I cannot solve this question, please help me. My whole class struggling...
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
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?
And here's my answers, hope this can help you understand how to do yours.
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!
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