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02 May 2022, 01:30
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B
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Difficulty:
15%
(low)
Question Stats:
77% (00:56) correct
23%
(01:16)
wrong
based on 100
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If r > 0, is \(\sqrt{r}\) an integer?
(1) r^2 is an integer.
(2) r = m^2, where m is an integer.
(1) r^2 is an integer.
(2) r = m^2, where m is an integer.
02 May 2022, 08:22
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Step 1: Analyse Question Stem
We have been given that r > 0.
We have to find if \(\sqrt{r}\) = integer
Squaring both sides, we have,
Is r = \((integer)^2\)?
The square of an integer is always a perfect square. Therefore, we are trying to find if ‘r’ is a perfect square.
Step 2: Analyse Statements Independently (And eliminate options) – AD / BCE
Statement 1: \(r^2\) is an integer.
This is a trap.
\(r^2\) = integer can be satisfied by r = 4 or r = 2.
In one case, r is a perfect square and in the other, it is not.
The data in statement 1 is insufficient to answer the question with a definite YES or NO.
Statement 1 alone is insufficient. Answer options A and D can be eliminated.
Statement 2: r = \(m^2\), where m is an integer.
Since m is an integer, \(m^2\) is the square of an integer; in other words, \(m^2\) is a perfect square.
Since r = \(m^2\), r is a perfect square. Since r is a perfect square, \(\sqrt{r}\) is an integer.
The data in statement 2 is sufficient to answer the question with a definite YES.
Statement 2 alone is sufficient. Answer option C and E can be eliminated.
The correct answer option is B.
We have been given that r > 0.
We have to find if \(\sqrt{r}\) = integer
Squaring both sides, we have,
Is r = \((integer)^2\)?
The square of an integer is always a perfect square. Therefore, we are trying to find if ‘r’ is a perfect square.
Step 2: Analyse Statements Independently (And eliminate options) – AD / BCE
Statement 1: \(r^2\) is an integer.
This is a trap.
\(r^2\) = integer can be satisfied by r = 4 or r = 2.
In one case, r is a perfect square and in the other, it is not.
The data in statement 1 is insufficient to answer the question with a definite YES or NO.
Statement 1 alone is insufficient. Answer options A and D can be eliminated.
Statement 2: r = \(m^2\), where m is an integer.
Since m is an integer, \(m^2\) is the square of an integer; in other words, \(m^2\) is a perfect square.
Since r = \(m^2\), r is a perfect square. Since r is a perfect square, \(\sqrt{r}\) is an integer.
The data in statement 2 is sufficient to answer the question with a definite YES.
Statement 2 alone is sufficient. Answer option C and E can be eliminated.
The correct answer option is B.
08 May 2022, 05:03
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Statement 1 : r^2 is an integer => r is an integer. But sq root(r) may or may not be an integer.
For Ex: r= 1(or any perfect square) => r^2 = 1 => sq root(r) = 1/-1. Yes !
but if r =2 => r^2 = 4 => sq root (r) => 1.414.. ( not an integer) . No!
Statement 2 : r = m^2 , where m is an integer.
therefore r is already a square of an integer. Therefore sq root of r is & will be an integer. Yes !
Ex: m =1 => r=1 => sq root(1) = 1/-1 => Yes !
m = 2 => r =4 => sq root(4) = 2/-2 => Integer. Yes!
Answer : B
For Ex: r= 1(or any perfect square) => r^2 = 1 => sq root(r) = 1/-1. Yes !
but if r =2 => r^2 = 4 => sq root (r) => 1.414.. ( not an integer) . No!
Statement 2 : r = m^2 , where m is an integer.
therefore r is already a square of an integer. Therefore sq root of r is & will be an integer. Yes !
Ex: m =1 => r=1 => sq root(1) = 1/-1 => Yes !
m = 2 => r =4 => sq root(4) = 2/-2 => Integer. Yes!
Answer : B
