Mind Warmups-3

Since two boats are travelling in opposite directions towards each other. Therefore their relative speed $= 15km/h=\frac 1 4 km/min.$ Hence before the collision, they must travel $\frac 1 4 km$ in last one minute.

time to collied two boat is given by, 5 T+10 T=20 which gives, T=80 min so , distance before 1 min of collison ie at 79th min is given by, distance= 20-(5 79/60 + 10 79/60) =1/4 km.

They are opposite in direction so 5+10=15 Total distance between them is 20 That is 15÷20=3/4 This time requires for cross the boat to each other. So remain time means 1-3/4=1/4 The ans.is 1/4

Suppose the boats start from the same position (collision point) and are travelling away from each other. After one minute has passed, the first boat will have travelled a distance of 5/60 Km and the second one 15/60 Km. Hence, the distance between them will be (5+15)/60 Km = 1/4 Km.

We can see this problem as two boats travel in opposite direction start from same point. The speed of each boats in minutes are 1/12 km/min and 1/6 km/min. In 1 minute, they travel 1/12 km and 1/6 km. Sum it and we get 1/4 km the distance of two boats.

The question is actually asking you "What is the combined distance moved by each of the boats in $One Minute$ ". No need to find where they collide since at collision the distance will be zero. lets call the distance moved by the left boat $d_l$ and the distance moved by the right boat $d_r$ then the sum is $d_f+d_r=5\times \frac{1}{60}+10\times\frac{1}{60}$ $=\frac{1}{12}+\frac{1}{6}$ $=\frac{1}{4}$

Since two boats are travelling towards each other, they will meet at the LCM of their speeds which is 10. One min before they collide is 9 min. Multiply their speed by 9 to get the distance covered in 9 min. 10/60 x 9 =9/6 km 5/60 x 9=9/12 km 9/6 - 9/12 =9/12 km = 3/4 km = distance covered in 9 min. So, distance between them is 1 - 3/4 = 1/4 km

by Aswathi 5th std.

( 5 + 10 ) / 60 = 0.25 (Opposite directions & regular units)