Aptitude and Reasoning Questions

Initial speed = 80km/hr
Total distance = 80 x 10 = 800km
New speed = 800/4 =200km/hr
Increase in speed = 200 - 80 = 120km/hr

Let the speed of current = 'C' km/hr

Given the speed of boat in still water = 6 kmph

Let 'd' kms be the distance it covers.

According to the given data,

Boat takes thrice as much time in going the same distance against the current than going with the current

i.e, d8 - C = 3 × d8 + C

$$\begin{gathered} ⇒ 24 - 3C = 8 + C \\ ⇒ 4C = 16 \\ ⇒ C = 4 kmph \end{gathered}$$

Hence, the speed of the current C = 4 kmph.

9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April May 

(31 + 28 + 31 + 30 + 9) = 129 days

129 days = (18 weeks + 3 day)= 3 odd day.

Total number of odd days = (0 + 0 + 3 ) = 3 odd day.

Given day is Wednesday.

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

So, required probability = 11!×3!13! = 39916800×66227020800 = 24625

The given series follows the rule,
+(5x1), +(5x2), +(5x3), +(5x4), +(5x5),....

So starting from 37, 37 + 5 = 42, 42 + 10 = 52, 52 + 15 = 67, 67 + 20 = 87,...
So wrong number is 47 which is to be 42.

Let x, x+2, x+4, x+6, x+8 and x+10 are six consecutive odd numbers.

Given that their average is 52

Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6

6x + 30 = 312

x = 47

So Product = 47 × 57 = 2679

Work done by the third pipe in 1 min = 1/50 - (1/60 + 1/75) = - 1/100.
[-ve sign means emptying]

The third pipe alone can empty the cistern in 100 min.