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# Test: Evens And Odds- 2

## 10 Questions MCQ Test Quantitative Aptitude for GMAT | Test: Evens And Odds- 2

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This mock test of Test: Evens And Odds- 2 for GMAT helps you for every GMAT entrance exam. This contains 10 Multiple Choice Questions for GMAT Test: Evens And Odds- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Evens And Odds- 2 quiz give you a good mix of easy questions and tough questions. GMAT students definitely take this Test: Evens And Odds- 2 exercise for a better result in the exam. You can find other Test: Evens And Odds- 2 extra questions, long questions & short questions for GMAT on EduRev as well by searching above.
QUESTION: 1

### If p and q are integers and p + q + p is odd, which of the following must be odd?

Solution:

First, let us simplify the original expression: p + q + p = 2p + q
Since the product of an even number and any other integer will always be even, the value of 2p must be even. If q were even, 2p + q would be the sum of two even integers and would thus have to be even. But the problem stem tells us that 2p + q is odd. Therefore, q cannot be even, and must be odd.
Alternatively, we can reach this same conclusion by testing numbers. We simply test even and odd values of p and q to see whether they meet our condition that p + q + p must be odd.
1) even + even + even = even (for example, 4 + 2 + 4 = 10). The combination (p even, q even) does not meet our condition.
2) odd + odd + odd = odd (for example, 5 + 3 + 5 = 13). The combination (p odd, q odd) does meet our condition.
3) even + odd + even = odd (for example, 4 + 3 + 4 = 11). The combination (p even, q odd) does meet our condition.
4) odd + even + odd = even (for example, 3 + 4 + 3 = 10). The combination (p odd, q even) does not meet our condition.
If we examine our results, we see that q has to be odd, while p can be either odd or even.  Our question asks us which answer must be odd; since q is an answer choice, we don't have to test the more complicated answer choices.

QUESTION: 2

### 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

Solution:

If ab2 were odd, the quotient would never be divisible by 2, regardless of what c is.  To prove this try to divide an odd number by any integer to come up with an even number; you can't.  If ab2  is even, either a is even or b is even.
(I) TRUE:  Since a or b is even, the product ab must be even
(II) NOT NECESSARILY:  For the quotient to be positive, a and c must have the same sign since b2 is definitely positive.  We know nothing about the sign of b.  The product of ab could be negative or positive.
(III) NOT NECESSARILY:  For the quotient to be even, ab2 must be even but c could be even or odd.  An even number divided by an odd number could be even (ex: 18/3), as could an even number divided by an even number (ex: 16/4).

QUESTION: 3

### If k and y are integers, and 10k + y is odd, which of the following must be true?

Solution:

(A) UNCERTAIN: k could be odd or even.
(B) UNCERTAIN: k could be odd or even.
(C) TRUE: If the sum of two integers is odd, one of them must be even and one of them must be odd.  Whether k is odd or even, 10k is going to be even; therefore, y must be odd.
(D) FALSE: If the sum of two integers is odd, one of them must be even and one of them must be odd.  Whether k is odd or even, 10k is going to be even; therefore, y must be odd.
(E) UNCERTAIN: k could be odd or even.

QUESTION: 4

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?

Solution:

Since the digits of G are halved to derive those of H, the digits of G must both be even. Therefore, there are only 16 possible values for G and H and we can quickly calculate the possible sums of G and H:  Alternately, we can approach this problem algebraically.  Let's call x and y the tens digit and the units digit of G.  Thus H can be expressed as 5x + .5y.  And the sum of G and H can be written as 15x + 1.5y.
Since we know that x and y must be even, we can substitute 2a for x and 2b for y and can rewrite the expression for the sum of G and H as: 15(2a) + 1.5(2b) = 30a + 3b. This means that the sum of G and H must be divisible by 3, so we can eliminate C and E.

Additionally, since we know that the maximum value of G is less than 100, then the maximum value of H must be less than 50. Therefore the maximum value of G + H must be less than 150. This eliminates answer choices A and B. This leaves answer choice D, 129. This can be written as 86 + 43.

QUESTION: 5

If a is an even integer and b is an odd integer, which of the following cannot be an even integer?

Solution:

Let's look at each answer choice:
(A) EVEN: Since a is even, the product ab will always be even.  Ex: 2 × 7 = 14.

(B) UNCERTAIN: An even number divided by an odd number might be even if the the prime factors that make up the odd number are also in the prime box of the even number.  Ex: 6/3 =2.

(C) NOT EVEN: An odd number is never divisible by an even number.  By definition, an odd number is not divisible by 2 and an even number is.  The quotient of an odd number divided by an even number will not be an integer, let alone an even integer. Ex: 15/4 = 3.75

(D) EVEN: An even number raised to any integer power will always be even.  Ex: 21 = 2

(E) EVEN: An even number raised to any integer power will always be even.  Ex: 23 = 8

QUESTION: 6

If x and y are prime integers and x < y, which of the following cannot be true?

