Volume of water collected in the tank in 1 hour

⇒ (0.3 × 0.2 × 20km × 1000mts) = 1200 m cubic

If after t hours, the water is at height of 12m,

1200t=200×150×12 

⇒ t = 300 Hours.

Let us assume A is the faster and B is slower, A takes 10s to finish a lap of 50 m, B takes 17s(approx) to finish a lap of 50 m.
So at 11th sec A starts his 2nd lap by the time B will be in between so that is the first time they will meet each other, at 21st sec A starts his 3rd lap by the time B will be coming back to finish his 2nd lap. So that is the second time they meet in opposite. Then at 31st sec A will start his 4th lap by the time B would not even finish 2nd lap as he take 34(2*17) sec to finish 2 laps. So at 35th sec B will start his 3rd lap by the time A will be coming back to finish his 4th lap in the half way so they will meet 3rd time. At 41st sec A will start its fifth lap by the time B will be going to finish 3rd lap in same direction. At 50th second both will complete A's 5th lap and B's 3rd lap. So they will start next lap(A's 6th and B's 4th) from the same point in same direction. And A will finish the race after completing 300m(6*50) total distance and then they wont meet opposite to each other.

So totally they met 3 times in opposite direction.

Required Length = H.C.F of 700 cm, 385 cm and 1295 c

                             = 35 cm.

Area of the room=(544 * 374)

size of largest square tile= H.C.F of 544 & 374 = 34 cm

Area of 1 tile = (34  x  34) cm2

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

Area of 4 walls = 2(l+b)h
=2(10+7) x 5 = 170 sq m
Area of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m

Cost of painting = Rs. 158 x 3 = Rs. 474

area of the room = 544 x 374 sq.cm

size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm

area of 1 tile = 34x34 sq cm

no. of tiles required = (544 x 374) / (34 x 34) = 176

The JVM performs three main tasks:

• Loads code

• Verifies code

• Executes code