chandu.
y+0.75y+2y = 4410
3.75y = 4410
y = 1176
Sum of the amount obtained by Bhanu and Chandu = 0.75y+2y
= 2.75y
= $2.75\times1176$
= Rs. 3234
Hence, option b is the correct answer.

Question 14:Â In a mixture of 280 litres, the quantity of water is 25% less than the quantity of juice. If 20 litres and â€˜yâ€™ litres of juice and water are added into the initial mixture, then the quantity of juice and water in the new mixture will be 9:8 respectively. Find out the value of â€˜yâ€™.

a)Â 25

b)Â 40

c)Â 50

d)Â 35

e)Â None of the above

Solution:

In a mixture of 280 litres, the quantity of water is 25% less than the quantity of juice.
quantity of water = 75% of quantity of juice
quantity of juice : quantity of water â‡’ 4:3
quantity of juice = 280 of (4/7) = 160 litres
quantity of water = 280 of (3/7) = 120 litres
If 20 litres and â€˜yâ€™ litres of juice and water are added into the initial mixture, then the quantity of juice and water in the new mixture will be 9:8 respectively.
$\frac{160+20}{120+y} = \frac{9}{8}$

$\frac{180}{120+y} = \frac{9}{8}$

$\frac{20}{120+y} = \frac{1}{8}$

160 = 120+y
y = 160-120 = 40
Hence, option b is the correct answer.

Question 15:Â A tank contains a mixture of milk and water in the ratio of 9 : 7 respectively. 32 litres of mixture is taken out and the tank is filled with 90 litres of milk, the ratio between milk and water becomes 2 : 1 respectively. What is the initial quantity of water in the tank?

a)Â 210 litres

b)Â 180 litres

c)Â 280 litres

d)Â 140 litres

e)Â 150 litres

Solution:

Initial ratio of milk and water = 9 : 7

Let the initial quantity of milk and water is 9p and 7p

32 litres of mixture is taken out. The quantity of milk and water taken out will be in the same ratio 9 : 7.

Quantity of milk taken out = $\frac{9}{9+7}\times32$ = 18 litres

Remaining quantity of milk = 9p – 18

Quantity of water taken out = $\frac{7}{9+7}\times32$ = 14 litres

Remaining quantity of water = 7p – 14

According to the problem,

$\frac{9p-18+90}{7p-14}=\frac{2}{1}$

$9p+72=14p-28$

$5p=100$

$p=20$

$\therefore$ Initial quantity of water = 7p = 7 x 20 = 140

Hence, the correct answer is Option D

Question 16:Â In a mixture, the initial quantity of milk and water is 4:1 respectively. If ‘yâ€™ and (y-15) litres of milk and water are poured into the mixture respectively, then in the mixture the new ratio of milk and water will be 7:2 respectively. If 40 litres of mixture is taken out from the initial mixture and (y-3) and (y-7) litres of milk and water are poured into the mixture respectively, then in the mixture, the new ratio of milk and water will be 49:15 respectively. Find out the value of â€˜yâ€™.

a)Â 50

b)Â 45

c)Â 30

d)Â 40

e)Â Cannot be determined

Solution:

In a mixture, the initial quantity of milk and water is 4:1 respectively.
Letâ€™s assume the initial quantity of milk and water in the mixture is 4z and z respectively.
If â€˜yâ€™ and (y-15) litres of milk and water are poured into the mixture respectively, then in the mixture the new ratio of milk and water will be 7:2 respectively.
$\frac{4z+y}{z+(y-15)} = \frac{7}{2}$
8z+2y = 7z+7y-105
z-5y = -105
5y-z = 105
z = 5y-105 Eq.(i)
If 40 litres of mixture is taken out from the initial mixture and (y-3) and (y-7) itres of milk and water are poured into the mixture respectively, then in the mixture the new ratio of milk and water will be 49:15 respectively.
$\frac{4z-32+(y-3)}{z-8+(y-7)} = \frac{49}{15}$

$\frac{4z-35+y}{z-15+y} = \frac{49}{15}$
60z-525+15y = 49z-735+49y
11z-34y = -735+525
34y-11z = 210 Eq.(ii)
Put Eq.(i) in Eq.(ii).
34y-11(5y-105) = 210
34y-55y+1155 = 210
21y = 945
y = 45
Hence, option b is the correct answer.

Question 17:Â In bucket A, the quantity of milk and water is 4:3 respectively. If 20 litres of milk is poured into the bucket A, then the ratio between the quantity of milk and water will become 3:2 respectively. In bucket B, the quantity of water is 30 litres less than the quantity of water in bucket A. If the total initial quantity of mixture in both the buckets together is 520 litres, then find out the quantity of milk in bucket B.

a)Â 110 litres

b)Â 120 litres

c)Â 140 litres

d)Â 180 litres

e)Â None of the above

Solution:

In bucket A, the quantity of milk and water is 4:3 respectively. If 20 litres of milk is poured into the bucket A, then the ratio between the quantity of milk and water will become 3:2 respectively.
$\frac{4y+20}{3y} = \frac{3}{2}$
8y+40 = 9y
9y-8y = 40
y = 40
Initial quantity of milk in bucket A = 4y = 160
Quantity of water in bucket A = 3y = 120
In bucket B, the quantity of water is 30 litres less than the quantity of water in bucket A.
Quantity of water in bucket B = 120-30 = 90
If the total initial quantity of mixture in both the buckets together is 520 litres.
Initial quantity of milk in bucket A + Quantity of water in bucket A + Quantity of milk in bucket B + Quantity of water in bucket B = 520
160+120+Quantity of milk in bucket B+90 = 520
Quantity of milk in bucket B+370 = 520
Quantity of milk in bucket B = 520-370 = 150 litres
Hence, option e is the correct answer.

