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(1/33)+(1/34)+……+(1/63)+(1/64) is including in which of

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by Max@Math Revolution » Sun Jan 24, 2016 3:48 pm
(1/33)+(1/34)+......+(1/63)+(1/64) is including in which of the following ranges?


A. (1/6)~(1/5)
B. (1/5)~91/4)
C. (1/4)~(1/3)
D. (1/3)~(1/2)
E. (1/2)~1


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by Testtrainer » Tue Jan 26, 2016 10:52 pm
32 terms, 31 of which are a bit less than 1/33. So adding them together yields a bit less than 32/33, which is close to 1. So (E).

BTW: 31 out of 32 terms are a bit more than 1/64. So adding them together yields a bit more than 32/64, which is 1/2.
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by GMATGuruNY » Wed Jan 27, 2016 3:19 am
Max@Math Revolution wrote:(1/33)+(1/34)+......+(1/63)+(1/64) is including in which of the following ranges?


A. (1/6)~(1/5)
B. (1/5)~91/4)
C. (1/4)~(1/3)
D. (1/3)~(1/2)
E. (1/2)~1
Number of consecutive integers = biggest - smallest + 1.
Here, the denominators consist of the integers between 33 and 64, inclusive.
Thus:
Total number of terms = 64 - 33 + 1 = 32.

If all of the 32 values were 1/64, the result would be the following sum:
32(1/64) = 1/2.
Since all but one of the values are actually GREATER THAN 1/64, the sum must be GREATER THAN 1/2.

The correct answer is E.
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by Max@Math Revolution » Wed Jan 27, 2016 5:37 pm
(1/33)+(1/34)+......+(1/63)+(1/64) is including in which of the following ranges?


A. (1/6)~(1/5)
B. (1/5)~91/4)
C. (1/4)~(1/3)
D. (1/3)~(1/2)
E. (1/2)~1


==> The sum of consecutive reciprocal numbers is determined by the first and the last terms. Also, the number of franctions is 64-33+1=32. Then, if S=(1/33)+(1/34)+......+(1/63)+(1/64), it becomes 32/64<S<32/33, and 1/2=32/64<S<32/33<1 -> 1/2<S<1. Therefore, the answer is E.
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