BANK & INSURANCE (MIXTURE AND ALLIGATION) PART 3

Total Questions: 30

21. A jar contains 120 litres of pure acid. Using a mug, 20% of the acid is replaced with water. Find the minimum number of times, does this procedure have to be followed before the quantity of acid in the jar falls below 60 litres.

Correct Answer: (e) 4 times
Solution:

ATQ
120 × (0.8)ⁿ ≤ 60

Quantity of acid in the tank after:
1st replacement = 120 × 0.80 = 96 litre
2nd replacement = 96 × 0.80 = 76.8 litres
3rd replacement = 76.8 × 0.80 = 61.44 litres

It is obvious that the quantity of acid will fall below 60 litres in the tank after one more replacement.
Therefore, the procedure has to be followed a total of 4 times.

Hence, option e.

22. Two jars ‘A’ and ‘B’ having equal capacities are completely filled with solution of ethanol. Jar ‘A’ contains 12% ethanol while jar ‘B’ contains 47% ethanol. Some of the solution from the jar ‘A’ is withdrawn and replaced with same quantity of solution from jar ‘B’. If jar ‘A’ now contains 26% ethanol, then how much percent solution is left in jar ‘B’ now?

Correct Answer: (a) 60% 
Solution:

A → 12%
B → 47%

26%

47% – 26% = 21%
26% – 12% = 14%

So, ‘A’ and ‘B’ were mixed in ratio of 3:2, respectively.
Since, total both the jars contained equal quantity of the mixture
So, required percentage = {(5 – 2)/5 × 100} = 60%

Hence, option a.

23. A 324 litre mixture of milk, water and honey contains water and honey in the ratio 3:5, respectively and the quantity of milk in the mixture is 60 litres more than that of water. If 54 litres of the mixture is replaced with 12 litres of milk and 18 litres of water, then water would form how much percentage of the resultant mixture?

Correct Answer: (c) 26% 
Solution:

Let the quantity of water in the 324 litre mixture be ‘3y’ litres

Then, quantity of honey in the 324 litre mixture
= (5/3) × 3y = ‘5y’ litres

Quantity of milk in the 324 litre mixture = (3y + 60) litres

So, 3y + 5y + 3y + 60 = 324
11y + 60 = 324
y = (324 – 60) ÷ 11 = 24

So, quantity of milk, water and honey in the 324 litres mixture is 132 litres, 72 litres and 120 litres, respectively.

So ratio of quantity of milk, water and honey in the mixture, respectively = 132:72:120 = 11:6:10

After replacing 54 litres of the mixture with 12 litres milk and 18 litres water,

Quantity of milk in the resultant mixture = (324 – 54) × (11/27) + 12 = 122 litres

Quantity of water in the resultant mixture = (324 – 54) × (6/27) + 18 = 78 litres

Quantity of honey in the resultant mixture = (324 – 54) × (10/27) = 100 litres

So, percentage of water in the mixture = (78 ÷ (122 + 78 + 100)) × 100
= (78/300) × 100 = 26%.

24. A 180 litre mixture of milk and water, contains __ litres more milk than water. If 60 litres of the mixture is replaced with __ litres of water, then the ratio of quantity of milk to that of water in the resultant mixture would become 3:4.

The values given in which of the following options will fill the blanks in the same order in which it is given so as to make the statement true?
I. 60, 45  II. 36, 48  III. 72, 80

Correct Answer: (b) Only II 
Solution:

From I:
Let the quantity of water in the mixture = ‘x’ litres

Then, quantity of milk in the mixture = (x + 60) litres

So, x + x + 60 = 180
2x + 60 = 180
So, x = (180 – 60) ÷ 2 = 60

Therefore, ratio of quantities of milk and water, respectively, in the mixture = 120:60 = 2:1

After replacing 60 litres of the mixture with 45 litres of water,

Quantity of milk in the resultant mixture = (180 – 60) × (2/3) = 80 litres

Quantity of water in the resultant mixture = (180 – 60) × (1/3) + 45 = 85 litres

So, ratio of quantity of milk to that of water in the resultant mixture = 80:85 = 16:17 ≠ 3:4

So, I is false.

