To check if the charges are in direct proportion, we need to see if the ratio of charge to time is constant.
\[ \frac{60}{4} = 15, \quad \frac{100}{8} = 12.5, \quad \frac{140}{12} \approx 11.67, \quad \frac{180}{24} = 7.5 \]
The ratios are not equal (15 ≠ 12.5 ≠ 11.67 ≠ 7.5), so the parking charges are not in direct proportion to the parking time.
| Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
|---|---|---|---|---|---|
| Parts of base | 8 | ... | ... | ... | ... |
Since the mixture is in direct proportion (1 part red pigment : 8 parts base), we can find the missing values by multiplying the red pigment parts by 8.
\[ 4 \times 8 = 32, \quad 7 \times 8 = 56, \quad 12 \times 8 = 96, \quad 20 \times 8 = 160 \]
The completed table is:
| Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
|---|---|---|---|---|---|
| Parts of base | 8 | 32 | 56 | 96 | 160 |
This is a direct proportion problem. Let \(x\) be the parts of red pigment needed for 1800 mL of base.
\[ \frac{1}{75} = \frac{x}{1800} \]
Cross-multiplying:
\[ 75x = 1800 \Rightarrow x = \frac{1800}{75} = 24 \]
This is a direct proportion problem. Let \(x\) be the number of bottles filled in 5 hours.
\[ \frac{840}{6} = \frac{x}{5} \]
\[ 140 = \frac{x}{5} \Rightarrow x = 140 \times 5 = 700 \]
Part 1: Actual length of bacteria
\[ \text{Actual length} = \frac{5}{50000} = 0.0001 \text{ cm} = 1 \times 10^{-4} \text{ cm} \]
Part 2: Enlarged length when enlarged 20,000 times
\[ \text{Enlarged length} = \text{Actual length} \times \text{Enlargement factor} = 0.0001 \times 20000 = 2 \text{ cm} \]
This is a direct proportion problem. The scale of the model is consistent.
First, find the scale factor:
\[ \text{Scale factor} = \frac{\text{Model mast height}}{\text{Actual mast height}} = \frac{9}{1200} = \frac{3}{400} \]
Now, find the model ship length (\(x\)):
\[ \frac{x}{28} = \frac{3}{400} \Rightarrow x = 28 \times \frac{3}{400} = \frac{84}{400} = 0.21 \text{ m} = 21 \text{ cm} \]
This is a direct proportion problem.
(i) For 5 kg of sugar:
\[ \frac{9 \times 10^6}{2} = \frac{x}{5} \Rightarrow x = \frac{9 \times 10^6 \times 5}{2} = 22.5 \times 10^6 = 2.25 \times 10^7 \]
(ii) For 1.2 kg of sugar:
\[ \frac{9 \times 10^6}{2} = \frac{x}{1.2} \Rightarrow x = \frac{9 \times 10^6 \times 1.2}{2} = 5.4 \times 10^6 \]
This is a direct proportion problem. Let \(x\) be the distance on the map.
\[ \frac{1}{18} = \frac{x}{72} \Rightarrow x = \frac{72}{18} = 4 \]
Convert all measurements to the same unit (cm):
First pole: Height = 560 cm, Shadow = 320 cm
(i) For a pole 10 m 50 cm high (1050 cm):
Let \(x\) be the length of the shadow.
\[ \frac{560}{320} = \frac{1050}{x} \Rightarrow x = \frac{1050 \times 320}{560} = 600 \text{ cm} = 6 \text{ m} \]
(ii) For a shadow 5 m long (500 cm):
Let \(y\) be the height of the pole.
\[ \frac{560}{320} = \frac{y}{500} \Rightarrow y = \frac{560 \times 500}{320} = 875 \text{ cm} = 8 \text{ m} 75 \text{ cm} \]
First, convert 5 hours to minutes: 5 × 60 = 300 minutes
This is a direct proportion problem. Let \(x\) be the distance traveled in 300 minutes.
\[ \frac{14}{25} = \frac{x}{300} \Rightarrow x = \frac{14 \times 300}{25} = 168 \]
NCERT Class 8 Mathematics (2024-25 Edition)