INTERMEDIATE ALGEBRA

[Gail Price]

Four numbers have a sum of 9900. The second exceeds the first by one-seventh of the first. The third exceeds the sum of the first two by 300. The fourth exceeds the sum of the first three by 300. Find the four numbers.

 

[arthur ohlsten]

a+b+c+d=9900

1) b=a+a/7
2)c=a+b+300
3)d=a+b+c+300

substitute for b in 2
c=a+a+a/7 +300

substitute for b and c in 3
d=a+[a+a/7] +[a+a+a/7+300]+300

substitute into original
a+a+a/7 +a+a+a/7+300+ 4a+2a/7 +600= 9900
8a+ 4a/7=9000
a[60/7]=9000
a=6300/6
a=1050 answer
b=1050+ 150
b=1200 answer
c=2100+150+300
c=2550 answer
d=1050+1200+2550+300
d=5100 answer

a+b+c+d=1050+1200+2550+5100
a+b+c+d=9900 check

Arthur

 

[Gail Price]

I do not understand how you got the 1050, how did you get the a+a+a/7, there is a lot to this problem and I don't know how you got that answer. I have studied it and can not come up with half of what you have. Can you explain step by step please. I do not want to seem difficult, but I would like to understand a little more. pleasesssssssssssss, pretty please

 

[Denis]

"The second exceeds the first by one-seventh of the first."

If a = the first, then the second = a + a/7 = 8a / 7 ; are you ok with that?

 

[Gail Price]

Thanks you, so much