Abhishek start s to paint a f ence on o ne day On the seco nd day two mo re

sorry guys for the mistakes in calculation in both solutions given by me...
here is the correct solution
for abhishek it takes (1+(1+2)+(1+2+3)+...+(1+2+3+...20)) days
let n=20 then
that is sigma n(n+1)/2=sigma n^2/2+sigma n/2
=[n(n+1)(2n+1)/6]/2+[n(n+1)/2]/2
=n(n+1)/4*[((2n+1)/3)+1]
=n(n+1)/4*[(2n+4)/3]
=n(n+1)(n+2)/6
=20(21)(22)/6 [since n=20]
=10*7*22 days
for 1 boy it takes 70*22 days...
for 1 girl it takes 70*22/2 days
for 10 girls it takes 70*11/10=77 days...
12 years agoHelpfull: Yes(12) No(3)

pls ....pls send me tcs new question paper tcs will come within 2 days....
id shweta.vardhan@gmail.com
12 years agoHelpfull: Yes(3) No(3)
for abhishek it takes (1+(1+2)+(1+2+3)+...+(1+2+3+...20)) days
let n=20 then
that is sigma n(n+1)/2=sigma n^2/2+sigma n/2
=[n(n+1)(2n+1)/6]/2+[n(n+1)/2]/2
=n(n+1)/4*[((2n+1)/3)+1]
=n(n+1)/4*[(2n+4)/3]
=n(n+1)(n+2)/6
=20(21)(22)/6 [since n=20]
=10*7*22 days
for 1 boy it takes 70*22 days...
for 1 girl it takes 70*22/2 days
for 10 girls it takes 70*11/10=77 days..
12 years agoHelpfull: Yes(3) No(2)
@tanya check my final solution.... it's correct....
n(n+1)(n+2)/6 is the sum...
there is some calculation mistakes in my previous ones....
i dont know why you said no to that solution....
PLEASE CHECK IT BEFORE DECIDING....
12 years agoHelpfull: Yes(2) No(2)
Do u mean number of ways or number of days ?
12 years agoHelpfull: Yes(1) No(1)
for abhishek it takes (1+(1+2)+(1+2+3)+...+(1+2+3+...20)) days
let n=20 then
that is sigma n(n+1)/2=sigma n^2+sigma n/2
=[n(n+1)(2n+1)/6]+[n(n+1)/2]/2
=n(n+1)/2*[((2n+1)/3)+(1/2)]
=n(n+1)/2*[(4n+5)/6]
=n(n+1)(4n+5)/12
=20(21)(85)/12 [since n=20]
=5*7*85 days
for 1 boy it takes 35*85 days...
for 1 girl it takes 35*85/2 days
for 10 girls it takes 35*85/20=148.75 days...
12 years agoHelpfull: Yes(1) No(3)
for abhishek it takes (1+(1+2)+(1+2+3)+...+(1+2+3+...20)) days
let n=20 then
that is sigma n(n+1)/2=sigma n^2/2+sigma n/2
=[n(n+1)(2n+1)/6]/2+[n(n+1)/2]/2
=n(n+1)/4*[((2n+1)/3)+1]
=n(n+1)/4*[(2n+4)/6]
=n(n+1)(n+2)/12
=20(21)(22)/12 [since n=20]
=5*7*22 days
for 1 boy it takes 35*22 days...
for 1 girl it takes 35*22/2 days
for 10 girls it takes 35*11/10=38.5 days...
12 years agoHelpfull: Yes(1) No(4)
In this problem which is on time and work it is days... there is no permutations and combinations problem my frnds...
12 years agoHelpfull: Yes(1) No(2)
@ Sainath,

Pls check.

As per ur solutiuon

for abhishek it takes (1+(1+2)+(1+2+3)+...+(1+2+3+...20)) days
let n=20 then
that is sigma n(n+1)/2=sigma n^2+sigma n/2
The series will be 1+3+6+10 +..
If n=4,
1+3+6+10 =20
which u will not get when u put n=4 in
n(n+1)(4n+5)/12 where we get 4*5*21/12 = 35
12 years agoHelpfull: Yes(1) No(2)
(1+3+6+10+15+21+........+171+190+210)=1540 days if abhishek works alone to finish the painting.....and for girls it would be 1540/(10*2)=77 days..
12 years agoHelpfull: Yes(1) No(2)
WAT DOES IT MEAN BY SO ON IS THE NUMBER INCREASING LIKE 1 3 6 10 N ??
12 years agoHelpfull: Yes(0) No(2)
number of days...the question states that 1st day boy was working...2nd day 3, 3rd day 6...but they are asking the question in terms of time...girls work twice as faster as boys...so i am confused!
12 years agoHelpfull: Yes(0) No(1)
i was solving these questions on the site...and this one stumped me...if the question is wrong please upload the correct one...or if it has a solution...please tell me how do we get the answer
12 years agoHelpfull: Yes(0) No(0)
SEE I HAVE GOT SO FAR 1+3+6 AND SO ON IF U ADD TILL 20TH DAY U WILL GET I CALCULATED THAT TOTAL MAN DAYS REQUIRED FOR THIS JOB IS 1553 AS GALS EFFICIENCY IS DOUBLE SO BY 2 IS 776 + 1 GAL POWER IS REQ SO NOW WE HAVE TO ARRANGE THE 10 GALS DOING IT COMPLETELY
12 years agoHelpfull: Yes(0) No(1)
if no days are asked then ans is 200.....
12 years agoHelpfull: Yes(0) No(3)

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