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03 Sep 2018, 03:27
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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:
25%
(medium)
Question Stats:
82% (01:35) correct
18%
(02:00)
wrong
based on 17
sessions
History
Date
Time
Result
Not Attempted Yet
A is the sum of the reciprocals of each of the integers from 3 to 7. B is the sum of the reciprocals of the even integers from 3 to 7. What is A – B?
A. \(\frac{10}{24}\)
B. \(\frac{61}{105}\)
C. \(\frac{71}{105}\)
D. \(\frac{2}{35}\)
E. \(\frac{7}{50}\)
A. \(\frac{10}{24}\)
B. \(\frac{61}{105}\)
C. \(\frac{71}{105}\)
D. \(\frac{2}{35}\)
E. \(\frac{7}{50}\)
03 Sep 2018, 03:28
Kudos
Bookmarks
Official Solution:
A is the sum of the reciprocals of each of the integers from 3 to 7. B is the sum of the reciprocals of the even integers from 3 to 7. What is A – B?
A. \(\frac{10}{24}\)
B. \(\frac{61}{105}\)
C. \(\frac{71}{105}\)
D. \(\frac{2}{35}\)
E. \(\frac{7}{50}\)
We’ll go for PRECISE because all the information we need is in the question.
A is the sum of the reciprocals of each of the integers from 3 to 7, and B is the sum of the reciprocals of the even integers only. Written mathematically:
\(A = \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\), and \(B = \frac{1}{4}+\frac{1}{6}\)
Therefore, \(A - B= \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}- (\frac{1}{4}+\frac{1}{6})=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\)
We’ll find 3, 5 and 7’s lowest common denominator: 3 x 5 x 7 = 105.
Therefore, \(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}=\frac{35+21+15}{105}=\frac{71}{105}\).
Answer: C
A is the sum of the reciprocals of each of the integers from 3 to 7. B is the sum of the reciprocals of the even integers from 3 to 7. What is A – B?
A. \(\frac{10}{24}\)
B. \(\frac{61}{105}\)
C. \(\frac{71}{105}\)
D. \(\frac{2}{35}\)
E. \(\frac{7}{50}\)
We’ll go for PRECISE because all the information we need is in the question.
A is the sum of the reciprocals of each of the integers from 3 to 7, and B is the sum of the reciprocals of the even integers only. Written mathematically:
\(A = \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\), and \(B = \frac{1}{4}+\frac{1}{6}\)
Therefore, \(A - B= \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}- (\frac{1}{4}+\frac{1}{6})=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\)
We’ll find 3, 5 and 7’s lowest common denominator: 3 x 5 x 7 = 105.
Therefore, \(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}=\frac{35+21+15}{105}=\frac{71}{105}\).
Answer: C
Moderator:
