​The product of two numbers is 64 and their sum is 20. If the first number is greater than the second, then the ratio of the two numbers is: 

Correct option is C

Given:

The product of two numbers is 64.

The sum of the two numbers is 20.

The first number is greater than the second, 

Concept Used:

​$$(x+y)^2-(x-y)^2 = 4 xy$$ 

Solution:

Let two numbers are x and y (Where x > y). such that:

x + y = 20 ........(Equation1)

x.y = 64 ..........(Equation2) 

Using identity $$(x+y)^2-(x-y)^2 = 4 xy$$​ 

Substitute values in identity from eq1 and eq 2 

​$$(20)^2-(x-y)^2 = 4 ×64$$ 

$$(x-y)^2 = 400- 256$$ 

$$(x-y)^2 = 144$$ 

$$x-y = \sqrt144 = 12$$ .......(Equation3)

By adding Equation1 and Equation3 

​$$x+y+x-y= 20+12$$ 

​$$2x = 32$$ 

$$x = \frac{32}{2} = 16$$ 

Substitute x = 16 in Equation1 

​$$16+ y = 20$$​​

​$$y = 20 - 16 = 4$$ 

The ratio of the two numbers is: 

​$$\frac{x}{y} = \frac{16}{4} = 4 : 1$$ 

Thus the ratio of the two numbers is 4:1 .