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

• 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

Answer 2

Answer:

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

A.T.Q.

6x + 4x = 5x + 55

=> 10x = 5x + 55

=> 10x − 5x = 55

=> 5x = 55

=> x = $\frac{55}{5}$

x = 11

Therefore-:

First no. =4×11=44

Second no. =5×11=55

Third no. =6×11=66

Hence the three numbers are 44,55 and 66.