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Mixture and Alligation Test 9
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Mixture and Alligation Test 9
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  • Question 1/20
    1 / -0

    A mixture contains milk and water in the ratio 2 : 3 and the other contains them in the ratio 3 : 1 respectively. What weight of 2nd mixture must be taken so as to make a third mixture 7 litres in weight with 70% milk?

    Solutions

    Explanation: 

    70/100 = 7/10

    Milk in first = 2/(2+3) = 2/5, milk in second = 3/(3+1) = 3/4

    By method of allegation:

    2/5                                3/4

    .                  7/10

    3/4 – 7/10               7/10 – 2/5

    1/20 : 3/10

    1 : 6

    So in 3rd mixture, 2nd mixture is [6/(1+6)] * 7 = 6 litres

     

  • Question 2/20
    1 / -0

    How many kg of wheat costing Rs 8 per kg must be mixed with 8 Kg of wheat costing Rs 12 per Kg, so that there may be gain of 20% by selling the mixture of Rs 11.52 per Kg? 

    Solutions

    Explanation: 

    With 20% gain and SP = 11.52, CP = (100/120)*11.52 = 9.6

    By method of allegation:

    1st wheat (x kg)                2nd wheat (8 kg)

    8                                                12

    .                             9.6

    12-9.6                                        9.6-8

    2.4 1.6

    3 : 2

    So x/8 = 3/2

     

  • Question 3/20
    1 / -0

    15 litres of the milk is drawn out of a jar and filled with water. This operation is performed 1 more times. If the ratio of the quantity of milk left in jar to that of water in jar is 16 : 9, what was the initial quantity of milk present in the jar? 

    Solutions

    Explanation: 

    Let initial quantity of milk = x litres

    After two times, quantity of milk left in jar = x [1 – 15/x]2

    So x [1 – 15/x]2 / x = 16/16+9

    [1 – 15/x]2 = 16/25

    Square root both sides, so [1 – 15/x] = 4/5

    Solve, x = 75

     

  • Question 4/20
    1 / -0

    Wheat worth Rs 30 per kg and Rs 42 per kg are mixed with a third variety of wheat in the ratio 1 : 1 : 2 respectively. The mixture is worth Rs 42 per kg. Find the price (per kg) of the third variety of wheat. 

    Solutions

     

    Explanation: 

    Since 1st and 2nd wheat mixed in equal ratio, their average price = (30+42)/2 = 72/2 = Rs 36

    Let 3rd variety of wheat be Rs x per kg

    So they are mixed as : (1+1) : 2 = 1 : 1

    So

    36                              x

    .                 42

    x-42                     42-36=6

    so x-42/6 = 1/1

    solve, x = 48

     

  • Question 5/20
    1 / -0

    A mixture contains A and B in the ratio 5 : 9. 14 litres of this mixture is taken out and 14 litres of B is poured in. Now the ratio of A to B becomes 2 : 5. Find the amount of B originally present in the mixture. 

    Solutions

    Explanation: 

    Total = 5x+9x+14 = 14x+14

    So 5x/9x+14 = 2/5

    Solve, x = 4

    So total = 14*4 + 14 = 70 litres

    So B = 9/(5+9) * 70 = 45

     

  • Question 6/20
    1 / -0

    A mixture contains A and B in the ratio 5 : 7. 24 litres of this mixture is taken out and 15 litres of A is poured in. Now the ratio of A to B becomes 10 : 7. Find the amount of B originally present in the mixture.

    Solutions

    Explanation: 

    Total = 5x+7x+24 = 12x+24

    So 5x+15/7x = 10/7

    Solve, x = 3

    So total = 12*3 + 24 = 60 litres

    So B = 7/(5+7) * 60 = 35

     

  • Question 7/20
    1 / -0

    Milk contains 20% water. What quantity of pure milk should be added to 75 litres of this milk to reduce the quantity of water to 15%? 

