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Test: Even And Odds- 1 - GMAT MCQ


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25 Questions MCQ Test Quantitative for GMAT - Test: Even And Odds- 1

Test: Even And Odds- 1 for GMAT 2025 is part of Quantitative for GMAT preparation. The Test: Even And Odds- 1 questions and answers have been prepared according to the GMAT exam syllabus.The Test: Even And Odds- 1 MCQs are made for GMAT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Even And Odds- 1 below.
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Test: Even And Odds- 1 - Question 1
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Is z even?

1) z/2 is even
2) 3z is even

Detailed Solution for Test: Even And Odds- 1 - Question 1

(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

Test: Even And Odds- 1 - Question 2
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If x is an integer, is x/2 an even integer?  

1) x is divisible by 2

2) x is divisible by 4

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Test: Even And Odds- 1 - Question 3
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What is the ratio of number of odd integers to the number of even integers between -10.5 and 10.5?

Detailed Solution for Test: Even And Odds- 1 - Question 3

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

 

Test: Even And Odds- 1 - Question 4
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If m is an integer, is m odd?

1) m/2 is not an even integer

2) m – 3 is not an even integer
Detailed Solution for Test: Even And Odds- 1 - Question 4

  • Statement 1: “m/2 is not an even integer.”

    • If m is odd, m/2 isn’t an integer (so certainly not an even integer).

    • If m ≡ 2 (mod 4), then m/2 is an odd integer (so again not even).

    • Only if m ≡ 0 (mod 4) would m/2 be even.

    • ⇒ Statement 1 holds for both odd m and for m ≡ 2 (mod 4). You cannot tell if m is odd. Not sufficient.

  • Statement 2: “m – 3 is not an even integer.”

    • If m is odd, m – 3 is even (so that would contradict the statement).

    • If m is even, m – 3 is odd (so not an even integer).

    • ⇒ The statement can only be true if m is even, so you conclude m is not odd. Sufficient.

Answer: A: Exactly one of the statements can answer the question.

Test: Even And Odds- 1 - Question 5
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 If n is a positive integer, then n(n + 1)(n + 2) is
Detailed Solution for Test: Even And Odds- 1 - Question 5

Answer: D: divisible by 4 whenever n is even

Reasoning without LaTeX or KaTeX:

  • Suppose n is even. Then we can write n = 2 k for some integer k.

  • The three consecutive numbers are:
    • n = 2 k
    • n+1 = 2 k + 1 (odd)
    • n+2 = 2 k + 2 = 2 (k+1)

  • In the product n × (n+1) × (n+2), two factors are even: 2 k and 2 (k+1).

  • Multiplying those two gives 2 k × 2 (k+1) = 4 [k × (k+1)], which is clearly a multiple of 4.

  • Therefore, whenever n is even, the entire product is divisible by 4.

Test: Even And Odds- 1 - Question 6
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The product of integers x, y, and z is even. Is z even?

 1) x/y = z

2) z = xy
Detailed Solution for Test: Even And Odds- 1 - Question 6

  • Statement 1 (“x / y = z”)
    ⇒ x = y·z, so xyz = (y·z)·y·z = y²·z².
    If y is even but z is odd, y²·z² is still even.
    We cannot force z to be even. ⇒ Not sufficient.

  • Statement 2 (“z = x·y”) ⇒ xyz = x·y·(x·y) = x²·y².
    For this to be even, either x or y must be even, which makes z = x·y even.
    ⇒ Sufficient.

Answer: A: Exactly one of the statements can answer the question.

Test: Even And Odds- 1 - Question 7
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The product of the units, tens, and hundreds digits of the positive 3-digit integer x is 42. Is x even?

(1)  x is less than 300. ?

(2)  The tens digit of x is 7. ?
Detailed Solution for Test: Even And Odds- 1 - Question 7

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

Test: Even And Odds- 1 - Question 8
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If m, n, and p are integers, is m+n odd??
(1) m = p2 + 4p + 4
(2) n = p2 + 2m + 1
Detailed Solution for Test: Even And Odds- 1 - Question 8

So both are required

Test: Even And Odds- 1 - Question 9
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Is the positive integer p even?

 (1) p2 + p is even.

(2) 4p + 2 is even. 
Detailed Solution for Test: Even And Odds- 1 - Question 9

  • Statement (1): p² + p = p·(p+1) is always even for any integer p (product of two consecutive numbers), so it gives no information about p’s parity.

  • Statement (2): 4p + 2 is always even for any integer p (4p is even, plus 2 remains even), so it also gives no information about p’s parity.

  • Even using both statements together, we still cannot determine whether p is even or odd.

Answer: D: More information is required as the information provided is insufficient to answer the question.

Test: Even And Odds- 1 - Question 10
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If p and q are integers and p + q + p is odd, which of the following must be odd? 
Detailed Solution for Test: Even And Odds- 1 - Question 10

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.

Test: Even And Odds- 1 - Question 11
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If a , b, and c are integers and ab2 / c is a positive even integer, which of the following must be true?

I. ab is even                

II. ab > 0          

III. c is even?
Detailed Solution for Test: Even And Odds- 1 - Question 11

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.

