Maths Question?

Question

Please help :

Andy, Benny, Catherine & Danny have a sum of money each. The amt of money Andy has is 1/3 of the total amt of money Benny, Catherine & Danny have. The amt of money Benny has is 1/4 of the total amt of money Andy, Catherine & Danny have. The amt of money Catherine has is 1/5 of the total amt of money Andy, Benny & Danny have. If Danny has $92, how much do the four children have?

Thanks.

Answer 1

This is another method to solve this question:

The amt of money Andy has is 1/3 of the total amt of money Benny, Catherine & Danny have
=> Andy has 1unit, Benny, Catherine & Danny have 3units
=> total amount of money = 4units
=> Andy has 1/4 of the total amt of money

Similarly,
Benny has 1/5 of the total amount of money
Catherine has 1/6 of the total amount of money

Thus Danny has (1-1/4-1/5-1/6)=23/60 of the total amount of money.

23/60 -----> $92
1 -----> $240
Thus,
Andy has ($240x1/4)=$60
Benny has ($240x1/5)=$48
Catherine has ($240x1/6)=40

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Answer 2

Let them have a, b, c, d resp.
a = 1/3 (b+c+d) b+c+d - 3a = 0
b = 1/4 (a+c+d) a+c+d - 4b = 0
c = 1/5 (a+b+d) a+b+d - 5c = 0
d = 92

5b - 4a = 0 ( Eqn 2 - Eqn 1)
6c - 5b = 0 ( Eqn 3 - Eqn 2)
4a = 5b = 6c
c = 4/6 a b = 4/5 a
b+c = 44/30 a

44/30 a + $92 - 3a = 0 (putting values in Eqn 1)
46/30 a = $92

a = $60
b = $48
c = $40
d = $92

Answer 3

Let A, B, C, and D be the amount of money each of them has (named after their first initials). Then we can write an equation for each sentence of information we're told:

A = (B+C+D)/3
B = (A+C+D)/4
C = (A+B+D)/5
D = 92

Multiply the first equation by 3, the second by 4, and the third by 5, and substitute the D=92, and you get:

3A = B+C+92
4B = A+C+92
5C = A+B+92

You've gotten this down to three equations with three unkowns. So solve the system to find A, B, and C, and put these answers with D.