Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 15 Jul 2019, 22:11

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

In a bag there are three kinds of marbles.

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
User avatar
G
Joined: 16 Jan 2019
Posts: 269
Location: India
Concentration: General Management
WE: Sales (Other)
In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post 15 Jun 2019, 00:12
1
4
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

63% (02:42) correct 38% (03:08) wrong based on 32 sessions

HideShow timer Statistics


In a bag there are three kinds of marbles. There are red marbles, blue marbles and green marbles. Initially, the ratio of red, blue and green marbles in the bag is 2:3:7. Now, a few green marbles are taken out from the bag and some red marbles and some blue marbles are put into the bag such that the new ratio of red, blue and green marbles is 4:5:9. What is the minimum possible number of red marbles which were put into the bag?

A. 1

B. 3

C. 9

D. 8

E. 2
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7764
In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post 15 Jun 2019, 01:18
1
firas92 wrote:
In a bag there are three kinds of marbles. There are red marbles, blue marbles and green marbles. Initially, the ratio of red, blue and green marbles in the bag is 2:3:7. Now, a few green marbles are taken out from the bag and some red marbles and some blue marbles are put into the bag such that the new ratio of red, blue and green marbles is 4:5:9. What is the minimum possible number of red marbles which were put into the bag?

A. 1

B. 3

C. 9

D. 8

E. 2



The ratios are
Colours-----R:B:G
Initial-------2:3:7
Final--------4:5:9

Some Inference..

1) We are adding red such that it becomes 4y from 2x..
This means the number of RED ball being added is EVEN, as 2x+A=4y, so A has to be even
All odd can be eliminated, ONLY D and E left.

2) Let us check the least out of two - 2
So 2x+2=4y or \(x+1=2y....x=2y-1\)
\(7x-B=9y.....7(2y-1)-B=9y...14y-7-B=9y....5y=7+B\), so least value of B is 3 and y becomes 2
Thus x=2y-1=2*2-1=3..
We can check too
If x is 3, the number is \(2*3:3*3:7*3=6:9:21\)
If we add 2 red, 1 blue and take out 3 green, we get \(6+2:9+1:21-3=8:10:18=4:5:9\)=Final ratio

Since 2 gives us integer values for x and y, it is our answer. Otherwise we would have taken 8 as our answer.

SO ans is E
_________________
Director
Director
avatar
P
Joined: 19 Oct 2018
Posts: 675
Location: India
Re: In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post 15 Jun 2019, 03:46
Before Exchange
Number of red Marbles= 2x
Number of blue marbles=3x
Number of Green Marbles=7x

After exchange
Number of red Marbles= 4y
Number of blue marbles=5y
Number of Green Marbles=9y

As we didn't exchange blue marbles, number of blue marbles will remain same
3x=5y
y=3x/5

Number of red Marbles= 4y= 12x/5
Number of red Marbles is an integral value; Minimum number of red marbles possible after exchange when x=5

Minimum Number of red Marbles before exchange= 2x=2*5=10
Minimum Number of red Marbles after exchange= (12*5)/5=12
Difference= 12-10=2




firas92 wrote:
In a bag there are three kinds of marbles. There are red marbles, blue marbles and green marbles. Initially, the ratio of red, blue and green marbles in the bag is 2:3:7. Now, a few green marbles are taken out from the bag and some red marbles and some blue marbles are put into the bag such that the new ratio of red, blue and green marbles is 4:5:9. What is the minimum possible number of red marbles which were put into the bag?

A. 1

B. 3

C. 9

D. 8

E. 2
Manager
Manager
avatar
S
Joined: 09 Nov 2015
Posts: 91
In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post Updated on: 15 Jun 2019, 23:54
1
We have an Initial Collection(IC) of red(R), blue(B) and green(G) marbles in the ratio of 2:3:7 respectively. Some of G have to be taken out and some R and B added in a manner that results in a Final Collection(FC) with a ratio of R:B:G=4:5:9. Also, this has to be done so that the minimum number of R is added.The possible total number of marbles in IC are (a) 12 (R-2, B-3 & G-7), (b) 24 (R-4, B-6, G-14), (c) 36 (R-6,B-9 & G-21)..... and so on. Similarly, the possible total numbers of FC are (a)18 (R-4, B-5 &G-9), (b) 36 (R-8, B-10 & G-18)... and so on. Now the trick is to choose a total number of IC that will enable us to take out a certain number of G and put in some R and B so that the ratio of FC is R:B:G=4:5:9. You will find that the IC quantity that will enable us to to do this (i.e. get the required ratio in FC) and, at the same time, ensure that the minimum quantity of R is put in is 36 (the LCM of the minimum combinations of IC and FC: 12 and 18). So IC quantity is 36: R-6, B-9 & G-21). We can then take out 3 green marbles and put in 1 blue marble and 2 red marbles which will give us an FC of R-8, B-10 and G-18 which is in the required ratio of 4:5:9. So answer is E.

P.S. You can also start with multiples of 36 as the IC quantity and do the needed removal and addition to get FC in the required ratio but in that case you would end up having to add more that 2 red marbles. For example, if you start with an IC quantity of 72 (R-12, B-18 & G-42) and remove 6 green marbles you would have to add 2 blue marbles and 4 red marbles to get the required FC ratio. So, in order to find the minimum possible number of red marbles to be added you have to assume that IC quantity is the LCM of the minimum possible quantities of IC and FC as dictated by their respective ratios.

SUMMARY:
In order to solve this and similar problems simply figure out the minimum quantities for IC and FC, calculate the LCM of these two numbers, take that as the IC and work from there.

