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Source: — Data Sufficiency |
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by 4meonly » Thu Nov 20, 2008 12:18 pm
If mn is not equal to zero, what is the ratio of m to n^2?

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


2. The ratio of m^2 to n is 7/5
m^2/n =7/5
7n = 5m^2
7 and 5 are primes, so n should be a multiply of 5 and m^2 - multiply of 7
m=sqrt7 if m>0
n=5, and ratio m/n^2 = sqrt7 / 25

but m can be <0

even combined there is no info about the sign of m

E

I think B
Last edited by 4meonly on Thu Nov 20, 2008 12:29 pm, edited 2 times in total.
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Re: ratio

by logitech » Thu Nov 20, 2008 12:20 pm
muzali wrote:If mn is not equal to zero, 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

Please explain your approach.
Interesting question:

1) insuf nothing about n
2) m^2 to n is 7/5

so m^2 = 7X

n=5x

we are asked

m to n^2? ; so we need M but since we don't know whether M is + or - we will have two different answers so

Insuf

1+2) still does not give us ( me ) anything:

E

What is OA ?
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by cramya » Thu Nov 20, 2008 12:30 pm
Let me the add to the confusion. I think its C)
Muzali, whats the OA?
Last edited by cramya on Thu Nov 20, 2008 2:10 pm, edited 1 time in total.
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by 4meonly » Thu Nov 20, 2008 12:34 pm
cramya,
what do u think about the sigh of m?

Actually, at 1st attempt i got B)))
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by logitech » Thu Nov 20, 2008 12:53 pm
cramya wrote:Let me the add to the confusion. I think its C) and ratio of m / n^2 being 7/5

Muzali, whats the OA?
Cramya:

M=+(7/5)^1/2 & N=1

M=-(7/5)^1/2 & N=1

What do you say ?
Last edited by logitech on Thu Nov 20, 2008 2:01 pm, edited 1 time in total.
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by rmpaes » Thu Nov 20, 2008 1:11 pm
If mn is not equal to zero, what is the ratio of m to n²?

1. The ratio of m² to 1 is 7/5
2. The ratio of m² to n is 7/5

Please explain your approach.
The easiest way to solve this problem is to pick numbers.

Statement 1:
Not sufficient as it gives no information about n.

Statement 2:
m²/n = 7/5
n = 5m²/7
Can we tell what
m:n² is?
let's see
if m = 1
n = 5/7
n² = 25/49
m:n² = 1/(25/49) = 49/25
if m = 2
n= 20/7
n² = 400/49
m:n² = 2/(400/49) = 49/200
We get different ratios so B is not sufficient

Now we combine both results.
from eq 1:
We do know that
m² can equal: 7x/5x, i.e 7/5, 14/10, 21/15 . . .
let m² = 7/5
We plug this into statement 2:
n = 5m²/7
n = 5(7/5)
n = 1
m:n² = root(7/5):1
let m² = 14/10
We plug this into statement 2:
n = 5m²/7
n = 5(14/10)
n = 1

m:n² = root(7/5):1
It is now easy to see that both statements will be sufficient.
When we combine both statements its easy to see that n = 1.
My answer is C.
Last edited by rmpaes on Thu Nov 20, 2008 2:01 pm, edited 1 time in total.
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by muzali » Thu Nov 20, 2008 1:25 pm
OA is C
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by logitech » Thu Nov 20, 2008 2:03 pm
rmpaes wrote:
m:n² = root(7/5):1
It is now easy to see that both statements will be sufficient.
When we combine both statements its easy to see that n = 1.
My answer is C.
M=+(7/5)^1/2 & N=1

M=-(7/5)^1/2 & N=1

What do you say ?

Why do you think that root (7/5) is positive ?
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by cramya » Thu Nov 20, 2008 2:10 pm
Stmt I

m^2/1 = 7/5 i.e m^2=7/5 i.e m = sqrt(7/5) (1)

No info about n

INSUFF

Stmt II

The ratio of m² to n is 7/5

m^2/n = 7/5

n=5 * m^2/7 (2)

Cant find m/n^2 ratio still as m terms will still be hanging around even when we substitute n =5 * m^2/7 in the expression m/n^2

INSUFF

Stmt I and II together

m/n^2

= m / (5/7*m^2) ^2 [From (2)] substituting n=5 * m^2/7]



= m / (5/7*7/5 ) ^2

= m/1 ^2

= sqrt(7/5):1

C)

I agree with rmpaes the ratio is sqrt(7/5) : 1


Muzali,
Whats the soucre of this nice question?
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by logitech » Thu Nov 20, 2008 2:20 pm
I think the sign of the ratio confuses me

Can we have negative ratios ? Because if we can the answer is E

so A:B is 3:1 that's cool but is there anything like A:B -3:1 ?
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by cramya » Thu Nov 20, 2008 2:22 pm
I thought about it too but since we are taking square root we would always consider the prinicpal square root of a number (i.e. non negative real number for the square root).
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by logitech » Thu Nov 20, 2008 2:24 pm
cramya wrote:I thought about it too but since we are taking square root we would always consider the prinicpal square root of a number (i.e. non negative real number for the square root).
come on, when you see X^2 in a DS question, do you only consider the positive root ?

Is X/Y < 0

1) Y=2
2) X^2 =1

What is your answer Cramya ?
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by cramya » Thu Nov 20, 2008 2:28 pm
Are u saying why not consider the case m = -sqrt(7/5) in the ratio?
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by logitech » Thu Nov 20, 2008 2:29 pm
cramya wrote:Are u saying why not consider the case m = -sqrt(7/5) in the ratio?
YES YESS YESSS!!!
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