Formulate the following problems as a part of equations, and hence find their solutions.
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days.Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately

Ans.(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,

\({{20} \over {x + y}} = 2\) and \({{4} \over {x - y}} = 2\)
\(2x + 2y = 20\) and \(2x - 2y = 4\)
\(x + y = 10\) … (1) and \(x – y = 2\) … (2)

Adding (1) and (2),
\(2x = 12\) => \(x = 6\)

Putting x = 6 in (1),
\(6 + y = 10\)
\(y = 10 – 6 = 4\)

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

So, 1 woman’s 1-day work = (\({{1} \over {x}}\))th part of the work

And, 1 man’s 1-day work = (\({{1} \over {y}}\))th part of the work

So, 2 women’s 1-day work = (\({{2} \over {x}}\))th part of the work

And, 5 men’s 1-day work = (\({{5} \over {y}}\))th part of the work

Therefore, 2 women and 5 men’s 1-day work = (\({{2} \over {x}} + {{5} \over {y}}\))th part of the work....… (1)

It is given that 2 women and 5 men complete work in = 4 days

It means that in 1 day, they will be completing (\({{1} \over {4 }}\))th part of the work ….... (2)

Since, (1) = (2)
\({{2} \over {x}} + {{5} \over {y}} = {{1} \over {4 }}\)….... (3)

Similarly, \({{3} \over {x}} + {{6} \over {y}} = {{1} \over {3}}\) .....… (4)

Let \({{1} \over {x}} = p\) and \( {{1} \over {y}} = q \)

Putting this in (3) and (4),

\(2p + 5q = {{1} \over {4 }}\) and \(3p + 6q = {{1} \over {3}}\)
\(8p + 20q = 1\) … (5) and \(9p + 18q = 1\) … (6)

Multiplying (5) by 9 and (6) by 8,
\(72p + 180q = 9\) … (7)
\(72p + 144q = 8\) … (8)

Subtracting (8) from (7), we get
\(36q = 1\) => \(q = {{1} \over {36}}\)

Putting this in (6), we get
\(9p + 18 ({{1} \over {36}}) = 1\)
\(9p = {{1} \over {2}}\) => \(p = {{1} \over {18}}\)

Putting values of p and q in , we get x = 18 and y = 36
Therefore, 1 woman completes work in = 18 days

And, 1 man completes work in = 36 days


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

According to given conditions,
\({{60} \over {x}} + {{240} \over {y}} = 4\) and \({{100} \over {x}} + {{200} \over {y}} = 4 + {{10} \over {60}}\)
Let \({{1} \over {x}} = p\) and \( {{1} \over {y}} = q \)

Putting this in the above equations, we get
\(60p + 240q = 4\) … (1)
And \(100p + 200q = {{25} \over {6}}\) … (2)

Multiplying (1) by 5 and (2) by 3,
\(300p + 1200q = 20 \)… (3)
\(300p + 600q = {{25} \over {2}}\) … (4)
Subtracting (4) from (3),
\(600q = 20 - {{25} \over {2}} = 7.5\)
\(q = {{7.5} \over {600}}\)

Putting value of q in (2),
\(100p + 200 ({{7.5} \over {600}}) = {{25} \over {6}}\)
\(100p + 2.5 = {{25} \over {6}}\)
\(100p = {{25} \over {6}} – 2.5\)
\(p = {{10} \over {600}}\)

But \({{1} \over {x}} = p\) and \( {{1} \over {y}} = q \)

Therefore, \(x = {{600} \over {10}} = 60\) km/h and \(y = {{600} \over {7.5}} = 80\) km/h

Therefore, speed of train = 60 km/h

And, speed of bus = 80 km/h