Direct and Inverse Proportion — Questions and Answers
Welcome to HSLC Guru. This page gives the complete, step-by-step solution to ASSEB (Assam State School Education Board) Class 8 General Mathematics Chapter 13, Direct and Inverse Proportion — every Activity table, worked Example, Exercise 13.1, Exercise 13.2 and the Multiple Choice Questions, each with clear working.
Summary
Two quantities $x$ and $y$ are in direct proportion when an increase (or decrease) in one causes a proportional increase (or decrease) in the other. Then the ratio $\frac{x}{y}$ is always a constant $k$, so $x = ky$ and $x \propto y$. For example, as the number of lemons increases, their total price increases in the same ratio.
Two quantities $x$ and $y$ are in inverse (indirect) proportion when an increase in $x$ causes a proportional decrease in $y$ (and vice versa). Then the product $xy$ is always a constant $k$, so $xy = k$. For example, as speed increases, the time to cover a fixed distance decreases.
For two pairs of values $(x_1, y_1)$ and $(x_2, y_2)$ — in direct proportion $\frac{x_1}{y_1} = \frac{x_2}{y_2}$, and in inverse proportion $x_1 y_1 = x_2 y_2$. Using these relations (or the unitary method) any unknown value can be found easily.
Summary: This ASSEB Class 8 General Mathematics Chapter 13 solution covers Direct and Inverse Proportion. Two quantities are in direct proportion when x/y stays constant (x = ky), and in inverse proportion when the product xy stays constant. The page fully solves every Activity table, worked Example, Exercise 13.1, Exercise 13.2 and the Multiple Choice Questions with clear step-by-step working.
Textbook Questions and Answers
Opening Activity (Completing the tables)
Activity 1: Complete the table based on example (a). (A pump draws 100 litres of water in 10 minutes, i.e. 10 litres per minute.)
Answer: Time and quantity of water are in direct proportion because $\frac{\text{water}}{\text{time}} = 10$ (constant), so water $= 10 \times$ time. Filling the blanks —
| Time (in minutes) | 1 | 2 | 3 | 5 | 10 | 18 | 20 |
|---|---|---|---|---|---|---|---|
| Quantity of water (in litres) | 10 | 20 | 30 | 50 | 100 | 180 | 200 |
Activity 2: Complete the table based on example (b). (Price of one book is Rs. 150.)
Answer: Number of books and price are in direct proportion, since price $= 150 \times$ number of books. Hence —
| Number of books | 1 | 2 | 5 | 7 | 9 | 12 | 14 | 20 |
|---|---|---|---|---|---|---|---|---|
| Price (in Rupees) | 150 | 300 | 750 | 1050 | 1350 | 1800 | 2100 | 3000 |
Direct Proportion (13.1)
Two quantities in direct proportion, when plotted on a graph, lie on a single straight line passing through the origin (O). In the graph below, number of lemons $x = 2, 3, 7, 15$ and their prices $y = 10, 15, 35, 75$; in every pair $\frac{x}{y} = \frac{1}{5}$ (constant).
Example 1: A car consumes 20 litres of petrol to cover 240 km. Find the distance covered when it consumes 5, 8, 12 and 25 litres of petrol.
Answer: Let petrol $= x$ litres and distance $= y$ km. More distance needs more petrol, so it is direct proportion. Here $\frac{x_1}{y_1} = \frac{20}{240} = \frac{1}{12}$, so $y = 12x$.
$$y = 12 \times 5 = 60,\quad y = 12 \times 8 = 96,\quad y = 12 \times 12 = 144,\quad y = 12 \times 25 = 300$$
| Petrol (x litres) | 20 | 5 | 8 | 12 | 25 |
|---|---|---|---|---|---|
| Distance (y km) | 240 | 60 | 96 | 144 | 300 |
Example 2: Madhurjya covers 1 km in 10 minutes on a bicycle. At the same speed, what distance can he cover in 1 hour 20 minutes?
