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GRE Math Challenge #47- a= 13/15 + 7/8 + 3/4
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Updated on: 18 Feb 2020, 15:45
Updated on: 18 Feb 2020, 15:45
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a= \(\frac{13}{15} + \frac{7}{8} + \frac{3}{4}\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
\(\frac{a}{3}\) |
\(1\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: GRE Math Challenge #47- a= 13/15 + 7/8 + 3/4
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11 Mar 2018, 06:55
11 Mar 2018, 06:55
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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
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
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Re: GRE Math Challenge #47- a= 13/15 + 7/8 + 3/4
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20 Jan 2019, 09:17
20 Jan 2019, 09:17
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sandy wrote:
a= \(\frac{13}{15} + \frac{7}{8} + \frac{3}{4}\)
Quantity A: \(\frac{a}{3}\)
Quantity B: 1
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
Answer: B
Cheers,
Brent
