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### c) Find the number of boys given 18 girls and a class ratio of 3:2.
Given:
- Number of girls = 18
- Ratio of girls to boys = 3:2
To find:
- Number of boys
Steps:
1. The ratio tells us that for every 3 girls, there are 2 boys.
2. We can set up the ratio as follows:
[tex]\[ \frac{\text{Number of girls}}{\text{Number of boys}} = \frac{3}{2} \][/tex]
3. Let the number of boys be [tex]\( x \)[/tex]. Then,
[tex]\[ \frac{18}{x} = \frac{3}{2} \][/tex]
4. Cross-multiplying gives us:
[tex]\[ 18 \cdot 2 = 3 \cdot x \implies 36 = 3x \implies x = \frac{36}{3} = 12 \][/tex]
Hence, the number of boys is 12.
### d) (i) Find the monthly expenditure of the family.
Given:
- Monthly income = Rs 21,000
- Ratio of income to expenditure = 7:4
To find:
- Monthly expenditure
Steps:
1. The ratio tells us that for every 7 parts of income, 4 parts are expenditure.
2. Let the expenditure be [tex]\( x \)[/tex]. Then,
[tex]\[ \frac{\text{Income}}{\text{Expenditure}} = \frac{7}{4} \][/tex]
3. Substituting the income value:
[tex]\[ \frac{21000}{x} = \frac{7}{4} \][/tex]
4. Cross-multiplying gives us:
[tex]\[ 21000 \cdot 4 = 7 \cdot x \implies 84000 = 7x \implies x = \frac{84000}{7} = 12000 \][/tex]
Hence, the monthly expenditure is Rs 12,000.
### d) (ii) Find the yearly savings of the family.
Given:
- Monthly income = Rs 21,000
- Monthly expenditure = Rs 12,000
Steps:
1. Monthly savings = Income - Expenditure
[tex]\[ 21000 - 12000 = 9000 \][/tex]
2. Yearly savings = Monthly savings [tex]\(\times\)[/tex] 12
[tex]\[ 9000 \times 12 = 108000 \][/tex]
Hence, the yearly savings of the family is Rs 1,08,000.
### 11. a) Divide Rs 100 in the ratio of 2:3.
Given:
- Total sum = Rs 100
- Ratio = 2:3
Steps:
1. Sum of ratio parts = 2 + 3 = 5
2. Each part's value = [tex]\(\frac{\text{Total sum}}{\text{Sum of ratio parts}}\)[/tex]
[tex]\[ \frac{100}{5} = 20 \][/tex]
3. Share of each part:
- For 2 parts: [tex]\(2 \times 20 = 40\)[/tex]
- For 3 parts: [tex]\(3 \times 20 = 60\)[/tex]
Hence, the shares are Rs 40 and Rs 60 respectively.
### 11. b) Divide Rs 1,80,000 in the ratio of 4:5.
Given:
- Total sum = Rs 1,80,000
- Ratio = 4:5
Steps:
1. Sum of ratio parts = 4 + 5 = 9
2. Each part's value = [tex]\(\frac{\text{Total sum}}{\text{Sum of ratio parts}}\)[/tex]
[tex]\[ \frac{180000}{9} = 20000 \][/tex]
3. Share of each part:
- For 4 parts: [tex]\(4 \times 20000 = 80000\)[/tex]
- For 5 parts: [tex]\(5 \times 20000 = 100000\)[/tex]
Hence, the son and daughter get Rs 80,000 and Rs 1,00,000 respectively.
### 11. c) Find the mass of metals in a 960 g alloy with a ratio of 5:3.
Given:
- Total mass = 960 g
- Ratio of copper to zinc = 5:3
Steps:
1. Sum of ratio parts = 5 + 3 = 8
2. Each part's mass = [tex]\(\frac{\text{Total mass}}{\text{Sum of ratio parts}}\)[/tex]
[tex]\[ \frac{960}{8} = 120 \][/tex]
3. Mass of each metal:
- Copper: [tex]\(5 \times 120 = 600\)[/tex] g
- Zinc: [tex]\(3 \times 120 = 360\)[/tex] g
Hence, the masses are 600 g of copper and 360 g of zinc.
### 12. a) Find the actual distance between two places represented by 4.5 cm on a map with a scale of 1:100000.
Given:
- Map distance = 4.5 cm
- Scale = 1:100000
Steps:
1. Actual distance = Map distance [tex]\(\times\)[/tex] Scale factor
[tex]\[ 4.5 \times 100000 = 450000 \text{ cm} \][/tex]
2. Convert to km: [tex]\(450000 \text{ cm} \times \frac{1 \text{ km}}{100000 \text{ cm}} = 4.5 \text{ km}\)[/tex]
Hence, the actual distance between the two places is 4.5 km or 450,000 cm.
### 12. b) Find the map distance if the actual distance is 15 km and the map scale is 1:500000.
Given:
- Actual distance = 15 km
- Scale = 1:500000
Steps:
1. Map distance = Actual distance [tex]\(\times\)[/tex] Scale factor
[tex]\[ 15 \text{ km} \times \frac{1}{500000} \text{ cm/km} = \frac{15}{500000} \text{ km cm} \][/tex]
2. Convert to cm:
[tex]\[ \frac{15}{500000} \times 100000 \text{ (to cm)} = 3 \text{ cm} \][/tex]
Hence, the map distance between the two places is approximately 3 cm.
