In a bag there are three kinds of marbles.

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.

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)

Oops!! I read the question wrongly, still got the right answer. :P

nick1816

I don't think anyone's gonna complain about having that kind of luck on the big day

Hi! Is there another explanation? I am really confused by the inequality method.