A charity has Rs. 6000 to buy blankets and sweaters. A blanket costs Rs. 300 and a sweater costs Rs. 200, and at least 6 blankets must be bought. The greatest total number of blankets and sweaters it can buy is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 27
Let x blankets and y sweaters be bought. Then
Maximise \(Z = x + y\)
subject to
\(300x + 200y \le 6000\), that is \(3x + 2y \le 60\) (money)
\(x \ge 6\) (blankets)
\(y \ge 0\)
The corner points are (6, 0), (20, 0) and, where \(x = 6\) meets \(3x + 2y = 60\),
\(18 + 2y = 60\)
\(y = 21\)
that is (6, 21). Values of Z:
\(Z(6, 0) = 6\)
\(Z(20, 0) = 20\)
\(Z(6, 21) = 27\)
The greatest number is 27: 6 blankets and 21 sweaters.
Why the other options are incorrect
Option (1) 20 — Incorrect. This is 20 blankets only; the cheaper sweaters give more items.
Option (2) 25 — Incorrect. It is possible but not the greatest; (6, 21) gives 27.
Option (4) 30 — Incorrect. 30 sweaters alone would use all Rs. 6000 and leave no money for the 6 blankets.