Four numbers: their sum is 9900; let the second exceed the first by one-seventh of the first; let the third exceed the sum of the first two by 300; and let the fourth exceed the sum of the first three by 300. Find the four numbers.
Explain
2007-04-30 15:44:34 · 4 answers · asked by Dr. Richards 2 in Science & Mathematics ➔ Mathematics
x + 8/7x + (x+8/7x+300) + (x+8/7x+300+300) = 9900
45/7x + 900 = 9900
45/7x = 9000
x = 1400
1400, 1600, 3300, 3600
algebra
2007-04-30 15:50:34 · answer #1 · answered by David K 2 · 0⤊ 2⤋
let the numbers be a, b, c, d
b = 8/7 a
c = a + b + 300
d = a + b + c + 300
a+b+c+d
= 2*(a+b+c) + 300
= 2 [ 2(a+b) + 300] + 300
= 2 [ 2 (15/7)a + 300 ] + 300 = 9900
(60/7) a + 900 = 9900
a = 1050
b = 1200
c = 2550
d = 5100
The first 2 answers have given this answer
1400, 1600, 3300, 3600
But the 4th number does not exceed the sum of the first 3 by 300.
2007-04-30 15:59:49 · answer #2 · answered by Dr D 7 · 2⤊ 0⤋
x = first
x + x/7 = second
x + x+ x/7 + 300 = third
x + x + x/7 + x + x + x/7 + 300 + 300 = fourth
8x + 4x/7 + 900 = 9900
(60/7)x = 9000
x = 7*9000/60 = 7*150 = 1050
1050 = first
1200 = second
2550 = third
5100 = fourth
2007-04-30 15:57:07 · answer #3 · answered by Raymond 7 · 2⤊ 0⤋
Let x be number 1
Let 8/7 x be number 2
Let x+8/7x+300 = number 3
Let x+8/7x+600 = number 4
Let the sum be 9900.
Then 3x + 24/7 x + 900 = 9900
45/7 x = 9000
x = 63000/45 = 1400
The others are 1600, 3300 and 3600
2007-04-30 15:53:24 · answer #4 · answered by cattbarf 7 · 0⤊ 2⤋