Quantitative Aptitude - Arithmetic Ability Questions

Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days.
Number of odd days = 30/7 = 2 odd days

Then no. of odd days = 2 => tuesday and wednesday(from monday).

Hence on 3rd january it is wednesday.

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

Given X can do in 10 days 

=> 1 day work of X = 1/10

Y can do in 15 days 

=> 1 day work of Y = 1/15

1day work of (X + Y) = 1/10 + 1/15 = 1/6

Given they are hired for 5 days

 => 5 days work of (X + Y) = 5 x 1/6 = 5/6

 Therefore, Unfinished work = 1 - 5/6 = 1/6 

Let the number be n. Then,

50% of n - 35% of n = 24

50/100 n - 35/100 n = 24

n = (24 x 100)/15 = 160.

P ( neither A nor B) = PA and B  

=  PA∩B=  PA∪B= 1-PA∪B  

= 1-61100=39100

work done=total number of person x number of days;
half of work done = 140 x 66;
For half of remaining work 25 extra men are added.
Therefore, total men for half work remaining = 140 + 25 = 165;
Let 2nd half work will be completed in K days;
140 x 66 = 165 x K
K = 122;
Hence, extra days => 122-120 = 2days.