ALLIGATION AND MIXTURE QUESTIONS

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Question 1 :

What is the ratio in which the two types of wheat mixed where the price of the first type is $9.30 per kg and the second type is $10.80 per kg so the mixture is having worth $10 per kg?

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From the given information, we have

cost price of the cheaper (c) = $9.30
cost price of the dearer (d) = $10.80
cost price of the mixture (m) = $10

Rule to find the ratio for producing mixture is (d-m):(m-c)
(d-m):(m-c) = (10.8-10):(10-9.3) = 0.8:0.7 =8:7

Hence the ratio in which the first kind and second kind to be mixed is 8:7

Question 2 :

Find the ratio in which, water to be mixed with milk to gain 20% by selling the mixture at cost price

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Let the cost price of 1 ltr pure milk be $1

Now we take some quantity of milk (less than 1 ltr),add some water and make it to be 1 ltr mixture

Let "x" be the money we invest for milk in 1 ltr of milk-water mixture

Since the gain is 20%, selling price = x + 20%of x
Selling price = 120% of x = (120/100)x = (6/5)x

But the mixture is sold at the cost price of pure milk
So, we have (6/5)x = 1
x = 5/6
Therefore cost price of the milk in the mixture = $(5/6

Cost price of the water = $0

Rule to find the ratio for producing mixture = (d-m):(m-c)
(d-m):(m-c) = 1-5/6:5/6-0 = 1/6:5/6 = 1:5

So, water and milk to be mixed in the ratio to gain 20% is 1:5

Question 3 :

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?

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Let the cost price of 1 liter pure milk be $1

Milk in 1 liter of mixture in A = 4/7 liter
Milk in 1 liter of mixture in B = 2/5 liter
Milk in 1 liter of mixture in C = 1/2 liter

C.P of 1 liter mixture in A (c)= $4/7
C.P of 1 liter mixture in B (d)= $2/5
C.P of 1 liter mixture in C (m)= $1/2
(cost of water is $0)

Rule to find the ratio for producing mixture = (d-m):(m-c) (d-m):(m-c) = (2/5-1/2):(1/2-4/7) = 1/10:1/14 = 7:5

Hence the required ratio is 7:5

Question 4 :

A milk vender has two cans of milk. The first contains 25% water and the rest milk. The second contains 50% water and the rest milk. How many liters of milk should he mix from each of the cans so as to get 12 liters of milk such that that the ratio of water to milk is 3:5?

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Let the cost price of 1 liter pure milk be $1

Milk in 1 liter mixture in the 1st can = 3/4 (that is 75%)
Milk in 1 liter mixture in the 2nd can = 1/2 (that is 50%)
Milk in 1 liter mixture in the final mix = 5/8
(from the given ratio w:m = 3:5)

C.P of 1 liter mixture in the 1st can (c) = $3/4
C.P of 1 liter mixture in the 2nd can (d) = $1/2
C.P of 1 liter mixture in the final mix (m) = $5/8

Rule to find the ratio for producing mixture = (d-m):(m-c)
(d-m):(m-c) = 1/2-5/8:5/8-3/4 = 1/8:1/8 = 1:1

The above found ratio 1:1 says that equal quantity of milk should be taken from each of the cans.

Since he wants to get 12 liters of milk, he should take 6 liters of milk from each of the cans.

The correct answer is option (B) (6,6).

Question 5 :

A merchant has 1000 kg of sugar, the part of which he sells at 8% profit and the rest at 18% profit. He gets 14% profit on the whole. The quantity sold at 18% profit is

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From the given information, we have

Profit on the first part (c) = 8%
Profit on the second part (d) = 18%
Profit on the whole (mixture) (m) = 14%

Rule to find the ratio for producing mixture = (d-m):(m-c)
(d-m):(m-c) = (18-14):(14-8) = 4:6 = 2:3

Quantity of the 2nd kind = 1000X3/5 = 600 kg

The quantity sold at 18% profit is 600 kg

Question 6 :

A dealer mixes tea costing $6.92 per kg. with tea costing $7.77 per kg. and sells the mixture at $8.80 per kg. and earns a profit of 17.5% on his sale price. In what ratio does he have to mix them?

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From the given information, we have

Cost price of the cheaper (c) = 6.92
Cost price of the dearer (d) = 7.77
Cost price of the mixture = selling price - profit
= 8.80-17.5%of8.8 = 8.80-1.54 = 7.26

Therefore, cost price of the mixture (m) = 7.26

Rule to find the ratio for producing mixture = (d-m):(m-c)
(d-m):(m-c) = (7.77-7.26):(7.26-6.92) = 0.51:0.34 = 51:34
So, (d-m):(m-c) = 3:2

Hence the required ratio is 3:2

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ALLIGATION AND MIXTURE QUESTIONS

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