Class 8 NCERT Solutions Chapter 13 Direct and Inverse Proportions Exercise 13.1 (original) (raw)

Last Updated : 11 Sep, 2024

In Chapter 13: Direct and Inverse Proportions we will learn about two important mathematical relationships—Direct Proportion and Inverse Proportion. These concepts are frequently used in real-life scenarios where two quantities either increase or decrease together. Understanding these relationships helps in solving problems where variables are related to each other in proportional ways.

What are Direct and Inverse Proportions?

Direct Proportion:

The Two quantities are said to be in direct proportion if an increase in one quantity results in a proportional increase in the other or a decrease in one quantity results in a proportional decrease in the other. Mathematically, if x and y are in direct proportion then:

\frac{x}{y} = \text{constant} \quad \text{or} \quad x \propto y

Inverse Proportion:

The Two quantities are said to be in inverse proportion if an increase in one quantity results in a proportional decrease in the other and vice versa. Mathematically, if x and y are in inverse proportion then:

x \times y = \text{constant} \quad \text{or} \quad x \propto \frac{1}{y}

**Question 1. The following are the car parking charges near a railway station up to.

**4 hours — ₹ 60
**8 hours — ₹ 100
**12 hours — ₹ 140
**24 hours — ₹ 180
**Check if the parking charges are in direct proportion to the parking time.

**Solution:

Table for the given question -

**Number of hours	4	8	12	24
**Parking Cost	60	100	140	180

The ratio of time and respective parking charge can be calculated as:

4/60 = 1/15

8/100 = 2/25

12/140 = 3/35

24/180 = 2/15

We can clearly see that 1/15 ≠ 2/25 ≠ 3/35 ≠ 2/15

So, the parking charges are **not direct proportions to the parking time.

**Question 2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of the base. In the following table, find the parts of the base that need to be added.

**Solution:

The mixture is prepared by adding 1 part of red pigment to 8 parts of the base, if red pigment parts increase, then correspondingly parts of the base will also increase, so they are in Direct proportion.

**Parts of red pigment	1	4	7	12	20
**Parts of base	8	a1	a2	a3	a4

As the red pigment is in direct proportion with parts of the base,

i) 1/8 = 4/a1

So, **a1 = 32

ii) 1/8 = 7/a2

So, **a2 = 56

iii) 1/8 = 12/a3

So, **a3 = 96

iv) 1/8 = 20/a4

So, **a4 = 160

Hence, the required parts of the bases are

**Parts of red pigment	1	4	7	12	20
**Parts of base	8	32	56	96	160

**Question 3. In Question 2 above, if 1 part of a red pigment requires 75 mL of the base, how much red pigment should we mix with 1800 mL of the base?

**Solution:

Let the parts of red pigment required to mix with 1800 mL of the base = a

**Parts of red pigment	1	a
**Parts of base	75	1800

Since the red pigment is in Direct proportion with parts of the base,

So, 1/75 mL = a/1800 mL

a = 1800/75 = 24

Hence, 24 parts of red pigment are required to mix with 1800 mL of a base to form a mixture.

**Question 4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

**Solution:

Let the number of bottles filled = a

**Number of Bottles	840	a
**Time in hours	6	5

Since the time taken to fill the bottles is directly proportional to the number of bottles filled,

So, 840/6 = a/5

a = (840 × 5)/6

a = 700

Hence, 700 bottles will be filled in 5 hours.

**Question 5. A photograph of bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

**Solution:

Let the actual length of bacteria is a cm and enlarged length is b cm

**Length of Bacteria (cm)	5	a	b
**Number of times photograph is enlarged	50,000	1	20,000

Since the length of bacteria is directly proportional to the number of times the photograph is enlarged,

So, 5/50000 = a/1

a = 1/10000 = 10-4

**The actual length is 10 -4 cm

And, 5/50000 = b/20000

b = 2 cm

**The enlarged length is 2 cm

Hence, the actual length of Bacteria is 10-4 cm, and the length of bacteria is 2 cm when the photograph is enlarger 20,000 times.

**Question 6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

**Solution:

Let the required length of the model ship is a cm

| | **Mast Length | **Ship Length | | | ------------------- | ----------------- | ---- | | **Model Ship | 9 cm | a cm | | **Actual Ship | 12 m | 28 m |

Since the mast length and ship length of both model and actual ship are directly proportional

So, 9/12 = a/28

a = (28 X 9)/12 = 21 cm

Hence, the length of the model ship is 21 cm, if the length of the actual ship is 28 m.

**Question 7. Suppose 2 kg of sugar contains 9 × 10 6 Crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?

**Solution:

i) Let the number of sugar crystals is a

**Weight of Sugar (kg)	2	5
**Number of Sugar crystals	9 X 106	a

Since the weight of Sugar is directly proportional to the number of sugar crystals,

So, 2/9 X 106 = 5/a

**a = 2.25 X 10 7 **Crystals

Hence, the number of sugar crystals is 2.25 X 107 in 5 kg of sugar.

ii) Let the number of sugar crystals is b

**Weight of Sugar (kg)	2	1.2
**Number of Sugar crystals	9 X 106	b

Since the weight of Sugar is directly proportional to the number of sugar crystals,

So, So, 2/9 X 106 = 1.2/b

b = **5.4 X 10 6 Crystals

Hence, the number of sugar crystals is 5.4 X 106 in 1.2 kg of sugar.

**Question 8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered on the map?

**Solution:

Let the distance covered on the map is a cm

**Distance covered on the road (km)	18	72
**Distance covered on the map (cm)	1	a

Since distance covered on road is directly proportional to the distance covered on the map

So, 18/1 = 72/a

**a = 4 cm

Hence, Rashmi covers only 4 cm on the map, while she drives 72 km on the road.

**Question 9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (i) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long

**Solution:

i) Let the length of the pole's shadow is a m

**Height of Pole (m)	5.60	10.50
**Length of Pole's Shadow (m)	3.20	a

As we know more is the height of the pole, more will be the length of its shadow, it means the height of the pole is directly proportional to the length of the shadow.

So, 5.60/3.20 = 10.50/a

**a = 6 m

Hence, the length of the pole shadow is 6 m, when the height of the pole is 10 m 50 cm.

ii) Let the height of the pole is b m

**Height of Pole (m)	5.60	b
**Length of Pole's Shadow (m)	3.20	5

As we know more is the height of the pole, more will be the length of its shadow, it means the height of the pole is directly proportional to the length of the shadow.

So, 5.60/3.20 = b/5

**b = 8.75 m or 8 m 75 cm

Hence, the height of the pole is 8 m 75 cm.

**Question 10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

**Solution:

Let the truck travels a km

As 1 hour = 60 minutes

So, 5 hours = 300 minutes

**Distance travelled (km)	14	a
**Time taken (minutes)	25	300

Since the distance travelled by truck is directly proportional to the time taken,

So, 14/25 = a/300

**a = 168 km

Hence, the truck travels 168 km in 5 hours.

Conclusion

Understanding direct and inverse proportions is crucial for the solving real-world problems that involve relationships between the two quantities. These concepts help in making better predictions and decisions in the everyday situations such as the calculating speed, cost, time or other related variables. Mastering these proportions will strengthen the mathematical foundation for the solving problems efficiently.