2 units of a% alcohol is mixed with 3 units of b% alcohol to give 40%

There's a trick that I learnt during my high school which has been a lifesaver on such questions.

In an equation of the form ax + by = N, you just have to find the first solution for (x, y) by plugging in some values and then onwards all the subsequent values of x will move in steps of b (coefficient of y) and all the corresponding values of y will move in steps of a (coefficient of x)

Let me illustrate this with the above example itself.

2x + 3y = 100
First solution (50, 0) (identified by hit and trial)

Subsequent solutions
(47, 2)
(44, 4)
(41, 6)
(38, 8)
..... And so on

If you carefully observe all the subsequent solutions are actual solutions of the given equation that have not been arrived at by extensive calculations or hit and trial but merely derived from our original solution (50, 0) by moving x and y in the steps of the coefficient of the other variable.

x takes values 50, 47, 44, 41, 38.. which have bear a difference of 3 (coefficient of y)

y takes values 0, 2, 4, 6, 8.. which bear a difference of 2 (coefficient of x)

Hope this helps.

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Given: 2 units of a% alcohol is mixed with 3 units of b% alcohol to give 40% alcohol solution.
Asked: If a > b and both and b are integers, how many integer values can a take?

(2*a% + 3*b%)/5 = 40%
2*a% + 3*b% = 200%
a% = (200% - 3*b%)/2
Since both a & b are integers, b must be even integer for a to be an integer.

a% = (200% - 3*b%)/2 > b%
200% - 3*b% > 2b%
b% < 40%
Lowest value b can take = 0
Highest value b can take = 38
Common difference = 2
Number of terms = (38-0)/2 +1 = 20

IMO D

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