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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?
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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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