CAT 2020 Quant Question [Slot 2] with Solution 11

Question

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

  1. 30
  2. 33
  3. 27
  4. 36
Option: 2
Solution:

Let the price of each pencil be Rs. \(x\) and price of each sharpens be Rs. y

Aron \(y - x = 2\quad  \Rightarrow \quad \therefore x = y - 2\)

\(a(y - 2) + b(y) = 2a(y - 2) + (b - 10)y\)

\(10y = a(y - 2) \to (1)\)

Required value \( = 3a\)

From \((1)\quad a = \frac{{10y}}{{y - 2}}a \in {I^ + }\)

Its possible only when \(y = 22\)

\(\therefore a = 11\)

Required answer =33

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CAT 2020 Quant questions with Solutions

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