Direct and Inverse Variation — Exercise 9.3

A farmer can reap a field in 10 days while his wife can do it in 8 days. If they work together, in how much time can they reap the field?

Answer

We know that,

One day's work = $\dfrac{1}{\text{number of days to complete the work}}$

Farmer's one day's work = $\dfrac{1}{10}$

Wife's one day's work = $\dfrac{1}{8}$

$\text{(Farmer + Wife)'s one day's work} = \dfrac{1}{10} + \dfrac{1}{8}\\[1em] = \dfrac{4 + 5}{40}\\[1em] = \dfrac{9}{40}$

Number of days taken by them working together = $\dfrac{1}{\dfrac{9}{40}} = \dfrac{40}{9} = 4\dfrac{4}{9}$

Hence, working together, the farmer and his wife can reap the field in $\mathbf{4\dfrac{4}{9}}$ days.

A can do $\dfrac{1}{5}$th of a certain work in 2 days and B can do $\dfrac{2}{3}$rd of it in 8 days. In how much time can they together complete the work?

Answer

A can do $\dfrac{1}{5}$th of the work in 2 days.

$\text{A's one day's work} = \dfrac{1}{5 \times 2}\\[1em] = \dfrac{1}{10}$

B can do $\dfrac{2}{3}$rd of the work in 8 days.

$\text{B's one day's work} = \dfrac{2}{3 \times 8}\\[1em] = \dfrac{2}{24}\\[1em] = \dfrac{1}{12}\\[1em] \text{(A + B)'s one day's work} = \dfrac{1}{10} + \dfrac{1}{12}\\[1em] = \dfrac{6 + 5}{60}\\[1em] = \dfrac{11}{60}$

Number of days taken by A and B working together = $\dfrac{1}{\dfrac{11}{60}} = \dfrac{60}{11} = 5\dfrac{5}{11}$

Hence, A and B together can complete the work in $\mathbf{5\dfrac{5}{11}}$ days.

One tap fills a tank in 20 minutes and another tap fills it in 12 minutes. The tank being empty and if both taps are opened together, in how many minutes will the tank be full?

Answer

First tap's one minute work = $\dfrac{1}{20}$

Second tap's one minute work = $\dfrac{1}{12}$

$\text{Both taps' one minute work} = \dfrac{1}{20} + \dfrac{1}{12}\\[1em] = \dfrac{3 + 5}{60}\\[1em] = \dfrac{8}{60}\\[1em] = \dfrac{2}{15}$

Number of minutes taken by both taps together = $\dfrac{1}{\dfrac{2}{15}} = \dfrac{15}{2} = 7\dfrac{1}{2}$

Hence, the tank will be full in $\mathbf{7\dfrac{1}{2}}$ minutes.

A can do a work in 6 days and B can do it in 8 days. They worked together for 2 days and then B left the work. How many days will A require to finish the work?

Answer

A's one day's work = $\dfrac{1}{6}$

B's one day's work = $\dfrac{1}{8}$

$\text{(A + B)'s one day's work} = \dfrac{1}{6} + \dfrac{1}{8}\\[1em] = \dfrac{4 + 3}{24}\\[1em] = \dfrac{7}{24}\\[1em] \text{(A + B)'s 2 days' work} = \dfrac{7}{24} \times 2\\[1em] = \dfrac{7}{12}\\[1em] \text{Remaining work} = 1 - \dfrac{7}{12}\\[1em] = \dfrac{12 - 7}{12}\\[1em] = \dfrac{5}{12}\\[1em] \text{Number of days taken by A to finish the remaining work} = \dfrac{\text{Remaining work}}{\text{A's one day's work}} = \dfrac{\dfrac{5}{12}}{\dfrac{1}{6}}\\[1em] = \dfrac{5 \times 6}{12 \times 1}\\[1em] = \dfrac{30}{12}\\[1em] = \dfrac{5}{2}\\[1em] = 2\dfrac{1}{2}$

Hence, A will require $\mathbf{2\dfrac{1}{2}}$ days to finish the work.

A can do a piece of work in 40 days. He works at it for 8 days and then B finishes the remaining work in 16 days. How long would they have taken to complete the work if they had worked together from the beginning?

Answer

A's one day's work = $\dfrac{1}{40}$

$\text{A's 8 days' work} = \dfrac{1}{40} \times 8\\[1em] = \dfrac{1}{5}\\[1em] \text{Remaining work} = 1 - \dfrac{1}{5}\\[1em] = \dfrac{5 - 1}{5}\\[1em] = \dfrac{4}{5}$

B finishes $\dfrac{4}{5}$ of the work in 16 days.

$\text{B's one day's work} = \dfrac{4}{5 \times 16}\\[1em] = \dfrac{4}{80}\\[1em] = \dfrac{1}{20}\\[1em] \text{(A + B)'s one day's work} = \dfrac{1}{40} + \dfrac{1}{20}\\[1em] = \dfrac{1 + 2}{40}\\[1em] = \dfrac{3}{40}$

Number of days taken by A and B working together = $\dfrac{1}{\dfrac{3}{40}} = \dfrac{40}{3} = 13\dfrac{1}{3}$

Hence, working together from the beginning, they would have taken $\mathbf{13\dfrac{1}{3}}$ days to complete the work.

