Question 212175This question is from textbook Elementary And Intermediate Algebra
: Ancient history. 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.
This question is from textbook Elementary And Intermediate Algebra
Found 2 solutions by checkley77, ankor@dixie-net.com:
Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website! A+B+C+D=9,900 OR D=9,900-(A+B+C)
B=A+A/7
C=(A+B)+300
D=(A+B+C)+300
D=-(A+B+C)+9,900 NOW ADD.
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2D=300+9,900
2D=10,200
D=10,200/2
D=5,100 ANS.
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C=A+B+300
-[5,100=A+B+C+300] SUBTRACT
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C-5,100=-C
2C=5,100
2C=5,100
C=5,100/2
C=2,550 ANS.
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B=A+A/7
C=(A+B)+300 OR C=(A+A+A/7)+300 OR C=(7A+7A+A)/7+300 OR C=15A/7+300
2,550=15A/7+300
15A/7=2,550-300
15A/7=2,250
15A=2,250*7
15A=15,750
A=15,750/15
A=1,050 ANS
B=1,050+1,050/7
B=(7*1,050+1,050)/7
B=(7,350+1,050)/7)
B=8,400/7
B=1,200
PROOF:
1,050+1,200+2,550+5,100=9,900
Answer by ankor@dixie-net.com(22740)  (Show Source):
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