@zuloo correct ans is 30 sec :(:(
@Logrhythm said:bhai i guess the answer here wld be 7002100*2/3*1/2 + 2100*4/5*2/3 + 2100*6/7*4/5 - 2((2100*2/3*4/5*1/2 - 2100*6/7*2/3*4/5*1/2) + (2100*4/5*6/7*2/3 - 2100*6/7*2/3*4/5*1/2)) - 3*2100*6/7*2/3*4/5*1/2 = 700but why do questions like these? this, i am sure won;t feature in any of the MBA entrance exams, IIT mein bhi nahi aayega
ya you are right.. but i found this question here itself on PG quant thread. 
@gupanki2 said:@zuloo correct ans is 30 sec
could you re-write the question. there is some confusion.
@gupanki2 said:In a race A beats by 10 sec, B beats C by 20 seconds. By how many seconds did A beats c?
time taken by A be t
=> time taken by B = t + 10
so time taken by C = (t + 10) + 20 = t + 30
so A beats C by 30 seconds
@zuloo said:could you re-write the question. there is some confusion.
yes, you are right it is 30 secs. was a big calculation mistake from my side.
@gupanki2 said:In a race A beats by 10 sec, B beats C by 20 seconds. By how many seconds did A beats c?
A cover D in t sec
B cover D in (t + 10) sec
C cover D in (t + 10) + 20 = t + 30 sec
so A beats C by 30 secs
What is the sum of all digits of first 1000 positive integers?
@techgeek2050 said:What is the sum of all digits of first 1000 positive integers?
1*300+2*300+9*300+1
13501??
@techgeek2050 said:What is the sum of all digits of first 1000 positive integers?
13501?
45*10*10(units digit) + 45*10*10(tens digit) + 45*100(hundred digit) + 1 = 13501
@techgeek2050 said:bhai i don't have the oa. plz share ur approach.
1 appears 20 times in every hundred
and 100 times extra in 100-199
so every no appears 100+20*10=300 times in first 999 numbers
1*300+2*300+9*300+1
13501
13501
what is the maximum no of different ways in which 8 identical balls can be placed in boxes 1,2,3,4 such that each box contain atleast one ball
@IIM-A2013 said:what is the maximum no of different ways in which 8 identical balls can be placed in boxes 1,2,3,4 such that each box contain atleast one ball
35?
@IIM-A2013 said:what is the maximum no of different ways in which 8 identical balls can be placed in boxes 1,2,3,4 such that each box contain atleast one ball
165
11C3
11C3
@IIM-A2013 said:@iLoveTorres@hedonistajay i was also doing like that but answer is 35 waiting for @ilovetorres ans
@hedonistajay said:cannot we apply n-1Cr-1 approach here
a + b + c + d = 8
n-1 C r -1 = 7C3 = 35