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13 Jun 2017, 01:41
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
67% (01:23) correct
33%
(01:38)
wrong
based on 69
sessions
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sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
Posts: 608
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Location: India
Concentration: Technology, Strategy
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
WE:Information Technology (Consulting)
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 608
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MathRevolution
1) y=3x+5
When x is odd; ==> y = 3(odd) + 5 --------- (odd x odd = odd)
y = odd + 5 = odd + odd = even
Therefore y would be even. == > x + y = odd + even = odd.
When x is even; ==> y = 3 (even) + 5 ----------- (odd x even = even)
y = even + 5 = even + odd = odd
Therefore y would be odd. ==> x + y = even + odd = odd.
When x is 0. ==> y = 3(0) + 5 = 5. Y would be odd. x + y = 0 + odd = Odd. ---------- (1) is Sufficient.
2) y=2x-3
When x is odd; ==> y = 2(odd) - 3 --------- (even x odd = even)
y = even - 3 = even - odd = odd
Therefore y would be odd. ==> x + y = odd + odd = even. ---------- (2) is Insufficient.
When x is even; ==> y = 2 (even) - 3 ----------- (even x odd = even)
y = even - 3 = even - odd = odd
Therefore y would be odd. ==> x + y = even + odd = odd.
When x is 0; ==> y = 2(0) - 3 ==> y = -3. Y would be odd. x + y = 0 + odd = odd.
In (2) there are more than one possibilities. Hence (2) is Insufficient.
Only (1) alone is Sufficient. Answer A...
13 Jun 2017, 02:11
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Since we need to find if x+y is ODD?
1) y=3x+5
There are 3 possibilities for value of x(odd, 0 or even)
If x=odd eg(x=3) y = 14, x+y = ODD
If x=EVEN eg(x=-4) y = -7, x+y = ODD
Even if x=0,y=5 making the sum ODD. Hence, sufficient
2) y=2x-3
There are 3 possibilities for value of x(odd, 0 or even)
If x=odd eg(x=3) y = 3, x+y = EVEN
if x=0,y= -3 making the sum ODD.
Since we have one case in each, this statement alone is insufficient (Option A)
1) y=3x+5
There are 3 possibilities for value of x(odd, 0 or even)
If x=odd eg(x=3) y = 14, x+y = ODD
If x=EVEN eg(x=-4) y = -7, x+y = ODD
Even if x=0,y=5 making the sum ODD. Hence, sufficient
2) y=2x-3
There are 3 possibilities for value of x(odd, 0 or even)
If x=odd eg(x=3) y = 3, x+y = EVEN
if x=0,y= -3 making the sum ODD.
Since we have one case in each, this statement alone is insufficient (Option A)
