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Question

Four friends, A,B,C,D each have a whole number of dollars. A has 1/7 of what B,C,D have combined. B has 1/9 of what A,C,D have combined. C has 1/11 of what A,B,D have combined. What is the least amount of money they could each have?

Answer

This is ratio problem and constant total problem.

A : B+C+D : Total => 1 : 7 : 8
B : A+C+D : Total => 1 : 9 : 10
C : A+B+D : Total => 1 : 11 : 12

Note Total is A+B+C+D, therefore we can make all the above ratios share the same unit ‘u’, by making the total common. Constant Total. LCM for the total is 120

A : B+C+D : Total => 15 : 105 : 120
B : A+C+D : Total => 12 : 108 : 120
C : A+B+D : Total => 10 : 110 : 120

D = A+B+D – (A+B) = 110 – (15 + 12) = 83

Lowest amount of money is when u =1 and I used LCM for the total, so
A = $15
B = $12
C = $10
D = $83

1 Reply 2 Likes

How to explain the 1st part? How to derive that ratio of 1:7:8 and so forth?

Very rusty with ratio. Thanks a million of the help.

1 Reply 1 Like

From A is 1/7 of B+C+D is the same as A:(B+C+D) => 1:7 in ratio.

Total is simply 1+7=8. Which is just A + ( B+C+D).

Fraction can covert to ratio simply. E.g A has 3/5 of B money is A :B => 3:5 . If you want to include total is A:B:total => 3:5:8.

 

1 Reply 1 Like

Ok. Understood. Again thank you very much!

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