The sum of two natural numbers, x and y, is 320 and the HCF of x and y is 20 . If x>y, how many such possible pairs of x and y are there?

Correct option is B

Given:

HCF = $$(x , y)= 20$$ 

The sum of $$x$$ and $$y$$ is = $$x+y = 320$$

Solution:

Let $$x= 20a\$$and $$y= 20b$$ (Where $$a,b$$ is the coprime integer.) HCF ($$a,b$$) = 1

The sum of $$x$$ and $$y$$ is  = $$x+y = 320$$  => $$20a+20b= 320$$ => $$a+b = 16$$ 

The integer pair $$(a,b)$$ such that $$a+b = 16$$ and HCF ($$a,b$$) = 1 are:

(1,15), (3,13), (5,11), (7,9)

For each pair calculate $$x= 20a\$$and $$y= 20b$$ : Where $$x>y$$​​

​$$(1, 15): x = 20 × 15 = 300, y = 20 × 1 = 20$$ 

​$$(3, 13): x = 20 × 13 = 260, y = 20 × 3 = 60$$

$$(5, 11): x = 20 × 11 = 220, y = 20 × 5 = 100$$

$$(7, 9): x = 20 × 9 = 180, y = 20 × 7 = 140$$ 

So the valid pairs of $$(x, y)$$ are:​

$$(300, 20), (260, 60), (220, 100), (180, 140)$$

Thus the correct answer is (B)