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Question:

Three cooks together have to make 80 samosas. While working together, they can make 20 pieces per minute. The first cook starts the work alone and makes 20 pieces while working for more than three minutes. The remaining part of the work is done by the second cook and the third cook, working together. If they take a total of 8 minutes to prepare 80 samosas, then how many minutes will the first cook take alone to cook 160 samosas?

Explanation:

Suppose,
the 1st cook makes A samosas per minute
the 2nd cook makes B samosas per minute
the 3rd cook makes C samosas per minute.

Given: A + B + C = 20 … (i)

Let the 1st cook take (3 + x) minutes to make 20 samosas.

∴ (3 + x)A = 20
A = $\frac{20}{3 + x}$ … (ii)

So, the 2nd cook and the 3rd cook take (5 - x) minutes to make 60 samosas.
∴ (5 - x)(B + C) = 60
B + C = $\frac{60}{5 - x}$ … (iii)

Substituting the values from (ii) and (iii), we get
⇒ $\frac{20}{3 + x}$ + $\frac{60}{5 - x}$ = 20
⇒ $\frac{1}{3 + x}$ + $\frac{3}{5 - x}$ = 1
⇒ 5 - x + 9 + 3x = 15 + 2x - x²
⇒ 14 - 2x = 15 + 2x - x²
⇒ 14 = 15 - x2
⇒ x = 1

So, the 1st cook takes 4 minutes to make 20 samosas.

∴ He will take 4 × 8 = 32 minutes to make 160 samosas.

Hence, option (c).

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