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Mixture and Alligation Test 11
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Mixture and Alligation Test 11
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  • Question 1/20
    1 / -0

    An alloy contains iron, copper and zinc in the ratio of 3:4:2. Another alloy contains copper, zinc and tin in the ratio of 10:5:3. If equal quantities of both alloys are melted, then weight of tin per kg in the new alloy.

    Solutions

    Explanation :

    I:C:Z = 3:4:2 (in first alloy) and C:Z:T = 10:5:3

    Equal quantities is taken. So, I:C:Z = 6:8:4 in first alloy and C:Z:T = 10:5:3

    I = 6

    C = 8 + 10 = 18

    Z = 4+5 = 9

    T = 3

    So weight of tin = 3/36 = 1/12

     

  • Question 2/20
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    8 litres are drawn from a flask containing milk and then filled with water. The operation is performed 3 more times. The ratio of the quantity of milk left and total solution is 81/625. How much milk the flask initially holds?

    Solutions

    Explanation :

    let initial quantity be Q, and final quantity be F

    F = Q*(1 – 8/Q)^4

    81/625 = (1-8/Q)^4

    3/5 = 1 – 8/Q

    Q = 20

     

  • Question 3/20
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    A 40 litre mixture contains milk and water in the ratio of 3:2. 20 litres of the mixture is drawn of and filled with pure milk. This operation is repeated one more time. At the end what is the ratio of milk and water in the resulting mixture?

    Solutions

    Explanation :

    milk = 40*3/5 = 24 and water  = 16 litres initially

    milk = 24 – 20*3/5 + 20 = 32 – 20*4/5 + 20 = 36

    water = 16 – 20*2/5 = 8 – 20*1/5 = 4

     

  • Question 4/20
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    Two vessels contain milk and water in the ratio of 7:3 and 2:3 respectively. Find the ratio in which the contents of both the vessels must be mixed to get a new mixture containing milk and water  in the ratio 3:2.

    Solutions

    Explanation :

    Let the ratio be k:1

    then in first mixture, milk = 7k/10 and water = 3k/10

    and in second mixture, milk = 2/5 and water = 3/5

    [7k/10 + 2/5]/[3k/10 3/5] = 3/2

    K = 2, so ratio will be 2:1

     

  • Question 5/20
    1 / -0

    How many Kgs of rice A costing rupees 20 per kg must be mixed with 20 kg of rice B costing rupees 32 per kg, so that after selling them at 35 rupees per kg, he gets a profit of 25%.

    Solutions

    Explanation :

    by rule of alligation,

    20           32

    …….28………..

    4            8

    So, x = 10

     

  • Question 6/20
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    How many litres of water must be added to 60 litre mixture that contains milk and water in the 7:3 such that the resulting mixture has 50% water in it?

    Solutions

    Explanation :

    milk = (7/10)*60 = 42 and water = 18

    so water must be added = 42 – 18 = 24

     

  • Question 7/20
    1 / -0

    A sample of x litre is replaced from a container containing milk and water in the ratio 2:3 by pure milk. If the container hold 30 litres of the mixture, and after the operation proportion of milk and water is same. Find the value of X?

    Solutions

    Explanation :

    milk = 30*2/5 = 12 and water = 30*3/5 = 18

    milk = 12 – x*2/5 + x and water = 18 – x*3/5

    equate both equation and we get x = 5

     

  • Question 8/20
    1 / -0

    Two cans P and Q contains milk and water in the ratio of 3:2 and 7:3 respectively. The ratio in which these two cans be mixed so as to get a new mixture containing milk and water in the ratio 7:4.

    Solutions

    Explanation :

    Milk in 1st can = 3/5 and water = 2/5. Similarly in second can milk = 7/10 and water = 3/10.

    Take the ratio = K:1

    (3k/5 + 7/10)/(2k/5 + 3/10) = 7/4

    Solve for k, we get k = 7/4. So the ratio is 7:4

     

  • Question 9/20
    1 / -0

    A trader mixes 6ltr of milk costing 5000 rupees with 7ltr of milk costing 6000 rupees per litre. The trader also mixes some quantity of water to the mixture so as to bring the price to 4800 per litre. How many litres of water is added.

    Solutions

    Explanation :

    (6*5000 + 7*6000)/(13 + w) = 4800 (w is the amount of water added)

     

  • Question 10/20
    1 / -0

    How many kilograms of rice costing Rs. 9 per kg must be mixed with 27kg of rice costing Rs.7 per kg so that there may be gain of 10% by selling the mixture at Rs.9.24 per kg?

    Solutions

    Explanation :

    By rule of allegation: –

    9        7

    ….8.4…..

    1.4     0.6

    So, x/27 = 1.4/0.6, we get x = 63 kg

     

  • Question 11/20
    1 / -0

    Two vessels A and B contain a mixture of Milk and Water. In the first vessel (i.e) Vessel A has the ratio of Milk to water is 8 : 3 and in the second vessel, Vessel B has the ratio of 5 : 1. A 35 litre capacity vessel is filled from these two vessels so as to contain a mixture of Milk and water in ratio of 4 : 1. Then how many litres should be taken from the first vessel, Vessel “A”.

