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:

2016-06-17_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

\(4+16+0.67x= \frac{1}{3} (37+32+x)\)

\(20+0.67x = \frac{1}{3}x+23\)

\(20-23 = \frac{1}{3}x-\frac{67}{100}\)

\(-3 = \frac{33}{300}x\)

\(-3 = \frac{11}{100}x\)

the above solution is incorrect, i kept 1/3 as fraction, so i wonder what did i do wrong

dave13The 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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