Question

Two numbers are in the ratio 3:4. The difference between their square is 28. Find the greater number

Answer 1

Step-by-step explanation:

lets us take two number be x,y

then, x/y=3/4

x=3y/4 ---(1)

since the difference between their squares is 28,

y^2-x^2=28 ---(2)

substitute (1) in (2)

y^2-(3y/4)^2=28

y^2-(9y^2/16)=28

(16y^2-9y^2)/16=28

7y^2=28*16=448

y^2=448/7=64

y=8

substitute y in (2)

y^2-x^2=28

(8)^2-x^2=28

64-x^2=28

x^2=64-28

x^2=36

x=6

therefore sum of square of the two numbers 6^2+8^2=36+64=100.

Answer 2

Answer:

The greater number is $100$

Step-by-step explanation:

Let us assume the two numbers$=x,y$

then

$\frac{x}{y} =\frac{3}{4}\\ x=\frac{3y}{4} ...(1)$

Now it is given that difference between their square$=28$

So, $y^2-x^2=28...(2)$

By substituting (2) in (1), we have

$y^2-(\frac{3y}{4} )^2=28\\y^2-\frac{9y^2}{16} =28\\16y^2-9y^2=28\times16\\7y^2=448$

Further solving,

$y^2=\frac{448}{7} \\y^2=64\\y=8$

Now plug this value in equation (2)

$y^2-x^2=28\\8^2-x^2=28\\64-x^2=28\\x^2=64-28$

By simplifying, we have

$x^2=36\\x=6$

Therefore, the sum of squares of two numbers is

$6^2+8^2=36+84\\=100$