If 1/3 of the total number of marbles in the three bags listed in the

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 .

Video solution from Quant Reasoning:

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

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.

Hi Bunuel - is 10.8 % (Bag P) the percentage that has been ALREADY ESTIMATED to the nearest tenth ?

I thought 10.8 % was NOT estimated and thus went with 11 %

It doesnt matter in a problem like this but I was wondering if you could figure out if the 10.8 % (Bag P) was already estimated to the nearest tenth or not

Thank you !

Since a bit more than 10% of the 37 marbles in P are blue, the number of blue marbles in P = a bit more than 3.7 = 4.
Since 1/2 of the 32 marbles in R are blue, the number of blue marbles in R = 16.

We can PLUG IN THE ANSWERS, which represent the total number of marbles in Q.

Rule:
MULTIPLE OF X + MULTIPLE OF X = MULTIPLE OF X
MULTIPLE OF X + NONMULTIPLE OF X = NONMULTIPLE OF X

Since 1/3 of all the marbles are blue, the total number of marbles must be a MULTIPLE OF 3.
Number of marbles in P and R combined = 37+32 = 69
Since 69 is a multiple of 3, the rule above indicates that the number of marbles in Q must also be a multiple of 3.
Eliminate A, D and E.

The chart indicates that 2/3 of the marbles in Q are blue.
When the correct answer is plugged in, \(\frac{total-blue}{total-marbles} = \frac{1}{3}\).

B: 9, implying 6 blue marbles in Q
\(\frac{total-blue}{total-marbles} = \frac{4+16+6}{37+32+9} = \frac{26}{78} = \frac{1}{3}\)
Success!

This is actually very simple. As with the majority of high level GMAT questions, you're thrown A LOT of data and are asked to process it into a form that answers the question at hand. First thing's first: remind yourself that every GMAT problem is designed to be solved within 2 minutes - you can do this!

We are asked to find the amount of blue marbles in bag Q. To make things complicated we're given the number of marbles in each bag and the percentage of each bag that is blue. GMAT makes this a party by throwing a bunch of nasty non integer percents at us to shake us up. But remember marbles are like people...you can only have a whole number of each!

Therefore when you see that 10.8% of 37 marbles in Bag P are blue...don't bother calculating. 10% of 37 is 3.7 so that 0.8% MUST take us up to 4. Bag P has 4 blue marbles. Bag R is easier...50% of 32 marbles are blue..that's 16 blue marbles. Easy! Thus far we have 20 blue marbles in total.

Now for the fun part. Bag Q has x amount of marbles...66.7% of which are blue. Time to throw in the towel? Not at all this is a piece of cake. 66.7% is the same as \(\frac{2}{3}\).

Now what do we do with this? Well we have that 20 blue marbles plus \(\frac{2}{3}\) of the marbles in Bag Q is \(\frac{1}{3}\) of the total number of marbles. Let's set this up as a formula to what we've described above and wrap this party up.

To restate what we have and what we're being asked to find: "The 20 blue marbles in bags R and P plus \(\frac{2}{3}\) of all of the marbles in Bag Q are equal to \(\frac{1}{3}\) of the 37 marbles in Bag P plus the 32 marbles in Bag R plus the x amount of marbles in Bag Q." Time to transcribe this into numbers and solve this problem!

20 + \(\frac{2}{3}\)x = \(\frac{1}{3}\)(69+x) .This looks tricky, doesn't it? It's actually easier than it looks. Multiply both sides by three to get rid of that disgusting denominator of 3.
60+2x=69+x
x=9

Now take a moment to look back at what you were thinking when you first saw this problem as opposed to now. It's conceptually extremely simple but the key to answering this question is knowing how to efficiently use all of the data that is being presented to us.

Correct option : B : 9

Say Bag Q has X marbles total. Then Total number of marbles in all bags = 37 + X + 32

Bag P has 37 marbles of which 10.8% are Blue. 10% of 37 is 3.7 so Blue marbles in the bag must be 4.
4/37 would be about 10.8%.

Bag R has 32 marbles, of which 50% are Blue. So it has 16 blue marbles.

Total Blue marbles = (1/3)*Total Marbles
4 + (2/3)X + 16 = (1/3)*(37 + X + 32)
X = 9

Answer (B)

GMATNinja, Bunuel, what may be the fastest way to multiply 10.8% by 37? (Looking for the most efficient approach to multiply by decimals in the exam, and avoid careless mistakes.

The better question: would it be more efficient to avoid multiplying decimals entirely? Frankly, I almost never do it on the GMAT, because there's almost always a better way -- either estimation or fractions, usually.

In this case? Well, the number of marbles has to be an integer. And 10% of 37 is 3.7, so if we round up a bit, there have to be 4 blue marbles in bag P. And that's it.

Think of this as a feature of the GMAT: the test very very rarely forces you into long, awkward, "brute force" calculations. Instead, the test rewards flexible thinking, and gives you opportunities to avoid messy things like multiplying or dividing decimals. More on that in this video.

I hope that helps a bit!

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