In a bag there are three kinds of marbles.

Question

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

A. 1

B. 3

C. 9

D. 8

E. 2

chetan2u · Answer

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

nick1816 · Answer

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

effatara · Answer

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.

firas92 · Answer

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}

nick1816 · Answer

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

firas92 · Answer

nick1816 I don't think anyone's gonna complain about having that kind of luck on the big day

SPatel1992 · Answer

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

bhm1972 · Answer

🧠 The Smart Trick — 10 Seconds Solution

Setup:

Red added = 4k - 2x = 2(2k - x)

💡 Key Insight:

🎯 Answer:  E. 2

In a bag there are three kinds of marbles.

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