A and B separately do a work in 10 and 15 days respectively. They worked together for some days and then A completed the remaining work in 5 days. For how many days had A and B worked together?

Answer

A's one day's work = $\dfrac{1}{10}$

B's one day's work = $\dfrac{1}{15}$

$\text{(A + B)'s one day's work} = \dfrac{1}{10} + \dfrac{1}{15}\\[1em] = \dfrac{3 + 2}{30}\\[1em] = \dfrac{5}{30}\\[1em] = \dfrac{1}{6}\\[1em] \text{A's 5 days' work} = \dfrac{1}{10} \times 5\\[1em] = \dfrac{1}{2}$

Let A and B work together for $x$ days.

Hence, (A + B)'s $x$ days' work + A's 5 days' work = 1

$\Rightarrow \dfrac{1}{6} \times x + \dfrac{1}{2} = 1\\[1em] \Rightarrow \dfrac{x}{6} = 1 - \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{x}{6} = \dfrac{1}{2}\\[1em] \Rightarrow x = \dfrac{1}{2} \times 6\\[1em] \Rightarrow x = 3$

Hence, A and B had worked together for 3 days.

If 3 women or 5 girls take 17 days to complete a piece of work, how long will 7 women and 11 girls working together take to complete the work?

Answer

Given, 3 women's work = 5 girls' work

⇒ 1 woman's work = $\dfrac{5}{3}$ girls' work

⇒ 7 women's work = $\left(\dfrac{5}{3} \times 7\right)$ girls' work = $\dfrac{35}{3}$ girls' work

⇒ 7 women and 11 girls' work = $\left(\dfrac{35}{3} + 11\right)$ girls' work

= $\dfrac{35 + 33}{3}$ girls' work

= $\dfrac{68}{3}$ girls' work

Since 5 girls take 17 days to complete the work,

Hence, 1 girl takes 5 × 17 = 85 days to complete the work.

$\text{Number of days taken by }\dfrac{68}{3} \text{ girls} = \dfrac{85}{\dfrac{68}{3}}\\[1em] = \dfrac{85 \times 3}{68}\\[1em] = \dfrac{255}{68}\\[1em] = \dfrac{15}{4}\\[1em] = 3\dfrac{3}{4}$

Hence, 7 women and 11 girls working together will complete the work in $\mathbf{3\dfrac{3}{4}}$ days.

A can do a job in 10 days while B can do it in 15 days. If they work together and earn ₹7000, how should they share the money?

Answer

We know that,

Remuneration is in proportion of the work done.

A's one day's work = $\dfrac{1}{10}$

B's one day's work = $\dfrac{1}{15}$

Hence, A and B should share the money in the ratio $\dfrac{1}{10} : \dfrac{1}{15}$

Multiplying both the terms of the ratio by 30, we get :

$\dfrac{1}{10} \times 30 : \dfrac{1}{15} \times 30 = 3 : 2$

Sum of the terms of the ratio = 3 + 2 = 5

A's share = $\dfrac{3}{5}$ of ₹ 7,000 = ₹ $\left(\dfrac{3}{5} \times 7{,}000\right) = 4{,}200$

B's share = $\dfrac{2}{5}$ of ₹ 7,000 = ₹ $\left(\dfrac{2}{5} \times 7{,}000\right) = 2{,}800$

Hence, A gets ₹ 4,200 and B gets ₹ 2,800.

A, B and C can separately do a work in 2, 6 and 3 days respectively. Working together, how much time would they require to do it? If the work earns them ₹1920, how should they divide the money?

Answer

A's one day's work = $\dfrac{1}{2}$

B's one day's work = $\dfrac{1}{6}$

C's one day's work = $\dfrac{1}{3}$

$\text{(A + B + C)'s one day's work} = \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{3}\\[1em] = \dfrac{3 + 1 + 2}{6}\\[1em] = \dfrac{6}{6} \\[1em] = 1.$

Number of days taken by A, B and C working together = $\dfrac{1}{1} = 1$

We know that,

Remuneration is in proportion of the work done.

Hence, they should divide the money in the ratio $\dfrac{1}{2} : \dfrac{1}{6} : \dfrac{1}{3}$

Multiplying all the terms of the ratio by 6, we get :

$\dfrac{1}{2} \times 6 : \dfrac{1}{6} \times 6 : \dfrac{1}{3} \times 6 = 3 : 1 : 2$

Sum of the terms of the ratio = 3 + 1 + 2 = 6

A's share = $\dfrac{3}{6}$ of ₹ 1,920 = ₹ $\left(\dfrac{3}{6} \times 1{,}920\right) = 960$

B's share = $\dfrac{1}{6}$ of ₹ 1,920 = ₹ $\left(\dfrac{1}{6} \times 1{,}920\right) = 320$

C's share = $\dfrac{2}{6}$ of ₹ 1,920 = ₹ $\left(\dfrac{2}{6} \times 1{,}920\right) = 640$

Hence, working together they require 1 day, and A gets ₹ 960, B gets ₹ 320 and C gets ₹ 640.