Originally posted by effatara on 15 Jun 2019, 08:14.
Last edited by effatara on 15 Jun 2019, 23:54, edited 1 time in total.
Senior Manager
Senior Manager
User avatar
G
Joined: 16 Jan 2019
Posts: 269
Location: India
Concentration: General Management
WE: Sales (Other)
In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post Updated on: 15 Jun 2019, 13:27
OFFICIAL EXPLANATION


Let the initial number of red, blue and green marbles be \(2x, 3x\) and \(7x\) respectively and the final number of red, blue and green marbles be \(4y, 5y\) and \(9y\) respectively

We know that the number of green marbles has decreased and so, \(7x>9y\) or \(x>\frac{9y}{7}\)

We also know that the number of red and blue marbles have increased. So, \(4y>2x\) or \(x<2y\) and \(5y>3x\) or \(x<\frac{5y}{3}\)

Combining the three equations we get \(\frac{9y}{7}<x<\frac{5y}{3}\)

For minimizing the number of red marbles, we have to minimize \(4y-2x\) or \(2(2y-x)\)

Hence, we have to minimize \(2y-x\) subject to the constraints \(\frac{9y}{7}<x<\frac{5y}{3}\)

We know that \(x\) and \(y\) have to be integers.

For \(y=1\), we have \(\frac{9}{7}<x<\frac{5}{3}\)

There is no integer value for \(x\) in this range

For \(y=2\), we have \(\frac{18}{7}<x<\frac{10}{3}\) and so \(x\) can be \(3\)

Therefore \(y=2, x=3\) are the values at which the number of additional red marbles is minimized.

Hence the minimum number of red marbles that were added \(= (4*2)-(2*3) = 2\)

Answer is (E)

Originally posted by firas92 on 15 Jun 2019, 13:07.
Last edited by firas92 on 15 Jun 2019, 13:27, edited 1 time in total.
Director
Director
avatar
P
Joined: 19 Oct 2018
Posts: 675
Location: India
Re: In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post 15 Jun 2019, 13:14
Oops!! I read the question wrongly, still got the right answer. :P

firas92 wrote:
OFFICIAL EXPLANATION


Let the initial number of red, blue and green marbles be \(2x, 3x\) and \(7x\) respectively and the final number of red, blue and green marbles be \(4y, 5y\) and \(9y\) respectively

We know that the number of green marbles has decreased and so, \(7x>9y\) or \(x>\frac{9y}{7}\)

We also know that the number of red and blue marbles have increased. So, \(4y>2x\) or \(x<2y\) and \(5y>3x\) or \(x<\frac{5y}{3}\)

Combining the three equations we get \(\frac{9y}{7}<x<\frac{5y}{3}\)

For minimizing the number of red marbles, we have to minimize \(4y-2x\) or \(2(2y-x)\)

Hence, we have to minimize \(2y-x\) subject to the constraints \(\frac{9y}{7}<x<\frac{5y}{3}\)

We know that \(x\) and \(y\) have to be integers.

For \(y=1\), we have \(\frac{9}{7}<x<\frac{5}{3}\)

There is no integer value for \(x\) in this range

For \(y=2\), we have \(\frac{18}{7}<x<\frac{10}{3}\) and so \(x\) can be \(3\)

Therefore \(y=2, x=3\) are the values at which the number of additional red marbles is minimized.

Hence the minimum number of red marbles that were added \(= (4*2)-(2*3) = 2\)

Answer is (E)
Senior Manager
Senior Manager
User avatar
G
Joined: 16 Jan 2019
Posts: 269
Location: India
Concentration: General Management
WE: Sales (Other)
Re: In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post 15 Jun 2019, 13:22
1
nick1816 wrote:
Oops!! I read the question wrongly, still got the right answer. :P

firas92 wrote:
OFFICIAL EXPLANATION


Let the initial number of red, blue and green marbles be \(2x, 3x\) and \(7x\) respectively and the final number of red, blue and green marbles be \(4y, 5y\) and \(9y\) respectively

We know that the number of green marbles has decreased and so, \(7x>9y\) or \(x>\frac{9y}{7}\)

We also know that the number of red and blue marbles have increased. So, \(4y>2x\) or \(x<2y\) and \(5y>3x\) or \(x<\frac{5y}{3}\)

Combining the three equations we get \(\frac{9y}{7}<x<\frac{5y}{3}\)

For minimizing the number of red marbles, we have to minimize \(4y-2x\) or \(2(2y-x)\)

Hence, we have to minimize \(2y-x\) subject to the constraints \(\frac{9y}{7}<x<\frac{5y}{3}\)

We know that \(x\) and \(y\) have to be integers.

For \(y=1\), we have \(\frac{9}{7}<x<\frac{5}{3}\)

There is no integer value for \(x\) in this range

For \(y=2\), we have \(\frac{18}{7}<x<\frac{10}{3}\) and so \(x\) can be \(3\)

Therefore \(y=2, x=3\) are the values at which the number of additional red marbles is minimized.

Hence the minimum number of red marbles that were added \(= (4*2)-(2*3) = 2\)

Answer is (E)


nick1816

I don't think anyone's gonna complain about having that kind of luck on the big day :lol:
Manager
Manager
avatar
B
Joined: 23 Jul 2015
Posts: 55
Re: In a bag there are three kinds of marbles.  [#permalink]

Show Tags

New post 15 Jun 2019, 15:35
Hi! Is there another explanation? I am really confused by the inequality method.
GMAT Club Bot
Re: In a bag there are three kinds of marbles.   [#permalink] 15 Jun 2019, 15:35
Display posts from previous: Sort by

In a bag there are three kinds of marbles.

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne