MIXTURE & ALLIGATION CONCEPT
Concept  Practice set 1  Practice set 2
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No. of Questions 
10 
Time 
10 min 
Medium 
english 
Marks 
10 
Positive Marks 
+1 
Negative Marks 
0 
INSTRUCTION : All question carry 1 mark and there is only single option correct for every questionMark review to see the question later 
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Question 1 of 10
1. Question
1 pointsIn what ratio must wheat A at Rs. 10.50 per kg be mixed with wheat B at Rs. 12.30 per kg, so that the mixture be worth of Rs. 11 per kg?
Correct
Correct option :(a)
Convert Rs into paise, to make the calculation easy
Ratio =(B – M)/(M – A)
The required ratio = 130 : 50 = 13 : 5Incorrect
Correct option :(a)
Convert Rs into paise, to make the calculation easy
Ratio =(B – M)/(M – A)
The required ratio = 130 : 50 = 13 : 5 
Question 2 of 10
2. Question
1 pointsIn what ratio must a shopkeeper mix Peas and Soybean of Rs. 16 and Rs. 25 per kg respectively, so as to obtain a mixture of Rs. 19.50 ?
Correct
Correct option: (c)
Use rule of alligation, to determine the ratio
The required ratio of Soybean and Peas = 5.50 : 3.50 = 11 : 7Incorrect
Correct option: (c)
Use rule of alligation, to determine the ratio
The required ratio of Soybean and Peas = 5.50 : 3.50 = 11 : 7 
Question 3 of 10
3. Question
1 points10 gallons are drawn from a container full of alcohol and filled with water again. 10 gallons of mixture are again drawn and the container is filled with water again. If the ratio of alcohol and water left in the container is 49 : 32, then find how much quantity does the container hold?
Correct
Correct option : (b)
Initially the container contains only wine. 10 gallons of alcohol was removed and same quantity of water was added.
This process is again repeated by replacing the mixture( alcohol + water) of 10 gallons with same quantity of water. Hence, the initial quantity of wine and the final quantity of water and alcohol is the same.
1) First assume that the initial quantity of alcohol is ‘A’ .
2) We are given that, the ratio of alcohol and water is 49 : 32
3) Assume initial quantity of alcohol in the container = 49 + 32 = 81 —– (This is because we have assumed that initial quantity of alcohol = final quantity of water and alcohol)
4) Subtract the quantity of alcohol replaced by water from the initial quantity of alcohol (A – B). As this operation is repeated n times, therefore (A – B)^{n}Therefore,
(Quantity of alcohol left after n^{th} operation)/ = (A – B)^{n} (Initial quantity of alcohol) or (Volume of flask) A 49 = (A – 10)^{2} 81 A Solving, we can find the value of A (initial quantity of alcohol)
A = 45 gallonsIncorrect
Correct option : (b)
Initially the container contains only wine. 10 gallons of alcohol was removed and same quantity of water was added.
This process is again repeated by replacing the mixture( alcohol + water) of 10 gallons with same quantity of water. Hence, the initial quantity of wine and the final quantity of water and alcohol is the same.
1) First assume that the initial quantity of alcohol is ‘A’ .
2) We are given that, the ratio of alcohol and water is 49 : 32
3) Assume initial quantity of alcohol in the container = 49 + 32 = 81 —– (This is because we have assumed that initial quantity of alcohol = final quantity of water and alcohol)
4) Subtract the quantity of alcohol replaced by water from the initial quantity of alcohol (A – B). As this operation is repeated n times, therefore (A – B)^{n}Therefore,
(Quantity of alcohol left after n^{th} operation)/ = (A – B)^{n} (Initial quantity of alcohol) or (Volume of flask) A 49 = (A – 10)^{2} 81 A Solving, we can find the value of A (initial quantity of alcohol)
A = 45 gallons 
Question 4 of 10
4. Question
1 pointsA container is filled with a mixture of water and milk in the ratio of 3 : 5. Find the quantity of mixture to be drawn off and replaced with water, in order to get the mixture as half milk and half water.
Correct
Correct option : (c)
A container contains milk and water in the ratio of 3 : 5. This means that the vessel contains 8 litres of mixture.
