BANK & INSURANCE (MIXTURE AND ALLIGATION) PART 3

Total Questions: 30

11. A mixture of milk, water and honey contains 32 litres more water than milk and __% as much honey as water. If half of the mixture is replaced with __ litres of honey, then the sum of quantities of milk and honey in the resultant mixture will be 20% more than that of water. The original quantity of milk in the mixture was __ litres.

The values given in which of the following options will fill the blanks in the same order so as to make the above statement true?
I. 60, 5, 78  II. 40, 6, 75  III. 55, 2, 208

Correct Answer: (e) None of these
Solution:For I:

Original quantity of milk in the mixture = 78 litres
So, original quantity of water in the mixture = 78 + 32 = 110 litres
And so, original quantity of honey in the mixture = 110 × 0.6 = 66 litres

After replacing half of the mixture with 5 litres of honey,
Quantity of milk in the mixture = 78 ÷ 2 = 39 litres
Quantity of water in the mixture = 110 ÷ 2 = 55 litres
Quantity of honey in the mixture = 66 ÷ 2 + 5 = 38 litres

So, sum of quantities of milk and honey = 39 + 38 = 77 ≠ 1.2 × 55
So, ‘I’ is false.

For II:
Original quantity of milk in the mixture = 75 litres
So, original quantity of water in the mixture = 75 + 32 = 107 litres
And so, original quantity of honey in the mixture = 107 × 0.4 = 42.8 litres

After replacing half of the mixture with 6 litres of honey,
Quantity of milk in the mixture = 75 ÷ 2 = 37.5 litres
Quantity of water in the mixture = 107 ÷ 2 = 53.5 litres
Quantity of honey in the mixture = 42.8 ÷ 2 + 6 = 27.4 litres

So, sum of quantities of milk and honey = 37.5 + 27.4 = 64.9 ≠ 53.5 × 1.2
So, ‘II’ is false.

For III:
Original quantity of milk in the mixture = 208 litres
So, original quantity of water in the mixture = 208 + 32 = 240 litres

And so, original quantity of honey in the mixture = 240 × 0.55 = 132 litres

After replacing half of the mixture with 6 litres of honey,
Quantity of milk in the mixture = 208 ÷ 2 = 104 litres
Quantity of water in the mixture = 240 ÷ 2 = 120 litres
Quantity of honey in the mixture = 132 ÷ 2 + 2 = 68 litres

So, sum of quantities of milk and honey = 104 + 68 = 172
But, 120 × 1.2 = 144 ≠ 172

So, ‘III’ is false.

Hence, option a.

12. Mixture ‘A’ and ‘B’ contain milk and water in them. Mixture ‘A’ contains 40% water while mixture ‘B’ contains same quantity of water as in mixture ‘A’. Both the mixtures are completely emptied in a 600 litre capacity container along with 50 litres water to make sure that the container is filled to the brim. If the ratio of quantity of milk to that of water in the container is 7:5, then find the quantity of milk in mixture ‘B’.

Correct Answer: (a) 200 litres 
Solution:

Quantity of water in the container
= 600 × (5/12) = 250 litres

So, quantity of water in the two mixtures together
= 250 - 50 = 200 litres

So, quantity of water in each mixture = 200 ÷ 2
= 100 litres

Let the total quantity of mixture ‘A’ be ‘x’ litres.

ATQ:
0.4x = 100

So, x = 250

So, quantity of mixture ‘B’ = 600 - 50 - 250 = 300 litres

So, quantity of milk in mixture ‘B’ = 300 - 100 = 200 litres

Hence, option a.

13. 360 ml of mixture ‘A’ containing ‘x%’ more water than milk in it is mixed with 480 ml of mixture ‘B’ which contains 16.x% more water than milk in it. Total quantity of mixture ‘C’ is 240 ml and it contains (x + 5)% of milk and 168 ml of water in it. Which of the following statement(s) is/are true?

I. x = 25
II. Quantity of milk in mixture ‘A’ is 200 ml
III. Quantity of milk in mixture ‘B’ is 160 ml

Correct Answer: (a) Only I 
Solution:

Let quantity of milk in mixture ‘A’ is ‘a’ ml

So, quantity of water in mixture ‘A’ = a(1 + x/100) ml

So, a + a(1 + x/100) = 360

a(1 + 1 + x/100) = 360

a(2 + x/100) = 360 .......... (1)

Let quantity of milk in mixture ‘B’ is ‘b’ ml

So, quantity of water in mixture ‘B’ = b(1 + 1.6x/100) ml

So, b + b(1 + 1.6x/100) = 480

b(1 + 1 + 1.6x/100) = 480

b(2 + 1.6x/100) = 480 .......... (2)

((x + 5)/100) × 240 = 72

So, x + 5 = 30

x = 25

Solving equation ‘I’ and ‘II’, we get
a = 160, b = 200

For ‘I’:
Since, x = 25
So, ‘I’ is true.

