Quantitative Aptitude - Arithmetic Ability Questions

Out of 26 alphabets two distinct letters can be chosen in 26P2 ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. 

Combined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.

The amount of petrol left after 4 operations  

= 200 × 1-402004  

= 200 × 454  

= 200 × 256625  

= 81.92 litres

Hence the amount of kerosene = 200 - 81.92 = 118. 08 litres

0.127 expressed as a fraction = 127/1000

1/8 can also be expressed as (1 x 125) / (8 x 125) = 125/1000

The difference is 2/1000, which is 1/500

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= 24-1 ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=23-1........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= 25..............(C)

From (A), (B) and (C), required number of ways
=  2524-123-1

The work to be done = Capacity of reservoir  = 30 litres.

 1^st Minute ---> inlet pipe opened ---> 5 lit filled

 2^nd minute ---> inlet pipe closed; outlet pipe opened ---> 4 lit emptied

In 2 minutes (5 litres - 4 litres = 1lit) is filled into the reservoir.

It takes 2 minutes to fill 1lit ---> it takes 50 minutes to fill 25 litres into the reservoir.  

In the 51^st minute inlet pipe is opened and the reservoir is filled.

Let the total capacity of tank be 90 liters.
Capacity of tank filled in 1 minute by K = 3 liters.
Capacity of tank filled in 1 minute by L = 15 liters.
Therefore, capacity of the tank filled by both K and L in 1 minute = 18 liters.
Hence, time taken by both the pipes to overflow the tank = 90/18 = 5 minutes.

Let us consider Karthik as K1 and Kalyan as K2.

Given k18+k1 and K2 - (K1 + 5) = 3 

K2 - K1 = 8

K2 = 8 + K1

Now, 

k18+k1= 45

=> K1 = 32 years

Therefore K2 = 40 years.

K1 + K2 = 72 years.

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.