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If 1/3 of the total number of marbles in the three bags listed in the
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17 Jun 2016, 04:19
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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 Attachment:
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Re: If 1/3 of the total number of marbles in the three bags listed in the
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19 Dec 2017, 06:42
Bunuel wrote: 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 Attachment: 20160617_1618.png 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
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Re: If 1/3 of the total number of marbles in the three bags listed in the
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17 Jun 2016, 12:56
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.




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Re: If 1/3 of the total number of marbles in the three bags listed in the
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20 Aug 2016, 08:37
FacelessMan wrote: 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. I have approximated 10.8% as 1/9, which is 11.1%



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If 1/3 of the total number of marbles in the three bags listed in the
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20 Aug 2016, 23:17
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.



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Re: If 1/3 of the total number of marbles in the three bags listed in the
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21 Aug 2016, 04:38
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)
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Re: If 1/3 of the total number of marbles in the three bags listed in the
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13 Jun 2017, 03:45
manlog wrote: I have approximated 10.8% as 1/9, which is 11.1% 10.8% of 37 > 10% of 37 = 3.7. 3.7 is a little low and the number of balls must be an integer; can round up to 4.



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Re: If 1/3 of the total number of marbles in the three bags listed in the
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22 Apr 2018, 09:44
Bunuel wrote: 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 Attachment: 20160617_1618.png No. of blue marbles in each bag is Bag P: 10.8% of 37 = 4 Bag Q:66.7% of X = 2/3 X Bag R: 50% of 32 = 16 Total Blue marbles = 4 + 2/3 X + 16 = 20 + 2/3 X Also Total marbles count = 37 + X + 32 = 69 + X So, 1/3 (69+ X ) = 20 + 2/3 X 23 + 1/3 X = 20 + 2/3 X 1/3 X = 3 X = 9 Answer B



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If 1/3 of the total number of marbles in the three bags listed in the
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Updated on: 29 Apr 2018, 02:17
many thanks generis and niks18 yeah working with fractions is easier , also good to know that \(\frac{2}{3}\)= 66% \(4+16+\frac{2}{3}x= \frac{1}{3} (37+32+x)\) \(20+\frac{2}{3}x = \frac{1}{3}x+23\) \(\frac{2}{3}x\frac{1}{3}x = 3\) \(\frac{1}{3}x =3\) \(x =9\) have a great weekend
Originally posted by dave13 on 28 Apr 2018, 05:41.
Last edited by dave13 on 29 Apr 2018, 02:17, edited 1 time in total.



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Re: If 1/3 of the total number of marbles in the three bags listed in the
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28 Apr 2018, 06:20
dave13 wrote: Bunuel wrote: 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 Attachment: 20160617_1618.png generis, niks18, pushpitkc, hello here is my unique solution two solutions one is incorrect and next one seems more accurate unfortunately i cant solve 700 level q under 2 min, takes me more time befire i figure out how the problem can be approached \(4+16+0.67x= \frac{1}{3} ( 37+32+x)\) \(20+0.67x = \frac{1}{3}x+23\)\(2023 = \frac{1}{3}x\frac{67}{100}\) \(3 = \frac{33}{300}x\) \(3 = \frac{11}{100}x\) the above solution is incorrect, i kept 1/3 as fraction, so i wonder what did i do wrong in the next solution I converted 1/3 to decimal = 0.33% \(4+16+0.67x= 0.33 (37+32+x)\)\(20+0.67x = 0.33x+23\)\(0.67x  0.33x = 2320\) \(3 = 0.34x\) \(x = 8.8\) round to the tenth \(8.8\) and get \(9\) have a great weekend hi dave13Pls check the calculations in the highlighted part above. How come 37+32 become 23? and when you are opening the bracket then each element of the bracket has to be multiplied by 1/3 or 0.33



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If 1/3 of the total number of marbles in the three bags listed in the
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28 Apr 2018, 06:30
niks18 wrote: dave13 wrote: Bunuel wrote: 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 Attachment: 20160617_1618.png generis, niks18, pushpitkc, hello here is my unique solution two solutions one is incorrect and next one seems more accurate unfortunately i cant solve 700 level q under 2 min, takes me more time befire i figure out how the problem can be approached \(4+16+0.67x= \frac{1}{3} ( 37+32+x)\) \(20+0.67x = \frac{1}{3}x+23\)\(2023 = \frac{1}{3}x\frac{67}{100}\) \(3 = \frac{33}{300}x\) \(3 = \frac{11}{100}x\) the above solution is incorrect, i kept 1/3 as fraction, so i wonder what did i do wrong in the next solution I converted 1/3 to decimal = 0.33% \(4+16+0.67x= 0.33 (37+32+x)\)\(20+0.67x = 0.33x+23\)\(0.67x  0.33x = 2320\) \(3 = 0.34x\) \(x = 8.8\) round to the tenth \(8.8\) and get \(9\) have a great weekend hi dave13Pls check the calculations in the highlighted part above. How come 37+32 become 23? and when you are opening the bracket then each element of the bracket has to be multiplied by 1/3 or 0.33 niks18, hello there to answer your question  How come 37+32 become 23?\(37+32 =69\) so\(\frac{1}{3}*69 =23\) so I get \(23+0,33x\) \(20+0.67x = 0.33x+23\) isnt it correct ?



