These NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1
Question 1.
Following are the car parking charges near a railway station upto
| 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.
Answer:
| Parking time | Parking charges | Ratio |
| 4 hours | ₹ 60 | \(\frac { 60 }{ 4 }\) = 15 |
| 8 hours | ₹ 100 | \(\frac{100}{8}=\frac{25}{2}\) |
| 12 hours | ₹ 140 | \(\frac{140}{12}=\frac{35}{3}\) |
| 24 hours | ₹ 180 | \(\frac{180}{24}=\frac{15}{2}\) |
Since \(\frac{15}{1} \neq \frac{25}{2} \neq \frac{35}{3} \neq \frac{15}{2}\)
∴ The parking charges are not in direct proportion with the parking time.
Question 2.
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Answer:
Let the number of parts of red pigment be x1, x2, x3, x4 , x5 and the number of parts of
the base be y1, y2, y3, y4, y5
As the number of parts of red pigment increases, number of parts of the base also increases in the same ratio It is a direct proportion
The required parts of bases are 32, 56, 96 and 160.
Question 3.
In question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Answer:
We have \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{1}{8}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x1 = 1, y1 = 75 and y2 = 1800
∴ \(\frac{1}{75}=\frac{x_{2}}{1800}\)
x2 = \(\frac{1 \times 1800}{75}\) = 24
Thus, the required no. of red pigments = 24.
Question 4.
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Answer:
| Number of bottles filled | Number of hours |
| 840 | 6 |
| X | 5 |
More the number of hours, more the number of bottles that would be filled. Thus, given quantities vary directly.
∴ \(\frac{840}{x}=\frac{6}{5}\)
x = \(\frac{840 \times 5}{6}\) = 140 × 5 = 700
∴ The required number of bottles = 700
Question 5.
A photograph of a 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?
Answer:
Actual length = \(\frac{5}{50000}=\frac{1}{10000}\) = 10-4 cm
| Number of times photograph enlarged | Length (in cm) |
| 50,000 | 5 |
| 20,000 | x |
The length increases with an increment in the number of times the photograph enlarged.
∴ It is a case of direct proportion.
\(\frac{50000}{20000}=\frac{5}{x}\)
50,000 × x = 5 × 20,000
x = \(\frac{5 \times 20,000}{50,000}\) = 2
∴The enlarged length = 2 cm.
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?
Answer:
Let the required length of the model ship be ‘x’ cm
| Length of the ship | Height of the mast |
| 28 | 12 |
| X | 9 |
Since, more the length of the ship, more would be the length of its mast.
∴ It is a direct variation \(\frac{28}{x}=\frac{12}{9}\)
12 × x = 28 × 9
x = \(\frac{28 \times 9}{12}=\frac{7 \times 9}{3}\) = 7 × 3 = 21
The required length of the model = 21 cm.
Question 7.
Suppose 2 kg of sugar contains 9 x 106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
Answer:
(i) Let the required number of sugar crystals be Y.
Since more the amount of sugar, more would be the number of sugar crystals.
∴ It is a direct variation
\(\frac{2}{5}=\frac{9 \times 10^{6}}{\mathrm{x}}\)
2x = 5 × 9 × 106
x = \(\frac{5 \times 9 \times 10^{6}}{2}=\frac{45 \times 10^{6}}{2}\) = 22.5 × 106
= 2.25 × 10 × 106
= 2.25 × 107
∴ Required number of sugar crystals = 2.25 x 107
(ii) Let the number of sugar crystals in a sugar be ‘y’
It is a direct variation
\(\frac{2}{1.2}=\frac{9 \times 10^{6}}{\mathrm{y}}\)
2y = 9 × 106 × 1.2
y = \(\frac{9 \times 10^{6} \times 1.2}{2}\) = 9 × 106 × 0.6
= 5.4 × 106
The required number of sugar crystals = 5.4 × 106
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 in the map?
Answer:
Let the required distance covered in the map be ‘x’ cm.
| Distance covered on the Road (in km) | Distance represented on the map (in cm) |
| 18 | 1 |
| 72 | x |
It is a direct variation
\(\frac{18}{72}=\frac{1}{x}\)
18 × x = 72 × 1
x = \(\frac{72 \times 1}{18}\) = 4 × 1 = 4
∴ The required distance on the map is 4 cm
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.
Answer:
Let the required length of shadow be ‘x’ cm
| Height of the pole | Length of the shadow |
| 5 m 60 cm = 560 cm | 3 m 20 cm = 320 cm |
| 10 m 50 cm = 1050 cm | cm |
As the height of the pole increases, the length of the shadow also increases in the same ratio
It is a direct variation.
\(\frac{560}{1050}=\frac{320}{\mathrm{x}}\)
560 × x = 1050 × 320
x = \(\frac{1050 \times 320}{560}\) = 600 cm
∴ Required length of the shadow = 600 cm (6 m)
(ii) Let the required height of the pole be y’ cm
| Height of the pole | Length of the shadow |
| 560 cm | 320 cm |
| y | 5 m = 500 cm |
It is a direct variation.
\(\frac{560}{y}=\frac{320}{500}\)
y × 320 = 560 × 500
y = \(\frac{560 \times 500}{320}\) = 125 × 7 = 875
7 320
∴ The height of the pole = 875 cm (or) 8 m 75 cm.
Question 10.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how for can it travel in 5 hours?
Answer:
Let the distance travelled in 5 hours be ‘x’ km.
| Distance (km) | Time (minutes) |
| 14 | 25 |
| x | 300 |
If the time increases, the distance also increases.
∴ It is a direct variation.
\(\frac{14}{x}=\frac{25}{300}\)
25 × x = 300 × 14
x = \(\frac{300 \times 14}{25}\) = 12 × 14 = 168
∴ The required distance =168 km.