Find three numbers such that the sum of any pair exceeds the

Find three numbers such that the sum of any pair exceeds the third by a given amount. Say the given excesses are 20, 30, and 40. Hint: Let the sum of all three numbers be 2x . Add the \'third number\' to \'largest number\' + \'second number\' = \'third number + 20.

Solution

Given excesses 20, 30, 40.

2x the sum of all three.

We have (1) + (2) = (3) + 20.

Adding (3) to each side,

we have: twice(3) + 20 = 2x,

so (3) = x-10.

Similarly the numbers (1) and (2) are x-15 and x-20 respectively. Therefore

3x – 45 = 2x

x = 45.

The required numbers are 30, 25, 35.