Direct and Inverse Variation — Check Your Progress

Match each of the entries in column I with the appropriate entry in column II.

Answer

(i) Two quantities $x$ and $y$ vary inversely to each other if $xy$ is constant.

Hence, (i) → (c).

(ii) Two quantities $m$ and $n$ vary directly if $\dfrac{m}{n}$ is constant, i.e. $m \propto n$.

Hence, (ii) → (f).

(iii) Two quantities $p$ and $q$ vary inversely if $pq$ is constant, i.e. $p \propto \dfrac{1}{q}$.

Hence, (iii) → (d).

(iv) If $x$ and $y$ vary inversely, then $xy$ is constant. So, on decreasing $x$, the value of $y$ will increase in proportion.

Hence, (iv) → (b).

(v) Two quantities $x$ and $y$ vary directly to each other if $\dfrac{x}{y}$ is constant.

Hence, (v) → (a).

(vi) If $x$ and $y$ vary directly, then $\dfrac{x}{y}$ is constant. So, on decreasing $x$, the value of $y$ will decrease in proportion.

Hence, (vi) → (e).

Thus, we have :

From the following table, determine $p$ and $q$ if $x$ and $y$ vary directly:

Answer

Two quantities $x$ and $y$ are said to be in direct variation if $\dfrac{x}{y}$ is constant.

$\Rightarrow \dfrac{6}{18} = \dfrac{p}{27} = \dfrac{15}{q}\\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{p}{27} \text{ and } \dfrac{1}{3} = \dfrac{15}{q}\\[1em] \Rightarrow p = \dfrac{27}{3} \text{ and } q = 15 \times 3\\[1em] \Rightarrow p = 9 \text{ and } q = 45$

Hence, $\bm{p} = 9$ and $\bm{q} = 45$.

From the following table, determine $a$ and $b$ if $x$ and $y$ vary inversely:

Answer

Two quantities $x$ and $y$ are said to be in inverse variation if $xy$ is constant.

$\Rightarrow 2 \times 45 = a \times 12.5 = 6 \times b\\[1em] \Rightarrow 90 = 12.5a = 6b\\[1em] \Rightarrow a = \dfrac{90}{12.5} \text{ and } b = \dfrac{90}{6}\\[1em] \Rightarrow a = 7.2 \text{ and } b = 15$

Hence, $\bm{a = 7.2}$ and $\bm{b = 15}$.

It rained 80 mm in first 20 days of April. What would be the total rainfall in April?

Answer

The rainfall in a month does not increase or decrease in a fixed ratio with the number of days. It may rain heavily on some days and may not rain at all on the other days.

So, the rainfall and the number of days are neither in direct variation nor in inverse variation. Hence, the total rainfall in April cannot be found from the rainfall of the first 20 days.

Hence, we can't say what the total rainfall in April would be.

Mamta earns ₹3,240 for a working week of 48 hours. If she was absent for 6 hours, how much did she earn?

Answer

Number of hours for which Mamta worked = 48 - 6 = 42

Let Mamta earn ₹ $x$ for 42 hours of work.

Less the number of hours worked, less will be the earning.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{48}{3{,}240} = \dfrac{42}{x}\\[1em] \Rightarrow x = \dfrac{3{,}240 \times 42}{48}\\[1em] \Rightarrow x = \dfrac{1{,}36{,}080}{48}\\[1em] \Rightarrow x = 2{,}835$

Hence, Mamta earned ₹ 2,835.

Navjot can do a piece of work in 6 days working 10 hours per day. In how many days can he do the same work if he increases his working hours by 2 hours per day?

Answer

New working hours per day = 10 + 2 = 12

Let Navjot do the same work in $x$ days working 12 hours per day.

More the working hours per day, less will be the number of days required.

Hence, it is a case of inverse variation.

$\Rightarrow 10 \times 6 = 12 \times x\\[1em] \Rightarrow 60 = 12x\\[1em] \Rightarrow x = \dfrac{60}{12}\\[1em] \Rightarrow x = 5$

Hence, Navjot can do the same work in 5 days.

Sharmila has enough money to buy 24 bananas at the rate of ₹1.50 per banana. How many bananas she can buy if the price of each banana is decreased by 30 paise?

Answer

Converting 30 paise into rupees,

30 paise = ₹ $\dfrac{30}{100} = 0.30$

Hence, the new price of each banana = ₹ (1.50 - 0.30) = ₹ 1.20

Let Sharmila buy $x$ bananas at ₹ 1.20 per banana.

For the same amount of money, less the price of each banana, more will be the number of bananas bought.

Hence, it is a case of inverse variation.

$\Rightarrow 1.50 \times 24 = 1.20 \times x\\[1em] \Rightarrow 36 = 1.20x\\[1em] \Rightarrow x = \dfrac{36}{1.20}\\[1em] \Rightarrow x = 30$

Hence, Sharmila can buy 30 bananas.

A fort has enough food for 180 soldiers for 40 days. After 10 days, 30 soldiers leave the fort. Find the total number of days for which the food will last.

Answer

After 10 days, the remaining food is sufficient for 180 soldiers for (40 - 10) days, i.e. 30 days.

