Mixtures and Alligations - Vikas Saini



  • A 30% solution of alcohol is mixed with a 50% solution of alcohol to form a 10 litre solution of 45% alcohol. How much of 30% solution was used.
    (a) 2 ltr
    (b) 2.5 ltr
    (c) 2.7 ltr
    (d) 3.2 ltr
    (e) 3.7 ltr

    By allegation
    30 50
    45
    (50-45 ) : (45-30)
    5 : 15 = 1 : 3.
    Ratio = 1 : 3
    30% solution was added = 10 x 1 / (1 + 3) = 2.5 ltr.

    The percentage volumes of milk in three solutions A,B and C form a geometric progression in that order. If we mix the first, second and third solutions in the ratio 2:3:4, by volume we obtain solution containing 32% milk. If we mix them in the ratio 3:2:1, by volume, we obtain a solution containing 22% milk. What is the percentage of milk in A?
    a) 6%
    b) 12%
    c) 18%
    d) 24%

    Suppose volumes of milk percentage in A,B,C are a, ar, ar^2 respectively.
    2a + 3ar + 4ar^2 = 32 x 9.
    3a + 2ar + ar^2 = 22 X 6.
    a= 12, r = 2.
    The percentage of milk in A = 12%.

    One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight ?

    10 X
    16
    X-16 : 16 - 10
    1 : 1/3

    X – 16 / 6 = 3 / 1
    X = 34.

    From a solution that has milk and water in the ratio 5 : 3, ‘x’ percent is removed and replaced with water. The concentration of milk in the resulting solution lies between 30% and 50%. Which of the following best describes the value of ‘x’?
    a) 25 < x < 30
    b) 20 < x < 52
    c) 20 < x < 48
    d) 25 < x < 60

    Suppose total unit is 8. Milk = 5, water = 3.
    X% of 8 is removed = 0.08x.
    Where milk part is 0.05x and water part is 0.03x.
    0.30 < 5 – 0.05x / 8 < 0.50
    2.40 < 5 – 0.05 x < 4.0
    -2.60 < -0.05x < -1.
    1 < 0.05x < 2.6
    20 < x < 52.
    Option B

    The concentration of milk in 60L milk and water solution is 30%. If 6L of the solution is replaced by water then 5L solution is again replaced by water, then find the concentration of milk in the final solution.
    a) 24.50 %
    b) 25.00%
    c) 24.75 %
    d) 25.50%

    The concentration in final solution = (3/10) ( 1 - 6/60) (1 – 5/60)
    = 24.75%.

    Rice of two different qualities are mixed and the mixture is sold at rs. X per kg, giving 25% profit. If higher quality rice is sold at rs X per kg, then there will be a loss of 100/11 %. If the ratio of the quantities of the lower quality rice and the higher quality rice in the mixture is 8 : 3, then what is the percentage profit when lower quality rice is sold at X per/kg ?
    Cost price of low quality rice per kg is l and cost price of high quality rice per kg is h.
    (8l + 3h / 11)(5/4) = h (10/11)
    8l = 5h.
    Cost price of lower quality rice is 5h/8.
    Selling price of mixture = 10h/11.
    Profit percentage = (10h/11 – 5h/8) x 100 / (5h/8) = 500/11 %.

    Two containers X and Y, equal quantities of water and acid. The concentration of acid is same in both the containers X and Y. If 4 litres of solution from X is replaced by acid and the concentration of acid becomes twice what it was initially. If 8 litres of solution is from Y is replaced with pure acid then what is the ratio of final concentration of acid in solution Y to the initial concentration of acid in solution Y ?

    Suppose quantity is q, concentration of acid = a, suppose quantity of water.
    In container X, 4 litre acid is replaced then concentration becomes twice.
    In container Y, 8 litre acid is when replaced then concentration becomes two times of twice.
    Hence increases by thrice.
    Ratio = 3 : 1.

    A and B are two perfumes which have oil and essence in the ratio of 11:3 and 2:5 respectively.
    New perfume is made by mixing A and B in the ratio of a:b such that the percentage of oil in the new perfume is 50%. Find a+b.

    Oil in A = 11 / 3+11 = 11/14.
    Oil in B = 2 / 2+5 = 2/7.
    (11a/14) + (2b/7) = (a+b)/2
    11a + 4b = 7a + 7b
    a/b = ¾
    a+b = 7.


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