If a, b and c are integers, is abc an even integer?
(1) b is halfway between a and c
(2) a=b−c
I chose E. But the answer is B. The explanation given is
a=b−c . Re-arrange: a+c=b. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.
Is xy≤1/2?
(1) x2+y2=1
(2) x2−y2=0
here also I chose E but the answer is A
Explanation given is (1) x2+y2=1. Recall that (x−y)2≥0(x−y)2≥0 (square of any number is more than or equal to zero). Expand: x2−2xy+y2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x2−y2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.
Please help
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Number Properties- Hi I need help on these two DS questions
This topic has expert replies
-
riteshpatnaik
- Junior | Next Rank: 30 Posts
- Posts: 29
- Joined: Wed Mar 09, 2016 12:55 am
-
fiza gupta
- Master | Next Rank: 500 Posts
- Posts: 216
- Joined: Sun Jul 31, 2016 9:55 pm
- Location: Punjab
- Thanked: 31 times
- Followed by:7 members
1) b is halfway between a and c
abc can be either odd or even
example a=7,c=9, so b=8,abc = even
a=7,c=11,so b=9, abc=odd
INSUFFICIENT
2) a=b-c
a+c=b
now a and c can be either odd or even
we can have 4 scenarios:
(i)let a and c are odd
odd + odd = even, s b= even
abc = even
(ii) let a be odd and c be even
odd + even = odd or vice versa even +odd = odd
abc = even
(iii) a and c are even
even + even = even
abc = even
in all cases product will be even
SUFFICIENT
abc can be either odd or even
example a=7,c=9, so b=8,abc = even
a=7,c=11,so b=9, abc=odd
INSUFFICIENT
2) a=b-c
a+c=b
now a and c can be either odd or even
we can have 4 scenarios:
(i)let a and c are odd
odd + odd = even, s b= even
abc = even
(ii) let a be odd and c be even
odd + even = odd or vice versa even +odd = odd
abc = even
(iii) a and c are even
even + even = even
abc = even
in all cases product will be even
SUFFICIENT
Fiza Gupta
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi riteshpatnaik,
In the future, you post just one question per thread. In that way, you can keep each conversation focused on just one question and avoid the confusion of criss-crossing conversations.
GMAT assassins aren't born, they're made,
Rich
In the future, you post just one question per thread. In that way, you can keep each conversation focused on just one question and avoid the confusion of criss-crossing conversations.
GMAT assassins aren't born, they're made,
Rich
GMAT/MBA Expert
-
Jay@ManhattanReview
- GMAT Instructor
- Posts: 3008
- Joined: Mon Aug 22, 2016 6:19 am
- Location: Grand Central / New York
- Thanked: 470 times
- Followed by:34 members
So, we have a situation that a + c = b necessarily implies that a*b*c is EVEN.riteshpatnaik wrote:If a, b and c are integers, is abc an even integer?
(1) b is halfway between a and c
(2) a=b−c
I chose E. But the answer is B. The explanation given is
a=b−c . Re-arrange: a+c=b. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.
Please help
Let's test this.
- 1. a and c are both ODD => ODD + ODD = EVEN => c is EVEN => ODD*ODD*EVEN = EVEN
2. a and c are both EVEN => EVEN + EVEN = EVEN => c is EVEN => EVEN*EVEN*EVEN = EVEN
3. One of a and c is ODD and the other is EVEN => ODD + EVEN = ODD => c is ODD => ODD*EVEN*ODD = EVEN
Hope this helps!
-Jay
_________________
Manhattan Review GMAT Prep
Locations: New York | Jakarta | Nanjing | Berlin | and many more...
Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
GMAT/MBA Expert
-
Jay@ManhattanReview
- GMAT Instructor
- Posts: 3008
- Joined: Mon Aug 22, 2016 6:19 am
- Location: Grand Central / New York
- Thanked: 470 times
- Followed by:34 members
We have to check if xy ≤ 1.riteshpatnaik wrote:
Is xy≤1/2?
(1) x2+y2=1
(2) x2−y2=0
here also I chose E but the answer is A
Explanation given is (1) x2+y2=1. Recall that (x−y)2≥0(x−y)2≥0 (square of any number is more than or equal to zero). Expand: x2−2xy+y2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x2−y2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.
Please help
S1: Given that x^2 + y^2 = 1
Adding 2xy on both the sides to make it a perfect square.
x^2 + y^2 + 2xy = 1 + 2xy
=> (x+y)^2 = 1 + 2xy
Since (x+y)^2 is a non-negative term, 0 ≤ 1 + 2xy => -1 ≤ 2xy => 2xy ≤ 1 [pay attention to the change of sign] => xy ≤ 1/2. Suffiicient!
S2: x^2 - y^2 = 0 => x^2 = y^2 => x = y or x = -y
If x = y, xy ≤ 1 becomes x^2 ≤ 1, but we do not know the value of x.
Similarly, if x = -y, xy ≤ 1 becomes -x^2 ≤ 1, again, we do not know the value of x. Insufficient.
Hope this helps!
-Jay
_________________
Manhattan Review GMAT Prep
Locations: New York | Jakarta | Nanjing | Berlin | and many more...
Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