Answer: 1 hour 20 minutes $= (60 + 20)$ minutes $= 80$ minutes. Time ($x$) and distance ($y$) are in direct proportion.
$$\frac{x_1}{y_1} = \frac{x_2}{y_2} \Rightarrow \frac{10}{1} = \frac{80}{y_2} \Rightarrow y_2 = \frac{80}{10} = 8$$
By the unitary method: in 10 minutes $= 1$ km; in 1 minute $= \frac{1}{10}$ km; in 80 minutes $= \frac{1}{10} \times 80 = 8$ km. So Madhurjya covers 8 kilometres.
Example 3: A train covers 720 km in 18 hours. At the same speed, how much time will it need to cover 200 km?
Answer: Let time $= x$ and distance $= y$. More distance needs more time, so it is direct proportion.
$$\frac{x_1}{y_1} = \frac{x_2}{y_2} \Rightarrow \frac{18}{720} = \frac{x_2}{200} \Rightarrow x_2 = \frac{18 \times 200}{720} = 5$$
So the train needs 5 hours to cover 200 km.
Example 4: The weight of 12 pages of thick paper is 40 grams. How many such pages weigh $2\frac{1}{2}$ kg?
Answer: $2\frac{1}{2}$ kg $= (2000 + 500)$ g $= 2500$ g. Let the number of pages $= x$; more pages means more weight, so direct proportion.
$$\frac{12}{40} = \frac{x}{2500} \Rightarrow x = \frac{12 \times 2500}{40} = 750$$
So the required number of pages is 750.
Example 5: The scale of a map (in centimetres) is $1 : 300000$. On this map the distance between two cities is shown as 4 cm. Find their actual distance.
Answer: Let map distance $= x$ cm and actual distance $= y$ cm. Given $\frac{x}{y} = \frac{1}{300000}$.
$$\frac{1}{300000} = \frac{4}{y} \Rightarrow y = 4 \times 300000 = 1200000 \text{ cm}$$
Since $100000$ cm $= 1$ km, actual distance $= \frac{1200000}{100000} = 12$ km. So the actual distance is 12 kilometres.
Exercise 13.1
1. Observe the following tables and find whether $x$ and $y$ are in direct proportion or not.
(i) $x$: 20, 17, 14, 11, 8, 5, 2 and $y$: 40, 34, 28, 22, 16, 10, 4
Answer: Taking $\frac{x}{y}$ for each pair, $\frac{20}{40} = \frac{17}{34} = \frac{14}{28} = \frac{11}{22} = \frac{8}{16} = \frac{5}{10} = \frac{2}{4} = \frac{1}{2}$ (constant). The ratio is always the same, so $x$ and $y$ are in direct proportion.
(ii) $x$: 6, 10, 14, 18, 22, 26, 30 and $y$: 4, 8, 12, 16, 20, 24, 28
Answer: $\frac{6}{4} = \frac{3}{2}$ but $\frac{10}{8} = \frac{5}{4}$ — the ratio is not the same, so $x$ and $y$ are not in direct proportion.
(iii) $x$: 5, 8, 12, 15, 18, 20 and $y$: 15, 24, 36, 60, 72, 100
Answer: $\frac{5}{15} = \frac{8}{24} = \frac{12}{36} = \frac{1}{3}$, but $\frac{15}{60} = \frac{1}{4}$ — the ratio is not the same, so $x$ and $y$ are not in direct proportion.
2. If the Principal is Rs. 1000 and the rate of interest is 8% per annum, fill up the following table and find in which case it is direct proportion.
Answer: Simple interest $= \frac{P \times R \times T}{100} = \frac{1000 \times 8 \times T}{100} = 80T$. Compound interest $= P\left(1 + \frac{R}{100}\right)^{T} – P$.
| Time | 1 year | 2 years | 3 years |
|---|---|---|---|
| Simple interest (Rupees) | 80 | 160 | 240 |
| Compound interest (Rupees) | 80 | 166.40 | 259.71 |
For simple interest $\frac{80}{1} = \frac{160}{2} = \frac{240}{3} = 80$ (constant), so simple interest is in direct proportion with time. For compound interest $\frac{80}{1}, \frac{166.40}{2}, \frac{259.71}{3}$ are not equal, so compound interest is not in direct proportion.