A, B and C together can do a piece of work in 15 days, B alone can do it in 30 days and C alone can do it in 40 days. In how many days will A alone do the work?

Answer

(A + B + C)'s one day's work = $\dfrac{1}{15}$

B's one day's work = $\dfrac{1}{30}$

C's one day's work = $\dfrac{1}{40}$

A's one day's work = (A + B + C)'s one day's work - B's one day's work - C's one day's work

$=\dfrac{1}{15} - \dfrac{1}{30} - \dfrac{1}{40}\\[1em] = \dfrac{8 - 4 - 3}{120}\\[1em] = \dfrac{1}{120}$

Number of days taken by A alone = $\dfrac{1}{\dfrac{1}{120}} = 120$

Hence, A alone will do the work in 120 days.

A, B and C working together can plough a field in $4\dfrac{4}{5}$ days. A and C together can do it in 8 days. How long would B working alone take to plough the field?

Answer

Converting the mixed fraction into an improper fraction,

A, B and C together can plough the field in $4\dfrac{4}{5} = \dfrac{24}{5}$ days.

(A + B + C)'s one day's work = $\dfrac{1}{\dfrac{24}{5}} = \dfrac{5}{24}$

(A + C)'s one day's work = $\dfrac{1}{8}$

B's one day's work = (A + B + C)'s one day's work - (A + C)'s one day's work

$= \dfrac{5}{24} - \dfrac{1}{8}\\[1em] = \dfrac{5 - 3}{24}\\[1em] = \dfrac{2}{24}\\[1em] = \dfrac{1}{12}$

Number of days taken by B alone = $\dfrac{1}{\dfrac{1}{12}} = 12$

Hence, B working alone will take 12 days to plough the field.

A and B together can build a wall in 10 days; B and C working together can do it in 15 days; C and A together can do it in 12 days. How long will they take to finish the work, working all together? Also find the number of days taken by each to do the same work, working alone.

Answer

(A + B)'s one day's work = $\dfrac{1}{10}$

(B + C)'s one day's work = $\dfrac{1}{15}$

(C + A)'s one day's work = $\dfrac{1}{12}$

Adding these three, we get :

$\text{2(A + B + C)'s one day's work} = \dfrac{1}{10} + \dfrac{1}{15} + \dfrac{1}{12}\\[1em] = \dfrac{6 + 4 + 5}{60}\\[1em] = \dfrac{15}{60}\\[1em] = \dfrac{1}{4}\\[1em] \text{(A + B + C)'s one day's work} = \dfrac{1}{4 \times 2}\\[1em] = \dfrac{1}{8}$

Number of days taken by A, B and C working together = $\dfrac{1}{\dfrac{1}{8}} = 8$

A's one day's work = (A + B + C)'s one day's work - (B + C)'s one day's work

$= \dfrac{1}{8} - \dfrac{1}{15}\\[1em] = \dfrac{15 - 8}{120}\\[1em] = \dfrac{7}{120}$

Number of days taken by A alone = $\dfrac{120}{7} = 17\dfrac{1}{7}$

B's one day's work = (A + B + C)'s one day's work - (C + A)'s one day's work

$= \dfrac{1}{8} - \dfrac{1}{12}\\[1em] = \dfrac{3 - 2}{24}\\[1em] = \dfrac{1}{24}$

Number of days taken by B alone = $\dfrac{1}{\dfrac{1}{24}} = 24$

C's one day's work = (A + B + C)'s one day's work - (A + B)'s one day's work

$= \dfrac{1}{8} - \dfrac{1}{10}\\[1em] = \dfrac{5 - 4}{40}\\[1em] = \dfrac{1}{40}$

Number of days taken by C alone = $\dfrac{1}{\dfrac{1}{40}} = 40$

Hence, working all together they will finish the work in 8 days, and working alone A takes $\mathbf{17\dfrac{1}{7}}$ days, B takes 24 days and C takes 40 days.

A pipe can fill a tank in 12 hours. By mistake, a waste pipe at the bottom is left opened and the tank is filled in 16 hours. If the tank is full, how much time will the waste pipe take to empty it?

Answer

Portion of the tank filled by the pipe in one hour = $\dfrac{1}{12}$

Portion of the tank filled in one hour when the waste pipe is also open = $\dfrac{1}{16}$

$\text{Portion of the tank emptied by the waste pipe in one hour} = \dfrac{1}{12} - \dfrac{1}{16}\\[1em] = \dfrac{4 - 3}{48}\\[1em] = \dfrac{1}{48}$

Time taken by the waste pipe to empty the full tank = $\dfrac{1}{\dfrac{1}{48}} = 48$

Hence, the waste pipe will take 48 hours to empty the full tank.