Question

The HCF and LCM of 2 numbers are 2 and 60 respectively. If one of the numbers is 14 more than the other, find the smaller number.

Answer 1

Given that:

HCF and LCM of two numbers are 2 and 60 respectively.

One number is 14 greater than the other.

To find:

The smaller number = ?

Solution:

First of all, let us understand one property of HCF (Highest Common Factor) and LCM (Least Common Multiple) of two numbers.

If P and Q are two numbers then the product of their HCF and LCM is equal to the product of numbers themselves.

$LCM\times HCF = P \times Q$

Let the given number be $x$ and $y$.

$y$ being the smaller number.

Using the Above Property:

$60\times 2= x \times y$ ...... (1)

Given that one number is 14 more than the other.

$x=y+14$

Putting value of $x$ in equation (1):

$60\times 2= (y+14)\times y\\\Rightarrow 120 = y^2+14y\\\Rightarrow y^2+14y -120 =0\\\Rightarrow y^2+20y-6y-120 =0\\\Rightarrow y(y+20)-6(y+20) =0\\\Rightarrow (y+20)(y-6) =0\\\Rightarrow y = 6, -20$

But HCF and LCM are in positive so value is 6.

Therefore, the smaller number = 6

Answer 2

Answer:

answer-6

Step-by-step explanation:

HCF and LCM of two numbers are 2 and 60 respectively.

One number is 14 greater than the other.

To find:

The smaller number = ?

Solution:

First of all, let us understand one property of HCF (Highest Common Factor) and LCM (Least Common Multiple) of two numbers.

If P and Q are two numbers then the product of their HCF and LCM is equal to the product of numbers themselves.

LCM\times HCF = P \times QLCM×HCF=P×Q

Let the given number be xx and yy .

yy being the smaller number.

Using the Above Property:

60\times 2= x \times y60×2=x×y ...... (1)

Given that one number is 14 more than the other.

x=y+14x=y+14

Putting value of xx in equation (1):

\begin{gathered}60\times 2= (y+14)\times y\\\Rightarrow 120 = y^2+14y\\\Rightarrow y^2+14y -120 =0\\\Rightarrow y^2+20y-6y-120 =0\\\Rightarrow y(y+20)-6(y+20) =0\\\Rightarrow (y+20)(y-6) =0\\\Rightarrow y = 6, -20\end{gathered}

60×2=(y+14)×y

⇒120=y

2

+14y

⇒y

2

+14y−120=0

⇒y

2

+20y−6y−120=0

⇒y(y+20)−6(y+20)=0

⇒(y+20)(y−6)=0

⇒y=6,−20

But HCF and LCM are in positive so value is 6.

Therefore, the smaller number = 6