(1/11)+(1/12)+……+(1/19)+(1/20) is including in which of the following

Question

(1/11)+(1/12)+……+(1/19)+(1/20) is including in which of
the following ranges?

A. 0 ~ (1/5) 
B. (1/5) ~ (1/4) 
C. (1/4)~(1/3) 
D. (1/3)~(1/2) 
E. (1/2)~1

chetan2u · Accepted Answer

Hi..
You don't have to start Adding..
 Just find the ATLEAST value..
Here the SUM has to be GREATER than 10 times lowest value= 10*1/20=1/2
So sum >1/2 
Only E fits in..
E

pushpitkc · Answer

Expression that we need to find the range is  {(\frac{1}{11})+(\frac{1}{12})+……+(\frac{1}{19})+(\frac{1}{20})}

The minimum range of that expression is  \frac{1}{20}+\frac{1}{20} ... total 10 times
Therefore the value of that minimum expression is greater than  \frac{10}{20}  or  \frac{1}{2}

Similarly the range for maximum value is  \frac{10}{11} = 0.91 (approximately)

Hence the range for the expression must be  \frac{1}{2}  < expression < 1 (Option E)

kunalsinghNS · Answer

Hi,

How are you deriving the minimum range as 1/20+1/20... ?

pushpitkc · Answer

We have a total of 10 elements in the expression. 
The minimum range can be got by adding the smallest number
10 times.

Since 1/20 is the smallest number in the expression,
adding the number to the itself 10 times will give you the 
smallest number possible(1/20 = 0.05)
When added 10 times to itself, 0.05*10 = 0.5 = 1/2!

Hope that helps!

kunalsinghNS · Answer

Thank you for your response !

yogeshmisthi · Answer

This explained method is not bad ..we can extract result from by multiply 10/20 for higher range and 10/11 for lower range...but it is a tricky solution

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MathRevolution · Answer

=> (1/20) < (1/11), (1/20) < (1/12), …, (1/20) < (1/19).
(1/20) + (1/20) + … + (1/20) < (1/11)+(1/12)+……+(1/19)+(1/20)
(1/20)*10 < (1/11)+(1/12)+……+(1/19)+(1/20)
(1/2) < (1/11)+(1/12)+……+(1/19)+(1/20)

(1/11) < (1/10), (1/12) < (1/10), … , (1/20) < (1/10)
(1/11)+(1/12)+……+(1/19)+(1/20) < (1/10) + (1/10) + … + (1/10) = 1

Thus, 1/2 < (1/11)+(1/12)+……+(1/19)+(1/20) < 1

Ans: E

shashankism · Answer

(1/20)+(1/20)+......+(1/20)+(1/20) < (1/11)+(1/12)+……+(1/19)+(1/20) < (1/10)+(1/10)+......+(1/10)+(1/10) -> 10/20 < (1/11)+(1/12)+……+(1/19)+(1/20) < 10/10 -> 1/2 <(1/11)+(1/12)+……+(1/19)+(1/20) < 1 Answer E

Mumthahina · Answer

Can anybody tell me why am I adding 1/20 by itself to get the minimum value? don't get the point anyhow

Bunuel · Answer

The idea is to find the minimum possible sum. If you assume every term in the sequence is 1/20 (the smallest value), the sum of ten such terms is 1/2. Since the actual terms are greater than 1/20, the true sum will be more than 1/2. Using 1/20 ten times is a shortcut to quickly find this minimum limit.

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(1/11)+(1/12)+……+(1/19)+(1/20) is including in which of the following

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(1/11)+(1/12)+……+(1/19)+(1/20) is including in which of the following

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