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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:
If 1/3 of the total number of marbles in the three bags listed in the
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21 Aug 2016, 00:17
7
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.
Re: If 1/3 of the total number of marbles in the three bags listed in the
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21 Aug 2016, 05:38
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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
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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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
\(4+16+0.67x= \frac{1}{3} (37+32+x)\)
\(20+0.67x =\frac{1}{3}x+23\)
\(20-23 = \frac{1}{3}x-\frac{67}{100}\)
\(-3 = \frac{33}{300}x\)
\(-3 = \frac{11}{100}x\)
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%
Pls 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
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
\(4+16+0.67x= \frac{1}{3} (37+32+x)\)
\(20+0.67x =\frac{1}{3}x+23\)
\(20-23 = \frac{1}{3}x-\frac{67}{100}\)
\(-3 = \frac{33}{300}x\)
\(-3 = \frac{11}{100}x\)
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%
Pls 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
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
\(4+16+0.67x= \frac{1}{3} (37+32+x)\)
\(20+0.67x = \frac{1}{3}x+23\)
\(20-23 = \frac{1}{3}x-\frac{67}{100}\)
\(-3 = \frac{33}{300}x\)
\(-3 = \frac{11}{100}x\)
the above solution is incorrect, i kept 1/3 as fraction, so i wonder what did i do wrong
dave13 The equation is correct. The arithmetic is off.
(EDIT As niks18 points out), check your math here:
Quote:
\(20-23 = \frac{1}{3}x-\frac{67}{100}\)
\(-3 = \frac{33}{300}x\)
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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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
\(4+16+0.67x= \frac{1}{3} (37+32+x)\)
\(20+0.67x =\frac{1}{3}x+23\)
\(20-23 = \frac{1}{3}x-\frac{67}{100}\)
\(-3 = \frac{33}{300}x\)
\(-3 = \frac{11}{100}x\)
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%
Pls 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
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 dave13 yes 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 1/3x-67x/100. Rest is ok
If 1/3 of the total number of marbles in the three bags listed in the
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20 Sep 2018, 20:24
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 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
Re: If 1/3 of the total number of marbles in the three bags listed in the
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05 Jul 2019, 01:31
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.
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
gmatclubot
Re: If 1/3 of the total number of marbles in the three bags listed in the
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25 Sep 2019, 21:03
close
67% (02:43) correct 
my mistake I got confused here. So in your method 1 the mistake is in highlighted part above. kindly check your calculation of 1/3x-67x/100. Rest is ok