    Solutions

    Explanation :

    [8/11(x) + 5/6(35-x)]/[3/11(x) + 1/6(35-x)] = 4/1

    x = 11

     

  • Question 12/20
    1 / -0

    When one litre of water is added to a mixture of milk and water, the new mixture contains 25% of milk. When one litre of milk is added to the new mixture, then the resulting mixture contains 40% milk. What is the percentage of milk in the original mixture?

    Solutions

    Explanation :

    Original Mixture = x L

    In (x + 1) Mixture, quantity of milk = (x + 1)* (25/100) = (x + 1)/4

    one litre of milk is added to the new mixture

    [((x + 1)/4 )+ 1 ]/ x + 2 = 40%

    x = 3 ; quantity of milk = (3 + 1)/4 = 1L

    percentage of milk in the original mixture = 1/3 * 100 = 100/3 %

     

  • Question 13/20
    1 / -0

    The price of a box and a pen is Rs.60. The box was sold at a 40% profit and the pen at a loss of 10%. If the Shop keeper gains Rs.4 in the whole transaction, then how much is the cost price of Box?

    Solutions

    Explanation :

    40x/100 – 10(60-x)/100 = 4

    40x + 10x = 400 + 600

    x = 20

     

  • Question 14/20
    1 / -0

    A vessel contains a mixture of diesel and petrol in which there is 20% diesel. Five litres are drawn off and then the vessel is filled with petrol. If the diesel present in the mixture is now 15% then how much does the vessel hold?

    Solutions

    Explanation :

    20/15 = x/x-5

    x = 20

     

  • Question 15/20
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    In a lab, two chemical solutions Acid “A” with 90% purity and Acid “B” with 96% purity are mixed resulting in 24 litres of mixture of 92% purity. How much is the quantity of the first solution, Acid “A” in the resulting mixture?

    Solutions

    Explanation :

    90x + (24 – x)* 96 = 24 * 92

    x = 16

     

  • Question 16/20
    1 / -0

    60 kg of a certain variety of Sugar at Rs.32 per kg is mixed with 48 kg of another variety of sugar and the mixture is sold at the average price of Rs.28 per kg. If there be no profit or no loss due to the new selling price, then what is the price of second variety of Sugar? 

    Solutions

    Explanation :

    Total CP of first variety = 60 * 32 = 1920

    Total CP of second variety = 48 * x = 48x

    SP of Mixture = 1920 + (108 * 28) = 3024

    1920 + 48x = 3024 => x = 23

     

  • Question 17/20
    1 / -0

    Six litre of milk was taken out from a vessel and is then filled with water. This operation is performed two more times. The ratio of the quantity of milk now left in vessel to that of the water is 8 : 27. How much is the quantity of the milk contained by the vessel originally?

    Solutions

    Explanation :

    [x(1-6/x)/x]³ = (8/27)

    [x(1-6/x)/x]³ = (2/3)³

    (x-6)/x = 2/3

    x = 18

     

  • Question 18/20
    1 / -0

    A vessel is filled with 120 litres of Chemical solution, Acid “A”. Some quantity of Acid “A” was taken out and replaced with 23 litres of Acid “B” in such a way that the resultant ratio of the quantity of Acid “A” to Acid “B” is 4:1. Again 23 litres of the mixture was taken out and replaced with 28 litre of Acid “B”. What is the ratio of the Acid “A” to Acid “B” in the resultant mixture?

    Solutions

    Explanation :

    In 23 litre mixture, Quantity of Acid “B” = 23 * 1/5 = 4.6 litre

    Acid “A” in the mixture = 23 – 4.6 = 18.4 litre

    120 – x / 23 = 4 / 1

    x = 28

    Ratio = 92-18.4 : 18.4 + 28

    Ratio = 46 : 29

     

  • Question 19/20
    1 / -0

    18 litres of Petrol was added to a vessel containing 80 litres of Kerosene. 49 litres of the resultant mixture was taken out and some more quantity of petrol and kerosene was added to the vessel in the ratio 2:1. If the respective ratio of kerosene and petrol in the vessel was 4:1, what was the quantity of kerosene added in the vessel?

    Solutions

    Explanation :

    Total quantity of the mixture = 18+80 = 98 litre

    quantity of petrol remaining = 18/2 = 9

    quantity of kerosene remaining = 80/2 = 40

    (40 + 2x) / (9 + x) = 4 / 1

    x = 2

    Quantity of kerosene added in the vessel = 2x = 4 litre

     

  • Question 20/20
    1 / -0

    A vessel which contains a mixture of acid and water in ratio 13:4. 25.5 litres of mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of acid is added to the mixture. If resultant mixture contains 25% water, what was the initial quantity of mixture in the vessel before the replacement in litres?

    Solutions

    Explanation :

    Quantity of Acid = 13x

    Quantity of water = 4x

    Total = 17x

    Resultant Mixture = 17x – 25.5 + 2.5 + 5 = 17x – 18

    Resultant water = 4x – 25.5 * (4/17) + 2.5 = 4x – 3.5

    Resultant mixture contains 25% water

    (17x – 18)*25/100 = 4x – 3.5

    x = 4

    Initial quantity = 17*4 = 68

     

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