Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
28 Oct 2021, 07:00
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
72% (02:05) correct
28%
(02:29)
wrong
based on 159
sessions
History
Date
Time
Result
Not Attempted Yet
A set consists of 7 positive integers, 4 are odd and 3 are even. Three numbers are selected at random and multiplied to each other, and the product is represented with A; three numbers are selected at random from the remained numbers and multiplied to each other, the product is represented with B. Which of the following must be true?
I. at least one of a and b is even.
II. at least one of a and b is odd.
III. a+b is odd.
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
I. at least one of a and b is even.
II. at least one of a and b is odd.
III. a+b is odd.
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
PyjamaScientist
Admitted - Which School Forum Moderator
Joined: 25 Oct 2020
Last visit: 04 Apr 2026
Posts: 1,126
Own Kudos:
Given Kudos: 633
Schools: Ross '25 (M$)
GMAT 1: 740 Q49 V42 (Online)
Products:
28 Oct 2021, 07:11
Kudos
Bookmarks
Option (A) is the correct choice.
There can be a case when either A or B have only odd numbers. In that case the other one will deifnitely have an even number and that would make it even (product by an even no. results in an even no.)
So either A or B will be even. This WILL be true always.
But, the same can not be said about the oddity of either two.
This is possible but not must true. Because, if one even number goes into product of A and B, both will be even. But, there can be a case, when no odd no. is in the product to make the product eventually odd. Anyway, it is possible, but not a must situation for sure.
Similarly, for the third condition, since there can be a case where both A and B are even, it is not necessary to have their sum always = odd! Hence, this can not also be a must be true condition.
There can be a case when either A or B have only odd numbers. In that case the other one will deifnitely have an even number and that would make it even (product by an even no. results in an even no.)
So either A or B will be even. This WILL be true always.
But, the same can not be said about the oddity of either two.
This is possible but not must true. Because, if one even number goes into product of A and B, both will be even. But, there can be a case, when no odd no. is in the product to make the product eventually odd. Anyway, it is possible, but not a must situation for sure.
Similarly, for the third condition, since there can be a case where both A and B are even, it is not necessary to have their sum always = odd! Hence, this can not also be a must be true condition.
26 Mar 2026, 00:50
Kudos
Bookmarks
Let's break this down step by step.
We have 7 numbers: 4 odd and 3 even. We pick 3 for product A, then 3 from the remaining 4 for product B. That means 6 numbers are used and 1 is left out.
Key rule: A product is EVEN if it contains at least 1 even number. A product is ODD only if ALL its factors are odd.
Now, how many even numbers are among the 6 used?
- If the leftover number is odd → 3 even numbers among the 6
- If the leftover number is even → 2 even numbers among the 6
Either way, there are at least 2 even numbers spread across the two groups.
Statement I: At least one of A and B is even.
For BOTH products to be odd, BOTH groups would need 0 even numbers. But we have at least 2 even numbers floating around — even if both land in the same group, that group's product is still even. So it's impossible for both A and B to be odd. Statement I is ALWAYS true. ✓
Statement II: At least one of A and B is odd.
Can both be even? Yes! Example: if the 2 or 3 even numbers split so each group gets at least 1 even number (say 1 even + 2 odd in group A, and 1 even + 2 odd in group B), both products are even. Statement II is NOT necessarily true. ✗
Statement III: A + B is odd.
A + B is odd only when one is even and the other is odd. Since both CAN be even (as shown above), A + B can be even. Statement III is NOT necessarily true. ✗
Key Insight: Only Statement I must be true — with at least 2 even numbers among the 6 chosen, it is impossible for both groups to consist entirely of odd numbers.
Answer: A
We have 7 numbers: 4 odd and 3 even. We pick 3 for product A, then 3 from the remaining 4 for product B. That means 6 numbers are used and 1 is left out.
Key rule: A product is EVEN if it contains at least 1 even number. A product is ODD only if ALL its factors are odd.
Now, how many even numbers are among the 6 used?
- If the leftover number is odd → 3 even numbers among the 6
- If the leftover number is even → 2 even numbers among the 6
Either way, there are at least 2 even numbers spread across the two groups.
Statement I: At least one of A and B is even.
For BOTH products to be odd, BOTH groups would need 0 even numbers. But we have at least 2 even numbers floating around — even if both land in the same group, that group's product is still even. So it's impossible for both A and B to be odd. Statement I is ALWAYS true. ✓
Statement II: At least one of A and B is odd.
Can both be even? Yes! Example: if the 2 or 3 even numbers split so each group gets at least 1 even number (say 1 even + 2 odd in group A, and 1 even + 2 odd in group B), both products are even. Statement II is NOT necessarily true. ✗
Statement III: A + B is odd.
A + B is odd only when one is even and the other is odd. Since both CAN be even (as shown above), A + B can be even. Statement III is NOT necessarily true. ✗
Key Insight: Only Statement I must be true — with at least 2 even numbers among the 6 chosen, it is impossible for both groups to consist entirely of odd numbers.
Answer: A
