Question

LCM of \(x\) and \(y\) is 64800 and HCF of \(x\) and \(y\) is 1080. If \(x>y\), then the least possible value of \(x\) is

• A. 10800
• B. 12960
• C. 16200
• D. 21600

Hint:

If HCF \(=d\), write the numbers as \(dm\) and \(dn\), where \(m\) and \(n\) are coprime. Then use \(\frac{LCM}{HCF}=mn\).

Answer 1

The Correct Option is B

Concept: If HCF of two numbers is \(d\), then the numbers can be written as \[ x=dm,\qquad y=dn \] where \(m\) and \(n\) are coprime. Also, \[ LCM \times HCF=x \times y \]

Step 1: Finding the product \(mn\).
\[ mn=\frac{LCM}{HCF} \] \[ =\frac{64800}{1080} \] \[ =60 \] Thus, \(m\) and \(n\) are coprime factors of 60.

Step 2: Finding the closest coprime factor pair.
Factor pairs of 60: \[ 60 \times 1,\quad 20 \times 3,\quad 15 \times 4,\quad 12 \times 5 \] Among these, the pair with minimum larger factor is \[ 10 \times 6 \] but they are not coprime. The valid coprime pair nearest to each other is \[ 12 \times 5 \] Hence, \[ x=1080 \times 12 \] \[ =12960 \] {12960}