Quantitative Aptitude - Arithmetic Ability Questions

55 min. spaces are covered in 60 min.
60 min. spaces are covered in ((60/55) x 60) min = 720/11 min.
Loss in 65 min. = (720/11 - 65 ) = 5/11
Loss in 24 hrs = (5/11 x 1/65 x 24 x 60)min = 1440/143 min.

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= 8640 ways.

Given

      125% ---- 3400

=> 100% ---- ?

=> ? = 3400x100/125 = 2720

=> Cost price of the article = Rs. 2720

Profit when article sold at Rs. 3265 = 3265 - 2720 = 545

Hence, Profit% = Gain x 100/cost price

=> P% = 545 x 100/2720

=> P% = 20%

S.I=PNR/100

Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4

Since order does not matter it is a combination. 

The word AND means multiply. 

Given 4 basketball, 3 volleyball, 2 soccer. 

We want 2 basketball games and 1 other event. There are 5 choices left. 

C(n,r) 

C(How many do you have, How many do you want) 

C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1) 

C(4,2) x C(5,1) = (6)(5) = 30

Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.

276+35961+50 = 10111011 = 1

Cash realised= Rs. (106.25 - 0.25)
                      = Rs. 106.