One number is 4 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 418​, find the numbers.

Question

One number is

44

times a first number. A third number is 100 more than the first number. If the sum of the three numbers is

418418​,

find the numbers.

Answer 1

To work this kind of problem, must choose a variable to represent the first number. How about "N".

Let N be the first number.

2nd number = 4 * the first number, so

2nd number = 4N

3rd number = 100 more than the 1st number, so

3rd number = N + 100

The sum of all three numbers is 418, so we add all of these

N + 4N + N+100 = 418

Combine like terms

6N + 100 = 418

Subtract 100 from both sides

6N = 318

Divide both sides by 6

N = 53

So...

1st number = 53

2nd number = 4 * 53 = 212

3rd number = 100 + 53 = 153

And to check, add these to be sure they total 418 (and they do!)