Quantitative Aptitude - Arithmetic Ability Questions

As the ratio of their present ages is 3 : 4 , let their present ages be 3X and 4X.

So, 5 years ago, as the ratio of their ages was 5 : 7, we can write  (3x-5) : (4x-5) = 5 : 7. Solving, we get X = 10. Hence, their present ages are 3X = 30 and 4X = 40

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.

 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 distance = x km.
Time taken at 3 kmph : dist/speed = x/3 = 20 min late.
time taken at 4 kmph : x/4 = 30 min earlier
difference between time taken : 30-(-20) = 50 mins = 50/60 hours.
x/3- x/4 = 50/60
x/12 = 5/6
x = 10 km.

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 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.