1. ## Numbers

The sum of 4 numbers is 80. If you add 3 to the first number, subtract 3 from the second number, multiply the third number by 3, the result will be equal. What is the difference between the largest and the smallest of the original numbers?

Thank you

2. Originally Posted by gwen
[snip]If you add 3 to the first number, subtract 3 from the second number, multiply the third number by 3, the result will be equal.
[snip]
This makes no sense. What result are you talking about?

3. I did a similar problem to this a while ago, this is what is was:

The sum of four numbers is 80. If you add 3 to the first number, subtract 3 from the second number, multiply the third number by 3 and divide the fourth number by 3, the result will be equal. What is the difference between the largest and the smallest of the original numbers?

Let the numbers be a, b, c & d.
a+b+c+d=80
a+3=x ; a=x-3
b-3=x; b=x+3
3c=x; c= (x/3)
(d/3) =x; d= 3x

Now substitute a,b,c & d in terms of x in the 1st equation.

(x-3) + (x+3) + (x/3) + 3x = 80
2x + (x/3) + 3x = 80

Multiply all terms by 3

6x + x + 9x = 240
16x = 240
x = (240/16) = 15

Then replace the value of x to obtain the 4 numbers.

a=15-3 =12
b=15+3 =18
c=(15/3) =5
d=3X15 =45.

So the difference between largest and smallest will be = 45-5 =40.
There you go, hope it helped!

4. Originally Posted by gwen
The sum of 4 numbers is 80. If you add 3 to the first number, subtract 3 from the second number, multiply the third number by 3, the result will be equal.
Probably means "the results will be equal and also equal the 4th number".

a = 1st, b = 2nd, c = 3rd; then 4th = 80 - a - b - c

a + 3 = b - 3 ; b = a + 6 [1]
a + 3 = 3c ; c = (a + b) / 3 [2]

a + 3 = 80 - a - b - c
2a + b + c = 77
Substitute [1] and [2] :
2a + a+6 + (a+b)/3 = 77 ; solve for a:
a = 21
[1]: b = 27
[2]: c = 8
4th = 80 - 21 - 27 - 8 = 24
21 + 3 = 24, 27 - 3 = 24, 8 * 3 = 24

5. Hi,

Thanks very much wilmer and jumba.