Problem Solving

Thanks.

Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A?


A. 1
B. 3
C. 4
D. 6
E. 8

skyjuice wrote:Thanks.

Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A?


A. 1
B. 3
C. 4
D. 6
E. 8

For every white marble we have 1/3 red marbles in A
For every white marble we have 1/4 red marbles in B

num red in A + num red in B = some integer, imples ...
1/3*w + 1/4*(30-w) = some integer

Only possible values for w are (A=6, B=24) or (A=18, B=12)

So, red marbles in A can be either 2 or 6 (i.e. 1/3 of white marbles)

My vote is for D

A clumsy way of doing it (backsolving)

A----------------B
R --- W --- R --- W
1 --- 3 --- 17/4 --- 17
3 --- 9 --- 21/4 --- 21
4 --- 12 --- 18/4 --- 18
6 --- 18 --- 3 --- 12
8 --- 24 --- 6/4 --- 6

Hence 6 implies D.

Regards

Sorry, but the editing is lost in the posting.
Take first option, 1 red marbel in Box A, use the ratio relation to find there will be three times as many white marbles.

So we get the second column. Subtract from 20 to get the last column, white marbles in Box B, since their sum is 20.

Finally, use the last ratio to get the number of red balls in Box B. (the third column).

If it is a integer, then bingo. I hope it is clearer now.

Regards

can u explain the second method of solving the marbel number or if there is some other method ....
thanks

Thanks guys

In Bag A - Ratio of Red : White : Blue = 2:6:9 (work it out using the 2 different ratios given) [let x be total no. of balls]
In Bag B - Ratio of Red : White = 1:4 [ let x be total number of balls]
All we know is 6x + 4y = 30 or 3x + 2y = 30
Work using the options (substitution)
Option C - Red Balls in Bag A = 4
2x = 4 ; x = 2 ; 6x = 12 ; Thus 4y = 18 and y = 18/4 (Not possible)
Option D - Red Balls in Bag A = 6
2x = 6 ; x = 3 ; 6x = 18 ; Thus 4y = 12 or y = 3 (Possible)

If you try it with any other choice, the ratios and the condition 3x + 2y won't work simultaneously. Hence D