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Answer :
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