Question 925611: Please Help!!! Q)Four friends A,B,C and D together fave 120 marbles. The number of marbles with A is 1/4 the total number of marbles with the other three. Marbles with B is 1/3 of the total marbles with the other three. C has 3/7 of the marbles with others. Who among A,B,C and D has the most of marbles?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We are given 4 equations,
A +B +C +D = 120
A = (1/4) * (B+C+D)
B = (1/3) * (A+C+D)
C = 120 - (A+B+C)
from the second equation, we have
4A = (B+C+D)
substitute in first equation,
A +4A = 120
5A = 120
A = 24
from the third equation, we have
3B = (A+C+D)
substitute in first equation
B +3B = 120
4B = 120
B = 30
from fourth equation, we have
C = 120 - (54+D)
C = 66 -D
C+D = 66
C =54
to find D, we have
24 +30 +54 +D = 120
108 +D = 120
D = 12
therefore,
A has 24 marbles, B has 30 marbles, C has 54 marbles, D has 12 marbles
C has the most marbles which is 54
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