We are given a table with the following information:

Bag P has 37 marbles and 10.8% of those marbles are blue. So there are 0.108(37) = 3.996, or 4, blue marbles in bag P.

Bag Q has X marble and 66.7% of those marbles are blue. Recall that the fraction 2/3 is approximately 66.7%, so there are (2/3)X blue marbles in bag Q.

Bag R has 32 marbles and 50% of those marbles are blue. So, there are 0.5(32) = 16 blue marbles in bag R.

We are also given that 1/3 of the total marbles in the 3 bags are blue. Thus, we can create the following equation:

1/3(37 + X + 32) = 4 + (2/3)X + 16

Multiplying the equation by 3, we have:

37 + X + 32 = 12 + 2X + 48

X + 69 = 2X + 60

X = 9

There are 9 marbles in bag Q.

Answer: B
_________________

Video solution from Quant Reasoning:

_________________

Bag P : 10.8% of 37 = 4
Bag Q : 66.7% of x = 2x/3
Bag R : 50% of 32 = 16.

Given, 4 + 2x/3 + 16 = 1/3 ( 37+x+32) => x = 9. Hence B.

This might look like it contains hard calculations, but the thing to notice here is that the number of balls are always going to be integers.

So, 10.8% of 37 will be 4, we know 50% of x is 32 and we can keep 66.7% as 2/3rds. Once you do this, the question can be solved to arrive at ~9 quickly.

If 1/3 of the total number of marbles in the three bags listed in the table above are blue, how many marbles are there in bag Q?

A) 5
B) 9
C) 12
D) 23
E) 46

Total has to be multiple of 3 since 1/3rd of them are blue and 1st and 3rd bag have total 69 marbles so (69 + x)/3 is blue total hence x also must be divisible by 3

hence narrow down to B) 9 or C) 12

66.67 % is nothing but 2/3 rd (ie 33.33 * 2 ie 1/3 * 2)

so if x is 9 then blue marbles in 2nd bag is 6 (2/3rd of 9 = 6)
or if x is 12 then blue marbles in 3rd bag is 8 (2/3rd of 12 = 8)

do that math : if x = 9 then total = 69 + 9 = 78
and total blue = 16 + 4 + 6 = 26
26/78 = 1/3 -----> fits (most likely that B is the ans)

cross check
if x = 12 then total = 81
total blue = 28
28/81 != 1/3 hence (B)
_________________

If we let \(x\) be the number of marbles in Bag \(Q\),

To start, the total of number of marbles in the three bags is \((37+ 32+ x) = 69 + x\) marbles.

Next, we need to find the number of blue marbles in the 3 bags. To do so, we multiply each percentage by the number of marbles in each bag (converting each percentage to decimal as needed):

The number of blue marbles present in each bag would be:

Bag P: \(0.108 \times 37 =3.996\) or \(4\) since we want the nearest whole number marbles
Bag R:\(0.50 \times 32 =\)\(16\) marbles
Bag Q:\(0.667(x) = 0.667x\) marbles or \(\frac{2x}{3}\) marbles (Note that 0.667 is equivalent to 2/3 in fractions)

The prompt states that \(\frac{1}{3}\) of the total marbles equals the total number of blue marbles. We can now form the equation below.

1/3 x Total Number of Marbles = Total Number of Blue Marbles

\(1/3 (69 + x) = (4 + 16 + 2x/3)\)
Distribute \(\frac{1}{3}\) into the parenthesis,
\(23 + \frac{x}{3} = 20 + \frac{2x}{3}\)
\(23 – 20= \frac{2x}{3} – \frac{x}{3}\)
\(3 = \frac{x}{3}\)
\(x = 3(3) = 9\)

Hence, there are 9 marbles in Bag Q. The final answer is .

Bag P: Number of blue marbles: 10.8% of 37 = 4

Bag Q: Number of blue marbles: 66.7% of X = ?

Bag R: Number of blue marbles: 50.0% of 32 = 16

=> \(\frac{1}{3}\) * (37 + X + 32) = 4 + ? + 16

=> \(\frac{1}{3}\) * (69 + X) = 20 + ?

=> 23 + \(\frac{X}{3}\) = 20 + 66.7% of X

=> 3 = 0.667X - 0.333X

=> 3 = 0.334X

=> X = \(\frac{3}{0.334}\)

=> X = 8.9 = 9

Answer B
_________________

Answer: Option B

Video solution by GMATinsight



Get TOPICWISE: Concept Videos | Practice Qns 100+ | Official Qns 50+ | 100% Video solution CLICK.
Two MUST join YouTube channels : GMATinsight (1000+ FREE Videos) and GMATclub
_________________

In the exam, if we are short on time we can quickly eliminate options A, D, and E using the below technique and guess from amongst B or C

Since the number of blue balls will be an integer
1/3 (37+32+X) = integer
1/3 (69+X) = integer
69 is divisible by 3 hence from amongst the answer options the only possible value of X (divisible by 3) is either (B) 9 or (C) 12.