3. Karim covers 15 km per hour by cycling. How much distance will he cover in (i) 3 hours (ii) 5 hours (iii) 1 hour 20 minutes?
Answer: The speed is constant, so distance is in direct proportion to time; distance $= 15 \times$ time (in hours).
$$\text{(i) } 15 \times 3 = 45 \text{ km} \qquad \text{(ii) } 15 \times 5 = 75 \text{ km}$$
(iii) 1 hour 20 minutes $= \frac{4}{3}$ hours, so distance $= 15 \times \frac{4}{3} = 20$ km. Answer: 45 km, 75 km, 20 km.
4. Julumi reaches Kaziranga in 3 hours 15 minutes by a car travelling at an average speed of 50 km/hr. From what distance does she come to Kaziranga?
Answer: 3 hours 15 minutes $= 3\frac{15}{60} = \frac{13}{4}$ hours. Distance $=$ speed $\times$ time.
$$\text{Distance} = 50 \times \frac{13}{4} = \frac{650}{4} = 162.5 \text{ km}$$
So Julumi comes from a distance of 162.5 kilometres.
5. What distance will a plane cover in 2 hours 20 minutes flying at a speed of 510 km per hour?
Answer: 2 hours 20 minutes $= 2\frac{20}{60} = \frac{7}{3}$ hours.
$$\text{Distance} = 510 \times \frac{7}{3} = \frac{3570}{3} = 1190 \text{ km}$$
So the plane covers 1190 kilometres.
6. Mary can run at 4.5 km per hour and Gurpreet can run 600 metres in 9 minutes. Who can run faster in an hour?
Answer: Find Gurpreet’s speed per hour. In 9 minutes she runs 600 m, so in 60 minutes $= 600 \times \frac{60}{9} = 4000$ m $= 4$ km. Mary’s speed is 4.5 km/hr, which is more than 4 km/hr. So Mary runs faster.
7. In a soft drink factory, a machine fills 840 bottles in 6 hours. How many bottles will be filled in 5 hours?
Answer: More time means more bottles — direct proportion.
$$\frac{840}{6} = \frac{y}{5} \Rightarrow y = \frac{840 \times 5}{6} = 700$$
So 700 bottles will be filled in 5 hours.
8. The height of the mast post of a model ship is 9 cm; that of the actual ship is 12 metres. If the length of the actual ship is 28 metres, what is the length of the model ship?
Answer: The model and actual ship are in direct proportion. Ratio of mast heights $= \frac{9 \text{ cm}}{12 \text{ m}} = \frac{9}{1200} = \frac{3}{400}$.
$$\text{Model length} = 28 \times \frac{3}{400} = \frac{84}{400} = 0.21 \text{ m}$$
So the length of the model ship is 0.21 metre, i.e. 21 cm.
9. A vertical tower of height 5 m 60 cm casts a shadow of length 3 m 20 cm. At the same time, (i) what will be the length of the shadow cast by a tower of height 10 m 50 cm? (ii) What is the height of the tower whose shadow is 5 m long?
Answer: At the same time, height and shadow are in direct proportion. Here height : shadow $= 5\text{ m }60\text{ cm} : 3\text{ m }20\text{ cm} = 560 : 320 = 7 : 4$.
(i) Tower height $10$ m $50$ cm $= 1050$ cm. Shadow $=$ height $\times \frac{4}{7}$.
$$\text{Shadow} = 1050 \times \frac{4}{7} = 600 \text{ cm} = 6 \text{ m}$$
(ii) Shadow $5$ m $= 500$ cm. Height $=$ shadow $\times \frac{7}{4}$.
$$\text{Height} = 500 \times \frac{7}{4} = 875 \text{ cm} = 8 \text{ m } 75 \text{ cm}$$
So (i) the shadow is 6 metres and (ii) the tower’s height is 8 m 75 cm (8.75 m).
10. Limsing has a road map. The ratio of the distance shown on the map to the actual distance is 1 cm : 18 km. If he drives 72 km, what will be the distance on the map?
Answer: Direct proportion: $\frac{1 \text{ cm}}{18 \text{ km}} = \frac{x}{72 \text{ km}}$.
$$x = \frac{72}{18} \times 1 = 4 \text{ cm}$$
So the distance on the map will be 4 cm.