Test: Even And Odds- 1 - Question 12
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If k and y are integers, and 10k + y is odd, which of the following must be true? 
Detailed Solution for Test: Even And Odds- 1 - Question 12

Answer: C: y is odd.

Explanation:

  • 10 × k is always an even number (because 10 is even).

  • An even number plus y is odd only when y itself is odd.

  • Therefore, for 10k + y to be odd, y must be odd.

Test: Even And Odds- 1 - Question 13
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Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H? 
Detailed Solution for Test: Even And Odds- 1 - Question 13

G + G/2 = 3G/2 --> the sum is a multiple of 3.

G is a two-digit number --> G < 100 --> 3G/2 < 150.

Among the answer choices the only multiple of 3 which is less than 150 is 129.

Answer: D.

Test: Even And Odds- 1 - Question 14
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If positive integers x and y are not both odd, which of the following must be even?
Detailed Solution for Test: Even And Odds- 1 - Question 14

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.

Test: Even And Odds- 1 - Question 15
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If i and j are integers, is i + j even?

1) i < 10

2) i = j
Detailed Solution for Test: Even And Odds- 1 - Question 15

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

Test: Even And Odds- 1 - Question 16
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If n is an integer, is n even?

1) n2 – 1 is an odd integer

2) 3n + 4 is an even integer
Detailed Solution for Test: Even And Odds- 1 - Question 16

Each statement alone tells us that n must be even:

  • From (1): if n² – 1 is odd, then n² is even, so n is even.

  • From (2): if 3n + 4 is even, then 3n is even, so n is even.

Thus each statement by itself is sufficient.
Answer: C: Each statement can answer the question individually.

Test: Even And Odds- 1 - Question 17
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If x and y are both integers, is xy even?

1) x + y is odd

2) x is even
Detailed Solution for Test: Even And Odds- 1 - Question 17

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

Test: Even And Odds- 1 - Question 18
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If x and y are integers, is xy even?

1) x = y + 1

2) x/y is an even integer
Detailed Solution for Test: Even And Odds- 1 - Question 18

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).

 

Test: Even And Odds- 1 - Question 19
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Is p2 an odd integer?

1) p is an odd integer

2) √p is an odd integer
Detailed Solution for Test: Even And Odds- 1 - Question 19

Answer: C: Each statement can answer the question individually.

Explanation:

  • Statement 1 says p is odd. Any odd integer squared is odd, so p² is odd.

  • Statement 2 says √p is an odd integer, so p = (odd)², which is odd; hence p² is also odd.

Each statement alone guarantees that p² is odd.

Test: Even And Odds- 1 - Question 20
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If x and y are prime integers and x < y, which of the following cannot be true??
Detailed Solution for Test: Even And Odds- 1 - Question 20

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

Test: Even And Odds- 1 - Question 21
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If x and y are positive integers, is the product xy even?

1) 5x - 4y is even

2) 6x + 7y is even
Detailed Solution for Test: Even And Odds- 1 - Question 21

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.

Test: Even And Odds- 1 - Question 22
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If x and y are integers, is x (y + 1) an even number?

1) x and y are prime numbers.

2) y > 7 ?
Detailed Solution for Test: Even And Odds- 1 - Question 22

  • From Statement 1 alone: “x and y are prime.” x could be 2 or an odd prime; y could be 2 or an odd prime.
    We can find cases where x(y+1) is odd
    (e.g. x=3, y=2 ⇒ y+1=3 ⇒ 3·3=9 odd) and
    cases where it’s even (e.g. x=2, y=3
    ⇒ y+1=4
    ⇒ 2·4=8).
    So Statement 1 is not enough.

  • From Statement 2 alone: “y>7.” y could be odd or even,
    and we have no information about x, so x(y+1) could be odd or even.
    Statement 2 by itself is also not enough.

  • Taken together: x is prime and y>7 ⇒ y is an odd prime (since the only even prime is 2, which is not >7), so y+1 is even. Hence x(y+1) is even regardless of x. Combined, they determine that the product is even.

Answer: B: Both statements are required to answer the question.

Test: Even And Odds- 1 - Question 23
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For all positive integers m, (m) = 3m when m is odd and (m) = (1⁄2) m when m is even, which of the following is equivalent to (9)*(6)? 
Detailed Solution for Test: Even And Odds- 1 - Question 23

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.

Test: Even And Odds- 1 - Question 24
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If m and n are integers, is m odd??

1) m + n is odd

2) m + n = n2 + 5
Detailed Solution for Test: Even And Odds- 1 - Question 24

(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.

Test: Even And Odds- 1 - Question 25
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For non-negative integers x, y, and z, is xz odd?

1) xz = odd

2) x = 2y 
Detailed Solution for Test: Even And Odds- 1 - Question 25

Statement 1 says “x times z is odd.” That immediately tells us the answer is “yes” (an odd product means xz is odd).

Statement 2 says  x = 2y  That only tells us x is a power of 2 (so either 1 if y=0, or an even number if y≥1). Without knowing z, we can’t tell if xz is odd or even (for example, if x=2 and z=1, xz=2 is even; if x=1 and z=any number, xz=z could be odd or even).

Therefore exactly one statement suffices.

Answer: (a) Exactly one of the statements can answer the question.

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