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anuramm
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DS - mn 29 Aug 2004, 04:54
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DS - mn 30 Aug 2004, 00:22
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1) m^2 : 1 = 7:5 does not tells us anything about n. So (1) is not sufficient.
2) m^2:n = 7:5, so we can write, m^2/n = 7/5, 5m^2 = 7n, n = 5m^2/7

Substituting to m/n^2 = (m)/(5m^2/7) does not solve our problem, as we will still get a ratio that is in the form of m

Using (1) and (2), we know n = 1, so we can find m and solve m/n^2.
(C) is the answer
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DS - mn 30 Aug 2004, 05:10
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I think the answer is B

M^2/N = 7/5 ==> sqrt 7 * sgrt 7 / 5

M/N^2 = sgrt 7/5^2


Correct me if I am wrong.

Regards,

Alex
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DS - mn 30 Aug 2004, 11:31
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My answer choice is E.

A is insufficient asit just says m^2=7/5
B is also insufficient as m^2/n=7/5
This rules out D option as well

From A m=+ or - sqrt 7/5
Using A and B n=1

But we need to find m/n^2. We don't know if it is +7/5 or -7/5.
Hence E
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DS - mn 31 Aug 2004, 01:14
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The answer given is C and even i got E ( explanation similar to dushver's).

Alex_NL, we cant say that if m^2/n = 7/5, then m/n^2 = sqrt7/25.
We can prove this by plugging some values.
m=6,n=8.
m^2/n=36/8=9/2
m/n^2=6/64 (actual value)
m/n^2 (per your method) = 3/4.
these two are not equal.
Thanks.
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DS - mn 31 Aug 2004, 01:17
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Anuramm,

You are right.
I realized that last night, but did not have the energy to correct myself.
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DS - mn 31 Aug 2004, 01:22
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It can't be E because if you use both equation, you can tell that n is actually 1. See my working above.
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DS - mn 31 Aug 2004, 01:32
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If mn!=0, what is the ratio of m to n^2?

1. The ratio of m^2 to 1 is 7/5
2. The ratio of m^2 to n is 7/5

From 1, nothing is said about n, so it is insufficient.

From 2, it can be written as m^2/n = 7/5, 5m^2 = 7n, n = (5m^2)/7
So using this, m/n^2 = 7m/25m^4 = 7/25m^3. Still we can’t figure out what the ratio is.

Using 1 and 2, we can see that n is actually 1, so with this in hand, we can get 5m^2=7n,
5m^2=7, m = sqrt(7/5). So m:n^2 = sqrt(7/5): 1. Sufficient.



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