From II:
Let the quantity of water in the mixture = ‘x’ litres

Then, quantity of milk in the mixture = (x + 36) litres

So, x + x + 36 = 180
Or, 2x + 36 = 180
So, x = (180 – 36) ÷ 2 = 72

Therefore, ratio of quantities of milk and water, respectively, in the mixture = 108:72 = 3:2

After replacing 60 litres of the mixture with 48 litres of water,

Quantity of milk in the resultant mixture = (180 – 60) × (3/5) = 72 litres

Quantity of water in the resultant mixture = (180 – 60) × (2/5) + 48 = 96 litres

So, ratio of quantity of milk to that of water in the resultant mixture = 72:96 = 3:4
So, II is true.

From III:
Let the quantity of water in the mixture = ‘x’ litres
Then, quantity of milk in the mixture = (x + 72) litres
So, x + x + 72 = 180
Or, 2x + 72 = 180
So, x = (180 – 72) ÷ 2 = 54

Therefore, ratio of quantities of milk and water, respectively, in the mixture = 126:54 = 7:3

After replacing 60 litres of the mixture with 80 litres of water,
Quantity of milk in the resultant mixture = (180 – 60) × (7/10) = 84 litres

Quantity of water in the resultant mixture = (180 – 60) × (3/10) + 80 = 116 litres

So, ratio of quantity of milk to that of water in the resultant mixture = 84:116 = 21:29 ≠ 3:4

So, III is false.
Hence, option b.

25. Jar ‘A’ and ‘B’ contains mixture of salt and water such that quantity of salt in jars ‘A’ and ‘B’ is 40% and 66%, respectively. Some of the mixture is taken out from both jars and poured into an empty container. If the container now contains 260 litres of mixture such that ratio of quantity of salt and water in it is 13:12, respectively, then how much mixture was withdrawn from jar ‘A’?

Correct Answer: (a) 140 litres 
Solution:

Percentage of quantity of salt in the container = 100 × (13/25) = 52%

Using alligation, we have;

Jar ‘A’ → 40%
Jar ‘B’ → 66%

52%

66% – 52% = 14%
52% – 40% = 12%

Required ratio = 14:12 = 7:6

So, Quantity of mixture withdrawn from jar ‘A’ = 260 × (7/13) = 140 litres

26. A container contains 960 litres of pure milk. A shopkeeper took out ‘x’ litres of milk and replaced it with the same quantity of water and repeated the same process two more time such that quantity of milk in the resultant mixture becomes 823.08 litres. Find the value of ‘p’ if one of the roots of the quadratic equation represented as (x – 39)y² + py – 8 = 0 is -4.

Correct Answer: (e) None of these
Solution:

ATQ;
960 × {1 – (x/960)}³ = 823.08

{1 – (x/960)}³ = (6859/8000) = (19/20)³

1 – (x/960) = (19/20)

(x/960) = 1 – (19/20) = (1/20)

x = 48

Given equation is:
(48 – 39)y² + py – 8 = 0
9y² + py – 8 = 0

Let another root of the equation be ‘r’
So, (-4) × r = (-8/9)
r = (2/9)

Also, (2/9) – 4 = (-p/9)
(-34/9) = (-p/9)
p = 34

27. Jugs ‘A’ and ‘B’ having capacity of 300 litres each are completely filled with a mixture containing 180 litres milk and 120 litres water. Half of the mixture in jug ‘A’ is replaced with an equal quantity of water and this process is repeated one more time. Now total mixture in jug ‘A’ must be mixed with how many litres of mixture in jug ‘B’ such that the resultant mixture has 33% of milk in it?

