Alice's take home pay is the same each month

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courtesy of user 'karenmeow'

alice's take home pay is the same each month. she saves same fraction each month. the total amt she saved at the end of the year was 3 times the amt of the portion she did NOT save. if all the money she saved last yr was from her pay, what fraction of her pay did she save each month?
Ron has been teaching various standardized tests for 20 years.

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by lunarpower » Sat Jul 12, 2008 1:50 am
alice's total take home per month = x
alice's saving per month = y
alice's saving for the entire year = 12y
what alice did not save = x-y
per the problem statement, alice's annual saving = 3x what she did not save per month
i.e., 12y = 3(x - y)
4y = x - y
y = x/5

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you can also do this problem by picking numbers, because you don't have a figure for the monthly pay (and, apparently, it doesn't matter, because the problem would be impossible to solve if it did matter).

so just let the monthly pay be '1'.
let the monthly savings be 'S', which is both a fraction and an actual amount of money. (it's both because the monthly pay is '1', so the fraction is the same as the dollar amount)
then she has saved 12S at the end of the year.
each month, she didn't save (1 - S)
so
12S = 3(1 - S)
12S = 3 - 3S
15S = 3
S = 3/15 = 1/5

faster, and probably less confusing, with only one variable
Ron has been teaching various standardized tests for 20 years.

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by parallel_chase » Sat Jul 12, 2008 2:38 pm
lunarpower wrote:alice's total take home per month = x
alice's saving per month = y
alice's saving for the entire year = 12y
what alice did not save = x-y
per the problem statement, alice's annual saving = 3x what she did not save per month
i.e., 12y = 3(x - y)
4y = x - y
y = x/5

--

you can also do this problem by picking numbers, because you don't have a figure for the monthly pay (and, apparently, it doesn't matter, because the problem would be impossible to solve if it did matter).

so just let the monthly pay be '1'.
let the monthly savings be 'S', which is both a fraction and an actual amount of money. (it's both because the monthly pay is '1', so the fraction is the same as the dollar amount)
then she has saved 12S at the end of the year.
each month, she didn't save (1 - S)
so
12S = 3(1 - S)
12S = 3 - 3S
15S = 3
S = 3/15 = 1/5

faster, and probably less confusing, with only one variable

I solved the problem exactly the way you did. But i could not understand the following part:

12y = 3 (x-y)

whereas, I did 12y= 3(12x-12y) since we are calculating for the entire year.

"the total amt she saved at the end of the year was 3 times the amt of the portion she did NOT save."

Since "month" is missing at the end of the statement. It would be nice if you could elucidate on this part of the question.


Thanks

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by lunarpower » Sat Jul 12, 2008 3:40 pm
ah yes, the original poster forgot to include the word 'monthly' in the original problem statement.

i had already answered this question, in fact more than once without noticing, on manhattangmat's own forums (see here and here), so i just copied and pasted the question from karenmeow's other post, without really examining it in detail.

here's the problem statement from our forum (please excuse the capital letters):
ALICE'S TAKE-HOME PAY LAST YEAR WAS THE SAME EACH MONTH, AND SHE SAVED THE SAME FRACTION OF HER TAKE-HOME PAY EACH MONTH. THE TOTAL AMOUNT OF MONEY THAT SHE HAD SAVED AT THE END OF THE YEAR WAS 3 TIMES THE AMOUNT OF THAT PORTION OF HER MONTHLY TAKE-HOME PAY THAT SHE DID NOT SAVE. IF ALL THE MONEY SHE SAVED LAST YEAR WAS HER TAKE-HOME PAY, WHAT FRACTION OF HER TAKE-HOME PAY DID SHE SAVE EACH MONTH?
Ron has been teaching various standardized tests for 20 years.

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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

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by parallel_chase » Sun Jul 13, 2008 2:19 am
lunarpower wrote:ah yes, the original poster forgot to include the word 'monthly' in the original problem statement.

i had already answered this question, in fact more than once without noticing, on manhattangmat's own forums (see here and here), so i just copied and pasted the question from karenmeow's other post, without really examining it in detail.

here's the problem statement from our forum (please excuse the capital letters):
ALICE'S TAKE-HOME PAY LAST YEAR WAS THE SAME EACH MONTH, AND SHE SAVED THE SAME FRACTION OF HER TAKE-HOME PAY EACH MONTH. THE TOTAL AMOUNT OF MONEY THAT SHE HAD SAVED AT THE END OF THE YEAR WAS 3 TIMES THE AMOUNT OF THAT PORTION OF HER MONTHLY TAKE-HOME PAY THAT SHE DID NOT SAVE. IF ALL THE MONEY SHE SAVED LAST YEAR WAS HER TAKE-HOME PAY, WHAT FRACTION OF HER TAKE-HOME PAY DID SHE SAVE EACH MONTH?
Thanks for this post. Everything is sound & clear.

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by alexdallas » Sun Jul 05, 2009 11:15 am
thanks lunarPower

great explanation

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by ghacker » Sun Jul 05, 2009 12:10 pm
No need to use equations

Logic will solve the question in a fraction of the time

ALICE'S TAKE-HOME PAY LAST YEAR WAS THE SAME EACH MONTH[/size], AND SHE SAVED THE SAME FRACTION OF HER TAKE-HOME PAY EACH MONTH. THE TOTAL AMOUNT OF MONEY THAT SHE HAD SAVED AT THE END OF THE YEAR WAS 3 TIMES THE AMOUNT OF THAT PORTION OF HER MONTHLY TAKE-HOME PAY THAT SHE DID NOT SAVE. IF ALL THE MONEY SHE SAVED LAST YEAR WAS HER TAKE-HOME PAY, WHAT FRACTION OF HER TAKE-HOME PAY DID SHE SAVE EACH MONTH?

So if u understand the marked portions you can see that

the amount she did not save = 4 times what she saved

or saved/ not saved = 1/4 hence

the answer is 1/5

Why use equations and complicate things when one can use simple logic and get the answer more quickly and efficiently

and who says that you have to solve / write equations for every simple sum you meet !!!!!!!!!!!!!