Solution:

Let's look at each answer choice:
(A) UNCERTAIN:  x could be the prime number 2.
(B) UNCERTAIN:  x could be the prime number 2, which when added to another prime number (odd) would yield an odd result. Ex: 2 + 3 = 5
(C) UNCERTAIN:  Since x could be the prime number 2, the product xy could be even.
(D) UNCERTAIN:  y > x and they are both prime so y must be odd.  If x is another odd prime number, the expression will be: (odd) + (odd)(odd), which equals an even (O + O = E).
(E) FALSE:  2x must be even and y must be odd (since it cannot be the smallest prime number 2, which is also the only even prime). The result is even + odd, which must be odd.

QUESTION: 7

If q, r, and s are consecutive even integers and q < r < s, which of the following CANNOT be the value of s2 – r2 – q2?

Solution:

If q, r, and s are consecutive even integers and q < r < s, then r = s – 2 and q = s – 4. The expression  s2– r2– q2 can be written as s2– (s –2)2 – (s – 4)2. If we multiply this out, we get:
s2– (s –2)2 – (s – 4)2 = s2– (s2 – 4s + 4) – (s2 – 8s + 16) = s2– s2 + 4s – 4 – s2 + 8s – 16 =
-s2 + 12s – 20
The question asks which of the choices CANNOT be the value of the expression -s2 + 12s – 20 so we can test each answer choice to see which one violates what we know to be true about s, namely that s is an even integer.
Testing (E) we get:
-s2 + 12s – 20 =16
-s2 + 12s – 36 = 0
s2+ 12s – 36 = 0
(s – 6)(s – 6) = 0
s = 6. This is an even integer so this works.
Testing (D) we get:
-s2 + 12s – 20 =12
-s2 + 12s – 32 = 0
s2+ 12s – 32 = 0
(s – 4)(s – 8) = 0
s = 4 or 8. These are even integers so this works.

Testing (C) we get:
-s2 + 12s – 20 = 8
-s2 + 12s – 28 = 0
s2+ 12s – 28 = 0
Since there are no integer solutions to this quadratic (meaning there are no solutions where s is an integer), 8 is not a possible value for the expression.

Alternately, we could choose values for q, r, and s and then look for a pattern with our results. Since the answer choices are all within twenty units of zero, choosing integer values close to zero is logical. For example, if q = 0, r = 2, and s = 4, we get 42 – 22 – 02 which equals
16 – 4 – 0 = 12. Eliminate answer choice D.

Since there is only one value greater than 12 in our answer choices, it makes sense to next test q = 2, r = 4, s = 6. With these values, we get 62 – 42 – 22 which equals 36 – 16 – 4 = 16. Eliminate answer choice E.

We have now eliminated the two greatest answer choices, so we must test smaller values for q, r, and s. If q = -2, r = 0, and s = 2, we get 22 – 02 – 22 which equals 4 – 0 – 4 = 0. Eliminate answer choice B.

At this point, you might notice that as you choose smaller (more negative) values for q, r, and s, the value of s2 < r2 < q2. Thus, any additional answers will yield a negative value. If not, simply choose the next logical values for q, r, and s: q = -4, r = -2, and s = 0. With these values we get 02 – (-2)2 – (-4)2 = 0 – 4 – 16 = -20. Eliminate answer choice A.

QUESTION: 8

For all positive integers m, (m) = 3m when m is odd and (m) = ½ m when m is even, which of the following is equivalent to (9)*(6)?

Solution:

(9)=27, (6)=3, so, (9)*(6)=81
Only (27) equals to 81.

QUESTION: 9

If a is an even integer and b is an odd integer, which of the following cannot be an even integer?

Solution:

Let's look at each answer choice:
(A) EVEN: Since a is even, the product ab will always be even.  Ex: 2 × 7 = 14.

(B) UNCERTAIN: An even number divided by an odd number might be even if the the prime factors that make up the odd number are also in the prime box of the even number.  Ex: 6/3 =2.

(C) NOT EVEN: An odd number is never divisible by an even number.  By definition, an odd number is not divisible by 2 and an even number is.  The quotient of an odd number divided by an even number will not be an integer, let alone an even integer. Ex: 15/4 = 3.75

(D) EVEN: An even number raised to any integer power will always be even.  Ex: 21 = 2

(E) EVEN: An even number raised to any integer power will always be even.  Ex: 23 = 8

QUESTION: 10

If x and y are prime integers and x < y, which of the following cannot be true?

Solution:

Let's look at each answer choice:

(A) UNCERTAIN:  x could be the prime number 2.
(B) UNCERTAIN:  x could be the prime number 2, which when added to another prime number (odd) would yield an odd result.  Ex: 2 + 3 = 5
(C) UNCERTAIN:  Since x could be the prime number 2, the product xy could be even.
(D) UNCERTAIN:  y > x and they are both prime so y must be odd.  If x is another odd prime number, the expression will be: (odd) + (odd)(odd), which equals an even (O + O = E).
(E) FALSE:  2x must be even and y must be odd (since it cannot be the smallest prime number 2, which is also the only even prime). The result is even + odd, which must be odd.