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09 Jul 2019, 00:41
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
72% (02:32) correct
28%
(02:45)
wrong
based on 480
sessions
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[GMAT math practice question]
What is \(\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56}?\)
\(A. \frac{3}{5}\)
\(B. \frac{5}{8}\)
\(C. \frac{7}{8}\)
\(D. \frac{8}{9}\)
\(E. \frac{9}{10}\)
What is \(\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56}?\)
\(A. \frac{3}{5}\)
\(B. \frac{5}{8}\)
\(C. \frac{7}{8}\)
\(D. \frac{8}{9}\)
\(E. \frac{9}{10}\)
09 Jul 2019, 01:28
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This is a trick based question.
We know that: \(\frac{1}{(a*b)}\) = \(\frac{1}{(b-a)}\) * \([\frac{1}{(a)} - \frac{1}{(b)}]\)
As per the question:
we can split every denominator in the above form:
\(\frac{1}{(2)}\) = \(\frac{1}{(2-1)}\) * \(((1) - \frac{1}{(2)})\)
As, 6 = 2*3 , 12 = 3*4 , 20 = 4*5, 30 = 5*6, 42 = 6*7, 56 = 7*8
=> as every two numbers have difference as 1, so we can use that for all.
We can have:
1/(1) * \([1 - \frac{1}{2} + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + .... + (\frac{1}{7} - \frac{1}{8}) ]\)
=> as we can see some of the terms will get cancel out.
=> 1 - 1/8
=> 7/8
The answer is 7/8.
Please hit kudos if you like the solution.
We know that: \(\frac{1}{(a*b)}\) = \(\frac{1}{(b-a)}\) * \([\frac{1}{(a)} - \frac{1}{(b)}]\)
As per the question:
we can split every denominator in the above form:
\(\frac{1}{(2)}\) = \(\frac{1}{(2-1)}\) * \(((1) - \frac{1}{(2)})\)
As, 6 = 2*3 , 12 = 3*4 , 20 = 4*5, 30 = 5*6, 42 = 6*7, 56 = 7*8
=> as every two numbers have difference as 1, so we can use that for all.
We can have:
1/(1) * \([1 - \frac{1}{2} + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + .... + (\frac{1}{7} - \frac{1}{8}) ]\)
=> as we can see some of the terms will get cancel out.
=> 1 - 1/8
=> 7/8
The answer is 7/8.
Please hit kudos if you like the solution.
09 Jul 2019, 00:54
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MathRevolution
\(\frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{9}{12}\)
\(\frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}\)
\(\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} = \frac{9}{12} + \frac{1}{12} = \frac{10}{12}\)
\(\frac{1}{42} + \frac{1}{56} = \frac{1}{7} * [ \frac{1}{6} + \frac{1}{8} ] = \frac{1}{7} * \frac{7}{24} = \frac{1}{24}\)
\(\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} = \frac{10}{12} + \frac{1}{24} = \frac{21}{24} = \frac{7}{8}\)
IMO C
