Quantitative Aptitude - Arithmetic Ability Questions

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x. 

S.P. of x articles = Rs. 20. 

Profit = Rs. (20 - x). 

20-xx*100=25 

=> 2000 - 100x = 25x 

=> 125x=2000 

=> x=16

Let cp= 100,
35 % increase in sp=135
10 % discount in 135((135*10)/100)=13.5
so 1st sp=(135-13.5)=121.5, again 15 % discount in 1st sp((121.5*15)/100)=18.225
2nd sp=(121.5-18.225)=103.275,

so finally cp=100,sp=103.275 ,gain by 3.27%

55min spaces are covered in 60min

60 min. spaces are covered in (60/55 x 60) min= 65+5/11 min.

loss in 64min=(65+5/11)- 64 =16/11

Loss in 24 hrs (16/11 x 1/64 x 24 x 60) min= 32 8/11.

clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.

Let the length of the rope be r mts 

then, πr2=9856 ⇒ r2=9856×722=3136 ⇒r =56m

Here n(S) = (6 x 6) = 36

Let E = event of getting a total more than 7
        = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.