working backward with percents....

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working backward with percents....

by wawatan » Wed May 21, 2008 1:44 pm
If today the price of an item is 3600, what was the price of the item exactly 2 years ago?

1) The price of the item increased by 10% per year during this 2 year period.

this is a data suff. problem but I want to know how to solve the answer.

any help would be greatly appreciated.

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by punit.kaur.mba » Wed May 21, 2008 1:58 pm
Let the price 2 yrs ago be x.
Given 10% increase per year -

End of first yr price = x + 10%x=x+(10/100)x=110x/100

End of second yr, price increased by 10%

price = (110x/100) + 10% of (110x/100)

Hence,

3600 = (110x/100) + (10/100)(110x/100)

Solve for x

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Re: working backward with percents....

by lunarpower » Sat May 24, 2008 2:39 am
wawatan wrote:If today the price of an item is 3600, what was the price of the item exactly 2 years ago?

1) The price of the item increased by 10% per year during this 2 year period.

this is a data suff. problem but I want to know how to solve the answer.

any help would be greatly appreciated.
ok, i think you already know this from the tenor of your comments, but you don't need to solve this problem in the context of the DS question you've presented. instead, it's sufficient (heh) to realize that there is exactly one starting price that, when increased by 10% twice, will yield 3600 (any lower starting price would yield a lower result, and any higher starting price would yield a higher result). that means 'sufficient'.

--

IMPORTANT BACKGROUND KNOWLEDGE that will save you lots and lots of time on the exam:

make sure you know how to use MULTIPLIERS to achieve percent changes in one step.

for instance, x increased by 14% is x + 0.14x, or 1.14x. this means that you can increase a quantity by 14% merely by multiplying it by 1.14.
by the same token,
to increase something by 40%, multiply by 1.4
to increase someting by 210%, multiply by 1 + 2.1 = 3.1
etc.

also, x decreased by 14% is x - 0.14x, or 0.86x. this means that you can decrease a quantity by 14% merely by multiplying it by 0.86. (this should make intuitive sense, because if you reduce a quantity by 14% then you have 86% of the original quantity remaining.)
by the same token,
to decrease something by 40%, multiply by 0.6
to decrease something by 8.1%, multiply by 0.919
etc.

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if you know the above background information, then statement (1) becomes
x(1.1)(1.1) = 3600
or
1.21x = 3600
which is a lot friendlier than the method in the above post (correct as that post may be).
learn the multipliers! like microwave ovens and cell phones, they will make your life easier, and you will never want to be without them again!
Ron has been teaching various standardized tests for 20 years.

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by Stuart@KaplanGMAT » Sat May 24, 2008 10:03 am
Even though it may not appear to be so at first glance, this is really a compound interest problem.

We have a fixed rate of growth and we have increases on the increases - the exact definition of compount interest.

So, another way we could have solved would be to use the compound interest formula:

Total of Principal + interest = P ((1 + r)^t)

in which P = principal, r = the rate expressed as a decimal and t = number of compound periods.

Here, we have:

3600 = P(1.1)^2
3600 = 1.21P

which is exactly the same as Ron's final answer (of course) - just another way to look at it.
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by netigen » Sat May 24, 2008 11:45 am
Compound interest is the way to go but you can also use the following formula:

first increase will be X%
Second year increase will be Y%+XY%/100

where X and Y are the percentages for the corresponding years, in this case x=y=10%

Assuming, the original price was p then by the formula:

3600 = p+px/100+py/100+pxy/10000 = p(1+0.1+0.1+0.01) = p(1.21)

or p = 3600/1.21

you can use this formula when you come across cases where multiple percentages are involved.

then