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In a bag there are three kinds of marbles.
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15 Jun 2019, 00:12
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63% (02:42) correct 38% (03:08) wrong based on 32 sessions
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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
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In a bag there are three kinds of marbles.
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15 Jun 2019, 01:18
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 ColoursR:B:G Initial2:3:7 Final4:5:9 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  2So 2x+2=4y or \(x+1=2y....x=2y1\) \(7xB=9y.....7(2y1)B=9y...14y7B=9y....5y=7+B\), so least value of B is 3 and y becomes 2 Thus x=2y1=2*21=3.. We can check tooIf 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 \(6+2:9+1:213=8:10:18=4:5:9\)=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
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Re: In a bag there are three kinds of marbles.
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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= 1210=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



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In a bag there are three kinds of marbles.
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Updated on: 15 Jun 2019, 23:54
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 (R2, B3 & G7), (b) 24 (R4, B6, G14), (c) 36 (R6,B9 & G21)..... and so on. Similarly, the possible total numbers of FC are (a)18 (R4, B5 &G9), (b) 36 (R8, B10 & G18)... 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: R6, B9 & G21). 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 R8, B10 and G18 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 (R12, B18 & G42) 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.



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In a bag there are three kinds of marbles.
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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 \(4y2x\) or \(2(2yx)\)
Hence, we have to minimize \(2yx\) 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.



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Re: In a bag there are three kinds of marbles.
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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 \(4y2x\) or \(2(2yx)\)
Hence, we have to minimize \(2yx\) 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)



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Re: In a bag there are three kinds of marbles.
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15 Jun 2019, 13:22
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 \(4y2x\) or \(2(2yx)\)
Hence, we have to minimize \(2yx\) 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) nick1816I don't think anyone's gonna complain about having that kind of luck on the big day



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Re: In a bag there are three kinds of marbles.
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15 Jun 2019, 15:35
Hi! Is there another explanation? I am really confused by the inequality method.




Re: In a bag there are three kinds of marbles.
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15 Jun 2019, 15:35






