Quantitative Aptitude - Arithmetic Ability Questions

The average age of a couple = 26 years.

Total age of couples = 26 x 2 = 52 years

 Total age of couple after 11 years = (52 + 2 * 11 ) = 74 years

Suppose average age of 3 children after 11 years is 3x years.

The average age of children is 7 years.

Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.

The first train covers 180 kms in 5 hrs

=> Speed = 180/5 = 36 kmph

Now the second train covers the same distance in 1 hour less than the first train => 4 hrs

=> Speed of the second train = 180/4 = 45 kmph

Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.

The number of words beign with A is 4! 

The number of words beign with E is 4! 

The number of words beign with M is 4! 

The number of words beign with R is 4!

Number of words beign with VA is 3! 

Words beign with VE are VEAMR 

VEARM

VEMAR

VEMRA

VERAM

VERMA 

Therefore, The Rank of the word VERMA = 4 x 4! + 3! + 6 = 96 + 6 + 6 =108             

60(x + 1) = 64x

X = 15

60 x 16 = 960 km

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. 

Let required fraction be  then 

           ............. (i)

and          ............ (ii)

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

 Required fraction = 

Speed on return trip = 150% of 50 = 75 km/hr.

Average speed = (2 x 50 x 75)/(50 + 75) = 60 km/hr.