Reciprocals Question

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Reciprocals Question

by jfranco23 » Wed Jan 28, 2009 7:10 am
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I Only
C. III Only
D. II and III only
E. I, II and III
Source: — Problem Solving |

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by DanaJ » Wed Jan 28, 2009 7:36 am
We are talking about 10 numbers from 91 to 100. What you can immediately notice is that, since 100 is the greatest of them, that means that 1/100 will be the smallest. So you have:
1/91 > 1/100
1/92 > 1/100
......
1/100 >= 1/100
Adding these babies (=equations for quant addicts like myself) up will give you that S > 10*1/10 = 1/10. So S is greater than 1/10. However, we don't konw the relationship between S and 1/8 and 1/9.
So here's part two: since 91 is the smallest number of the series 91 - 100, this means that 1/91 will be the biggest of them all, compared to 1/92 ... 1/100. So you get that:

1/91 >= 1/91
1/91 > 1/92
.....
1/91 > 1/100
Adding it all up leads us to the conclusion that 10*1/91 > S, or that 10/91 > S. Now all we have to do is compare 10/91 with 1/9. Since 1/9 = 10/90 and 91 < 90, this means that 10/91 < 1/9. But since 10/91 > S, this will definately mean that 1/9 > S. And again, since 1/8 > 1/9, 1/8 is obviously greater than S.

Conclusion: only 1/10 is smaller than S.

Answer: C.

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by jfranco23 » Wed Jan 28, 2009 7:53 am
Thanks Dana, but i don´t understand your method. Can you explain it again?

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by DanaJ » Wed Jan 28, 2009 8:07 am
Sure... My method relies on writing the equations described above and just adding them together... The first set of equations is smth like :
1/91 > 1/100
1/92 > 1/100
1/93 > 1/100
........ [meaning so on snd so forth]
1/99 > 1/100
1/100 >= 1/100

If you add all the numbers on the left side of the ">" you get S, if you add everything on the right side you get ten times 1/100 or 10*1/100.

The same goes for the other set of equations, just the other way around: on the left side you get 1/91 ten times or 10*1/91 and on the right side you get S.

Hope this makes it a bit more clear. IMO, just try to write the whole thing on paper and you'll be able to tell how I did it...

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by krisraam » Wed Jan 28, 2009 8:21 am
Answer is C

1/91, 1/92,......1/100

1/100 <1/99 <1/98 ..... <1/91

The sum will be greater than 10( total numbers) times the lowest value(1/100) and less than 10 times higher value(1/91).

10(1/100) < S < 10(1/91)


so 1/10 is the less than S

Thanks
Raama