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M70-10

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Math Expert
Joined: 02 Sep 2009
Posts: 58453

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03 Sep 2018, 03:27
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Difficulty:

35% (medium)

Question Stats:

67% (01:13) correct 33% (02:00) wrong based on 6 sessions

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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. $$\frac{7}{50}$$

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Math Expert
Joined: 02 Sep 2009
Posts: 58453

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03 Sep 2018, 03:28
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, $$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}=\frac{35+21+15}{105}=\frac{71}{105}$$.

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Re M70-10   [#permalink] 03 Sep 2018, 03:28
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