Question 18:Â In 2024 litres of the mixture, the ratio between the quantity of milk and water is 5:3 respectively. If (y+44) litres of milk and â€˜yâ€™ litres of water is poured into the mixture, then the ratio between the quantity of milk and water will be 27:16 respectively. Find out the value of (y+8).

a)Â 49

b)Â 41

c)Â 45

d)Â 47

e)Â None of the above

Solution:

In 2024 litres of the mixture, the ratio between the quantity of milk and water is 5:3 respectively.
Initial quantity of milk = $\frac{5}{8}\times2024$
= 1265
Initial quantity of water = $\frac{3}{8}\times2024$
= 759
If (y+44) litres of milk and â€˜yâ€™ litres of water is poured into the mixture, then the ratio between the quantity of milk and water will be 27:16 respectively.

$\frac{1265+(y+44)}{759+y} = \frac{27}{16}$

$\frac{1309+y}{759+y} = \frac{27}{16}$

20944+16y = 20493+27y
27y-16y = 20944-20493
11y = 451
y = 41
value of (y+8) = 41+8 = 49
Hence, option a is the correct answer.

Question 19:Â A certain amount of money was divided among three persons P, Q, and R in the ratio 6:4:3 respectively. The amount obtained by P was further divided between his two sons in the ratio 7:5 respectively such that the elder son got more amount than the younger son. If the difference between the amount obtained by the elder son and that of the younger son of P is Rs.1786, then find the difference between the amount obtained by Q and that of R.

a)Â Rs.3572

b)Â Rs.4218

c)Â Rs.1786

d)Â Rs.993

e)Â None of these

Solution:

Let the amount obtained by P, Q and R be Rs.6x, Rs.4x and Rs.3x respectively.

Amount obtained by the elder son of P = $\dfrac{7}{12}\times6x = Rs.3.5x$

Amount obtained by the younger son of P = 6x – 3.5x = Rs.2.5x.

Given, 3.5x – 2.5x = 1786
x = 1786.

Hence, The required difference = 4x – 3x = Rs.1786.

Question 20:Â In a jar P, (y-35) litres of mixture is available, where the quantity of water is 63.63% less than the quantity of juice. In another jar Q, â€˜yâ€™ litres of mixture is available, where the quantity of juice is $\frac{3}{7}$ times more than the quantity of water. If the mixture of both the jars is mixed together, then the quantity of water will be 452 litres. Find out the value of â€˜yâ€™.

a)Â 520

b)Â 560

c)Â 640

d)Â 680

e)Â None of the above

Solution:

In a jar P, (y-35) litres of mixture is available, where the quantity of water is 63.63% less than the quantity of juice.
In a jar P, the ratio between the quantities of juice and water is 11:4 respectively.
In another jar Q, â€˜yâ€™ litres of mixture is available, where the quantity of juice is $\frac{3}{7}$ times more than the quantity of water.
In a jar Q, the ratio between the quantities of juice and water is 10:7 respectively.
If the mixture of both the jars is mixed together, then the quantity of water will be 452 litres.
(4/15) of (y-35) + (7/17) of y = 452

68y-2380+105y = 115260
173y = 115260+2380 = 117640
y = 680
Hence, option d is the correct answer.

Question 21:Â In a â€˜yâ€™ liters of mixture, the quantity of milk is 37.5% more than the quantity of water. If 10 litres of water is added into the mixture, then the ratio between the quantity of milk to water will become 4:3. Find out the difference between the quantity of milk and water in the initial mixture.

a)Â 125 liters

b)Â 175 liters

c)Â 180 liters

d)Â 120 liters

e)Â None of the above

Solution:

In a â€˜yâ€™ liters of mixture, the quantity of milk is 37.5% more than the quantity of water.
The ratio between the initial quantity of milk and water is 11:8 respectively.
If 10 litres of water is added into the mixture, then the ratio between the quantity of milk to water will become 4:3.
$\frac{\frac{11}{19}y}{\frac{8}{19}y + 10} = \frac{4}{3}$

$\frac{11y}{8y + 190} = \frac{4}{3}$

33y = 32y + 760
y = 760
Difference between the quantity of milk and water in the initial mixture = $\frac{11}{19}y – \frac{8}{19}y$
= $\frac{3}{19}y$

= $\frac{3}{19}\times760$

= 120 liters
Hence, option d is the correct answer.