Question: 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?
- 1
- 3
- 4
- 6
- 8
“Bag A contains red, white and blue marbles such that”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.
Solution and Explanation:
Approach Solution 1:
Bag A:
Red/White=1/3=2/6;
White/Blue=2/3=6/9;
So Red/White/Blue=2/6/9 --> # of marbles in bag A would be 2x, 6x and 9x for some positive integer multiple x, where 6x corresponds to the # of white marbles and 2x corresponds to the # of red marbles (so # of red marbles must be a multiple of 2, so answers A and B are out at this stage);
Bag B:
Red/White=1/4 --> # of marbles in bag B would be y and 4y, for some positive integer multiple y where 4y corresponds to the # of white marbles;
Given:
->6x+4y = 30
->3x+2y = 15
there are two positive integer solutions for this equation:
- x=3 and y=3
--> in this case # of red marbles equals to 2x=6
- x=1 and y=6
--> in this case # of red marbles equals to 2x=2
Only 6 is in the answer choices.
Correct Answer: D
Approach Solution 2:
Explanation:
Bag A:
Red:White = 1:3
White:Blue = 2:3
W is the common one here so make it equal i.e. Red:White = 2:6 and White:Blue = 6:9 (the ratios remain the same). So Red:White:Blue = 2:6:9
Since number of marbles has to be an integer, number of red marbles in this bag must be 2 or a multiple of 2 and number of white marbles must be 6 or a multiple of 6.
Bag B:
Red:White = 1:4
The number of white marbles must be 4 or a multiple of 4.
To make 30 white marbles, one could mix white marbles from Bag A and Bag B in many ways.
BagA: 6 + BagB: 24 (No. of red marbles in BagA = 2)
BagA: 12 + BagB: 18 - Not possible because 18 is not a multiple of 4
BagA: 18 + BagB: 12 (No. of red marbles in BagA = 6)
BagA: 24 + BagB: 6 - Not possible because 6 is not a multiple of 4
No of red marbles in bag A can be both 2 and 6.
Correct Answer: D
Approach Solution 3:
White/Blue=2/3=6/9, hence Red/White/Blue=2/6/9 in Bag A. Answers A and B are now ruled out since the number of marbles in bag A would be 2x, 6x, and 9x for any positive integer multiple x, where 6x corresponds to the number of white marbles and 2x relates to the number of red marbles;
Red/White = 1/4 in Bag B - If y were a positive integer multiple, then the number of marbles in bag B would be y and 4y, where y is the number of white marbles;
Given:
->6x+4y = 30\s->3x+2y = 15
For this equation, there are two positive integer solutions:
In this scenario, the number of red marbles equals 2x=6 since x=3 and y=3
In this scenario, the number of red marbles equals 2x=2 since x=1 and y=6.
There are just 6 options for answers.
Correct Answer: D
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