After 10 days, 30 soldiers leave the fort.

Hence, the number of soldiers in the fort = 180 - 30 = 150

Let the remaining food last for $x$ days when there are 150 soldiers.

Less the number of soldiers, more will be the number of days for which the food lasts.

Hence, it is a case of inverse variation.

$\Rightarrow 180 \times 30 = 150 \times x\\[1em] \Rightarrow 5400 = 150x\\[1em] \Rightarrow x = \dfrac{5400}{150}\\[1em] \Rightarrow x = 36$

Hence, the total number of days for which the food will last = 10 + 36 = 46

Hence, the food will last for a total of 46 days.

If 4 goats or 6 sheep can graze a field in 40 days, how many days will 4 goats and 14 sheep take to graze the same field?

Answer

Given, 4 goats' work = 6 sheep's work

⇒ 4 goats and 14 sheep's work = (6 + 14) sheep's work = 20 sheep's work

Let 20 sheep graze the same field in $x$ days.

More the number of sheep, less will be the number of days taken to graze the field.

Hence, it is a case of inverse variation.

$\Rightarrow 6 \times 40 = 20 \times x\\[1em] \Rightarrow 240 = 20x\\[1em] \Rightarrow x = \dfrac{240}{20}\\[1em] \Rightarrow x = 12$

Hence, 4 goats and 14 sheep will take 12 days to graze the same field.

A tap can fill a tank in 20 hours, while the other can empty it in 30 hours. The tank being empty and if both taps are opened together, how long will it take for the tank to be half full?

Answer

Portion of the tank filled by the first tap in one hour = $\dfrac{1}{20}$

Portion of the tank emptied by the second tap in one hour = $\dfrac{1}{30}$

$\text{Portion of the tank filled in one hour when both taps are opened together} = \dfrac{1}{20} - \dfrac{1}{30}\\[1em] = \dfrac{3 - 2}{60}\\[1em] = \dfrac{1}{60}$

Time taken to fill the tank completely = $\dfrac{1}{\dfrac{1}{60}} = 60$ hours

Hence, the time taken for the tank to be half full = $\dfrac{60}{2} = 30$ hours

Hence, it will take 30 hours for the tank to be half full.

Three ants separately can gobble a grasshopper in 3, 4 and 6 hours respectively. How much time will they take together to finish off the poor chap? If the grasshopper weighs 63 gram, find the share of each.

Answer

First ant's one hour work = $\dfrac{1}{3}$

Second ant's one hour work = $\dfrac{1}{4}$

Third ant's one hour work = $\dfrac{1}{6}$

$\text{All three ants' one hour work} = \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{6}\\[1em] = \dfrac{4 + 3 + 2}{12}\\[1em] = \dfrac{9}{12}\\[1em] = \dfrac{3}{4}$

Time taken by the three ants together = $\dfrac{1}{\dfrac{3}{4}} = \dfrac{4}{3} = 1\dfrac{1}{3}$ hours

The share of each ant is in proportion of the work done by it.

Hence, they should share the grasshopper in the ratio $\dfrac{1}{3} : \dfrac{1}{4} : \dfrac{1}{6}$

Multiplying all the terms of the ratio by 12, we get :

$\dfrac{1}{3} \times 12 : \dfrac{1}{4} \times 12 : \dfrac{1}{6} \times 12 = 4 : 3 : 2$

Sum of the terms of the ratio = 4 + 3 + 2 = 9

First ant's share = $\dfrac{4}{9}$ of 63 g = $\left(\dfrac{4}{9} \times 63\right)$ g = 28 g

Second ant's share = $\dfrac{3}{9}$ of 63 g = $\left(\dfrac{3}{9} \times 63\right)$ g = 21 g

Third ant's share = $\dfrac{2}{9}$ of 63 g = $\left(\dfrac{2}{9} \times 63\right)$ g = 14 g

Hence, the three ants together take $\mathbf{1\dfrac{1}{3}}$ hours, and their shares are 28 g, 21 g and 14 g respectively.

A and B together can do a piece of work in 12 days; B and C together can do it in 15 days. If A is twice as good a workman as C, in how many days A alone will do the same work?

Answer

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

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

Subtracting the second from the first, we get :

$\text{A's one day's work - C's one day's work} = \dfrac{1}{12} - \dfrac{1}{15}\\[1em] = \dfrac{5 - 4}{60}$

= $\dfrac{1}{60}$ ....(1)

Since A is twice as good a workman as C,

Hence, A's one day's work = 2 × C's one day's work ....(2)

Substituting the value from (2) in (1), we get :

$\Rightarrow 2 \times \text{C's one day's work} - \text{C's one day's work} = \dfrac{1}{60}\\[1em] \Rightarrow \text{C's one day's work} = \dfrac{1}{60}\\[1em] \text{Hence, A's one day's work} = 2 \times \dfrac{1}{60}\\[1em] = \dfrac{2}{60}\\[1em] = \dfrac{1}{30}$

Number of days taken by A alone = $\dfrac{1}{\dfrac{1}{30}} = 30$

Hence, A alone will do the same work in 30 days.