Assume that x litres of this mixture is replaced with water.
From the mixture containing water and milk of 3 : 5, x quantity of mixture is withdrawn and is replaced by water of the same quantity in the mixture. From the mixture (3x)/(8) part of water is removed and x quantity of water is added.
1) Quantity of water in the newly formed mixture = [3 – ( 3x ) + x] 8 — (3 is the quantity of water, x is the quantity of mixture replaced by water)
2) Similarly, quantity of milk in the newly formed mixture = [5 – ( 5x )] 8 — (Here x is not added because only water was added in the mixture and not the syrup)
Therefore,
[ 3 – ( 3x ) + x] = [5 – ( 5x )] 8 8 Solving 1 and 2, we get
5x + 24 = 40 5x
x = 8 / 5
So the part of mixture replaced from 8 litres = 8 * 1 = 1 5 8 5 Incorrect
Correct option : (c)
A container contains milk and water in the ratio of 3 : 5. This means that the vessel contains 8 litres of mixture.
Assume that x litres of this mixture is replaced with water.
From the mixture containing water and milk of 3 : 5, x quantity of mixture is withdrawn and is replaced by water of the same quantity in the mixture. From the mixture (3x)/(8) part of water is removed and x quantity of water is added.
1) Quantity of water in the newly formed mixture = [3 – ( 3x ) + x] 8 — (3 is the quantity of water, x is the quantity of mixture replaced by water)
2) Similarly, quantity of milk in the newly formed mixture = [5 – ( 5x )] 8 — (Here x is not added because only water was added in the mixture and not the syrup)
Therefore,
[ 3 – ( 3x ) + x] = [5 – ( 5x )] 8 8 Solving 1 and 2, we get
5x + 24 = 40 5x
x = 8 / 5
So the part of mixture replaced from 8 litres = 8 * 1 = 1 5 8 5 
Question 5 of 10
5. Question
1 pointsFind in what ratio must water be mixed with alcohol to gain 10% profit by selling the mixture at cost price.
Correct
Correct option : (b)
Assume, C.P. and S.P. of alcohol = Re. 1 per litre
The general formula to calculate C.P. in case of profit = 100 * S.P. (100 + Gain%) Here, 10 % profit is gained, therefore
Cost price of 1 litre of mixture = Rs. 100 *1 = Rs. 100 = 10 (100 + 10) 110 11 Rs. 10 / 11 is the cost price of mixture.
Now, use the rule of alligation to determine the ratio of water and alcohol.
Water is free of cost, hence C.P. of water is zero.
The ratio of milk and water = 1 : 10 = 1 : 10 11 11 Incorrect
Correct option : (b)
Assume, C.P. and S.P. of alcohol = Re. 1 per litre
The general formula to calculate C.P. in case of profit = 100 * S.P. (100 + Gain%) Here, 10 % profit is gained, therefore
Cost price of 1 litre of mixture = Rs. 100 *1 = Rs. 100 = 10 (100 + 10) 110 11 Rs. 10 / 11 is the cost price of mixture.
Now, use the rule of alligation to determine the ratio of water and alcohol.
Water is free of cost, hence C.P. of water is zero.
The ratio of milk and water = 1 : 10 = 1 : 10 11 11 
Question 6 of 10
6. Question
1 pointsA shopkeeper has 100 kg of salt. He sells part of the total quantity A at 7% profit and the rest B at 17 % profit. If he gains 10 % profit on the whole quantity, then find how much is sold at 7 % profit?
Correct
Correct option : (a)
Assume that A and B are two parts of the mixture. To determine the quantity A and B, first calculate ratio of A : B.
Given:
1) Selling price of mixture with 10% profit = Rs. 110
2) With 17 % profit, the selling price of A = Rs. 117
3) With 7 % profit, the selling price of B = Rs. 107Now, this question can be easily solved by using the rule of alligation
Now, the ratio of A : B = 3 : 7
Let the quantity of part A be 3x and part B be 7x in the total quantity of 100 kg.