Activity — Ages of Hima and Sima
Question: The table shows the ages of Hima ($x$) and Sima ($y$). As $x$ increases, $y$ also increases; are their ages in direct proportion?
| Age 5 years ago | Present age | Age 5 years later | |
|---|---|---|---|
| Age of Hima (x) | 9 | 14 | 19 |
| Age of Sima (y) | 10 | 15 | 20 |
Answer: Taking $\frac{x}{y}$: $\frac{9}{10}, \frac{14}{15}, \frac{19}{20}$ — the ratio is not the same. So although $x$ and $y$ increase together, they are not in direct proportion. This shows that two quantities increasing together are not necessarily proportional.
Inverse Proportion (13.2) — Complete the tables
The distance between Nagaon and Guwahati is 120 km; at 40 km/hr it takes 3 hours, at 60 km/hr only 2 hours (as speed increases, time decreases). Again, 5 men do a piece of work in 18 days, while 10 men do it in 9 days. In all these cases, as $x$ increases $y$ decreases and the product $xy$ stays constant — this is inverse proportion. Complete the tables below.
Table 1 (fixed distance 120 km; $xy = 120$):
| Speed (x km/hr) | 10 | 15 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|---|
| Time (y hours) | 12 | 8 | 6 | 3 | 2 | 1.5 |
| xy | 120 | 120 | 120 | 120 | 120 | 120 |
Table 2 (fixed work; $xy = 90$):
| Number of men (x) | 5 | 10 | 15 | 18 | 30 | 45 |
|---|---|---|---|---|---|---|
| Time (y days) | 18 | 9 | 6 | 5 | 3 | 2 |
| xy | 90 | 90 | 90 | 90 | 90 | 90 |
Table 3 (rectangle of fixed area 144 sq. m; $xy = 144$):
| Length of one side (x m) | 12 | 9 | 36 | — | 8 | 24 | 48 |
|---|---|---|---|---|---|---|---|
| Length of other side (y m) | 12 | 16 | 4 | 26 | 18 | 6 | 3 |
| xy | 144 | 144 | 144 | 144 | 144 | 144 | 144 |
Since the product is always constant ($120$, $90$ and $144$), all three tables show inverse proportion. For example, in Table 3, when $x = 8$, $y = \frac{144}{8} = 18$, and when $y = 3$, $x = \frac{144}{3} = 48$. (Note: the printed cell $y = 26$ does not fit the constant product 144 exactly — $\frac{144}{26} \approx 5.5$ — which appears to be a printing slip.)
Activity — Arranging rows
Question: In a school assembly, 400 students are arranged in 10 rows with 40 students in each row. Keeping the same number of students, change the number of rows and complete the table.
| Number of rows (x) | 10 | 16 | 20 | 8 |
|---|---|---|---|---|
| Students in each row (y) | 40 | 25 | 20 | 50 |
Answer: Total students $= 10 \times 40 = 400$; so $y = \frac{400}{x}$. Thus $x = 16 \Rightarrow y = 25$, $x = 20 \Rightarrow y = 20$, $x = 8 \Rightarrow y = 50$.
(i) As $x$ increases $y$ decreases (and as $x$ decreases $y$ increases) — Yes. (ii) $xy = 400$ in each case (same) — Yes. (iii) $x_1 y_1 = x_2 y_2$ (e.g. $10 \times 40 = 16 \times 25 = 400$) — Yes. (iv) $\frac{x_1}{x_2} = \frac{y_2}{y_1}$ — Yes. (v) $\frac{x_2}{x_1} = \frac{y_1}{y_2}$ — Yes. So $x$ and $y$ are in inverse proportion.
Inverse Proportion (13.2) — Worked Examples
Example 1: Observe the following tables and find those pairs of $x$ and $y$ that are inversely proportional.
(i) $x$: 50, 40, 30, 20 and $y$: 5, 6, 7, 8
Answer: $xy = 250, 240, 210, 160$ — not the same, so not inversely proportional.
