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

The correct answer is (B): \(33\)

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

Aron \(y-x=2 ⇒ ∴ x=y-2\)

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

\(\Rightarrow10y=a(y-2)→(1)\)

Required value = \(3a\)

from(1) \(a=\frac{10y}{y-2}a∈I^+\)

Its possible only when y = \(22\)

\(∴ a = 11\)

Required answer =\(33\)

Answer 2

Let "p" be the number of pencils Aron bought and "a" be the price per pencil.
Let "s" be the total number of sharpeners that Aron purchased, and "b" be the price per sharpener. 
Aron will now have spent \((pa)+(sb). \)
Aditya purchased pencils (2p) and sharpeners (s-10). 
The total amount will be \((2pa) + (s-10)b.\)
The same amount is spent in both cases\( (pa + sb = 2pa + (s-10)b\) or \(pa = 10b).\)
The question now states that the cost of a sharpener is two times that of a pencil, 
therefore b\(=a+2 pa\)
\(= 10a+20\) or 
\(a=\frac{20}{(p-10)} \)
Now that the pencil count must be as low as possible, we must determine the smallest "p" such that "p" and "a" are both integers. 
P=11 is the smallest of these values.
The total quantity of pencils bought will be \(p+2p=11+22=33.\)