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12 Mar 2018, 04:51
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
59% (01:47) correct
41%
(02:34)
wrong
based on 63
sessions
History
Date
Time
Result
Not Attempted Yet
If X and Y are non-zero integers, Is X/Y >= 5/7 ?
(1) X^3 >= 125 and Y^2 <= 49
(2) X*Y <= 0
(1) X^3 >= 125 and Y^2 <= 49
(2) X*Y <= 0
Mo2men
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12 Mar 2018, 05:08
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amanvermagmat
(2) \(X*Y\leq{0}\)
This means that x & y have DIFFERENT sings hence x/y is laways negative and can't be >= 5/7.
Sufficient
(1) \(x^3\geq{125}\) and \(Y^2\leq{49}\)
\(x\geq{5}\) , -7<= Y <= 7
Let x = 5 & Y =1........5/1 =5...............Answer is Yes
Let x = 5 & Y =-1........5/1 =5...............Answer is NO
Insufficient
Answer: B
19 Mar 2018, 12:32
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SOLUTION
We are given:
- • \(X\) and \(Y\) are non-zero integers.
We need to find:
- • If \(\frac{X}{Y}\) is greater than \(\frac{5}{7}\) or not?
Let us analyze both the statements one by one.
Statement-1: \(X^3 >= 125\) and \(Y^2 <= 49\).
From \(X^3>=125\), we can write:
- • \(X >=5\).
- • \(-7<= Y<= 7\)
For \(X=6\) and \(Y=7\), the value of \(\frac{X}{Y}\) is greater than \(\frac{5}{7}\).
However, for \(X=6\) and \(Y=-7\), the value of \(\frac{X}{Y}\) is less than \(\frac{5}{7}\).
Hence, Statement 1 alone is not sufficient to answer the question.
Statement-2 \(X*Y <= 0\).
Since \(X\) and \(Y\), both, are non-zero integers, \(X*Y\) can only be less than 0.
The product of \(X\) and \(Y\)is negative.
- • Hence, we can say that \(X\) and \(Y\) are of different sign.
- o Therefore, \(\frac{X}{Y}\) is also negative.
Hence, we can certainly say that \(\frac{X}{Y}\) is a negative value which is always less than \(\frac{5}{7}\).
Thus, Statement 2 alone is sufficient to answer the question.
Answer: B
