Question

There are two numbers

x

and

y

such that

x < y

. Their H.C.F. is

4

and L.C.M. is

48

. If

2

is added to

y

, the H.C.F. is increased by

2

. If

2

is subtracted from

x

, the H.C.F. is reduced by

2

. Find values of

x

and

y

.

Accepted Answer

There are two numbers $x$ and $y$ . Their H.C.F. is $4$ and L.C.M. is $48$ .

We know that HCF $\times$ LCM $=$ the product of two numbers.

Thus, we have, $x \times y = 4 \times 48$

$\Rightarrow x y = 192$ -----(i)

Now, if $2$ is added to $y$ , means $y$ becomes $(y + 2)$ . the H.C.F. becomes $4 + 2 = 6$ [H.C.F. is increased by $2$ ]

So, we have, $x (y + 2) = 6 \times 48$

$\Rightarrow x y + 2 x = 288$ -----(ii)

Again, if $2$ is subtracted from $x$ , means $x$ becomes $(x - 2)$ , the H.C.F. becomes $4 - 2 = 2$ [H.C.F. is reduced by $2$ ]

Thus, we have, $(x - 2) y = 2 \times 48$

$\Rightarrow x y - 2 y = 96$ -----(iii)

$\Rightarrow 192 - 2 y = 96$

$\Rightarrow 2 y = 192 - 96$

$\Rightarrow 2 y = 96$

$\Rightarrow y = \frac{96}{2} = 48$

From (i), we obtain, $x = \frac{192}{y} = \frac{192}{48} = 4$ .

Therefore, the required value of $x$ and $y$ are $4$ and $48$ , respectively.

Dr. Sanjeev Verma and R.K. Tyagi Solutions for Chapter: Playing With Whole Numbers, Exercise 13: HOTS

Dr. Sanjeev Verma Mathematics Solutions for Exercise - Dr. Sanjeev Verma and R.K. Tyagi Solutions for Chapter: Playing With Whole Numbers, Exercise 13: HOTS

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