### c) Find the number of boys given 18 girls and a class ratio of 3:2.
Given:
- Number of girls = 18
- Ratio of girls to boys = 3:2
To find:
- Number of boys
Steps:
1. The ratio tells us that for every 3 girls, there are 2 boys.
2. We can set up the ratio as follows:
[tex]\[ \frac{\text{Number of girls}}{\text{Number of boys}} = \frac{3}{2} \][/tex]
3. Let the number of boys be [tex]\( x \)[/tex]. Then,
[tex]\[ \frac{18}{x} = \frac{3}{2} \][/tex]
4. Cross-multiplying gives us:
[tex]\[ 18 \cdot 2 = 3 \cdot x \implies 36 = 3x \implies x = \frac{36}{3} = 12 \][/tex]
Hence, the number of boys is 12.
### d) (i) Find the monthly expenditure of the family.
Given:
- Monthly income = Rs 21,000
- Ratio of income to expenditure = 7:4
To find:
- Monthly expenditure
Steps:
1. The ratio tells us that for every 7 parts of income, 4 parts are expenditure.
2. Let the expenditure be [tex]\( x \)[/tex]. Then,
[tex]\[ \frac{\text{Income}}{\text{Expenditure}} = \frac{7}{4} \][/tex]
3. Substituting the income value:
[tex]\[ \frac{21000}{x} = \frac{7}{4} \][/tex]
4. Cross-multiplying gives us:
[tex]\[ 21000 \cdot 4 = 7 \cdot x \implies 84000 = 7x \implies x = \frac{84000}{7} = 12000 \][/tex]
Hence, the monthly expenditure is Rs 12,000.
### d) (ii) Find the yearly savings of the family.
Given:
- Monthly income = Rs 21,000
- Monthly expenditure = Rs 12,000
Steps:
1. Monthly savings = Income - Expenditure
[tex]\[ 21000 - 12000 = 9000 \][/tex]
2. Yearly savings = Monthly savings [tex]\(\times\)[/tex] 12
[tex]\[ 9000 \times 12 = 108000 \][/tex]
Hence, the yearly savings of the family is Rs 1,08,000.
### 11. a) Divide Rs 100 in the ratio of 2:3.
Given:
- Total sum = Rs 100
- Ratio = 2:3
Steps:
1. Sum of ratio parts = 2 + 3 = 5
2. Each part's value = [tex]\(\frac{\text{Total sum}}{\text{Sum of ratio parts}}\)[/tex]
[tex]\[ \frac{100}{5} = 20 \][/tex]
3. Share of each part:
- For 2 parts: [tex]\(2 \times 20 = 40\)[/tex]
- For 3 parts: [tex]\(3 \times 20 = 60\)[/tex]
Hence, the shares are Rs 40 and Rs 60 respectively.
### 11. b) Divide Rs 1,80,000 in the ratio of 4:5.
Given:
- Total sum = Rs 1,80,000
- Ratio = 4:5
Steps:
1. Sum of ratio parts = 4 + 5 = 9
2. Each part's value = [tex]\(\frac{\text{Total sum}}{\text{Sum of ratio parts}}\)[/tex]
[tex]\[ \frac{180000}{9} = 20000 \][/tex]
3. Share of each part:
- For 4 parts: [tex]\(4 \times 20000 = 80000\)[/tex]
- For 5 parts: [tex]\(5 \times 20000 = 100000\)[/tex]
Hence, the son and daughter get Rs 80,000 and Rs 1,00,000 respectively.
### 11. c) Find the mass of metals in a 960 g alloy with a ratio of 5:3.
Given:
- Total mass = 960 g
- Ratio of copper to zinc = 5:3
Steps:
1. Sum of ratio parts = 5 + 3 = 8
2. Each part's mass = [tex]\(\frac{\text{Total mass}}{\text{Sum of ratio parts}}\)[/tex]
[tex]\[ \frac{960}{8} = 120 \][/tex]
3. Mass of each metal:
- Copper: [tex]\(5 \times 120 = 600\)[/tex] g
- Zinc: [tex]\(3 \times 120 = 360\)[/tex] g
Hence, the masses are 600 g of copper and 360 g of zinc.
### 12. a) Find the actual distance between two places represented by 4.5 cm on a map with a scale of 1:100000.
Given:
- Map distance = 4.5 cm
- Scale = 1:100000
Steps:
1. Actual distance = Map distance [tex]\(\times\)[/tex] Scale factor
[tex]\[ 4.5 \times 100000 = 450000 \text{ cm} \][/tex]
2. Convert to km: [tex]\(450000 \text{ cm} \times \frac{1 \text{ km}}{100000 \text{ cm}} = 4.5 \text{ km}\)[/tex]
Hence, the actual distance between the two places is 4.5 km or 450,000 cm.
### 12. b) Find the map distance if the actual distance is 15 km and the map scale is 1:500000.
Given:
- Actual distance = 15 km
- Scale = 1:500000
Steps:
1. Map distance = Actual distance [tex]\(\times\)[/tex] Scale factor
[tex]\[ 15 \text{ km} \times \frac{1}{500000} \text{ cm/km} = \frac{15}{500000} \text{ km cm} \][/tex]
2. Convert to cm:
[tex]\[ \frac{15}{500000} \times 100000 \text{ (to cm)} = 3 \text{ cm} \][/tex]
Hence, the map distance between the two places is approximately 3 cm.
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