Formulate the following problems as a part of equations, and hence find their solutions.

(i) Let speed of rowing in still water =x km/h

Let speed of current =y km/h

So, speed of rowing downstream = (x +y) km/h

And, speed of rowing upstream = (x −y) km/h

According to given conditions,

2(x+y) = 20 and 2(x-y) = 4

⇒ 2x + 2y = 20 and 2x − 2y = 4

⇒x +y = 10 … (1) andx –y = 2 … (2)

Adding (1) and (2), we get

2x = 12⇒x = 6

Puttingx = 6 in (1), we get

6 +y = 10

⇒y = 10 – 6 = 4

Therefore, speed of rowing in still water = 6 km/h

Speed of current = 4 km/h

(ii) Let time taken by 1 woman alone to finish the work =x days

Let time taken by 1 man alone to finish the work =y days

Therefore,work done by a woman in 1 day =1/x

work done by a man in 1 day =1/y

According to the question,

putting 1/x = p and 1/y = q, in these equations
we obtain,

2p + 5q =1/4
8p + 20q = 1

3p + 6q = 1/3
= 9p + 18q = 1

By cross-multiplication,we obtain

x = 18, y = 36
Hence, number of days taken by a woman = 18 Number of days taken by a man = 36

(iii) Let speed of train =x km/h and let speed of bus =y km/h

According to given conditions,

putting in these equations,we obtain

Putting this in the above equations, we get

60p + 240q = 4 … (3)

And 100p + 200q = 25/6

600p + 1200q = 25 … (4)

Multiplying (3) by 10, we get

600p +2400q = 40 … (5)

substracting equation (4) from (5),we obtain
1200q =15

Substituting in equation(3),we obtain
60p + 3 = 4
60p = 1

u = 60km/h and v = 80km/h
Hence,speed of train = 60 km/h
speed of bus = 80km/h