    Solutions

    Explanation: 

    To have final mixture with water 15% means, milk = (100-15) = 85%

    Pure milk is 100% milk, and in 75 litres of milk there is 80% milk so

    Milk (75 litres)                        Pure milk (x litres)

    80                                                  100

    .                             85

    15                                                        5

    3 : 1

    So 75/x = 3/1

    Solve, x = 25

     

  • Question 8/20
    1 / -0

    An article is bought for Rs 560. Some of the part is sold at 20% profit and remaining at 15% loss giving a total of 10% profit. Find the cost price of part sold at 20% profit.

    Solutions

    Explanation: 

    20                         -15

    .                10

    10-(-15)                 20-10

    25 : 10

    5 : 2

    So at 20% profit = 5/(5+2) * 560

     

  • Question 9/20
    1 / -0

    A man travelled a distance of 75 km in 5 hours partly on foot at the rate of 6 km/hour and partly by scooter at 18 km/hour. Find the distance travelled by scooter? 

    Solutions

    Explanation: 

    Average speed = 75/5 = 15 km/hr

    By method of allegation:

    6                          18

    .             15

    3                           9

    1 : 3

    1 : 3 is the ratio of times.

    So time by scooter = 3/(1+3) * 5 = 3.75 hrs

    So distance by scooter = 3.75 * 18

     

  • Question 10/20
    1 / -0

    A 50 litres mixture of milk and water contains 10% water. 10 litres of this mixture is replaced by 10 litres of milk. What is the percentage of water in the final mixture? 

    Solutions

    Explanation: 

    When 10 litres taken out, quantity of mixture left =50-10=40, so

    milk left = 90/100 *40 = 36

    And water left = (10/100)*40 = 4

    now 10 litres of milk poured in, so there is no change in quantity of water, and total mixture again becomes 50l

    so water now is 4/50 * 100 = 8%

     

  • Question 11/20
    1 / -0

    Two varieties of rice costing Rs 25 and Rs 35 respectively are mixed in a certain ration and the resulting mixture is sold at a profit of 20% at Rs 34.8. What is the respective ratio in which they are mixed? 

    Solutions

    Explanation: 

    SP = 34.8, profit = 20%, so CP = (100/120)*34.8 = 29

    By method of allegation:

    25               35

    .        29

    6                4

    3 : 2

     

  • Question 12/20
    1 / -0

    How much quantity of water should be mixed with 10 l of milk costing Rs 50 per litre so that the resultant mixture is to be sold at Rs 44 per kg?

    Solutions

    Explanation: 

    CP of water = Rs 0

    By method of allegation:

    water (x kg)             milk (10 litres)

    0                                      50

    .                    44

    6                                      44

    6 : 44

    3 : 22

    So x/10 = 3/22

     

  • Question 13/20
    1 / -0

    In what ratio must the 3 varieties of wheat costing Rs 42, Rs 54 and Rs 65 respectively be mixed so that the resulting mixture is sold for Rs 63.8 at a profit of 10%? 

    Solutions

    Explanation: 

    SP = 63.8, Profit=10%, so CP = (100/110)*63.8 = 58

    Now 58 is greater than 42 and 54 and less than 65

    So

    42               65
    .        58

    7                 16

    And

    54               65

    .        58

    7                  4

    So 1 part of 1st wheat A, 1 part of 2nd wheat B and 2 parts of 3rd wheat C gives

    A : C = 7 : 16, and B : C = 7 : 4

    So A : B : C = 7 : 7 : (16+4)

    *we have taken 2 parts of C so it is added – when there are 3 varieties to be mixed

    it is not done like simple calculation of A : C and B : C

     

  • Question 14/20
    1 / -0

    A can containing 25 litres of mixture of milk and water has 80% milk in it. How much quantity of the mixture be drawn out and replaced with water such that the new ratio of water to milk becomes 1 : 3? 

