Test: Even And Odds-1 - GMAT MCQ

(1) Z = 2 * even
=> z is even
Sufficient

(2) 3z = even
3 is odd
So z is even if z is an integer
But if z = 8/3
Not Sufficient

Answer - A

Integers between -10.5 and 10.5 are 

-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Odd integers are -9, -7, -5, -3, -1, 1, 3, 5, 7, 9 (Total 10)

Even integers are -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10 (Total 11)

Hence, ratio of number of odd integers to the number of even integers between -10.5 and 10.5 is 10/11

(E) divisible by 4 whenever n is even

First, let's analyze the given expression n(n - 1)(n - 2).

Notice that the three factors are consecutive integers. This means that at least one of them must be even (divisible by 2), and at least one of them must be divisible by 3.

(A) is incorrect because if n is odd, then (n - 1) is even, so the product is even.
(B) is incorrect because if n is even, then n is even, so the product is even.
(C) is incorrect because if n is odd, then (n - 1) is even, so the product is even.
(D) is incorrect because if n is even, then either n or (n - 2) is divisible by 3, so the product is divisible by 3.

(E) is correct because if n is even, then n is divisible by 2. Furthermore, since the three factors are consecutive integers, one of the other two factors (n - 1) or (n - 2) must also be divisible by 2. Therefore, the product must be divisible by 4.

Correct Answer :- d

Explanation : If the product of the three digits is 42 then the possible set of three digits is 2,3,7 or 6,7,1

Numbers from 2,3,7 are 237,273,327,372,723,732

Numbers from 6,7,1 are 167,176,617,671,716,761

Numbers less than 300 are 237,273,167,176 so x can be both even or odd- Insufficient.

Tens digit is 7

Numbers can be 273,372,671,176.. x can be both even and odd -Insufficient

Taking both the statements together

Number <300 and tens digit is 7 nos are 273,176 both even and odd possible so Insufficient

So both are required

q must be odd, e.g. follwing patterns;

2+1+2 = 5
1+3+1 = 5
1+2+1 = 4
2+4+2 = 8

Hence all combinations with an even q would yield even results, thus q must be odd.

Given: = even > 0 ab² = c* even = even--> either a is even or b or both.

I. ab is even --> according to the above this must be true;

II. ab > 0 --> not necessarily true, bb could be positive as well as negative (for example a=1, c = 1 and b =−2);

III. c is even --> not necessarily true, see above example.

Positive integers x and y are NOT both odd, means that either both x and y are even or one is even and the other one is odd. In either case xy must be even.

Now as the statement of the question suggests i+ j will be even (when)
statement 1 : i < 10
but this doesn't fit as i can be 2 and j can be 3 and i + j = 5 which is not even.
 hence, fails the criteria
Statement 2 : i = j,
whenever i = j, the resultant will ALWAYS be even.
Hence, Option A is correct where Exactly on of the statements can answer the question

Correct Answer :- C

Explanation : In order the product of two integers to be even either (or both) of them must be even. So, the question basically asks whether either x or y is even.

(1) x = y + 1. If x is odd then y is even and vise-versa. Sufficient.

(2) x is even 

Let x = 2, y= 3

x*y = 2*3 = 6 (Even), Sufficient

If either x or y is zero, then xy=0=even, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

 

2X+Y===Even +odd= odd...Always False
(As x <y then x can be smallest prime 2 leaving only odd prime choices for y)

1) 4y will always be even. Then we have 5x−even=even ,

For this to be the case, 5x must be even. Since 5 can't be even, then x must be even. Thus the product xy will be even. Sufficient.

2) 6x will always be even. Then we have even+7y=even.

Thus 7y is even, and y is even, and xy is even. Sufficient.

Notice that [ ] is just some function such that "[m]=3m when m is odd and [m]=(1/2)*m when m is even".

So, [m]=3m when m is odd, [m]=(1/2)*m when m is even.
As 9 is odd then [9] equals to 3*9=27;
As 6 is even then [6] equals to 1/2*6=3;

So [9]*[6]=27*3=81. Note that numbers in the answer choices are also in boxes, so we have: [m]=81. m could be 27 (in this case as 27 is odd [27]=3*27=81) OR 162 (in this case as 162 is even [162]=162/2=81) --> only [27] is in the answer choices.

(1) n + m is odd

The sum of two integers is odd only if one is odd and another is even, hence m may or may not be odd. Not sufficient.

(2) n + m = n² + 5

-->  m−5=n²−n

=> m−5=n(n−1)

either n or n−1 is even hence n(n−1)=even

=> n(n−1)=even

=> m−5=m−odd=even

 --> m=odd. Sufficient.

(1) The product xz is odd.
=> x, z are odd => x^z  odd => SUF

(2) x = 2^y
=> x is even => x^z even => SUF