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If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1

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If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1  [#permalink]

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New post 12 May 2017, 00:10
5
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

78% (01:34) correct 22% (01:09) wrong based on 50 sessions

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If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1/30). What is the value of a?

(A) 0.15 < a < 0.25
(B) 0.25 < a < 0.35
(C) 0.35 < a < 0.45
(D) 0.45 < a < 0.55
(E) 0.55 < a < 0.65

\(\displaystyle\sum_{k=21}^{30} \frac{1}{k}>\frac{5}{25}+\frac{5}{30}=\frac{1}{5}+\frac{1}{6}=\frac{11}{30}>0.35\)

\(\displaystyle\sum_{k=21}^{30} \frac{1}{k}<\frac{5}{20}+\frac{5}{25}=\frac{1}{4}+\frac{1}{5}=\frac{9}{20}=0.45\)

The answer is C

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Re: If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1  [#permalink]

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New post 12 May 2017, 00:54
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2
C

As 1/30 is the smallest in the series.
So [(1/21) + (1/22)...... + (1/30)] > [(1/30) + (1/30)........+ (1/30)... 10 times]
a > 10/30
a > 0.33

Similarly as 1/21 is the greatest in the series
So a [(1/21) + (1/22)...... + (1/30)] < [(1/21) + (1/21)........+ (1/21)... 10 times]
a < 10/21
a < 0.47

This leads to 0.33 < a < 0.47
So as C lies within this limit, C is the answer

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Re: If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1  [#permalink]

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New post 12 May 2017, 01:01
ziyuen wrote:
If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1/30). What is the value of a?

(A) 0.15 < a < 0.25
(B) 0.25 < a < 0.35
(C) 0.35 < a < 0.45
(D) 0.45 < a < 0.55
(E) 0.55 < a < 0.65

a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1/30)

Now Lets start checking lowest and highest value of terms..by taking values which can be easily simplified.
So a>10/30 -> a>0.33
a<10/20 -> a<0.5

By looking at the option we can say that 0.35< a <0.45 lies in this range..
So Correct answer is C

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Re: If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1  [#permalink]

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New post 12 May 2017, 01:07
anugrahs wrote:
C

As 1/30 is the smallest in the series.
So [(1/21) + (1/22)...... + (1/30)] > [(1/30) + (1/30)........+ (1/30)... 10 times]
a > 10/30
a > 0.33

Similarly as 1/21 is the greatest in the series
So a [(1/21) + (1/22)...... + (1/30)] < [(1/21) + (1/21)........+ (1/21)... 10 times]
a < 10/21
a < 0.47

This leads to 0.33 < a < 0.47
So as C lies within this limit, C is the answer

Regards
Anugrah


If we want to go in more details of this solution...
We can use the method as suggested by nguyendinhtuong..

i.e. a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1/30)
Now Lets start checking lowest and highest value of terms..by taking values which can be easily simplified.
So a>5/30 + 5/25 -> a>1/6 + 0.2 -> a >0.166 +0.2 > 0.3666
a<5/20 +5/25 -> a< .25 +.2 -> a<0.45

By looking at the option we can say that 0.35< a <0.45 lies in this range..
So Correct answer is C

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Re: If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1  [#permalink]

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New post 12 May 2017, 01:11
anugrahs wrote:
C

As 1/30 is the smallest in the series.
So [(1/21) + (1/22)...... + (1/30)] > [(1/30) + (1/30)........+ (1/30)... 10 times]
a > 10/30
a > 0.33

Similarly as 1/21 is the greatest in the series
So a [(1/21) + (1/22)...... + (1/30)] < [(1/21) + (1/21)........+ (1/21)... 10 times]
a < 10/21
a < 0.47

This leads to 0.33 < a < 0.47
So as C lies within this limit, C is the answer

Regards
Anugrah


Don't take a<10/21
Either take a<10/20 --> a<0.5
or
a<5/20+5/25 --> a<0.25+0.2 --> a<0.45
This will make the process easy and you will save sometime..
I also learned to solve this type of question here only..
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Re: If a=(1/21)+(1/22)+(1/23)+(1/24)+(1/25)+(1/26)+(1/27)+(1/28)+(1/29)+(1 &nbs [#permalink] 12 May 2017, 01:11
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