Inverse proportion

More work­ers fin­ish a job in fewer days. A faster car takes less time for the same trip. Here, when one quan­tity goes up, the other comes down. This is inverse pro­por­tion. It is just as com­mon as direct pro­por­tion, and the work­ing rule is just as short.

A fort and its food#

A fort has enough food for 1000 men for 20 days. Some men leave, and now the food lasts 25 days. How many men left?

Fewer men eat the food more slowly, so it lasts longer. The total amount of food is fixed. It is enough for $1000 \times 20$ "man-days", that is, one man eat­ing for 20,000 days, or 1000 men eat­ing for 20 days.

Let $x$ men leave. Then $1000 - x$ men stay, and the food lasts 25 days.

The total food is the same both times:

$$\displaystyle \begin{aligned} 1000 \times 20 &= (1000 - x) \times 25 \\ 1000 - x &= \frac{1000 \times 20}{25} = 800 \\ x &= 1000 - 800 = 200 \end{aligned}$$

So 200 men left.

What inverse pro­por­tion is#

Inverse pro­por­tion. Two quan­ti­ties are in inverse pro­por­tion if an increase in one causes a decrease in the other in the same ratio, and a decrease in one causes an increase in the other in the same ratio.

If $x$ and $y$ vary inversely, we write

$$\displaystyle x \propto \frac{1}{y}$$

Read it as "$x$ is inversely pro­por­tional to $y$", or "$x$ is directly pro­por­tional to $\displaystyle \frac{1}{y}$".

Then $\displaystyle x = \frac{k}{y}$ for some con­stant $k$. Mul­ti­ply both sides by $y$:

$$xy = k$$

So the prod­uct of the two quan­ti­ties stays the same. If $y_1$ and $y_2$ are the val­ues of $y$ that go with $x_1$ and $x_2$, then

$$x_1 y_1 = x_2 y_2$$

Com­pare the two rules side by side.

Exam­ple 1: Men and days#

56 men can do a piece of work in 42 days. How many men will do the same work in 14 days?

Fewer days need more men, so $\displaystyle x \propto \frac{1}{y}$ and $x_1 y_1 = x_2 y_2$.

$$\displaystyle \begin{aligned} x \times 14 &= 56 \times 42 \\ x &= \frac{56 \times 42}{14} = 168 \end{aligned}$$

168 men are needed.

Exam­ple 2: Hours a day#

A man works 8 hours a day and takes 5 days to fin­ish a project. How many hours a day must he work to fin­ish it in 4 days?

$$\begin{aligned} x \times 4 &= 8 \times 5 \\ x &= 10 \end{aligned}$$

He must work 10 hours a day.

Exam­ple 3: Direct and inverse together#

$A$ varies directly as $B$ and inversely as $C$. $A$ is 12 when $B$ is 6 and $C$ is 2. Find $A$ when $B$ is 12 and $C$ is 3.

"Directly as $B$ and inversely as $C$" means

$$\displaystyle A \propto \frac{B}{C} \quad \text{so} \quad A = \frac{kB}{C}$$

Put in $A = 12$, $B = 6$ and $C = 2$:

$$\displaystyle 12 = \frac{k \times 6}{2} = 3k \quad \text{so} \quad k = 4$$

Now $\displaystyle A = \frac{4B}{C}$. Put in $B = 12$ and $C = 3$:

$$\displaystyle A = \frac{4 \times 12}{3} = 16$$

Exam­ple 4: Pumps#

20 pumps can fill a tank in 20 min­utes. If 4 pumps stop work­ing, how long will the other pumps take?

Pumps left $= 20 - 4 = 16$.

$$\displaystyle \begin{aligned} 16 \times x &= 20 \times 20 \\ x &= \frac{400}{16} = 25 \end{aligned}$$

The 16 pumps take 25 min­utes.

Exam­ple 5: Cycles#

A man has enough money to buy 25 cycles at ₹500 each. How many can he buy if each cycle costs ₹125 more?

New price $= 500 + 125 = 625$ rupees. His total money is the same, $25 \times 500 = 12500$ rupees.

$$\displaystyle \begin{aligned} x \times 625 &= 25 \times 500 \\ x &= \frac{12500}{625} = 20 \end{aligned}$$

He can buy 20 cycles.

Food prob­lems where peo­ple join later#

In these prob­lems, some of the food is eaten before the change. Work out how much food is left at the time of the change, then use inverse pro­por­tion on what is left.

Exam­ple 6#

There is food for 760 men for 22 days. After 2 days, some more men join. Now the food that is left lasts 19 days. How many men joined?

Mea­sure food in man-days.

Total food $= 760 \times 22 = 16720$ man-days.

Food eaten in the first 2 days $= 760 \times 2 = 1520$ man-days.

Food left $= 16720 - 1520 = 15200$ man-days.

This must feed all the men for 19 days. So the num­ber of men now is

$$\displaystyle \frac{15200}{19} = 800$$

Men who joined $= 800 - 760 = 40$.

Exam­ple 7#

A town of 2000 peo­ple has food for 30 days. After 10 days, 500 more peo­ple arrive. How long will the remain­ing food last, at the same rate?

Total food $= 2000 \times 30 = 60000$ per­son-days.

Eaten in 10 days $= 2000 \times 10 = 20000$ per­son-days.

Food left $= 60000 - 20000 = 40000$ per­son-days.

Peo­ple now $= 2000 + 500 = 2500$.

$$\displaystyle \text{days} = \frac{40000}{2500} = 16$$

The food will last 16 more days.

Prac­tice#

1. 12 work­ers can build a wall in 10 days. How many days will 15 work­ers take?
2. A car going at 60 km/h takes 4 hours for a trip. How long will it take at 80 km/h?
3. A hos­tel has food for 300 stu­dents for 12 days. If 60 more stu­dents join, how long will the food last?
4. 9 taps fill a tank in 16 hours. How long will 12 taps take?
5. $x$ is inversely pro­por­tional to $y$, and $x = 15$ when $y = 4$. Find $x$ when $y = 10$.
6. $A$ varies directly as $B$ and inversely as $C$. $A = 10$ when $B = 15$ and $C = 6$. Find $A$ when $B = 20$ and $C = 8$.
7. A hos­tel has food for 120 stu­dents for 25 days. After 5 days, 40 stu­dents leave. How long will the remain­ing food last?
8. A school has money to buy 36 chairs at ₹750 each. If the price goes up to ₹900, how many chairs can it buy?

Answers#