For ‘II’:
Since, quantity of milk in mixture ‘A’ = a = 160 ml
So, ‘II’ cannot be true.

For ‘III’:
Since, quantity of milk in mixture ‘B’ = b = 200 ml
So, ‘III’ cannot be true.

Hence, option a.

14. Directions (14-15): Answer the questions based on the information given below.

Mixture ‘A’ contains milk and water in the ratio 7:8, respectively and mixture ‘B’ contains milk and water in the ratio 5:4, respectively. If ‘P’ litres of mixture ‘A’ and ‘Q’ litres of mixture ‘B’ was taken out in an empty container ‘X’ and then R% of that mixture from container ‘X’ was taken out and was replaced with same quantity of juice, then the ratio of milk, water and juice in the container becomes 2:2:1, respectively. Roots of the equation, x² – 90x – 9000 = 0 are (2P) and -(Q + 15) respectively, (Where P > Q).

Ques. Find the value of (R + 25).

Correct Answer: (c) 45 
Solution:

x² – 90x – 9000 = 0
x² – 150x + 60x – 9000 = 0
x(x – 150) + 60(x – 150) = 0
(x + 60)(x – 150) = 0
x = -60, 150

As, P > Q
2P = 150 and – (Q + 15) = -60
P = 75 and Q = 45

Now,
Quantity of milk in 75 litres of mixture ‘A’ = 75 × 7/15
= 35 litres.

Quantity of water in 75 litres of mixture ‘B’ = 75 × 8/15 = 40 litres.

Quantity of milk in 45 litres of mixture ‘B’ = 45 × 5/9
= 25 litres.

Quantity of water in 45 litres of mixture ‘B’ = 45 × 4/9
= 20 litres.

Total quantity of milk in the container = 35 + 25 = 60 litres.

Total quantity of water in the container = 40 + 20 = 60 litres.

Ratio of milk to water in the container = 60:60
= 1:1

Total quantity of mixture in the container = 60 + 60 = 120 litres.

After the first replacement,
Quantity of milk = 60 × (120 – R)/100
= 3(100 – R)/5 litres.

Quantity of water = 60 × (100 – R)/100
= 3(100 – R)/5 litres.

Quantity of juice = 120 × R/100 = 6R/5 litres.

According to the statement,
{3(100 – R)/5}/{6R/5} = 2:1

(100 – R)/2R = 2

100 – R = 4R

R = 100/5 = 20

R = 20

Required equation:
(x – 20)(x – 45) = 0
x² – 65x + 900 = 0

Hence, 45 is answer .

15. Find the quadratic equation whose roots are ‘Q’ and ‘R’.

Correct Answer: (b) x² – 65x + 900 = 0
Solution:

x² − 90x − 9000 = 0
x² − 150x + 60x − 9000 = 0
x(x − 150) + 60(x − 150) = 0
(x + 60)(x − 150) = 0
x = −60, 150
As, P > Q
2P = 150 and − (Q + 15) = −60
Or, P = 75 and Q = 45
Now,
Quantity of milk in 75 litres of mixture ‘A’
= 75 × 7/15 = 35 litres.
Quantity of water in 75 litres of mixture ‘B’
= 75 × 8/15 = 40 litres.
Quantity of milk in 45 litres of mixture ‘B’
= 45 × 5/9 = 25 litres.
Quantity of water in 45 litres of mixture ‘B’
= 45 × 4/9 = 20 litres.
Total quantity of milk in the container
= 35 + 25 = 60 litres.

Total quantity of water in the container
= 40 + 20 = 60 litres.
Ratio of milk to water in the container = 60:60
= 1:1
Total quantity of mixture in the container = 60 + 60
= 120 litres.
After the first replacement,
Quantity of milk = 60 × (120 − R)/100
= 3(100 − R)/5 litres.
Quantity of water = 60 × (100 − R)/100
= 3(100 − R)/5 litres.
Quantity of juice = 120 × R/100 = 6R/5 litres.
According to the statement,
{3(100 − R)/5}/{6R/5} = 2:1
(100 − R)/2R = 2
100 − R = 4R
R = 100/5 = 20
R = 20
Required equation:
(x − 20)(x − 45) = 0
x² − 65x + 900 = 0
Hence, option b.

16. 504 ml of mixture ‘A’ contains milk and water in the ratio of 5:x, respectively; ‘63x’ ml of mixture ‘A’ is taken out and the remaining mixture is mixed with (12x + 40) ml of mixture ‘B’ containing milk and water in the ratio of (x + 1):(x + 2), respectively. If ratio of milk and water in the resultant mixture is 9:8, respectively, then find the value of ‘x’.

