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GRE Prep Club Legend  Joined: 07 Jun 2014
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GRE Math Challenge #47- a= 13/15 + 7/8 + 3/4 [#permalink]
Expert's post 00:00

Question Stats: 85% (01:03) correct 14% (00:00) wrong based on 7 sessions
a= $$\frac{13}{15} + \frac{7}{8} + \frac{3}{4}$$

Quantity A: $$\frac{a}{3}$$

Quantity B: 1
• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
[Reveal] Spoiler: OA

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Sandy
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Re: GRE Math Challenge #47- a= 13/15 + 7/8 + 3/4 [#permalink]
sandy wrote:
a= $$\frac{13}{15} + \frac{7}{8} + \frac{3}{4}$$

Quantity A: $$\frac{a}{3}$$

Quantity B: 1
• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined

Let's find the LCM of [15,8, 4] which is 120.

thus we got $$\frac{104}{120} + \frac{105}{120} + \frac{90}{120}$$

= $$\frac{299}{120} For Quant A [m]\frac{a}{3}$$= $$\frac{290}{360}$$ which is samller than 1 .

Thus, Ans B
GRE Instructor Joined: 10 Apr 2015
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Re: GRE Math Challenge #47- a= 13/15 + 7/8 + 3/4 [#permalink]
Expert's post
sandy wrote:
a= $$\frac{13}{15} + \frac{7}{8} + \frac{3}{4}$$

Quantity A: $$\frac{a}{3}$$

Quantity B: 1

There's no need to find the value of a. Instead, we can approximate

Instead, recognize the following:
13/15 is less than 1
7/8 is less than 1
3/4 is less than 1

So, 13/15 + 7/8 + 3/4 = some number that's less than 3.
In other words, a = some number that's less than 3.

We get:
QUANTITY A: (some number that's less than 3)/3
QUANTITY B: 1

Simplify to get:
QUANTITY A: Some number that's less than 1
QUANTITY B: 1

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com Sign up for our free GRE Question of the Day emails Re: GRE Math Challenge #47- a= 13/15 + 7/8 + 3/4   [#permalink] 20 Jan 2019, 09:17
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