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

The correct answer is 33.

Let's define our variables carefully before setting up the equation.

Let:
  • p = number of pencils Aron bought
  • s = number of sharpeners Aron bought
  • c = cost (in ₹) of one pencil
  • Cost of one sharpener = c + 2   (given: sharpener costs ₹2 more than a pencil)

Aditya, spending the same amount as Aron, bought:
  • Pencils: 2p (twice as many)
  • Sharpeners: s − 10 (10 less)

Step 1: Set up the equal-spending equation

Aron's total spend = Aditya's total spend:

$$p\cdot c+s\cdot (c+2)=2p\cdot c+(s-10)\cdot (c+2)$$

Step 2: Simplify the equation

Expand the right-hand side:

$$pc+s(c+2)=2pc+s(c+2)-10(c+2)$$

The s(c+2) term cancels from both sides:

$$pc=2pc-10(c+2)$$

Rearranging:

$$10(c+2)=2pc-pc=pc$$

So our key equation is:

$$pc=10c+20$$

Step 3: Solve for p

Dividing both sides by c:

$$p=10+\frac{20}{c}$$

Since p must be a positive integer, c must be a positive integer divisor of 20.

The divisors of 20 are: 1, 2, 4, 5, 10, 20

Step 4: Apply the constraint on sharpeners

Aditya bought s − 10 sharpeners. For this to be a valid (positive) quantity:

$$s-10\ge 1⟹s\ge 11$$

Aron needs to have bought at least 11 sharpeners. Since s is a free positive integer (not restricted by the main equation), this constraint can always be satisfied by choosing s ≥ 11 for any valid c.

Step 5: Minimize the total number of pencils

Total pencils bought by Aron and Aditya together:

$$p+2p=3p$$

To minimize 3p, we need to minimize p, which means we need to maximize c.

The largest valid divisor of 20 is c = 20. Substituting:

$$p=10+\frac{20}{20}=10+1=11$$

Step 6: Verify this is valid

With c = 20 (pencil costs ₹20) and p = 11:
  • Sharpener cost = ₹22
  • Take any valid s ≥ 11, say s = 11
  • Aron's spend = 11 × 20 + 11 × 22 = 220 + 242 = ₹462
  • Aditya's pencils = 22, Sharpeners = 1
  • Aditya's spend = 22 × 20 + 1 × 22 = 440 + 22 = ₹462 ✓

Step 7: Calculate the final answer

$$\text{Minimum total pencils}=3p=3\times 11=\mathbf{33}$$

The minimum possible number of pencils bought by Aron and Aditya together is 33.