What is the ratio of 5m to 7n?

Question

GMATBusters’ Quant Quiz Question -8 
What is the ratio of 5m to 7n?
1) The ratio of m^2 to n^2 = 81/36
2) The ratio of m^5 to n^3 >1

GMATBusters · Answer

Official Solution

We need 5m/7n or m/n

St 1) m^2/n^2 = 81/36
hence m/n = +/- 9/6
NOT Sufficient

St 2) ratio of m^5 to n^3 >1
Hence m/n >1 or m/n = +ve
NOT Sufficient

Combining St 1 & 2, we get 
m/n = 9/6
Sufficient

Answer C

shameekv1989 · Answer

What is the ratio of 5m to 7n?
1) The ratio of m^2 to n^2 = 81/36
2) The ratio of m^5 to n^3 >1

i)  \frac{m^2}{n^2} = \frac{81}{36} => \frac{|m|}{|n|} = \frac{9}{6} = \frac{3}{2}

Since we don't know the signs of m and n we can't determine the ratio i.e. \frac{m}{n} can be +-\frac{3}{2}

ii) The ratio of  \frac{m^5}{n^3} >1  => the ratio of m to n is positive => m and n have same signs

Insufficient

Combining we know m and n have same signs

thus  \frac{m}{n}  can only be  \frac{3}{2}

Sufficient

Answer - C

Kinshook · Answer

Asked: What is the ratio of 5m to 7n?

1) The ratio of m^2 to n^2 = 81/36
 \frac{m^2}{n^2} = \frac{81}{36} 
 \frac{|m|}{|n| }= \frac{9}{6} =\frac{3}{2} 
Since m & n can be of the same sign or of the opposite sign
NOT SUFFICIENT

2) The ratio of m^5 to n^3 >1 
 \frac{m^5}{n^3} > 1 >0 
m & n are of the same sign
Since magnitude of m & n are unknown
NOT SUFFICIENT

(1) + (2) 
1) The ratio of m^2 to n^2 = 81/36
 \frac{m^2}{n^2} = \frac{81}{36} 
 \frac{|m|}{|n| }= \frac{9}{6} =\frac{3}{2} 
2) The ratio of m^5 to n^3 >1 
 \frac{m^5}{n^3} > 1 >0 
m & n are of the same sign
Since m & n are of the same sign
 \frac{|m|}{|n|} = \frac{m}{n} = \frac{3}{2} 
 \frac{5m}{7n} = \frac{3}{2} * \frac{5}{7} = \frac{15}{14} 
SUFFICIENT

IMO C

Raxit85 · Answer

What is the ratio of 5m to 7n?
5m/7n = ?

1) The ratio of m^2 to n^2 = 81/36
(m/n)^2 = (9/6)^2 = (3/2)^2.
So, m = 3/-3 and n=2/-2. 
5m/7n could vary as per respective m and n values. Insufficient.

2) The ratio of m^5 to n^3 >1
m^5/n^3 >1
(m/n)^3 * m^2>1
So, m can be positive/negative and n can be positive only.
Again, 5m/7n could vary as per respective m and n values. Insufficient.

1) + 2) 
Based on both statements conditions, m can be positive/negative and n can be positive only.
Again, 5m/7n could vary as per respective m and n values. Insufficient.

Ans. E

Danchar42 · Answer

What is the ratio of 5m to 7n?
1) The ratio of m^2 to n^2 = 81/36
2) The ratio of m^5 to n^3 >1

Sqrt of statement 1 is tempting, but wouldn't tell us anything about the answer because doesn't capture negative values of m and n, which may have different ratios than 3:2 --> insufficient --> eliminate (a) and (d)

Statement 2 tells us that m is probably not a fraction or negative

Combined, we still don't know enough... for example:
m = 9, n = 6 satisfies 1 and 2
m = 9, n = -6 satisfied 1 and 2

however, they have different answers to the actual ratio being asked about --> insufficient together -->  answer is (e)

mohagar · Answer

we need sign and value to answer the question we get from both the statement so C

neharaj0001 · Answer

(C) Both together are sufficient
m^2/n^2=81/36=9/4
=>m/n=+/-sqrt(9/4)=+/-1.5
Hence 1 Alone isn't sufficient
Statement 2 implies that ratio of odd power of m to odd power of n is larger than 1 implying that both together are either positive or negative. Hence m/n will always he positive.
Therefore together combining statements we can say that m/n=1.5

Posted from my mobile device

AnirudhaS · Answer

(1)  m^2 : n^2= 81 : 36 
or, m : n = 9 : 6
Therefore,  m = \frac{9n}{6} 
So,  \frac{5m}{7n} = \frac{15}{14}
 
(2)  m^5 : n^3 >1 
Insufficient

Answer: A

What is the ratio of 5m to 7n?

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