Correct Answer: (a) 200 litres 
Solution:

After replacing half of the mixture in jug ‘A’ with equal quantity of water,

Quantity of milk in jug ‘A’ = 180 – 180 × 0.5 = 90 litres

Quantity of water in jug ‘A’ = 120 – 120 × 0.5 + (300/2) = 210 litres

After repeating this process again,

Quantity of milk in jug ‘A’ = 90 – 90 × 0.5 = 45 litres

Quantity of water in jug ‘A’ = 210 – 210 × 0.5 + (300/2) = 255 litres

So, percentage of quantity of milk in jug ‘A’
= (45/300) × 100 = 15%

Percentage of quantity of milk in jug ‘B’
= (180/300) × 100 = 60%

Using alligation rule,

Percentage of milk in jug ‘A’ = 15%
Percentage of milk in jug ‘B’ = 60%

Desired percentage of milk = 33%

60 – 33 = 27
33 – 15 = 18

Desired ratio = 27:18 = 3:2

So, quantity of mixture in jug ‘B’ to be mixed = 300 × (2/3) = 200 litres.

28. 456 ml of mixture contains milk and water in the ratio of 12:7, respectively. If ‘a’ ml of mixture is taken out and 285 ml of mixture (milk + water) containing 156(1/4)% more water than milk in it, is mixed with the remaining mixture then the ratio of milk to water in the resultant mixture becomes 7:12. Find the value of ‘a’.

Correct Answer: (b) 361 
Solution:

Quantity of milk in 456 ml of mixture = 456 × 12/19 = 288 ml

Quantity of water in 456 ml of mixture
= 456 – 288 = 168 ml

Let quantity of milk and water taken out be ‘12x’ ml and ‘7x’, respectively.

156(1/4)% = 25/16

Let quantity of milk in 285 ml of mixture is ‘b’ ml

So, quantity of water in 285 ml of mixture = (1 + 25/16) × b = 41b/16 ml

So, b + 41b/16 = 285
b = 80

So, quantity of milk in 285 ml of mixture = 80 ml

So, quantity of water in 285 ml of mixture = 285 – 80 = 205 ml

So, (288 – 12x + 80)/(168 – 7x + 205) = 7/12

4416 – 144x = 2611 – 49x

95x = 1805

x = 19

So, a = 19x = 19 × 19 = 361 ml.

29. Brass and Bronze are the two alloys which were made by mixing X and Y in the ratio 6:9 and 7:11 respectively. Scientists are planning to develop a new alloy which will help in reducing corrosion and have come up with a new technique to implement the same. For implementation, 40 grams of alloy Brass and 60 grams of alloy bronze are melted and mixed to form another alloy Z. What is the ratio of X and Y in the new alloy Z?

Correct Answer: (b) 59:91 
Solution:

Alloy Brass contains:
X = 16 grams
Y = 24 grams

Alloy Bronze contains:
X = 70/3 gram
Y = 110/3 gram

Alloy Z contains:
X = 16 + 70/3 = 118/3 gram
Y = 24 + 110/3 = 182/3 gram

The required ratio = 59 : 91.

30. A man mixes two kinds of pulses and mixes them in a certain ratio. He buys the first kind of pulses at Rs. 22 per kg and second one at Rs. 16 per kg. How much quantity of the first kind should be mixed with 25 kg of the second kind so that he can gain a profit of 20% if he is selling the mixture at Rs. 24 per kg?

Correct Answer: (b) 50kg 
Solution:

S.P of 1 kg of mixture = Rs. 24
Gain = 20%

C.P of 1 kg of mixture = (100/120) × 24 = Rs. 20

C.P of 1 kg of the 1st kind of pulses = 22
C.P of 1 kg of pulses of 2nd kind = 16

Mean price = 20

By the rule of allegation, we have:

C.P of 1kg of pulses of 1st kind (22)
C.P of 1 kg of pulses of 2nd kind (16)

Mean Price (20)

20 – 16 = 4
22 – 20 = 2

Required ratio = 4 : 2 = 2 : 1

Let x kg of pulses of the 1st kind be mixed with 25 kg of 2nd kind

2/1 = x/25
x = 50 kg