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ims quant

Expert replies
by angelic_devil » Tue Feb 17, 2009 6:36 am
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
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Source: — Problem Solving |

by tini » Tue Feb 17, 2009 7:54 am
5b=4a
or b/a=4/5

4a=6c
or a/c=6/4

b:a:c=6*4:6*5:5*4=24:30:20=12:15:10

let total amt=x

a's share from this new ratio=(15/37)x

a's original share=(4/15)x

(15/37)x-(4/15)x=154
solve for x,x=154*37*15/77=1110
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exp

by naaga » Tue Feb 17, 2009 10:04 am
tini , how did you get this step, i didn't understand the aproach
plz clear ..

b:a:c=6*4:6*5:5*4=24:30:20=12:15:10
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Re: ims quant

by billzhao » Wed Feb 18, 2009 3:03 am
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.
Yiliang
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Re: ims quant

by sureshbala » Wed Feb 18, 2009 3:16 am
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.
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