Good qn...

well take the last and firat,2nd last and 2nd 1st....
=(1/11+1/20)+(1/12+1/19)+...+(1/15+1/16)
now see the symmetry 20+11=12+19=..=15+16=31
so,
=31(1/11.20+1/12.19+1/13.18+...+1/15.16)
now again see a good pattern..an increasing AP with decreasing cd like:
11.20=220
12.19=228
13.18=234
14.17=238
15.16=240
...Please note no need to calculate these multiplication...i have just shown here for symmetry..
so where we were
31(1/11.20+1/12.19+1/13.18+...+1/15.16)
now little bit approx for quick calculation
31*5/230( as there are total 5 terms and 1st deno is 220 and last deno is 240,so for nearly approx we can take 230)
so 31*5/230 will give u near approximation...so now go for option...post if there is any doubt.

Sum = Average * # items

Average of a 10 number set is the average of digits 5 and digit 6.


Average = (1/15 + 1/16)/2 = (32/240)/2 = (16/240)

Sum = (16/240) * 10 = 16/24 = 2/3

Ashish8 wrote:Sum = Average * # items

Average of a 10 number set is the average of digits 5 and digit 6.


Average = (1/15 + 1/16)/2 = (32/240)/2 = (16/240)

Sum = (16/240) * 10 = 16/24 = 2/3

You can only apply this rule if you have a set of consecutive numbers, which is not the case with the fractions in this question. Raising the denominator by 1 does not create an equidistant (i.e. a constant distance between each pair of terms) set.

For example, let's simplify the question to:

what's the sum of 1/2, 1/3 and 1/4.

Well, using your approach, the average of the set is the middle term, 1/3 and the sum is 3*(1/3) = 1.

However, 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12.

If the question had been:

what's the sum of 1/12 + 2/12 + 3/12 + 4/12 +....+ 10/12, then we could have applied the formula, since each of those terms is 1/12 more than the previous term (i.e. it's an equidistant set).

ahh yes you're correct. My apologies.

kuntalkkc wrote: 11.20=220
12.19=228
13.18=234
14.17=238
15.16=240
...Please note no need to calculate these multiplication...i have just shown here for symmetry..

now little bit approx for quick calculation
1st deno is 220 and last deno is 240,so for nearly approx we can take 230).

i think if difference in answer choices not much (which is in this type of question) then the appoximation of number will make a huge difference in answer......
so is that calcuation really unnecessary ....
and if we do start caluculating i think 2-3 minutes may not be sufficient....(ok multipication number seems to be easier one in this case,but is not always the case....)

In my opinion,

reduction of fractions to a common denominator

(12*13*14*15*16*17*18*19*20 +
11*13*14*15*16*17*18*19*20 +
11*12*14*15*16*17*18*19*20 +
... 11*12*13*14*15*16*17*18*19)
/ 11*12*13*14*15*16*17*18*19*20

then, sum of the denominator is

(12*13*14*15*16*17*18*19*20 + 12*13*14*15*16*17*18*19*20) / 2 * 10
= 12*13*14*15*16*17*18*19 * 5 * (11+20)

then, equation is

12*13*14*15*16*17*18*19 * 5 * (11+20)
------------------------------------------------------------
11*12*13*14*15*16*17*18*19*20


5 * (11+20)
--------------------
11*20


5*31
-----------
220


so, my answer is 31/44

^0^...

Here's another approach if you are good with decimals:
1/11: <.1
1/12: .083 (half of 1/6 which is half of 1/3 which is .33)
1/13: .077
1/14: .072 (half 1/7 which is .14)
1/15: .068
1/16: 0.625 (half of 1/8 which is half of 1/4 which is .25)
1/17: .058
1/18: .055 (half of 1/9 which is 1/3 of 1/3 which is .33) So divide 1/3 (.33) by 3 to get .11, divide by 2 to get .055
1/19: .0525 (know that 1/20 is .05, so 1/19 should be between .05 and .055. So lets say .0525)
1:20: .050

Then add and see what you get. This method should be used if the answers choices are somewhat apart
Total: 0.6 approx which is 6/10 or 3/5

Again, you gotta be quick and dirty with some tough fractions. But like I got 1/19 (by guessing between 1/20 and 1/18), you should do that for 1/13,1/15,1/17 as well.