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10 Jun 2020, 02:58
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
81% (01:18) correct
19%
(01:37)
wrong
based on 268
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(\sqrt{x + y}\) an integer?
(1) \(x^3 = 64\)
(2) \(x^2 = y – 3\)
(1) \(x^3 = 64\)
(2) \(x^2 = y – 3\)
10 Jun 2020, 03:12
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Bunuel
Question: is x+y an integer?
Statement 1: x^3 = 64
No information about y hence
NOT SUFFICIENT
STatemrnt 2: x^2 = y – 3
for y = 7, x = 2 and √(x+y) is an Integer
for y = 5, x = √2 and √(x+y) is NOT an Integer
NOT SUFFICIENT
COmbining the statements
x^2 = y – 3 and x^3 = 64
i.e. x = 4 and y = 19 hence \(\sqrt{x+y} = √23\) NOT an integer
SUFFICIENT
ANswer: Option C
10 Jun 2020, 03:21
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St. 1:
x^3 = 64
=> x = 3
But, no information about y.
=> St. 1 is insufficient.
St.2:
x^2+3 = y
If x = 1, then y = 4 => root(5) is not an integer.
But, if, x = 2, then y = 7 => root (9) = 3 which is an integer.
=> St. 2 is insufficient.
Taking both St.1 and St.2
x = 4
=> y = 19
=> root(x+y) = root(23) = not an integer.
=> Option C is the answer
Posted from my mobile device
x^3 = 64
=> x = 3
But, no information about y.
=> St. 1 is insufficient.
St.2:
x^2+3 = y
If x = 1, then y = 4 => root(5) is not an integer.
But, if, x = 2, then y = 7 => root (9) = 3 which is an integer.
=> St. 2 is insufficient.
Taking both St.1 and St.2
x = 4
=> y = 19
=> root(x+y) = root(23) = not an integer.
=> Option C is the answer
Posted from my mobile device
