SOLUTION: Keleah 1 needs help to solve this Ancient history word problem. This problem is from the second century. Four numbers have a sum of 9900. The second exceeds the first by one-sev

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Question 212142: Keleah 1 needs help to solve this Ancient history word problem. This problem is from the second century. 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.
Answer by Theo(13342) About Me  (Show Source):
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let a = first number, b = second number, c = third number, d = fourth number
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a+b+c+d = 9900
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a = a = (7a/7)
b = a + (1/7)a = (8a/7)
c = a + b + 300 = 7a/7 + 8a/7 + 300 = (15a/7 + 300)
d = a + b + c + 300 = 7a/7 + 8a/7 + 15a/7 + 300 + 300 = (30a/7 + 600)
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sum of all 4 number is:
7a/7 + 8a/7 + (15a/7 + 300) + (30a/7 + 600)
combine like terms to get:
60a/7 + 900 = 9900
subtract 900 from both sides to get:
60a/7 = 9000
multiply both sides by 7 and divide both sides by 60 to get:
a = 7*9000/60 = 1050
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a = 1050
b = 8/7 * 1050 = 1200
c = a + b + 300 = 1050 + 1200 + 300 = 2550
d = a + b + c + 300 = 1050 + 1200 + 2550 + 300 = 5100
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a + b + c + d = 1050 + 1200 + 2550 + 5100 = 9900
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