Class 6 Ratio, Proportion and Unitary Method Exercise 9.3

\begin{array}{l} \text {Q1. Which of the following statements are true? } \\ \text {(i) 16: 24=20: 30 } \\ \text {(ii) 21: 6=35: 10 }\\ \text {(iii) 12: 18=28: 12 } \\ \text {(iv) 51: 58=85: 102 }\\ \text {(v) 40 men : 200 men = Rs. 5: Rs. 25 } \\ \text {(vi) 99 kg: 45 kg=Rs. 44: Rs. 20 } \\ \\\text {Sol. We know that two ratios are equal if they are proprotion } \\ \Rightarrow \text {Product of extremes} = \text {Product of means} \\ \\\text {(i) 16: 24=20: 30 } \\ \text {Product of extremes} = 16 \times 30 = 480 \\ \text {Product of means} = 24 \times 20 = 480 \\ \text {Clearly, Product of extremes} = \text {Product of means} \\ \text {Hence, } 16: 24 = 20: 30 \text { is true} \\ \\\text {(ii) 21: 6=35: 10 } \\\text {Product of extremes} = 21 \times 10 = 210 \\ \text {Product of means} = 6 \times 35 = 210 \\ \text {Clearly, Product of extremes} = \text {Product of means} \\ \text {Hence, } 21: 6 = 35 : 10 \text { is true} \\ \\\text {(iii) 12: 18=28: 12 } \\\text {Product of extremes} = 12 \times 12 = 144 \\ \text {Product of means} = 18 \times 28 = 504 \\ \text {Clearly, Product of extremes } \neq \text { Product of means} \\ \text {Hence, } 12: 18=28: 12 \text { is false} \\ \\\text {(iv) 51: 58=85: 102 } \\\text {Product of extremes} = 51 \times 102 = 5202 \\ \text {Product of means} = 58 \times 85 = 4930 \\ \text {Clearly, Product of extremes } \neq \text { Product of means} \\ \text {Hence, } 51: 58=85: 102 \text { is false} \\ \\\text {(v) 40 men : 200 men = Rs. 5: Rs. 25 } \\ 40 : 200 = 5: 25 \\ \text {Product of extremes} = 40 \times 25 = 1000 \\ \text {Product of means} = 200 \times 5 = 1000 \\ \text {Clearly, Product of extremes } = \text { Product of means} \\ \text {Hence, } 40 men : 200 men = Rs. 5: Rs. 25 \text { is true} \\ \\\text {(vi) 99 kg: 45 kg=Rs. 44: Rs. 20 } \\ 99 : 45 = 44: 20 \\ \text {Product of extremes} = 99 \times 20 = 1980 \\ \text {Product of means} = 45 \times 44 = 1980 \\ \text {Clearly, Product of extremes } = \text { Product of means} \\ \text {Hence, } 99 kg: 45 kg=Rs. 44: Rs. 20 \text { is true} \\ \\\text {Q2. Find which of the following are in proportion: } \\ \text {(i) } \quad 8,16,6,12 \\ \text {(ii) } \quad 6,2,4,3 \\ \text {(iii) } \quad 150, 250, 200, 300 \\ \\\text {Sol. We know that 4 numbers are in proprotion if } \\ \text {Product of extremes} = \text {Product of means} \\ \\\text {(i) } \quad 8,16,6,12 \\ \text {Product of extremes} = 8 \times 12 = 96 \\ \text {Product of means} = 16 \times 6 = 96 \\ \text {Clearly, Product of extremes} = \text {Product of means} \\ \text {Hence, 8,16,6,12 are in proportion.} \\ \\\text {(ii) } \quad 6,2,4,3 \\ \text {Product of extremes} = 6 \times 3 = 18 \\ \text {Product of means} = 2 \times 4 = 8 \\ \text {Clearly, Product of extremes} \neq \text {Product of means} \\ \text {Hence, 6,2,4,3 are not in proportion.} \\ \\\text {(iii) } \quad 150,250,200,300 \\ \text {Product of extremes} = 150 \times 300 = 45000 \\ \text {Product of means} = 250 \times 200 = 50000 \\ \text {Clearly, Product of extremes} \neq \text {Product of means} \\ \text {Hence, 150,250,200,300 are not in proportion.} \\ \\\text {Q3. Find x in the following proportions: } \\ \text {(i) } \quad x: 6=55: 11 \\ \text {(ii) } \quad 18: x=27: 3 \\ \text {(iii) } \quad 7: 14=15: x \\ \text {(iv) } \quad 16: 18=x: 96 \\ \\\text {Sol. (i) } \quad x: 6=55: 11 \\ \Rightarrow \frac {x}{6}= \frac {55}{11} \\ \Rightarrow x= \frac {55 \times 6}{11} \\ \Rightarrow x= 30 \\ \\\text {(ii) } \quad 18: x=27: 3 \\ \Rightarrow \frac {18}{x}= \frac {27}{3} \\ \Rightarrow x= \frac {18 \times 3}{27} \\ \Rightarrow x= 2 \\ \\\text {(iii) } \quad 7: 14=15: x \\ \Rightarrow \frac {7}{14}= \frac {15}{x} \\ \Rightarrow x= \frac {15 \times 14}{7} \\ \Rightarrow x= 30 \\ \\\text {(iv) } \quad 16: 18=x: 96 \\ \Rightarrow \frac {16}{18}= \frac {x}{96} \\ \Rightarrow x= \frac {16 \times 96}{18} \\ \Rightarrow