A man covers a certain distance on scooter. Had he moved 3 kmph faster, he would have taken 40 min. less. If he had moved 2 kmph slower, he would have taken 40 min. more. The distance (in km) is:
@Faruq said:A man covers a certain distance on scooter. Had he moved 3 kmph faster, he would have taken 40 min. less. If he had moved 2 kmph slower, he would have taken 40 min. more. The distance (in km) is:
12/7
@hippophobia said:(1034)n = n^3 + 3n + 4 this part is not clear....i dont think it is correct as you cannot write 1034 in this form
Why Nt.. (1034)n = 1*n^3 + 0*n^2 + 3*n^1 + 4*n^0= n^3 + 3n^2 + 4..
I dnt thnk yeh galat hai..
@pyashraj said:Why Nt.. (1034)n = 1*n^3 + 0*n^2 + 3*n^1 + 4*n^0= n^3 + 3n^2 + 4..I dnt thnk yeh galat hai..
nope it is fine i was talking bout approach....i got ur approach now....was making silly logical mistake..thnks
@Faruq said:A man covers a certain distance on scooter. Had he moved 3 kmph faster, he would have taken 40 min. less. If he had moved 2 kmph slower, he would have taken 40 min. more. The distance (in km) is:
40 km...
d/s - d/(s+3) = 40/60 -(1)
d/(s-2) - d/s = 40/60 -(2)
on solving s= 12kmph and d = 40km..
Find the sum of 7+77+777+...777...(99 times)?
a)(7/9)[10(10^99-1)-891]
b)(7/81)[100(100^99-99)-891]
c)(7/81)[10(10^99-1)-891]
d)(7/81)[100(10^99-1)-2673]
e)(7/81)[10(100^99-1)-2673]
a)(7/9)[10(10^99-1)-891]
b)(7/81)[100(100^99-99)-891]
c)(7/81)[10(10^99-1)-891]
d)(7/81)[100(10^99-1)-2673]
e)(7/81)[10(100^99-1)-2673]
Q-2:Find the solution:
@iLoveTorres said:@saurabhlumarrai is the answer right? i dont think its right.
actaully jahan se maine question liya hai-usme answer none of these diya hai to none of these can be 5/2 also...so mujhe laga ki sahi hai....anyways approach would be appreciated
@saurabhlumarrai said:Find the sum of 7+77+777+...777...(99 times)?a)(7/9)[10(10^99-1)-891]b)(7/81)[100(100^99-99)-891]c)(7/81)[10(10^99-1)-891]d)(7/81)[100(10^99-1)-2673]e)(7/81)[10(100^99-1)-2673]
Option C ???
An express train travelling at 80kmph overtakes a goods traain,twice as long and going at 40kmph on a parallel track,in 54 sec.How long will the express train take to cross a station 400m long?
a)36sec b) 45s c)27s d)none
@Todd said:Bhai wat are the options for above question ???
a)[(2^1/2)-1]^2
a)2*[(2^1/2)-1]^2
a)[[(2^1/2)-1]^2]/2
d)None of these
a)2*[(2^1/2)-1]^2
a)[[(2^1/2)-1]^2]/2
d)None of these
@saurabhlumarrai said:bhai kitni baar ek hi cheez post karogey....jara approach bhi samjha do..
Approach (striking off the options)
lets consider for
7 + 77 + 777 + .... 77(9 times)
7(1+ 11+ 111 + .... 1111 (9 times))
7( 123456789)
Now if you add one more number that is (7777 (10 times)) The sum will become
7(1234567890)
For entire series then the sum will be like
7(1234567890.... (10 times))
Multiply and divide 9 to the above number
7/9 * (11111111010 (10 times)
Again multiply and divide 9 then,
7/81 * (99999999090 (10 times))
Looking at the option I choose C as the close to answer don't know weather it's right or wrong.
Let me know if you have a better approach. I am also looking into a better and faster approach.
@IIM-A2013 said:An express train travelling at 80kmph overtakes a goods traain,twice as long and going at 40kmph on a parallel track,in 54 sec.How long will the express train take to cross a station 400m long?a)36sec b) 45s c)27s d)none
ans c)27?