Quantitative Aptitude - Arithmetic Ability Questions

Speed of lorry = 36012 = 30 kmph

Speed of van =250100 x 30 = 75 kmph

Speed of bike = 35 x 75 = 45 kmph

Therefore, now required average speed of bike and van  = 75 + 452= 60 kmph.

Let required fraction be  then 

           ............. (i)

and          ............ (ii)

Solving (i) and (ii) we get x=3, y=5

 Required fraction = 

It won't requires that much solving.

Here given the increase in weekly collection by Rs. 1,05,000

That is this increase denotes for all 7 days.

=> Gross Collection increase per day = Rs. 1,05,0007 = Rs. 15,000

Hence, the gross collection increase per day = Rs. 15,000.

According to the given data,

P + Q = 41 ......(1)

R - 1 = P + 2

R = P + 3

and

P + 4 = Q - 1

=> Q = P + 5 ....(2)

From (1)&(2)

P = 18

Q = 18 + 5 = 23

R = 18 + 3 = 21

=> P/S = 3/4

=> S = 4/3 x 18 = 24

Required Difference = S - R = 24 - 21 = 3.

Given exp . is like this 854*854*854-276*276*276854*854+854*276+276*276=?

so ans = 854 - 276 = 578

The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14. 

Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get 212. Since the case of no candy being selected is not an option, we have to subtract 1 from our answer.

Therefore,there are 212- 1 = 4095 ways of selecting one or more candies