x= \frac {256}{3} \\ \\\text {Q4. Set up all proportions from the numbers 9, 150, 105, 1750} \\ \\\text {Sol. We have, } 9 \times 1750 = 15750 \text { and } 150 \times 105 = 15750 \\ \therefore \quad 9 \times 1750 = 150 \times 105 \ldots (i) \\ \Rightarrow \text {First term } = 9 \text {, Second term } = 150 \text {, Third term } = 105,\text {Fourth term } = 1750 \\ \text {Hence, } 9 :150 :: 105 : 1750 \\ \text {We can also write (i) as } \\ 9 \times 1750 = 105 \times 150 \text {Hence, } 9 : 105 :: 150 : 1750 \\ \text { Again, (i) can be written as } \\ 1750 \times 9 = 150 \times 105 \\ \text {Hence, } 1750 : 150 :: 105 : 9 \\ \text {Finally (i) can be written as } \\ 1750 \times 9 = 105 \times 150 \\ \text {Hence, } 1750 : 105 :: 150 : 9 \\ \text {Thus, the required proportions are } \\ 9 :150 :: 105 : 1750 \\ 9 : 105 :: 150 : 1750 \\ 1750 : 150 :: 105 : 9 \\ 1750 : 105 :: 150 : 9 \\ \\\text {Q5. Find the other three proportions involving terms of each of the following: } \\ \text {(i) } \quad 45: 30=24: 16 \\ \text {(ii) } \quad 12: 18=14: 21 \\ \\\text {Sol. (i) Given } 45: 30=24: 16 \\ \Rightarrow 45 \times 16 = 24 \times 30 \\ \text {The other three proportions are } 45 \times 16 = 30 \times 24 \\ \Rightarrow \quad 45 : 30 :: 24 : 16 \\ 16 \times 45 = 30 \times 24 \\ \Rightarrow \quad 16 : 30 :: 24 : 45 \\ 16 \times 45 = 24 \times 30 \\ \Rightarrow \quad 16 : 24 :: 30 : 45 \\ \\\text {(ii) Given } 12: 18=14: 21 \\ \Rightarrow 12 \times 21 = 14 \times 18 \\ \text {The other three proportions are } 12 \times 21 = 18 \times 14 \\ \Rightarrow \quad 12 : 18 :: 14 : 21 \\ 21 \times 12 = 14 \times 18 \\ \Rightarrow \quad 21 : 14 :: 18 : 12 \\ 21 \times 12 = 18 \times 14 \\ \Rightarrow \quad 21 : 18 :: 14 : 12 \\ \\\text {Q6. If 4, x, 9 are in continued proportion, find the value of x } \\ \\\text {Sol. Since 4, x, 9 are in continued proportion, 4, x, x 9 are in proportion.} \\ \Rightarrow \text {Product of extreme terms } = \text {Product of mean terms} \\ \Rightarrow \quad 4 \times 9 = x \times x \\ \Rightarrow \quad x^{2} = 36 \\ \Rightarrow \quad x = 6 \\ \\\text {Q7. If in a proportion, the first, second and fourth terms are 32,112 and 217 } \\ \text { respectively, find the third term. } \\ \\\text {Sol. Given that the first, second and fourth terms are 32,112 and 217 respectively} \\ \text {Lets assume that the third term be x}\\\Rightarrow \quad 32, 112, x ,217 \text { are in proportion}. \\\Rightarrow \text {Product of extreme terms } = \text {Product of mean terms} \\ \Rightarrow \quad 32 \times 217 = 112 \times x \\ \Rightarrow \quad x = \frac {32 \times 217}{112} \\ \Rightarrow \quad x = 62 \\ \\\text {Q8. Show that the following numbers are in continued proportion: } \\ \text {(i) } \quad 36,90,225 \quad \text {(ii) } \quad 48,60,75 \quad \text {(iii) } \quad 16,84,441 \\ \\\text {Sol. (i) If 36,90,225 are in continued proportion then 36,90, 90, 225 are in proportion.} \\ \Rightarrow \text {Product of extreme terms } = \text {Product of mean terms} \\ \Rightarrow \quad 36 \times 225 = 90 \times 90 \\ \Rightarrow \quad 8100 = 8100 \\ \Rightarrow \quad LHS = RHS \\ \\ \text {Hence, 36,90,225 are in continued proportion} \\\text {(ii) If 48,60,75 are in continued proportion then 48,60, 60, 75 are in proportion.} \\ \Rightarrow \text {Product of extreme terms } = \text {Product of mean terms} \\ \Rightarrow \quad 48 \times 75 = 60 \times 60 \\ \Rightarrow \quad 3600 = 3600 \\ \Rightarrow \quad LHS = RHS \\ \\ \text {Hence, 48,60,75 are in continued proportion} \\\text {(iii) If 16,84,441 are in continued proportion then 16,84,84, 441 are in proportion.} \\ \Rightarrow \text {Product of extreme terms } = \text {Product of mean terms} \\ \Rightarrow \quad 16 \times 441 = 84 \times 84 \\ \Rightarrow \quad 7056 = 7056 \\ \Rightarrow \quad LHS = RHS \\ \\ \text {Hence, 16,84,441 are in continued proportion} \\ \\\text {Q9. The ratio of the length of a school ground to its width is 5: 2. } \\ \text {Find its length if the width is 40 metres.