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# Formulate the following problems as a pair 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

**To do:**

We have to formulate the given problems as a pair of equations, and hence find their solutions.

**Solution:**

(i) Let the speed of the current be $x\ km/hr$ and the speed of her rowing in still water be $y\ km/hr$.

Upstream speed $=y−x\ km/hr$

Downstream speed $=y+x\ km/hr$

$Time=\frac{speed}{distance}$

Ritu can row downstream 20 km in 2 hours.

Time taken $=\frac{20}{y+x}$

$2=\frac{20}{y+x}$

$2(y+x)=20$

$y+x=10$.....(i)

Ritu can row upstream 4 km in 2 hours.

Time taken $=\frac{4}{y-x}$

$2=\frac{4}{y-x}$

$2(y-x)=4$

$y-x=2$.....(ii)

Adding equations (i) and (ii), we get,

$y+x+y-x=10+2$

$2y=12$

$y=\frac{12}{2}$

$y=6\ km/hr$

This implies,

$6-x=2$

$x=6-2$

$x=4\ km/hr$

Therefore,

The speed of her rowing in still water is 6 km/hr and the speed of the current is 4 km/hr.

(ii) Let the number of days taken by one man alone to finish a piece of embroidery be $x$.

This implies,

The amount of work done by one man in a day $=\frac{1}{x}$.

Let the number of days taken by one woman alone to finish the embroidery be $y$.

This implies,

The amount of work done by one woman in a day $=\frac{1}{y}$.

In the first case, 2 women and 5 men can together finish the piece of embroidery in 4 days.

The amount of work done by 5 men in 1 day $=5\times\frac{1}{x}=\frac{5}{x}$.

The amount of work done by 2 women in 1 day $=2\times\frac{1}{y}=\frac{2}{y}$.

According to the question,

$4(\frac{5}{x}+\frac{2}{y})=1$

$\frac{20}{x}+\frac{8}{y}=1$....(i)

In the second case, 6 men and 3 women finish the work in 3 days.

The amount of work done by 6 men in 1 day $=6\times\frac{1}{x}=\frac{6}{x}$.

The amount of work done by 3 women in 1 day $=3\times\frac{1}{y}=\frac{3}{y}$.

According to the question,

$3(\frac{6}{x}+\frac{3}{y})=1$

$\frac{18}{x}+\frac{9}{y}=1$....(ii)

Multiplying equation (i) by 9 and equation (ii) by 8, we get,

$9(\frac{20}{x}+\frac{8}{y})=9(1)$

$\frac{180}{x}+\frac{72}{y}=9$.....(iii)

$8(\frac{18}{x}+\frac{9}{y})=8(1)$

$\frac{144}{x}+\frac{72}{y}=8$.....(iv)

Subtracting (iv) from (iii), we get,

$\frac{180}{x}+\frac{72}{y}-(\frac{144}{x}+\frac{72}{y})=9-8$

$\frac{180-144}{x}=1$

$x=36$

Substituting $x=36$ in (i), we get,

$\frac{20}{36}+\frac{8}{y}=1$

$\frac{8}{y}=1-\frac{5}{9}$

$\frac{8}{y}=\frac{9-5}{9}$

$\frac{8}{y}=\frac{4}{9}$

$y=\frac{8\times9}{4}$

$y=18$

Therefore, one man alone takes 36 days to finish the work and one woman alone takes 18 days to finish the work.

(iii) Total distance to the home $=300\ km$.

Let the speed of the train be $x$ km/hr and the speed of the bus be $y$ km/hr.

We know that,

Time $=$ Distance $\div$ Speed

In the first case, she takes 4 hours if she travels 60 km. by train and the remaining by bus.

Time taken $=\frac{60}{x}+\frac{300-60}{y}$

$\Rightarrow \frac{60}{x}+\frac{240}{y}=4$.....(i)

In the second case, she takes 10 minutes more if she travels 100 km by train and the remaining by bus.

Time taken $=\frac{100}{x}+\frac{300-100}{y}$

$\Rightarrow \frac{100}{x}+\frac{200}{y}=4+\frac{10}{60}$

$\Rightarrow \frac{100}{x}+\frac{200}{y}=4+\frac{1}{6}$

$\Rightarrow \frac{100}{x}+\frac{200}{y}=\frac{4\times6+1}{6}$

$\Rightarrow \frac{100}{x}+\frac{200}{y}=\frac{25}{6}$......(ii)

Multiplying equation (i) by 5 and equation (ii) by 3, we get,

$5(\frac{60}{x}+\frac{240}{y})=5(4)$

$\frac{300}{x}+\frac{1200}{y}=20$....(iii)

$3(\frac{100}{x}+\frac{200}{y})=3(\frac{25}{6})$

$\frac{300}{x}+\frac{600}{y}=\frac{25}{2}$.....(iv)

Subtracting equation (iv) from (iii), we get,

$\frac{300}{x}+\frac{1200}{y}-\frac{300}{x}-\frac{600}{y}=20-\frac{25}{2}$

$\frac{1200-600}{y}=\frac{20(2)-25}{2}$

$\frac{600}{y}=\frac{15}{2}$

$y=\frac{600\times2}{15}$

$y=40\times2=80$

Substituting $y=80$ in equation (i), we get,

$\frac{60}{x}+\frac{240}{80}=4$

$\frac{60}{x}+3=4$

$\frac{60}{x}=4-3=1$

$x=60(1)$

$x=60$

Therefore, the speed of the train is $60\ km/hr$ and the speed of the bus is $80\ km/hr$ respectively.

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