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If 1/3 of the total number of marbles in the three bags listed in the
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28 Apr 2018, 06:34
dave13 wrote: Bunuel wrote: 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 Attachment: 20160617_1618.png generis, niks18, pushpitkc, hello here is my unique solution two solutions one is incorrect and next one seems more accurate unfortunately i cant solve 700 level q under 2 min, takes me more time befire i figure out how the problem can be approached \(4+16+0.67x= \frac{1}{3} (37+32+x)\) \(20+0.67x = \frac{1}{3}x+23\) \(2023 = \frac{1}{3}x\frac{67}{100}\) \(3 = \frac{33}{300}x\) \(3 = \frac{11}{100}x\) the above solution is incorrect, i kept 1/3 as fraction, so i wonder what did i do wrong dave13The equation is correct. The arithmetic is off. ( EDIT As niks18 points out), check your math here: Quote: \(2023 = \frac{1}{3}x\frac{67}{100}\)
\(3 = \frac{33}{300}x\) Try using \(\frac{67}{100}=\frac{2}{3}\) \(\frac{1}{3}x  \frac{2}{3}x\) = ???? Calculate again and see what you get. Tip: use fractions with fractions and decimals with decimals. IMO fractions are easier. Another way to think about setup and solving of the equation . . . (I do not know whether either suggestion will be easier for you  they're just possibilities) After you replace \(.67\) with \(\frac{2}{3}\) Change the setup and cross multiply. (1) Setup: If blue marbles equal \(\frac{1}{3}\) of three bags' total marbles: \(\frac{Blue}{Total}=\frac{1}{3}\) \(\frac{16+4+\frac{2}{3}x}{37+32+x}=\frac{1}{3}\) (2) Solve. Cross multiply. Just one arithmetic mistake, in other words. Your second solution is correct. If you use decimals (IMO, harder here!) what you calculated is correct. Hope that helps.
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Re: If 1/3 of the total number of marbles in the three bags listed in the
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28 Apr 2018, 06:37
dave13 wrote: niks18 wrote: dave13 wrote: generis, niks18, pushpitkc, hello here is my unique solution two solutions one is incorrect and next one seems more accurate unfortunately i cant solve 700 level q under 2 min, takes me more time befire i figure out how the problem can be approached \(4+16+0.67x= \frac{1}{3} ( 37+32+x)\) \(20+0.67x = \frac{1}{3}x+23\)\(2023 = \frac{1}{3}x\frac{67}{100}\)
\(3 = \frac{33}{300}x\)
\(3 = \frac{11}{100}x\)the above solution is incorrect, i kept 1/3 as fraction, so i wonder what did i do wrong in the next solution I converted 1/3 to decimal = 0.33% \(4+16+0.67x= 0.33 (37+32+x)\)\(20+0.67x = 0.33x+23\)\(0.67x  0.33x = 2320\) \(3 = 0.34x\) \(x = 8.8\) round to the tenth \(8.8\) and get \(9\) have a great weekend hi dave13Pls check the calculations in the highlighted part above. How come 37+32 become 23? and when you are opening the bracket then each element of the bracket has to be multiplied by 1/3 or 0.33 niks18, hello there to answer your question  How come 37+32 become 23?\(37+32 =69\) so\(\frac{1}{3}*69 =23\) so I get \(23+0,33x\) \(20+0.67x = 0.33x+23\) isnt it correct ? hi dave13yes its correct. my mistake I got confused here. So in your method 1 the mistake is in highlighted part above. kindly check your calculation of 1/3x67x/100. Rest is ok



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If 1/3 of the total number of marbles in the three bags listed in the
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20 Sep 2018, 19:24
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 marblesBag 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 .



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Re: If 1/3 of the total number of marbles in the three bags listed in the
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05 Jul 2019, 00:31
Just want to clarify, should one always round up to the nearest integer if the variable is an actual object?
My error in solving this problem was assuming it was 3 marbles in Bag P, instead of 4. Due to the fact that, I didnt see it possible for there to be .994 of a marble, just like how in other problems 1.5 people is equivalent to just 1 person, as you cant have half a person.
Thanks!



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Re: If 1/3 of the total number of marbles in the three bags listed in the
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25 Sep 2019, 20:03
There are 2 ways to do this question quickly.
1. is detailed by several other ppl. 10.8% of 37 = 4 66.7% of x = 2x/3 50% of 31 = 16
4 + 16 + 2x/3 = 1/3(32 + 37 + x) 20 + 2x/3 = 1/3(69 +x) => x = 9
2. Alternatively, use the answer choices and information given. Since Bag Q has 66.7% (2/3) blue marbles, and you can't really have fractions of marbles, the answer must be a multiple of 3: either (B) 9 or (C) 12.
Plug in the answers to the question. 9: 10.8%(37) + 66.7%(9) + 50%(32) = 26 AND 37 + 9 + 32 = 78 => WORKS
12: 10.8%(37) + 66.7%(12) + 50%(32) = 28 AND 37 + 12 + 32 = 81 => DOESN'T WORK



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Re: If 1/3 of the total number of marbles in the three bags listed in the
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23 Apr 2020, 21:52
Bunuel wrote: 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 Attachment: 20160617_1618.png If x=9, the total blue marble is 4 (Bag P; 10.8% of 37)+6 (Bag Q; 66.7% of 9)+16 (Bag R; 50% of 32)= 26Total marbles in three bags=37+9+32=78. So, \(\frac{1}{3}\) of 78= 26Here, 66.7% is \(\frac{2}{3}\). A) 5>\(\frac{2}{3}\) of 5 is the broken marble  we need integer..So, out B) 9>correct C) 12 D) 23>\(\frac{2}{3}\) of 23 is the broken marble  we need integer..So, out E) 46>>\(\frac{2}{3}\) of 46 is the broken marble  we need integer..So, out Hope it helps..
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Re: If 1/3 of the total number of marbles in the three bags listed in the
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01 May 2020, 09:57
Bunuel wrote: 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 Attachment: 20160617_1618.png Total Marbles: 32+37+x, 69+x Q Blue: 2/3*x P,R Blue: 4+16, 20 Total Blue: 1/3*(69+x)=20+2/3*x …69+x=60+2x, x=9 Ans (B)




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