A river is flowing with a steady speed of 4 km/h. One rows his boat downstream in the river and then returns by rowing upstream in the same river. When he returns to the starting points, the total distance covered by him is 42 km. If the return journey takes 2 h more than his outward journey, then the speed of his rowing in still water must be?

∴ Required ratio = 4 : 3

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) ≡ 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan.  Feb.   March    April    May      June 
(31 +  28  +  31   +   30   +   31   +   17) = 168 days

∴ 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.

Let 1872/?N = 234.
Then ?N = 1872/234 = 8
? N = 8 x 8 = 64

Let distance covered by dog in 1 leap is x and
Distance covered by cat in 1 leap is y

Then, 3x = 4y
? x= 4/3y

Now, Ratio of speed of dog and cat = Ratio of distance covered by them in the same time = 4x : 5y
= 16/3y : 5y
= 16 : 15

Total volume of cuboid = (10 x 5 x 2) cm³
= 100 cm³
Volume curved = 1/3 x (22/7) x 3 x 3 x 7 cm³
= 66 cm³
% of wood wasted = (100 - 66)% = 34 %

Series pattern
1², 3², 5², 7², 9², 11²
So missing term = 11² = 121

Angle of 360° is covered by hour hand in 12 h .
So, angle of 135° is covered by hour hand in 12/360° x 135° = 4.5 h
Thus, required time = (3 + 4.5)h = 7.5 h = 7 : 30

Try yourself. It was Wednesday.