(ii) $x$: 100, 200, 300, 400 and $y$: 60, 30, 20, 15
Answer: $xy = 6000, 6000, 6000, 6000$ — always the same (constant), so inversely proportional.
(iii) $x$: 90, 60, 45, 30, 20, 5 and $y$: 10, 15, 20, 25, 30, 35
Answer: $xy = 900, 900, 900, 750, 600, 175$ — not the same, so not inversely proportional.
Example 2: The pencils in a box are shared among 15 students, and each gets 8 pencils. If the number of students increases by 5, how many pencils will each get?
Answer: As the number of students increases, the pencils each gets decreases — inverse proportion. New number of students $= 15 + 5 = 20$.
$$x_1 y_1 = x_2 y_2 \Rightarrow 15 \times 8 = 20 \times y_2 \Rightarrow y_2 = \frac{120}{20} = 6$$
So each student gets 6 pencils.
Example 3: A bus covers a distance in $4\frac{1}{2}$ hours at a speed of 40 km/hr. If it returns at 45 km/hr, how much time will it take to cover the same distance?
Answer: A higher speed takes less time — inverse proportion. Speed $x$, time $y$.
$$x_1 y_1 = x_2 y_2 \Rightarrow 40 \times \frac{9}{2} = 45 \times y_2 \Rightarrow 180 = 45\, y_2 \Rightarrow y_2 = 4$$
So it takes 4 hours on the return trip.
Example 4: One man can build a thatched house in 15 days. In how many days can 3 men build it?
Answer: More men means less time — inverse proportion.
$$x_1 y_1 = x_2 y_2 \Rightarrow 1 \times 15 = 3 \times y_2 \Rightarrow y_2 = \frac{15}{3} = 5$$
So 3 men can build the house in 5 days.
Example 5: 10 men can dig a pond in 17 days. In how many days can 34 men dig it?
Answer: More men means less time — inverse proportion.
$$x_1 y_1 = x_2 y_2 \Rightarrow 10 \times 17 = 34 \times y_2 \Rightarrow y_2 = \frac{170}{34} = 5$$
So 34 men can dig the pond in 5 days.
Exercise 13.2
1. Write whether the following quantities are inversely proportional or not.
(i) Number of workers and time taken to complete the work.
Answer: As the number of workers increases, the time decreases, so they are inversely proportional.
(ii) Speed of a vehicle and time taken to cover a distance.
Answer: As the speed increases, the time decreases, so they are inversely proportional.
(iii) Area of cultivable land and its production.
Answer: As the area increases, the production also increases (direct proportion), so they are not inversely proportional.
2. The people of a village decide to dig a public pond. If 30 villagers can dig it in 35 days, in how many days can 210 villagers dig it?
Answer: More people means less time — inverse proportion.
$$30 \times 35 = 210 \times y \Rightarrow y = \frac{1050}{210} = 5 \text{ days}$$
So 210 villagers can dig the pond in 5 days.
3. 15 women of a Self Help Group can weave 500 gamosas in 24 days. In how many days can 5, 8 or 24 women weave the same 500 gamosas?
Answer: More women means less time — inverse proportion; constant $= 15 \times 24 = 360$.
$$5 \text{ women: } \frac{360}{5} = 72 \text{ days},\quad 8 \text{ women: } \frac{360}{8} = 45 \text{ days},\quad 24 \text{ women: } \frac{360}{24} = 15 \text{ days}$$
4. An empty tank can be filled in 1 hour 20 minutes by 6 pipes. If 5 similar pipes are used, how much time will be needed to fill the tank?
Answer: Fewer pipes means more time — inverse proportion. 1 hour 20 minutes $= 80$ minutes.
$$6 \times 80 = 5 \times y \Rightarrow y = \frac{480}{5} = 96 \text{ minutes}$$
So 5 pipes will fill the tank in 96 minutes, i.e. 1 hour 36 minutes.
5. A dam 4 km long can be built in 42 days by 120 workers. How many workers are needed to build the same dam in 30 days?
Answer: Fewer days needs more workers — inverse proportion.
$$120 \times 42 = x \times 30 \Rightarrow x = \frac{5040}{30} = 168 \text{ workers}$$
So 168 workers are needed.