} \\ \\\text {Sol. Let’s assume that length of the ground be x metres.} \\ \text {Given that width of school ground is 40 metres.} \\ \Rightarrow \text {The ratio of length to the width of school ground} = x: 40 \\ \text {But its given that ratio of the length of a school ground to its width is 5: 2} \\ \therefore \quad x:40 = 5:2 \\ \Rightarrow \quad x \times 2 = 40 \times 5 \quad \quad [\text {Product of extremes = Product of means}] \\ \Rightarrow \quad x = \frac {40 \times 5}{2} \\ \Rightarrow \quad x = 100 \\ \text {Hence, the lenght of school ground is 100 metres.} \\ \\\text {Q10. The ratio of the sale of eggs on a Sunday to that of the whole week of a } \\ \text { grocery shop was 2: 9 If the total sale of eggs in the same week was Rs. 360, } \\ \text { find the sale of eggs on Sunday.} \\ \\\text {Sol. Let’s assume that sale of eggs on Sunday be Rs. x } \\\text {Given that total sale of eggs for full week is Rs. 360} \\ \Rightarrow \text {The ratio of eggs sale on sunday to full week } = x: 360 \\ \text {But its given that ratio of eggs sale on sunday to full week is 2: 9} \\ \therefore \quad x:360 = 2:9 \\ \Rightarrow \quad x \times 9 = 360 \times 2 \quad \quad [\text {Product of extremes = Product of means}] \\ \Rightarrow \quad x = \frac {360 \times 2 }{9} \\ \Rightarrow \quad x = \text {Rs. } 80 \\ \text {Hence, the sale of eggs on Sunday is Rs. 80} \\ \\\text {Q11. The ratio of copper and zinc in an alloy is 9 : 7. If the weight of zinc } \\ \text {in the alloy is 9.8 kg, find the weight of copper in the alloy.} \\ \\\text {Sol. Let’s assume that weight of copper in alloy be x kg} \\ \text {Given that weight of zinc in alloy is 9.8 kg.} \\ \Rightarrow \text {The ratio of copper to zinc in alloy} = x : 9.8 \\ \text {But its given that ratio of copper and zinc in an alloy is 9 : 7} \\ \therefore \quad x : 9.8 = 9 : 7 \\ \Rightarrow \quad x \times 7 = 9.8 \times 9 \quad \quad [\text {Product of extremes = Product of means}] \\ \Rightarrow \quad x = \frac {9.8 \times 9}{7} \\ \Rightarrow \quad x = 12.6 \text { kg}\\ \text {Hence, the weight of copper in the alloy is 12.6 kg.} \\ \\\text {Q12. The ratio of the income to the expenditure of a family is 7: 6.} \\ \text {Find the savings if the income is Rs. 1400} \\ \\\text {Sol. Lets assume that the expenditure be Rs. x} \\\text {Given that income is Rs. 1400} \\ \Rightarrow \text {The ratio of income to expenditure} = 1400 : x \\ \text {But its given that ratio of the income to the expenditure of the family is 7 : 6} \\ \therefore \quad 1400 : x = 7 : 6 \\ \Rightarrow \quad 1400 \times 6 = x \times 7 \quad \quad [\text {Product of extremes = Product of means}] \\ \Rightarrow \quad x = \frac {1400 \times 6}{7} \\ \Rightarrow \quad x = \text {Rs. } 1200\\ \text {We know that Saving = Income – Expenditure } \\ \Rightarrow \quad \text {Saving } =\text {Rs. } 1400 – 1200 =\text {Rs. } 200 \\ \text {Hence, saving of the family is Rs. 200} \\ \\\text {Q13. The ratio of story books in a library to other books is 1: 7. } \\ \text {The total number of story books is 800. } \\ \text {Find the total number of books in the library.} \\ \\\text {Sol. Lets assume that total number of other books in the library be x} \\\text {Given that total number of story books are 800} \\ \Rightarrow \text {The ratio of story books to total no. of other books} = 800 : x \\ \text {But its given that ratio of story books in a library to other books is 1 : 7} \\ \therefore \quad 800 : x = 1 : 7 \\ \Rightarrow \quad 800 \times 7 = x \times 1 \quad \quad [\text {Product of extremes = Product of means}] \\ \Rightarrow \quad x = 5600 \\ \therefore \text {Total no. of other books in the library are 5600.} \\ \Rightarrow \text {Total no. of books in library } \\ = \text {Story Books } + \text { Other books } \\ = 800 + 5600 = 6400 \\ \text {Hence, a total of 6400 books are there in the library.} \\ \\ \end{array}