Last visit was: 19 Nov 2025, 12:46 It is currently 19 Nov 2025, 12:46
Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,343
 [7]
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,343
 [7]
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
avatar
NKJJ1990
Joined: 22 Apr 2020
Last visit: 22 Sep 2022
Posts: 8
Own Kudos:
Given Kudos: 10
Location: United Arab Emirates
Concentration: General Management, Marketing
Posts: 8
Kudos: 12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
jain3120
Joined: 03 Jun 2020
Last visit: 06 Jun 2020
Posts: 10
Own Kudos:
18
 [3]
Given Kudos: 3
Posts: 10
Kudos: 18
 [3]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
hacorus
avatar
Current Student
Joined: 30 Apr 2020
Last visit: 24 Jul 2021
Posts: 53
Own Kudos:
50
 [1]
Given Kudos: 148
Location: United Kingdom
GMAT 1: 760 Q49 V44 (Online)
GMAT 1: 760 Q49 V44 (Online)
Posts: 53
Kudos: 50
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This is a long question to calculate with the following method (took me 3min with the online whiteboard), so there must be a better way.

Packages:
3C
4N
2G

Production:
750C
975N
575G
Total<3000 (no need to calculate as all of the terms are less than 1000) so the checking machine can handle it all.

Quality products = 80%
750-75*2=675
975-97-98=975-100-95=780
575-57-58=575-60-55=460

In packages this gives
675/3=225
780/4=195 we can stop here, this is the lowest possible answer choice.

Solution is A I only 195
User avatar
Lipun
Joined: 05 Jan 2020
Last visit: 08 Jan 2025
Posts: 144
Own Kudos:
Given Kudos: 291
Posts: 144
Kudos: 157
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ratio in each packet = C:N:G = 3:4:2
Production ratio = C:N:G = 750:975:575

Comparing all 3, we can find out 'N' is the limiting factor (there won't be sufficient quantity of N(975) for the respective quantities of C(750) and G(575)).
Max packets that can be produced = \(\frac{975}{4}\)

No. of packets that pass quality check = 80% of \(\frac{975}{4}\) = \(\frac{4}{5}*\frac{975}{4}\) = 195.

Ans: A
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 19 Nov 2025
Posts: 21,716
Own Kudos:
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,716
Kudos: 26,996
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
At a food processing and packaging plant, identical packets of confectionary items are prepared at a constant rate by a packaging machine. Each packet contains 3 chocolates, 4 candied nuts and 2 chewing gums. The production rate of chocolates, candied nuts and chewing gums at the plant is 750, 975 and 575 units per hour respectively. All the produced items are immediately sent to a common quality checking machine that can process 3000 units of confectionary items per hour. 20 percent each of chocolates, candied nuts and chewing gums fail the quality check. The remaining items are sent to the packaging plant without any holdup or delay. Which of the following can represent the number of packets of confectionary items prepared by the packaging machine in one hour?

I. 195
II. 230
III. 280

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III



Solution:

Let’s analyze each Roman numeral:

I. 195 packets prepared

If 195 packets were prepared, then 195 x 3 = 585 chocolates, 195 x 4 = 780 candied nuts and 195 x 2 = 390 chewing gums were sent to the packaging machine. This means that 585/0.8 ≈ 731 chocolates, 780/0.8 = 975 candied nuts and 390/0,8 ≈ 488 chewing gums were sent to the quality checking machine. Notice that since 731 + 975 + 488 = 2194 is less than 3000, the quality checking machine is able to process all the confectionary items within the hour. Further, notice that the number of chocolates, number of candied nuts and the number of chewing gums produced in one hour (which are 731, 975 and 488, respectively) are all within the limits of the production rate for each item (which are 750, 975 and 575, respectively). Thus, it is possible that 195 packets were produced.

II. 230 packets prepared

Preparing 230 packets would require 230 x 3 = 690 chocolates passing the quality check; which in turn would require producing 690/0.8 ≈ 862 chocolates per hour. Since the plant can only produce 750 chocolates per hour, preparing 230 packets is not possible.

III. 280 packets prepared

We already know that the plant does not produce chocolates fast enough to prepare 230 packets and preparing 280 packets would require even more chocolates. Thus, without further calculations, we can conclude that the plant cannot produce 280 packets either.

Answer: A

avatar
veron90
Joined: 13 Jan 2019
Last visit: 24 Mar 2022
Posts: 18
Own Kudos:
Given Kudos: 179
Location: India
Concentration: Marketing, Operations
GRE 1: Q161 V160
GPA: 2.9
GRE 1: Q161 V160
Posts: 18
Kudos: 8
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
At a food processing and packaging plant, identical packets of confectionary items are prepared at a constant rate by a packaging machine. Each packet contains 3 chocolates, 4 candied nuts and 2 chewing gums. The production rate of chocolates, candied nuts and chewing gums at the plant is 750, 975 and 575 units per hour respectively. All the produced items are immediately sent to a common quality checking machine that can process 3000 units of confectionary items per hour. 20 percent each of chocolates, candied nuts and chewing gums fail the quality check. The remaining items are sent to the packaging plant without any holdup or delay. Which of the following can represent the number of packets of confectionary items prepared by the packaging machine in one hour?

I. 195
II. 230
III. 280

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


Are You Up For the Challenge: 700 Level Questions
Hi,

Is this also the right way of solving this :

Each packet contains : 3 Choc + 4 Nuts + 2 Candied Gum = 9 per packet

Number of Chocolates, Nuts and Candied Gum produced = 750 + 975 + 575 = 2300

Thus, 80% of 2300 = 1840 (20% rejected)


Hence, Total packets per hour = 1840/9 = 204(approx.)

So as per the answer choices, 195 is the only viable option.


Is this flawed, or correct - would appreciate input...thanks! Because it appears that the question is asking how many units the packaging machine can pack in an hour.


Varun
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
Posts: 38,588
Kudos: 1,079
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
105390 posts
Tuck School Moderator
805 posts