6. A family of 6 members can run its expenses for 42 days on a fixed income. If one more member is added, for how many days can the family run on the same income?
Answer: More members means fewer days — inverse proportion. New number of members $= 6 + 1 = 7$.
$$6 \times 42 = 7 \times y \Rightarrow y = \frac{252}{7} = 36 \text{ days}$$
So the family can run for 36 days.
7. 35 women can complete a work in 160 days. In how many days can 28 women complete it?
Answer: Fewer women means more days — inverse proportion.
$$35 \times 160 = 28 \times y \Rightarrow y = \frac{5600}{28} = 200 \text{ days}$$
So 28 women can complete the work in 200 days.
8. Two carpenters can complete 4 new windows in 7 days. (i) A carpenter fell ill before starting; in how many days can the remaining carpenter complete the work? (ii) How many carpenters are needed to complete the work in a single day?
Answer: Number of carpenters and time are inversely proportional; constant $= 2 \times 7 = 14$.
(i) $$2 \times 7 = 1 \times y \Rightarrow y = 14 \text{ days}$$
(ii) $$2 \times 7 = x \times 1 \Rightarrow x = 14 \text{ carpenters}$$
So (i) one carpenter completes the work in 14 days, and (ii) 14 carpenters are needed to finish it in a single day.
Multiple Choice Questions
1. If $x$ and $y$ are directly proportional, which one is correct? (a) $\frac{x_1}{y_1} = \frac{x_2}{y_2}$ (b) $x_1 y_1 = x_2 y_2$ (c) $x_1 x_2 = y_1 y_2$ (d) $\frac{x_1}{x_2} = \frac{y_2}{y_1}$
Answer: (a) $\frac{x_1}{y_1} = \frac{x_2}{y_2}$.
2. If $x_1 = 4$, $y_1 = 10$, $x_2 = 2$ and $x, y$ are directly proportional, then $y_2 = $ ? (a) 20 (b) 5 (c) 25 (d) 10
Answer: (b) 5, since $\frac{4}{10} = \frac{2}{y_2} \Rightarrow y_2 = \frac{2 \times 10}{4} = 5$.
3. If $x$ and $y$ are inversely proportional, which is correct? (a) $y$ increases when $x$ increases (b) $y$ decreases when $x$ decreases (c) $y$ decreases when $x$ increases (d) None of the above
Answer: (c) $y$ decreases when $x$ increases.
4. If the weight of 5 books is 4 kg, what is the weight of 8 such books? (a) 5 kg (b) 6 kg (c) 6.4 kg (d) 10 kg
Answer: (c) 6.4 kg, since $\frac{4}{5} = \frac{y}{8} \Rightarrow y = \frac{4 \times 8}{5} = 6.4$.
5. 10 persons can dig a tank in 6 hours. How many persons can dig the same tank in 12 hours? (a) 20 (b) 5 (c) 7 (d) 15
Answer: (b) 5, since $10 \times 6 = x \times 12 \Rightarrow x = \frac{60}{12} = 5$ (inverse proportion).
6. A man can complete a work in 6 days. What portion of the work can 2 men complete in a single day? (a) $\frac{1}{3}$ (b) $\frac{1}{6}$ (c) $\frac{1}{2}$ (d) $\frac{1}{12}$
Answer: (a) $\frac{1}{3}$. One man does $\frac{1}{6}$ in a day, so 2 men do $2 \times \frac{1}{6} = \frac{1}{3}$.
7. 36 persons can do a piece of work in 20 days. In how many days can 12 persons do the same work? (a) $\frac{20}{3}$ days (b) 40 days (c) 60 days (d) 8 days
Answer: (c) 60 days, since $36 \times 20 = 12 \times y \Rightarrow y = \frac{720}{12} = 60$.
8. A canteen can provide food to 300 persons for 20 days. If 50 persons decrease, for how many days will the canteen run on the same food? (a) 120 days (b) 17 days (c) 25 days (d) 24 days
Answer: (d) 24 days. Persons $= 300 – 50 = 250$; $300 \times 20 = 250 \times y \Rightarrow y = \frac{6000}{250} = 24$.
