Quantitative Aptitude - Arithmetic Ability Questions

The year 2006 is an ordinary year. So, it has 1 odd day.

So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.

But, 8th Dec, 2007 is Saturday

S0, 8th Dec, 2006 is Friday.

Given year is divided by 4, and the quotient gives the number of leap years.

Here, 100%4 = 25.

But, as 100 is not a leap year => 25 - 1= 24 leap years.

Let , S -  sample space        E - event of selecting 1 girl and 2 boys. 

Then, n(S) = Number ways of selecting 3 students out of 25 

                = 25C3 

                = 2300.

n(E) = 10C1×15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.

1. Ans. A.
98 = 17 + 81
26 = 98 – 72
89 = 26 + 63
35 = 89 – 54
? = 35 + 45, i.e. ? = 80

 gain in 2 years =Rs.5000*254*2100-5000*4*2100  

= Rs. (625 - 400) 

= Rs. 225 

 gain in 1 year =Rs.2252=112.50

Let length of plot = L meters, then breadth = L - 20 meters

and perimeter = 2[L + L - 20] = [4L - 40] meters

[4L - 40]  * 26.50 = 5300

[4L - 40] = 5300 / 26.50 = 200

4L = 240

L = 240/4= 60 meters.