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GRE Math Challenge #99- (3/5) + (2/3) or 1

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GRE Math Challenge #99- (3/5) + (2/3) or 1 [#permalink] New post 09 May 2015, 11:17
Expert's post
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Question Stats:

94% (00:19) correct 5% (00:13) wrong based on 17 sessions
Quantity A: (3/5) + (2/3)
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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Re: GRE Math Challenge #99- (3/5) + (2/3) or 1 [#permalink] New post 27 Feb 2018, 08:16
Expert's post
sandy wrote:
Quantity A: (3/5) + (2/3)
Quantity B: 1


One option is to replace the fractions with decimal equivalents

GIVEN:
Quantity A: (3/5) + (2/3)
Quantity B: 1

3/5 = 0.6 and 2/3 ≈ 0.66

So, we get:
Quantity A: 0.6 + 0.66
Quantity B: 1

Evaluate:
Quantity A: 1.26
Quantity B: 1

Quantity A is greater

Answer: A

Cheers,
Brent
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Re: GRE Math Challenge #99- (3/5) + (2/3) or 1 [#permalink] New post 27 Feb 2018, 08:19
Expert's post
sandy wrote:
Quantity A: (3/5) + (2/3)
Quantity B: 1

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


Another approach is to ADD THE FRACTIONS

3/5 + 2/3 = 9/15 + 10/15 [I rewrote the fractions with common denominators]
= 19/15
= 1 4/15

So, we have:
Quantity A: 1 4/15
Quantity B: 1

Quantity A is greater

Answer: A

Cheers,
Brent
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Re: GRE Math Challenge #99- (3/5) + (2/3) or 1 [#permalink] New post 28 Feb 2018, 02:34
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Also common multiple approach could be used to solve this problem

Multiplying both sides by 15 as 15 = 3* 5

Quantity A = 3/5 * 15 + 2/3 * 15 = 9 + 10 = 19

Quantity B = 1* 15 = 15

Option A
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Re: GRE Math Challenge #99- (3/5) + (2/3) or 1   [#permalink] 28 Feb 2018, 02:34
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