SOLUTION: Hi ben and tom had a total of 144 sweets.ben gave tom 1/5 of what he had. Tom the gave ben 1/4 of what he had to ben. If both had an equal number of sweets in the end, how many di

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Question 1105576: Hi
ben and tom had a total of 144 sweets.ben gave tom 1/5 of what he had. Tom the gave ben 1/4 of what he had to ben. If both had an equal number of sweets in the end, how many did each of them have at first.
Thanks
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
RETRY---------------------

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ben gave tom 1/5 of what he had. Tom the gave ben 1/4 of what he had to ben.
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After that process is done,
Ben has 4b%2F5%2Bt%2F4%2Bb%2F20 and Tom has t%2Bb%2F5-t%2F4-b%2F20. These values are given as equal and their equality can be simplified.

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If both had an equal number of sweets in the end, how many did each of them have at first.
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4b%2F5%2Bt%2F4%2Bb%2F20=t%2Bb%2F5-t%2F4-b%2F20
LCD 20 so multiply both members by 20.

16b%2B5t%2Bb=20t%2B4b-5t-b
14b=10t
7b=5t
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Use b%2Bt=144 and substitute for t.
7b=5%28144-b%29
7b=5%2A144-5b
12b=5%2A144
b=5%2A12
highlight%28b=60%29--------------original amount Ben had.

t=144-60
highlight%28t=84%29------------original amount Tom had.






***********************(BELOW STILL CONTAINS UNFIXED MISTAKE)******************************

Follow the description step-wise literally.

b, Ben had originally
t, Tom had originally

b+t=144
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b-b%2F5 and t%2Bb%2F5, Ben and Tom
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b-b%2F5%2B%28t%2Bb%2F5%29%2F4 and t%2Bb%2F5-%28t%2Bb%2F5%29%2F4, Ben and Tom

Now these last two numbers are given as equal.
b-b%2F5%2B%28t%2Bb%2F5%29%2F4=t%2Bb%2F5-%28t%2Bb%2F5%29%2F4

b-b%2F5%2Bt%2F4%2Bb%2F20=t%2Bb%2F5-t%2F4-b%2F20

LCD is 20, so multiply both sides by 20.

20b-4b%2B5t%2Bb=20t%2B4b-5t-b

17b%2B5t=23b-5t

10t=6b

5t=3b

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Make substitution from b%2Bt=144
b=144-t
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5t=3%28144-t%29
5t=-3t%2B432
8t=432
cross%28t=54%29-------------Tom had originally.

cross%28b=90%29------------Ben had originally.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There was an error in the algebra in the solution provided by the other tutor; the given answer does not satisfy the conditions in the problem.

Let b and t represent the numbers Ben and Tom start out with, respectively:
Start:
Ben = b;
Tom = t.

Ben gives 1/5 of the number he has to Tom; after that:
Ben = (4/5)b;
Tom = t+(1/5)b.

Tom now gives 1/4 of what he has -- which is (1/4)t+(1/20)b -- to Ben; after that:
Ben = (4/5)b+(1/4)t+(1/20)b = (17/20)b+(1/4)t;
Tom = (3/4)t+(3/20)b.

At this point the two of them have the same number of sweets, 72.

Ben: %2817%2F20%29b%2B%281%2F4%29t+=+72
(1) 17b%2B5t=1440
Tom: %283%2F4%29t%2B%283%2F20%29b+=+72
(2) 15t%2B3b=1440

Solve (1) and (2) by elimination:
15t%2B51b+=+4320
15t%2B3b+=+1440
48b+=+2880
b+=+60

Ben started with 60 sweets; Tom with 84.

Check:
Start: Ben 60, Tom 84
After Ben gives 1/5 of his to Tom: Ben 60-12 = 48; Tom 84+12 = 96
After Tom gives 1/4 of his to Ben: Ben 48+24 = 72; Tom 96-24 = 72