Events & Promotions
| Last visit was: 20 Sep 2026, 13:35 |
It is currently 20 Sep 2026, 13:35 |
|
|
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.
Sep 1810:00 AM IST
-11:00 PM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.
Sep 2003:30 AM EDT
-04:30 AM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included!
Sep 2312:30 AM EDT
-01:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Sep 2601:00 PM EDT
-04:00 PM EDT
Top scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included. Class starts Saturdays Sept. 26, 2026 - Dec. 19, 2026, 1:00pm-4:00pm ES
Sep 2606:00 PM PDT
-07:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
22 Mar 2018, 08:15
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
48% (02:19) correct
52%
(02:14)
wrong
based on 401
sessions
History
Date
Time
Result
Not Attempted Yet
Samuel has a big drawer full of exactly 59 packets, each containing a marker which is either black or blue or red in colour. All 59 packets are sealed and are completely identical from outside. It is not possible to know which colour marker is inside unless the packet is fully opened. Out of 59, 23 packets contain a black marker each. How many packets contain a blue marker?
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
22 Mar 2018, 09:23
Kudos
Bookmarks
TOTAL = 59
Black = 23
Blue = x
Red = y
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
In that case, 23+ 18 + 1 = 42 and he has at least one marker of each color. Then Black = 23, Blue = 18, Red = 18. SUFFICIENT
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
In that case R/59 = Blue/59 --> R = B. Then 59 - 23 (Black) = 36.
As Blue = Red, 36/2 = 18 = Red = Blue. SUFFICIENT
Ans: (D)
Please correct me if I'm wrong
Black = 23
Blue = x
Red = y
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
In that case, 23+ 18 + 1 = 42 and he has at least one marker of each color. Then Black = 23, Blue = 18, Red = 18. SUFFICIENT
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
In that case R/59 = Blue/59 --> R = B. Then 59 - 23 (Black) = 36.
As Blue = Red, 36/2 = 18 = Red = Blue. SUFFICIENT
Ans: (D)
Please correct me if I'm wrong
22 May 2018, 18:40
Kudos
Bookmarks
Given:
59 total
23 black
36 blue or red
Find blue.
Statement 1
42 packets ensures that you'll have one of each color. This means that in any 42 packets you withdraw, you'll have:
- ALL of the packets of the color that has the most packets,
- ALL of the packets of the color that has the second most packets, and
- ONE of the packets of the color that has the least packets.
In any selection of 42 packets, 23 will be black, and 19 will be blue or red... so black has the most packets (23), and either blue or red has the second most (18). But note that in our initial analysis, blue+red=36. Thus, if one of the colors has 18 packets, the other will also have 18 packets.
Sufficient
Statement 2
36 packets of blue or red, half of them are red, thus half of them are blue.
Sufficient
Answer: D
59 total
23 black
36 blue or red
Find blue.
Statement 1
42 packets ensures that you'll have one of each color. This means that in any 42 packets you withdraw, you'll have:
- ALL of the packets of the color that has the most packets,
- ALL of the packets of the color that has the second most packets, and
- ONE of the packets of the color that has the least packets.
In any selection of 42 packets, 23 will be black, and 19 will be blue or red... so black has the most packets (23), and either blue or red has the second most (18). But note that in our initial analysis, blue+red=36. Thus, if one of the colors has 18 packets, the other will also have 18 packets.
Sufficient
Statement 2
36 packets of blue or red, half of them are red, thus half of them are blue.
Sufficient
Answer: D
23 May 2018, 00:58
Kudos
Bookmarks
chetan2u
Hi Chetan ,
Statement B is ok to understand .
But i have trouble understanding statement 1 .
Can you please explain statement 1 and how to interpret such statement in general
