Question

three numbers are in the ratio of 4:5:6. if the sum of the largest and smallest exceeds the third number by 55 find the number​

Answer 1

Explanation:

`\huge \color {brown} \boxed { \colorbox {black}{ ANSWER }}`

Let us assume the ratio's as 4x, 5x and 6x.

A.T.Q

6x+4x=5x+55

→ 10x = 5x + 55

→ 10x - 5x = 55

→ 5x = 55

→ x =
`(55)/(5)`

x = 11

Therefore,

First no. = 4x = 4 × 11 = 44

Second no. = 5x = 5 × 11 = 55

Third no. = 6x = 6 × 11 = 66

Hence the three numbers are 44,55 and 66.

Answer 2

• Let numbers be 4x,5x,6x

ATQ

`\\ \sf\longmapsto 4x+6x=5x+55`

`\\ \sf\longmapsto 10x=5x+55`

`\\ \sf\longmapsto 10x-5x=55`

`\\ \sf\longmapsto 5x=55`

`\\ \sf\longmapsto x=11`

• 5x=55