There are 764 marbles to be shared among 4 boys : a, b, c and d. If a has 20 marbles lesser, b has 30 marbles more, c has to halve his marbles and d has to double his marbles, all the boys will have equal marbles at the end. How many marbles does each boy have initially?
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2 comments:
A - 2 units +20 [20 less to be equal]
B - 2 units -30 [30 more to be equal]
C - 4 units [half to equal]
D - 1 unit [double to equal]
Total - 9 units -10
Thus, each unit is (764+10)/9 = 86
A has 192
B has 142
C has 344
D has 86
(I have a primary school aged daughter ^_^ quite used to doing these type of sums now)
Good job with the modelling methodology.
Let's solve this problem with simultaneous equation.
Let x be the equal amount for each person at the end.
(x+20) + (x-30) + 2x + 1/2x = 764
x = 172
Therefore,
A has to be 20 less to be equal = 172 + 20 = 192
B has to be 3o more to be equal = 172 - 30 = 142
C has to half to be equal = 172 x 2 = 344
D has to double to equal = 172 / 2 = 86
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