Question:hard

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

Updated On: Jan 15, 2026
  • 30
  • 33
  • 27
  • 36
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The Correct Option is B

Solution and Explanation

Paso 1: Definir Variables

  • Sea \( x \) el precio de un lápiz
  • Sea \( y \) el precio de un sacapuntas
  • Dado: \( y = x + 2 \Rightarrow x = y - 2 \)

Paso 2: Formular las Ecuaciones de Costo

Costo total para Aron: \( ax + by \)
Costo total para Aditya: \( 2a x + (b - 10) y \)

Dado que gastan lo mismo: \[ ax + by = 2ax + (b - 10)y \] Sustituir \( x = y - 2 \): \[ a(y - 2) + by = 2a(y - 2) + (b - 10)y \]

Paso 3: Expandir y Simplificar

\[ ay - 2a + by = 2ay - 4a + by - 10y \] Restar \( by \) de ambos lados: \[ ay - 2a = 2ay - 4a - 10y \] Mover todo a un lado: \[ ay - 2a - 2ay + 4a + 10y = 0 \Rightarrow -ay + 2a + 10y = 0 \] \[ \Rightarrow a(y - 2) = 10y \quad \text{(Ecuación 1)} \]

Paso 4: Resolver para \( a \) entero

\[ a = \frac{10y}{y - 2} \] Queremos el menor entero positivo \( a \), así que probamos valores de \( y \) tales que \( \frac{10y}{y - 2} \in \mathbb{Z} \). Intentamos con \( y = 22 \): \[ a = \frac{10 \times 22}{20} = \frac{220}{20} = 11 \quad \text{✓ Válido} \]

Paso 5: Respuesta Final

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

✅ Respuesta Final: 33

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