    Solutions

    Explanation: 

    Milk = (80/100)*25 = 20, so water = 5 l

    So water : milk = 5 : 20 = 1 : 4

    Let x litres drawn out

    So water left = 5 – (1/(1+4))*x = 5 – x/5

    Milk left = 20 – (4/(1+4))*x = 20 – 4x/5

    Now x litres of water is added too, so water becomes = 5 – x/5 + x = 5 + 4x/5

    So [5 + 4x/5] / [20 – 4x/5] = 1/3

    75x+12x = 100x-4x

    16x = 25

    x=25/16 = 1.5625

     

  • Question 15/20
    1 / -0

    A can contains 60 litres of milk. 4 litres of milk is drawn and replaced with water. This procedure is repeated once again. How much quantity of milk is remained in the can? 

    Solutions

    Explanation: 

    Performed 2 times, so

    Milk left = 60 [1 – 4/60]2

     

  • Question 16/20
    1 / -0

    A variety of wheat costing Rs 8.70 is mixed with another variety in a ratio 2 : 3. If the mixture is sold at Rs 8.10 making a loss of 10%, then what is the cost of 2nd variety of wheat?

    Solutions

    Explanation: 

    SP = 8.1, loss = 10%, so CP = (100/90)*8.1 = Rs 9

    1st wheat              2nd wheat

    8.70                                 x

    .                   9

    2                                    3

    So (x – 9)/(9 – 8.70) = 2/3

    Solve, x = 9.20

     

  • Question 17/20
    1 / -0

    An alloy contains 4 parts bronze and 6 parts copper. How much part of mixture should be drawn out and replaced with bronze so that the ratio of bronze to copper gets reversed in new mixture? 

    Solutions

    Explanation: 

    Total = 4+ 6 = 10

    The old ratio is 4/6 = 2/3, so new ratio of bronze to copper should be 3/2

    Let x kg of mixture is drawn out and then x kg of bronze added.

    So

    Bronze is now = 4 – (4/10)*x + x = 4 + (3x/5)

    Copper is now = 6 – (6/10)*x = 6 – (3x/5)

    Now [4 + (3x/5)] / [6 – (3x/5)] = 3/2

    Solve, x = 10/3

    So part of mixture drawn out is (10/3)*10 = 1/3

     

  • Question 18/20
    1 / -0

    Wheat worth Rs 50 per kg and Rs 56 per kg are mixed with a third variety of wheat in the ratio 2 : 2 : 3 respectively. If the mixture obtained is worth Rs 61 per kg. Find the price (per kg) of the third variety of wheat. 

    Solutions

    Explanation: 

    Since 1st and 2nd wheat mixed in equal ratio, their average price = (50+56)/2 = Rs 52

    Let 3rd variety of wheat be Rs x per kg

    So they are mixed as : (2+2) : 3 = 4 : 3

    So

    52                       x

    .           61

    x-61              61-52=9

    so x-61/9 = 4/3

    solve, x = 73

     

  • Question 19/20
    1 / -0

    A mixture of milk and water contains 25% water. 12 litres of this mixture is drawn out and replaced with 5 litres of water. If the new ratio of water to milk becomes 2 : 5, what is the amount of milk originally present in the mixture? 

    Solutions

    Explanation: 
    Milk : water = 75% : 25% = 3 : 1
    Total = 3x+x+12 = 4x+12
    So (x+5)/3x = 2/5
    Solve, x = 25
    So total = 4*25 + 12 = 112 litres
    So originally milk = 3/(3+1) * 112 = 84

  • Question 20/20
    1 / -0

    Mixture A contain water and milk in the ratio 2 : 5 and mixture B contain them in the ratio 3 : 4 respectively. Equal quantities from both the mixture are taken and mixed to form mixture C. What is the ratio of milk to water in mixture C? 

    Solutions

    Explanation: 

    Let x litres taken from both mixtures,

    Then new ratio of milk to water is

    [5/(5+2)] * x + [4/(3+4)] * x : [2/(5+2)] * x + [3/(3+4)] * x

     

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