billzhao wrote:angelic_devil wrote:a certain amount was to be divided among A, B and C in the ratio of 4:5:6 but by mistake, it was divided in such a manner that 4 time of A's share was equal to 5 times of B's share and 6 times of C's share. as a result A got $154 more than the excepted amount . what was the amount that was divided among them.
a)$370
b)$450
c)$560
d)$1000
e)$1110
answer is E)
plz help to understand
Assume the amount A is supposed to get is A and the amount A was given by mistake is A', And assume the similar variable used for B and C, The total amount is M.
From
a certain amount was to be divided among A, B and C in the ratio of 4:5:6 , we have A=4*k, B=5*k and C=6*k where k is a positive number.
And we have A+B+C=M =>(4+5+6)*k=M =>k=M/15 and thus
A=4*M/15
The second step is to find A':
We have two equations:
A'+B'+C'=M (note: M is still the same)..................(1)
4A'=5B'=6C'...................(2)
From (2), we have: B'=4/5*A' and C'=4/6*A'
Substitute B' and C' into (1), we have A'+4/5*A'+4/6*A'=M =>
A'=15/37*M
From
as a result A got $154 more than the excepted amount, we have A'-A=154 =>
15/37*M-4*M/15=154.
We can solve the above equation and M=1110.
Folks, you can conclude these retios in a quicker way and also using options you can avoid the calculation involved in the final stage here...
Given that money must be divided among A,B and C in the ratio 4:5:6 .
But they divided such that 4A=5B=6C=k
So A:B:C = 1/4 : 1/5 : 1/6
In order to convert these fractions into natural numbers take the lcm of the denominators and write the corresponding numerators.
So the ratio of A:B:C = 15:12:16
Now if you observe carefully, initially the money must be divided in the ratio 4:5:6. So A must get 4/15 (M) where M is total money
But he got 15/37(M).
Given M(15/37-4/15) = 154 (a natural number). which means M must be divisible by both 15 and 37 i.e M must be divisible by 111. Only E satisfies this.
I completely agree that the correct way to answer this is to solve the equation but since we are comfortable with the divisibility rules we can make use of them. Also I am lucky enough to have only one option which is divisible by 111.