Correct Answer: (c) 4 
Solution:

Quantity of milk in mixture ‘A’
= {5/(5 + x)} × 504 ml
Quantity of water in mixture ‘A’
= {x/(5 + x)} × 504 ml

Quantity of milk taken out = {5/(5 + x)} × 63 ml
Quantity of water taken out = {x/(5 + x)} × 63 ml

Quantity of milk in mixture ‘B’
= {(x + 1)/(2x + 3)} × (12x + 40) ml

Quantity of water in mixture ‘B’ = {(x + 2)/(2x + 3)} × (12x + 40) ml

According to question;
[{5/(5 + x)} × (504 − 63x) + {(x + 1)/(2x + 3)} × (12x + 40)] / [{x/(5 + x)} × (504 − 63x) + {(x + 2)/(2x + 3)} × (12x + 40)] = 9:8

The above equation will satisfy only when ‘x’ = 4
Hence, option c.

17. A mixture contains 120 litres milk and rest water. 40% of the mixture is taken out in container ‘A’ and 12 litres of water is added in container ‘A’. In the remaining mixture, 30 litres of milk and 18 litres of water is added. These two mixtures are again mixed together such that the ratio of milk to water in the resultant mixture becomes 3:7. Find the initial quantity of water in the mixture.

Correct Answer: (c) 320 litres
Solution:

Let the quantity of water in the initial mixture be ‘x’ litres

According to the question,
(120 + 30)/(0.6x + 0.4x + 12 + 18) = 3/7
Or, x = 350 − 30 = 320 litres

Hence, option c.

18. Each of the three mixtures ‘A’, ‘B’ and ‘C’ contain petrol and diesel, only. The ratios of amount of petrol to diesel in mixtures ‘A’, ‘B’ and ‘C’ are 3:5, 5:4 and 8:3, respectively. The quantity of petrol in mixtures ‘A’ and ‘B’ are equal. The sum of quantities of petrol in mixtures ‘B’ and ‘C’ is equal to 100 litres while mixture ‘C’ contains 33 litres less diesel than that in ‘B’. Find the ratio of quantity of diesel in mixtures ‘B’ and ‘C’.

Correct Answer: (a) 16:5 
Solution:

Let the quantity of petrol and diesel in mixture ‘A’ be 3x litres and 5x litres, respectively
Therefore, quantity of petrol in mixture ‘B’ = 3x litres
Quantity of diesel in mixture ‘B’ = 3x × (4/5)
= 2.4x litres

Quantity of petrol in mixture ‘C’ = (100 − 3x) litres
Quantity of diesel in mixture ‘C’ = (2.4x − 33) litres

According to the question,
(100 − 3x)/(2.4x − 33) = 8/3
300 − 9x = 19.2x − 264
28.2x = 564
x = 20

Therefore, quantity of diesel in mixture ‘B’ = 2.4x = 48 litres

Quantity of diesel in mixture ‘C’
= 2.4x – 33 = 48 – 33 = 15 litres
Required ratio = 48:15 = 16:5

19. A vessel contains 80 litres of pure alcohol. A jar is used to withdraw the alcohol and replace it with the same quantity of water. When the same process is repeated 2 more times, the quantity of alcohol left in the jar is 40.96 litres. Find the capacity of the jar.

Correct Answer: (b) 16 litres 
Solution:

Let the capacity of jar be ‘x’ litres
So, 80 × (1 – (x/80))³ = 40.96
(1 – (x/80))³ = 0.512
1 – (x/80) = 0.8
0.2 = (x/80)
x = 16

Therefore, the capacity of the jar is 16 litres.

20. A (x + 32) litres of a mixture (milk + water) contains 84 litres milk such that if (x – 64) litres of the mixture is replaced with 4 litres of milk, then quantity of milk would be 60% of the resultant mixture. Find the value of ‘x’.

Correct Answer: (e) 112
Solution:

Quantity of mixture (milk + water) remaining after removing (x – 64) litres = x + 32 – (x – 64)
= 96 litres

Let the quantity of milk and water in 96 litres of the mixture be ‘y’ litres and (96 – y) litres

According to question,
(y + 4)/(96 – y) = (60/40) = (3/2)

2y + 8 = 288 – 3y
5y = 280
So, y = (280/5) = 56

So, ratio of quantity of milk to that of water in the original mixture = 56:(96 – 56) = 56:40 = 7:5

So, quantity of water in the (x + 32) litre mixture = 84 × (5/7) = 60

So, 84 + 60 = x + 32
x + 32 = 144
So, x = 144 – 32 = 112