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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

- $30$
- $33$
- $27$
- $36$

1 vote

We can make a table for better understanding.

$\begin{array}{lcc} & \text{Pencil} & \text{Sharpeners} \\ \text{Aron:} & a & b \\ \text{Aditya:} & 2a & b-10 \\ \text{Cost of:} & ₹x & ₹(x+2) \end{array}$

The total amount of money spend by Aron is same as Aditya. Then,

$ax + b(x+2) = 2ax + (b – 10)(x+2)$

$ \Rightarrow ax + bx + 2b = 2ax + bx + 2b – 10x – 20$

$ \Rightarrow 10x + 20 = ax$

$ \Rightarrow \boxed{a = \frac{20}{x} + 10} \quad \longrightarrow (1)$

The minimum number of pencils bought by Aron and Aditya together $ = a + 2a = (3a)_{\text{min}}$

From equation $(1),$

$a = \frac{20}{x} + 10 $

$ \Rightarrow a_{\text{min}} = \frac{20}{20} + 10 \quad [\because \text{Maximum value of}\; x = 20]$

$ \Rightarrow \boxed{a_{\text{min}} = 11}$

$\therefore (3a)_{\text{min}} = 3 \times 11 = 33.$

Correct Answer $: \text{B}$

$\begin{array}{lcc} & \text{Pencil} & \text{Sharpeners} \\ \text{Aron:} & a & b \\ \text{Aditya:} & 2a & b-10 \\ \text{Cost of:} & ₹x & ₹(x+2) \end{array}$

The total amount of money spend by Aron is same as Aditya. Then,

$ax + b(x+2) = 2ax + (b – 10)(x+2)$

$ \Rightarrow ax + bx + 2b = 2ax + bx + 2b – 10x – 20$

$ \Rightarrow 10x + 20 = ax$

$ \Rightarrow \boxed{a = \frac{20}{x} + 10} \quad \longrightarrow (1)$

The minimum number of pencils bought by Aron and Aditya together $ = a + 2a = (3a)_{\text{min}}$

From equation $(1),$

$a = \frac{20}{x} + 10 $

$ \Rightarrow a_{\text{min}} = \frac{20}{20} + 10 \quad [\because \text{Maximum value of}\; x = 20]$

$ \Rightarrow \boxed{a_{\text{min}} = 11}$

$\therefore (3a)_{\text{min}} = 3 \times 11 = 33.$

Correct Answer $: \text{B}$