What We Have Learnt
- Two quantities $x$ and $y$ are directly proportional if their ratio of increase or decrease is constant, i.e. $\frac{x}{y} = k$. For two pairs, $\frac{x_1}{y_1} = \frac{x_2}{y_2}$.
- Two quantities $x$ and $y$ are inversely proportional if $y$ decreases proportionally as $x$ increases; then their product is constant, i.e. $xy = k$. For two pairs, $x_1 y_1 = x_2 y_2$.
Additional Questions and Answers
Multiple Choice Questions
1. In direct proportion, the value of $\frac{x}{y}$ is — (a) always constant (b) always increasing (c) always zero (d) always decreasing
Answer: (a) always constant.
2. In inverse proportion, which is always constant? (a) $x + y$ (b) $x – y$ (c) $xy$ (d) $\frac{x}{y}$
Answer: (c) $xy$.
3. If the price of 3 books is Rs. 90, what is the price of 7 books? (a) Rs. 180 (b) Rs. 210 (c) Rs. 270 (d) Rs. 630
Answer: (b) Rs. 210. ($\frac{90}{3} = 30$; $30 \times 7 = 210$)
4. 4 men do a work in 12 days. In how many days can 6 men do it? (a) 8 days (b) 18 days (c) 9 days (d) 6 days
Answer: (a) 8 days. ($4 \times 12 = 6 \times y \Rightarrow y = 8$)
5. What is the graph of two quantities in direct proportion like? (a) a straight line through the origin (b) a circle (c) a curved line (d) a line parallel to the $x$-axis
Answer: (a) a straight line through the origin.
Fill in the Blanks
1. In direct proportion, $x = k \_\_\_$ .
Answer: $y$ (that is, $x = ky$).
2. In inverse proportion, $xy = \_\_\_$ .
Answer: a constant ($k$).
3. As speed increases, the time to cover a fixed distance $\_\_\_$ .
Answer: decreases.
4. In direct proportion, $\frac{x_1}{y_1} = \_\_\_$ .
Answer: $\frac{x_2}{y_2}$.
5. Number of workers and time to complete a work are in $\_\_\_$ proportion.
Answer: inverse.
True or False
1. In direct proportion, when one quantity increases the other also increases.
Answer: True.
2. In inverse proportion, the sum of the two quantities is always constant.
Answer: False (the product is constant).
3. Any two quantities that increase together are in direct proportion.
Answer: False (only when the ratio is constant).
4. More workers complete a work in fewer days.
Answer: True.
5. Proportion problems can also be solved by the unitary method.
Answer: True.
Short Answer Questions
1. What is the difference between direct and inverse proportion?
Answer: In direct proportion, when $x$ increases $y$ also increases in the same ratio and $\frac{x}{y}$ stays constant. In inverse proportion, when $x$ increases $y$ decreases in the same ratio and the product $xy$ stays constant.
2. If 2 oranges cost Rs. 24, what is the cost of 5 oranges?
Answer: One orange costs $\frac{24}{2} = 12$ rupees; so 5 oranges cost $12 \times 5 = 60$ rupees (direct proportion).
3. 8 men do a work in 15 days. In how many days can 12 men do it?
Answer: Inverse proportion: $8 \times 15 = 12 \times y \Rightarrow y = \frac{120}{12} = 10$ days.
4. A car covers a distance in 4 hours at 60 km/hr. How much time is needed to cover the same distance at 80 km/hr?
Answer: Inverse proportion: $60 \times 4 = 80 \times y \Rightarrow y = \frac{240}{80} = 3$ hours.
Key Terms
| Term | Meaning |
|---|---|
| Direct proportion | When one quantity increases, the other increases in the same ratio |
| Inverse proportion | When one quantity increases, the other decreases in the same ratio |
| Constant (k) | A fixed, unchanging value |
| Ratio | A comparison of two quantities |
| Proportional | Related by a fixed proportion |
| Unitary method | Solving by first finding the value of one unit |
| Speed | Distance covered per unit time |
| Simple interest | Interest calculated only on the principal |
| Compound interest | Interest calculated on the amount including previous interest |
| Scale | Ratio of map distance to actual distance |