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

• A. 30
• B. 33
• C. 27
• D. 36

Answer 1

The Correct Option is B

Step 1: Define Variables 

• Let the price of a pencil be \( x \)
• Let the price of a sharpener be \( y \)
• Given: \( y = x + 2 \Rightarrow x = y - 2 \)

Step 2: Formulate the Cost Equations

Total cost for Aron: \( ax + by \)
Total cost for Aditya: \( 2a x + (b - 10) y \)

Given that they spend the same: \[ ax + by = 2ax + (b - 10)y \] Substitute \( x = y - 2 \): \[ a(y - 2) + by = 2a(y - 2) + (b - 10)y \]

Step 3: Expand and Simplify

\[ ay - 2a + by = 2ay - 4a + by - 10y \] Subtract \( by \) from both sides: \[ ay - 2a = 2ay - 4a - 10y \] Bring everything to one side: \[ ay - 2a - 2ay + 4a + 10y = 0 \Rightarrow -ay + 2a + 10y = 0 \] \[ \Rightarrow a(y - 2) = 10y \quad \text{(Equation 1)} \]

Step 4: Solve for Integer \( a \)

\[ a = \frac{10y}{y - 2} \] We want the smallest positive integer \( a \), so test values of \( y \) such that \( \frac{10y}{y - 2} \in \mathbb{Z} \) Try \( y = 22 \): \[ a = \frac{10 \times 22}{20} = \frac{220}{20} = 11 \quad \text{✓ Valid} \]

Step 5: Final Answer

\[ \text{Total pencils} = a + 2a = 3a = 3 \times 11 = \boxed{33} \]

✅ Final Answer: 33

Answer 2

Let: 

• \( p \) = number of pencils Aron bought
• \( a \) = price per pencil
• \( s \) = number of sharpeners Aron bought
• \( b \) = price per sharpener

 

Step 1: Equation from Aron's purchase

Total cost for Aron: \[ \text{Aron’s total cost} = pa + sb \]

Step 2: Equation from Aditya's purchase

Aditya buys:

• 2p pencils → cost: \( 2pa \)
• (s - 10) sharpeners → cost: \( (s - 10)b \)

So total cost: \[ \text{Aditya’s total cost} = 2pa + (s - 10)b \]

 

Step 3: Equating both totals

Given that both spend the same: \[ pa + sb = 2pa + (s - 10)b \] Simplifying: \[ pa + sb = 2pa + sb - 10b \Rightarrow pa = 10b \quad \text{(Equation 1)} \]

Step 4: Given relation between sharpener and pencil price

The sharpener costs ₹2 more than a pencil: \[ b = a + 2 \quad \text{(Equation 2)} \] Substitute in Equation 1: \[ pa = 10(a + 2) \Rightarrow pa = 10a + 20 \] Bring terms together: \[ pa - 10a = 20 \Rightarrow a(p - 10) = 20 \Rightarrow a = \frac{20}{p - 10} \]

Step 5: Find the smallest integer value of \( p \)

We need \( p \) and \( a \) both to be integers. Try smallest integer values of \( p > 10 \) such that \( a \) is also integer: \[ \begin{align*} p = 11 &\Rightarrow a = \frac{20}{1} = 20 \ (\text{✓}) \\ p = 12 &\Rightarrow a = \frac{20}{2} = 10 \ (\text{✓}) \\ p = 15 &\Rightarrow a = \frac{20}{5} = 4 \ (\text{✓}) \\ \end{align*} \] But we want the **smallest** value of \( p \), so choose \( p = 11 \).

Step 6: Total number of pencils bought

Total pencils = Aron's pencils + Aditya's pencils = \( p + 2p = 3p = 3 \times 11 = 33 \)

✅ Final Answer:

The total number of pencils bought is: \( \boxed{33} \)