Question

The sum of the squares of three numbers is 336 and the ratio of the first and the

second as also of the second and the third is 1 : 2. Then find the difference of third

and first number?

Answer 1

The difference of third and first number is 12.

Step-by-step explanation:

Let the three numbers be x, y and z.

By the given conditions,

x² + y² + z² = 336 ..... (1)

x : y = 1 : 2

or, x/y = 1/2

or, x = y/2 ..... (2)

y : z = 1 : 2

or, y/z = 1/2

or, z = 2y ..... (3)

Using (2) and (3), from (1), we get

(y/2)² + y² + (2y)² = 336

or, (1/4 + 1 + 4) y² = 336

or, 21/4 * y² = 336

or, y² = 336 * 4/21

or, y² = 64

i.e., y = 8

Solving we get

x = 4, y = 8, z = 16

Therefore the required difference of third and first number is

= |x - z|

= |4 - 16|

= |- 12|

= 12

Answer 2

Answer:

12

Step-by-step explanation:

x : y : z

1 : 2 : 2

1 : 1 : 2

overll ratio x : y : z = 1 : 2 : 4

(1x)^2 + (2x)^2 + (4x)^2 = 336

21x^2 = 336

x = 4

so

x = 4

y = 2x4 = 8

z = 4×4 = 16

ans z - x = 12