What is the ratio of 5m to 7n?

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

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

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)

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

(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

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(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