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 Post subject: If m = 9/25, w = 15/32, and m + w + c = 1, which of the foll
 Post Posted: Wed Oct 20, 2010 7:02 pm 
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Students


Posts: 4
If m = 9/25, w = 15/32, and m + w + c = 1, which of the following gives the values of m, w, and c in increasing order?

* c, m, w
* c, w, m
* m, w, c
* w, c, m
* w, m, c

The correct aswer choice is c, m, w

I would like to know if there is a short way to solve this. I took the long route, subsituting m and w on m + w + c = 1 and solving for c. This takes a long time and at the end I had to guess which fraction was bigger or smaller.

Thanks

tito


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 Post subject: Re: If m = 9/25, w = 15/32, and m + w + c = 1, which of the foll
 Post Posted: Fri Oct 22, 2010 12:10 am 
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Course Students


Posts: 1
I will give it a shot

I quickly cross multiplied m and w to find out which of them was greater than the other
m=9/25 and w=15/32

9/25 X 15/32
288 < 375

Hence m < w. Eliminate B,D and E

then I estimated the fractions as
m=9/25 ~= 9/27 ~=1/3
w=15/32~=15/30~=1/2

c~= 1-(1/3 + 1/2)
c~= 1/3

Hence c<w. The only option that satisfies this is (A). Might not be the most elegant solution but is quick.


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 Post subject: Re: If m = 9/25, w = 15/32, and m + w + c = 1, which of the foll
 Post Posted: Sat Oct 23, 2010 5:13 am 
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ManhattanGMAT Staff


Posts: 7146
titoestrada wrote:
If m = 9/25, w = 15/32, and m + w + c = 1, which of the following gives the values of m, w, and c in increasing order?

* c, m, w
* c, w, m
* m, w, c
* w, c, m
* w, m, c

The correct aswer choice is c, m, w

I would like to know if there is a short way to solve this. I took the long route, subsituting m and w on m + w + c = 1 and solving for c. This takes a long time and at the end I had to guess which fraction was bigger or smaller.

Thanks

tito


tito, how long is "a long time"?
ideally you should be able to perform the calculations in this problem within 1 minute or less, especially since the common denominator winds up being a very nice number (25 x 32 = 800).

if you couldn't do the calculations within about a minute, then you may actually want to pick up a 6th/7th grade pre-algebra textbook (with lots of arithmetic calculations) from the local library, and practice the homework problems.
this is a serious suggestion (it's not meant to be insulting, in case it might be perceived that way), which i've given to many tutoring students. many students today, especially younger ones, have depended on calculators and Excel their whole lives and are not able to perform arithmetic calculations quickly.

anyway, below are 2 solutions.

--

method 1 * do the arithmetic
make a common denominator.
m = 9/25
multiply by 32/32 --> m = 288/800

w = 15/32
multiply by 25/25 --> w = 375/800

subtract from 800/800 --> c = 137/800

there you go.

--

method 2 * fairly precise estimates

m = 9/25 you can calculate exactly, without much trouble. using long division (or memorization of the fact that 1/25 = 0.04, then multiplication by 9), you get
m = 0.36

w = 15/32
this is a little less than 1/2 (since 1/2 = 16/32).
so, just estimate it -- say it's about 0.45

if m = 0.36 and w is about 0.45, then that's a total of more than 0.8 already, so c is really small (less than 0.2) -- clearly smaller than the other two.

--

method 3 * crude estimates
w is close to 1/2.
m is nowhere close to 1/2, or at least not as close to 1/2 as w is.
therefore, m is smaller than w.
that's a humble observation, but it already eliminates all choices except (a) and (c).

now observe that, upon subtracting these numbers from 1, you are definitely going to get a number that is substantially less than 1/2. therefore, c is less than w.
that leaves choice (a), and you're done.

--

in any case -- you should be able to do the fraction arithmetic in a minute or less. if you can't, you should grab a pre-algebra book and practice!


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