Last visit was: 19 Nov 2025, 01:34 It is currently 19 Nov 2025, 01:34
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
avatar
mightymk
Joined: 15 Oct 2013
Last visit: 27 Jan 2020
Posts: 12
Own Kudos:
27
 [27]
Given Kudos: 15
Concentration: Marketing, Finance
Schools: McCombs '16
Schools: McCombs '16
Posts: 12
Kudos: 27
 [27]
2
Kudos
Add Kudos
25
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,379
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,379
Kudos: 778,166
 [16]
8
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
General Discussion
avatar
mightymk
Joined: 15 Oct 2013
Last visit: 27 Jan 2020
Posts: 12
Own Kudos:
Given Kudos: 15
Concentration: Marketing, Finance
Schools: McCombs '16
Schools: McCombs '16
Posts: 12
Kudos: 27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
kunals31
Joined: 08 Jan 2015
Last visit: 20 Oct 2016
Posts: 3
Own Kudos:
8
 [1]
Given Kudos: 6
Posts: 3
Kudos: 8
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Given that Red is to yellow is 2:5, Red is to green is 3:8 and Yellow is to blue is 9:2.

Therefore, the total number of red, yellow, blue and green balls will be 18x, 45x, 10x and 48x respectively, where x is a constant.

If the combined number of red and blue balls is fewer than 85, i.e. max 84 balls, then the maximum number of yellow and green balls will be 279.

(10x+18x) < 85.

28x < 85, i.e. 28x <= 84 (Since number of balls cannot be in fraction). Thus, x<=3.

(45x+48x) = 93x.

Max (93x) = 279.

Answer is D.
User avatar
VeritasPrepBrandon
User avatar
Veritas Prep GMAT Instructor
Joined: 23 Oct 2013
Last visit: 07 Jun 2016
Posts: 143
Own Kudos:
940
 [4]
Given Kudos: 9
Expert
Expert reply
Posts: 143
Kudos: 940
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Fun question. This problem is all about least common multiples. It helps to initially lay out the ratios individual, r:y = 2:5, r:g = 3:8, and y:b = 9:2. Then combine them. First combine the r:y:g by finding the least common multiple between the red in the r:y and r:g ratios. In the first ratio, the red is 2, and in the second, the red is 3, so the LCM is 6. Therefore multiply the first ratio by 3 and the second by 2 to get - r:y:g = 6:15:16. You do this in order to make the red a common number so that you can combine the ratios.

Then we have to use the yellow to incorporate the last ratio. The yellow in the ratio that we just did is 15, and in the y:b ratio is 9, therefore the LCM is 45. Multiply the ratio that we currently have by 3, and the y:b ratio by 5, to get r:y:g:b = 18:45:48:10. The r+b candies are thus a multiple of 28 (18+10). The highest multiple of 28 that is still below 85 is 84 (28*3). Therefore we must multiply the y+g multiplier (45+48 = 93) times 3 in order to get to 279 yellow and green candies, or answer choice D.

I hope this helps!
avatar
Economist2015
Joined: 18 Jan 2015
Last visit: 22 Jul 2015
Posts: 13
Own Kudos:
Given Kudos: 1
Posts: 13
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Many thanks Brandon!Didn't think it could be solved using LCMs!

How would you rate this question?Is this a 700+ question?
User avatar
Temurkhon
Joined: 23 Jan 2013
Last visit: 06 Apr 2019
Posts: 412
Own Kudos:
314
 [1]
Given Kudos: 43
Schools: Cambridge'16
Schools: Cambridge'16
Posts: 412
Kudos: 314
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
what is max of y+g if r+b<85?

need to find r+b, to do so multiply r:y and y:b in numbers which are in 9/5 ratio
r:y - 2:5 | *27
r:g - 3:8
y:b - 9:2 |*15

r+b=54+30=84<85, done

y+g=135+8*(54/3)=279

D
User avatar
mvictor
User avatar
Board of Directors
Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,124
Own Kudos:
1,263
 [1]
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
GMAT 1: 650 Q49 V30
Posts: 2,124
Kudos: 1,263
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
mightymk
In a candy dish the ratio of red to yellow candies is 2:5, the ratio of red to green candies is 3:8, and the ratio of yellow ot blue candies is 9:2.what is the maximum total number of yellow and green candies combined if the maximum combined number of red and blue candies is fewer than 85?

A. 144
B. 189
C. 234
D. 279
E. 309

we can rearrange to below:
R 18
Y 45
G 48
B 10

so 18x+10x<85. x must be an integer. 28x<85. x can be 3, and 28x=74, which is true. if x is 4, then the condition is not

then: 45*3 + 48*3 = 135+144 = 279
D
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 18 Nov 2025
Posts: 21,712
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: 21,712
Kudos: 26,995
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mightymk
In a candy dish the ratio of red to yellow candies is 2:5, the ratio of red to green candies is 3:8, and the ratio of yellow ot blue candies is 9:2.what is the maximum total number of yellow and green candies combined if the maximum combined number of red and blue candies is fewer than 85?

A. 144
B. 189
C. 234
D. 279
E. 309

Since the ratio of red to yellow candies is 2:5 and the ratio of yellow to blue candies is 9:2, we must multiply each ratio by a different constant such that the numbers for yellow candy are the same in each ratio:

Multiplying the first ratio by 9, we see that the ratio of red to yellow candies is 18:45. If we multiply the second ratio by 5, we obtain a ratio of yellow to blue candies of 45:10. Now that the ratio numbers for the yellow candies is the same (45), we can directly relate red to blue candies as a ratio of 18:10, or 9:5.

Thus, we can create the equation:

9x + 5x < 85

13x < 85

x < 6.54

Since x is an integer, the maximum integer value of x is 6. Therefore, we have a maximum of 9(6) = 54 red candies. Since the ratio of red to yellow candies is 2:5 and 2 x 27 = 54, we have a maximum of 5 x 27 = 135 yellow candies. Likewise, since the ratio of red to green candies is 3:8 and 3 x 18 = 54, we have a maximum of 8 x 18 = 144 green candies. Therefore, we have a maximum of 135 + 144 = 279 yellow and green candies.

Answer: D
User avatar
HarshavardhanR
Joined: 16 Mar 2023
Last visit: 18 Nov 2025
Posts: 425
Own Kudos:
Given Kudos: 59
Status:Independent GMAT Tutor
Affiliations: Ex - Director, Subject Matter Expertise at e-GMAT
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 425
Kudos: 461
Kudos
Add Kudos
Bookmarks
Bookmark this Post
­
___
Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com
 
Attachment:
GMAT-Club-Forum-jphjlsb0.png
GMAT-Club-Forum-jphjlsb0.png [ 244.92 KiB | Viewed 853 times ]
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,584
Own Kudos:
Posts: 38,584
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
105379 posts
Tuck School Moderator
805 posts