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28 Mar 2020, 19:47
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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
28 Mar 2020, 19:48
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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
28 Mar 2020, 20:04
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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
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
