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Question 999050: Four numbers, taken two at a time, give the sums 84, 88, 100, 100, 112 and 116. What are the four numbers?
Found 2 solutions by macston, ikleyn:
Answer by macston(5194)   (Show Source):
You can put this solution on YOUR website! .
A, B, C, and D are the four numbers
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GIVEN:
A+B=84
A+C=88
A+D=100
B+C=100
B+D=112
C+D=116
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SOLVE:
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A+B=84
A=84-B
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B+C=100
C=100-B
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B+D=112
D=112-B
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A+C=88
(84-B)+(100-B)=88
-2B=-96
B=48
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A+B=84
A+48=84
A=36
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B+C=100
48+C=100
C=52
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B+D=112
48+D=112
D=64
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ANSWER:
A=36
B=48
C=52
D=64
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CHECK:
A+B=84
36+48=84
84=84
.
A+C=88
36+52=88
88=88
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A+D=100
36+64=100
100=100
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B+C=100
48+52=100
100=100
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B+D=112
48+64=112
112=112
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C+D=116
52+64=116
116=116
Answer by ikleyn(52824)  (Show Source):
You can put this solution on YOUR website! .
Four numbers, taken two at a time, give the sums 84, 88, 100, 100, 112 and 116. What are the four numbers?
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GIVEN:
A+B=84
A+C=88
A+D=100
B+C=100
B+D=112
C+D=116
Let us consider those sums that relate to 3 numbers A, B and C:
A+B=84 (1)
A+C=88 (2)
B+C=100 (3)
Sum up the left and the right sides. You will get
2*(A + B + C) = 84 + 88 + 100 = 272. It gives
A + B + C =  = 136 (4).
Now, distract (1) from (4). You will get
C = 136 - 84 = 52.
Distract (2) from (4). You will get
B = 136 - 88 = 48.
Distract (3) from (4). You will get
A = 136 - 100 = 36.
Now you can easily determine D from A+D=100, for instance:
D = 100 - 36 = 64.
Answer. A=36, B=48, C=52, D=64.
Notice. We have 4 unknowns, therefore 4 equations is enough to find the unknowns. The rest of equations are excessive.
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