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# The sum 7/8+1/9 is between

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Math Expert
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The sum 7/8+1/9 is between [#permalink]  24 Sep 2012, 03:53
1
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Expert's post
00:00

Difficulty:

5% (low)

Question Stats:

89% (01:42) correct 11% (00:44) wrong based on 467 sessions
The sum $$\frac{7}{8}+\frac{1}{9}$$ is between

(A) $$\frac{1}{2}$$ and $$\frac{3}{4}$$

(B) $$\frac{3}{4}$$ and 1

(C) 1 and $$1\frac{1}{4}$$

(D) $$1\frac{1}{4}$$ and $$1\frac{1}{2}$$

(E) $$1\frac{1}{2}$$ and 2

Practice Questions
Question: 46
Page: 158
Difficulty: 550
[Reveal] Spoiler: OA

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Math Expert
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Kudos [?]: 42184 [3] , given: 5957

Re: The sum 7/8+1/9 is between [#permalink]  24 Sep 2012, 03:53
3
KUDOS
Expert's post
SOLUTION:

The sum $$\frac{7}{8}+\frac{1}{9}$$ is between

(A) $$\frac{1}{2}$$ and $$\frac{3}{4}$$

(B) $$\frac{3}{4}$$ and 1

(C) 1 and $$1\frac{1}{4}$$

(D) $$1\frac{1}{4}$$ and $$1\frac{1}{2}$$

(E) $$1\frac{1}{2}$$ and 2
________________________________

$$\frac{7}{8}+\frac{1}{9}=\frac{63}{72}+\frac{8}{72}=\frac{71}{72}$$.

Now, $$\frac{71}{72}$$ is less but very close to 1, so the answer is B.

Else, one can notice that since $$\frac{1}{9}$$ is less than $$\frac{1}{8}$$, thus $$\frac{7}{8}+\frac{1}{9}$$ is less than 1. Therefore we can eliminate answer choices C, D, and E. Next, since $$\frac{7}{8}>\frac{3}{4}=\frac{6}{8}$$, then answer choice A is also out. So, only answer choice B remains.

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Re: The sum 7/8+1/9 is between [#permalink]  24 Sep 2012, 04:24
1
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7/8 + 1/9 ---> .864+ .11--->.97
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Re: The sum 7/8+1/9 is between [#permalink]  24 Sep 2012, 04:52
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7/8 + 1/9 = 71/72, which is surely between 3/4 to 1.

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Re: The sum 7/8+1/9 is between [#permalink]  24 Sep 2012, 07:54
1
KUDOS
Bunuel wrote:

The sum $$\frac{7}{8}+\frac{1}{9}$$ is between

(A) $$\frac{1}{2}$$ and $$\frac{3}{4}$$

(B) $$\frac{3}{4}$$ and 1

(C) 1 and $$1\frac{1}{4}$$

(D) $$1\frac{1}{4}$$ and $$1\frac{1}{2}$$

(E) $$1\frac{1}{2}$$ and 2

taking LCM of denominator, the expression becomes
7/8 + 1/9 = (63+8)/72 = 71/72
This is slightly lesser than 1, which satisfies option B

Hence B
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Re: The sum 7/8+1/9 is between [#permalink]  24 Sep 2012, 09:45
1
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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

The sum $$\frac{7}{8}+\frac{1}{9}$$ is between

(A) $$\frac{1}{2}$$ and $$\frac{3}{4}$$

(B) $$\frac{3}{4}$$ and 1

(C) 1 and $$1\frac{1}{4}$$

(D) $$1\frac{1}{4}$$ and $$1\frac{1}{2}$$

(E) $$1\frac{1}{2}$$ and 2

Practice Questions
Question: 46
Page: 158
Difficulty: 600

GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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Thank you!

$$\frac{7}{8}+\frac{1}{9}<\frac{7}{8}+\frac{1}{8}=1$$ and $$\frac{7}{8}+\frac{1}{9}>\frac{6}{8}=\frac{3}{4}$$.

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Re: The sum 7/8+1/9 is between [#permalink]  24 Sep 2012, 10:44
1
KUDOS
Expert's post
LCM ----> 71/72 we can notice that 72/72 is 1 so 71/72 must be a bit less than 1 but is impossible to be under 3/4 (0.75). Must be between 3/4 and 1. So B
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Math Expert
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Kudos [?]: 42184 [0], given: 5957

Re: The sum 7/8+1/9 is between [#permalink]  28 Sep 2012, 03:25
Expert's post
SOLUTION:

The sum $$\frac{7}{8}+\frac{1}{9}$$ is between

(A) $$\frac{1}{2}$$ and $$\frac{3}{4}$$

(B) $$\frac{3}{4}$$ and 1

(C) 1 and $$1\frac{1}{4}$$

(D) $$1\frac{1}{4}$$ and $$1\frac{1}{2}$$

(E) $$1\frac{1}{2}$$ and 2
________________________________

$$\frac{7}{8}+\frac{1}{9}=\frac{63}{72}+\frac{8}{72}=\frac{71}{72}$$.

Now, $$\frac{71}{72}$$ is less but very close to 1, so the answer is B.

Else, one can notice that since $$\frac{1}{9}$$ is less than $$\frac{1}{8}$$, thus $$\frac{7}{8}+\frac{1}{9}$$ is less than 1. Therefore we can eliminate answer choices C, D, and E. Next, since $$\frac{7}{8}>\frac{3}{4}=\frac{6}{8}$$, then answer choice A is also out. So, only answer choice B remains.

Kudos points given to everyone with correct solution. Let me know if I missed someone.
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Re: The sum 7/8+1/9 is between [#permalink]  27 Dec 2013, 08:00
Bunuel wrote:
The sum $$\frac{7}{8}+\frac{1}{9}$$ is between

(A) $$\frac{1}{2}$$ and $$\frac{3}{4}$$

(B) $$\frac{3}{4}$$ and 1

(C) 1 and $$1\frac{1}{4}$$

(D) $$1\frac{1}{4}$$ and $$1\frac{1}{2}$$

(E) $$1\frac{1}{2}$$ and 2

Practice Questions
Question: 46
Page: 158
Difficulty: 550

If you know how to efficiently convert fractions to decimals, this question is pretty straightforward:

1/8 = 0.125 so 7/8 = 0.875... 1/9 = 0.111111...

So 0.875 + 0.111 = 0.986.. Only B works.
Manager
Joined: 07 Apr 2014
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Kudos [?]: 10 [0], given: 81

Re: The sum 7/8+1/9 is between [#permalink]  10 Sep 2014, 08:10
Bunuel wrote:
The sum $$\frac{7}{8}+\frac{1}{9}$$ is between

(A) $$\frac{1}{2}$$ and $$\frac{3}{4}$$

(B) $$\frac{3}{4}$$ and 1

(C) 1 and $$1\frac{1}{4}$$

(D) $$1\frac{1}{4}$$ and $$1\frac{1}{2}$$

(E) $$1\frac{1}{2}$$ and 2

Practice Questions
Question: 46
Page: 158
Difficulty: 550

71/72

approx== 0.9

c,d,e,- greater than 1 .
a- less than .75
hence B
Re: The sum 7/8+1/9 is between   [#permalink] 10 Sep 2014, 08:10
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