Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is
Explanation:
Let the price of a pencil = x and that of a sharpener = x + 2.
Suppose Aron bought p pencils and s sharpeners. Aditya buys 2x pencils and s – 10 sharpeners.
∴ px + s(x + 2) = 2px + (s – 10)(x + 2)
⇒ px + sx + 2s = 2px + sx + 2s – 10x – 20
⇒ 20 = px – 10x
⇒ x(p – 10) = 20
Here, the least value of p can be 11.
∴ Aron and Aditya together bought 3p pencils and least value of 3p = 3 × 11 = 33.
Hence, option (c).
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