Therefore, 3x + 7x = 100
10x = 100
x = 10Quantity of part A = 3x = 3 x 10 = 30 kg
Quantity of part B = 7x = 7 x 10 = 70 kgIncorrect
Correct option : (a)
Assume that A and B are two parts of the mixture. To determine the quantity A and B, first calculate ratio of A : B.
Given:
1) Selling price of mixture with 10% profit = Rs. 110
2) With 17 % profit, the selling price of A = Rs. 117
3) With 7 % profit, the selling price of B = Rs. 107Now, this question can be easily solved by using the rule of alligation
Now, the ratio of A : B = 3 : 7
Let the quantity of part A be 3x and part B be 7x in the total quantity of 100 kg.
Therefore, 3x + 7x = 100
10x = 100
x = 10Quantity of part A = 3x = 3 x 10 = 30 kg
Quantity of part B = 7x = 7 x 10 = 70 kg 
Question 7 of 10
7. Question
1 pointsSugar A worth Rs. 130/kg and B of Rs. 120/kg are mixed with a third variety C in the ratio of 1 : 1 : 2. If the mixture is worth Rs. 160, then find the price of third variety of sugar.
Correct
Correct option: (a)
1) First calculate the average of A and B variety of sugar: (120 + 130) = Rs. 125 2 2) Now, the mixture is formed by two varieties of sugar, one at Rs. 125 /kg and assume the cost of type C Rs. x. It is formed in the ratio of 2 : 2, i.e 1 : 1
Use the rule of alligation, to easily determine the unknown quantity.
Therefore,
(x – 160) = 1 35 x = 35 + 160 = Rs. 195
The cost of third variety of sugar C = Rs. 195
Incorrect
Correct option: (a)
1) First calculate the average of A and B variety of sugar: (120 + 130) = Rs. 125 2 2) Now, the mixture is formed by two varieties of sugar, one at Rs. 125 /kg and assume the cost of type C Rs. x. It is formed in the ratio of 2 : 2, i.e 1 : 1
Use the rule of alligation, to easily determine the unknown quantity.
Therefore,
(x – 160) = 1 35 x = 35 + 160 = Rs. 195
The cost of third variety of sugar C = Rs. 195

Question 8 of 10
8. Question
1 pointsTwo containers P and Q contain milk and water in the ratio of 5 : 2 and 7 : 6 respectively. Find the ratio in which these two mixtures can be mixed so that a new mixture formed in the container R is in the ratio of 8 : 5.
Correct
Correct Option:(c)
Let the cost price of milk be Re. 1 per litre.
Therefore, cost of milk in 1 litre of mixture in
Container A (Milk : Water = 5 : 2) = 5 * Re. 1 = Re. 5 7 7 Container B (Milk : Water = 7 : 6) = 7 *Re. 1 = Re. 7 13 13 Container C (Milk : Water = 8 : 5) = 8 * Re. 1 = Re. 8 13 13 Now use the rule of alligation, to find the required ratio
The required ratio of milk and water:
1 : 9 13 91 Simplifying, we get 7 : 9
Incorrect
Correct Option:(c)
Let the cost price of milk be Re. 1 per litre.
Therefore, cost of milk in 1 litre of mixture in
Container A (Milk : Water = 5 : 2) = 5 * Re. 1 = Re. 5 7 7 Container B (Milk : Water = 7 : 6) = 7 *Re. 1 = Re. 7 13 13 Container C (Milk : Water = 8 : 5) = 8 * Re. 1 = Re. 8 13 13 Now use the rule of alligation, to find the required ratio
The required ratio of milk and water:
1 : 9 13 91 Simplifying, we get 7 : 9

Question 9 of 10
9. Question
1 pointsThe cost of Type 1 rice is Rs. 20 per kg and Type 2 rice is Rs. 30 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2:3, then the price per kg of the mixed variety of rice is :
Correct
CORRECT ANSWER : a. Rs. 24
Incorrect
CORRECT ANSWER : a. Rs. 24

Question 10 of 10
10. Question
1 pointsIn what ratio must a shopkeeper mix two varieties of pulses costing Rs. 14 and Rs. 16 per kg respectively so as to get mixture worth Rs. 12.50?
Correct
CORRECT ANSWER : b. 7:3
Incorrect
CORRECT ANSWER : b. 7:3