1-(1/2-2/3)= : GMAT Problem Solving (PS)
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# 1-(1/2-2/3)=

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Math Expert
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24 Sep 2012, 03:59
Expert's post
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$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

Practice Questions
Question: 48
Page: 158
Difficulty: 550
[Reveal] Spoiler: OA

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24 Sep 2012, 03:59
SOLUTION

$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

$$1-(\frac{1}{2}-\frac{2}{3})=1-(\frac{3}{6}-\frac{4}{6})=1-(-\frac{1}{6})=1+\frac{1}{6}=\frac{7}{6}$$.

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24 Sep 2012, 04:11
1
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1-(-1/6) = 7/6

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24 Sep 2012, 07:36
1
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Here we should solve the expression within the bracket

1/2 - 2/3 = (3 - 4)/6 (by taking LCM of denominator) =-1/7

So, the overall expression becomes: 1 - (-1/7) = 1 + 1/6= 7/6

Hence B
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24 Sep 2012, 09:16
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The answer is B but is a 300-500 level question not 600, in my opinion.

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24 Sep 2012, 09:27
2
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Bunuel wrote:
$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

Practice Questions
Question: 48
Page: 158
Difficulty: 600

$$1-(\frac{1}{2}-\frac{2}{3})=1-\frac{1}{2}+\frac{2}{3}=\frac{1}{2}+\frac{2}{3}$$
The result must be greater than $$1$$, so we have to choose between A and B.
Since we have two fractions with $$2$$ and $$3$$ in the denominator, the final result should be an improper fraction with $$6$$ in the denominator.

NB: Even without a calculator on the test, I don't think it would be a problem for anybody. I would say a below 500 level question.
I am just trying to find any type of shortcut for such questions, taking advantage of the multiple choices...really boring to simply add/subtract fractions...
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24 Sep 2012, 09:47
1
KUDOS
1-(-1/6)
= 1+1/6
=7/6
Ans (B)
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28 Sep 2012, 03:36
SOLUTION

$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

$$1-(\frac{1}{2}-\frac{2}{3})=1-(\frac{3}{6}-\frac{4}{6})=1-(-\frac{1}{6})=1+\frac{1}{6}=\frac{7}{6}$$.

Kudos points given to everyone with correct solution. Let me know if I missed someone.
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22 Oct 2013, 13:59
EvaJager wrote:
Bunuel wrote:
$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

Practice Questions
Question: 48
Page: 158
Difficulty: 600

$$1-(\frac{1}{2}-\frac{2}{3})=1-\frac{1}{2}+\frac{2}{3}=\frac{1}{2}+\frac{2}{3}$$
The result must be greater than $$1$$, so we have to choose between A and B.
Since we have two fractions with $$2$$ and $$3$$ in the denominator, the final result should be an improper fraction with $$6$$ in the denominator.

NB: Even without a calculator on the test, I don't think it would be a problem for anybody. I would say a below 500 level question.
I am just trying to find any type of shortcut for such questions, taking advantage of the multiple choices...really boring to simply add/subtract fractions...

Eva shortcut is to quickly get rid of 1/2 from 1 which leaves half and 1/2 and 2/3 mentions that the sum is above 1. Now we just multiply denominators look fro 6 in denominator and bingo. only one option.
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10 Sep 2014, 08:20
Bunuel wrote:
$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

Practice Questions
Question: 48
Page: 158
Difficulty: 550

(1/2-2/3) = -1/6

1-(-1/6) =7/6
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10 Sep 2014, 09:19
1 -(1/2 -2/3)= 1 -(3-4)/6 = 1 + 1/6 =7/6
ans = B
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10 Sep 2014, 19:02
Bunuel wrote:
$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

$$1-(\frac{1}{2}-\frac{2}{3})= 1 + \frac{2}{3} - \frac{1}{2} = \frac{5}{3} - \frac{1}{2} = \frac{7}{6}$$

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30 May 2016, 04:18
Bunuel wrote:
$$1-(\frac{1}{2}-\frac{2}{3})=$$

(A) 6/5
(B) 7/6
(C) 6/7
(D) 5/6
(E) 0

We are given the expression 1 – (1/2 – 2/3), and we need to determine the value. We start by getting a common denominator of 6 for the fractions inside the parentheses. This gives us:

1 – (3/6 – 4/6)

1 – (–1/6) = 1 + 1/6 = 1 1/6 = 7/6

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Re: 1-(1/2-2/3)=   [#permalink] 30 May 2016, 04:18
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