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Algebra

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by ritula » Mon Aug 25, 2008 2:53 am
If Sam has 5 more marbles than Jack, Jack has 4 more marbles than Frank, and Frank has 5 more marbles than Ted, what is the least number of marbles that must change hands if the four boys are to have an equal number of marbles?
Philosophers have interpreted world in various ways, the point is to change it!
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by amitansu » Mon Aug 25, 2008 3:44 am
I think it's 9 altogether.
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by 4meonly » Mon Aug 25, 2008 8:23 am
total 9 marbles
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by parallel_chase » Mon Aug 25, 2008 9:20 am
S=T+14=J+5=F+9

14+5+9 = 28

28/4 = 7

The question is asking least number of marbles. Hence 7 is the answer.

whats the OA?
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by rishi235 » Mon Aug 25, 2008 9:46 am
Hi...
I also got 7...which is the average of the no. of marbles..
But i do not think its the right answer...

@ chase...
I guess we need the find the least no. of marbles 'exchanged'...

I just substituted the values as
0 , 5 , 9 & 14...

So all the boys to have 7 marbles... we need to exchange 7+2 = 9 marbles..
So ans is 9.

@ Ritula... wats the OA...
Last edited by rishi235 on Mon Aug 25, 2008 10:13 am, edited 2 times in total.
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by kris610 » Mon Aug 25, 2008 10:08 am
I agree with 9:

Total in terms of T: 4(T+7), which means each boy should've T+7. Sam needs to give up 7 and Jack nees to give up 2.
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by ritula » Mon Aug 25, 2008 9:54 pm
yes OA is 9 indeed. Thanks to all.
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