Class 7 Ratio and Proportion Exercise 9.3

\begin{array}{l} \text {Q1. Find which of the following are in proportion? } \\ \text {(i) } \quad 33,44,66,88 \quad \text {(ii) } \quad 46,69,69,46 \quad \text {(iii) } \quad 72,84,186,217 \end{array}\begin{array}{l} \text {Sol. We know that if four numbers are in proportion then } \\ \text {Product of the extreme terms } = \text {Product of mean/middle terms.} \\ \\\text {(i) } \quad 33,44,66,88 \\ \text {Product of extreme terms } =33 \times 88=2904 \\ \text {Product of mean terms }=44 \times 66=2904 \\ \because \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \text {Hence, given numbers are in proportion. } \\ \\\text {(ii) } \quad 46,69,69,46 \\ \text {Product of extreme terms } =46 \times 46=2116 \\ \text {Product of mean terms }=469 \times 69=4761 \\ \because \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \text {Hence, given numbers are in proportion. } \\ \\\text {(iii) } \quad 72,84,186,217 \\ \text {Product of extreme terms } =72 \times 217=15624 \\ \text {Product of mean terms }=84 \times 186=15624 \\ \because \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \text {Hence, given numbers are in proportion. } \\ \\\end{array}\begin{array}{l} \text {Q2. Find x in the following proportions: } \\ \text {(i) } \quad 16: 18=x: 96 \quad \text {(ii) } \quad x: 92=87: 116 \end{array}\begin{array}{l} \text {Sol. We know that if four numbers are in proportion then } \\ \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \\ \text {(i) } \quad 16: 18=x: 96 \\ \Rightarrow \quad \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \Rightarrow \quad 16 \times 96 = 18 \times x \\ \text {On dividing both side by 18, we get} \Rightarrow \quad \frac {x \times 18}{18}=\frac{16 \times 96}{18} \\ \Rightarrow \quad x=\frac{16 \times 96}{18} \\ \Rightarrow \quad x=\frac{256}{3} \\ \\\text {(ii) } \quad x: 92=87: 116 \\ \Rightarrow \quad \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \Rightarrow \quad x \times 116 = 92 \times 87 \\ \text {On dividing both side by 116, we get} \Rightarrow \quad \frac {x \times 116}{116}=\frac{92 \times 87}{116} \\ \Rightarrow \quad x=\frac{92 \times 87}{116} \\ \Rightarrow \quad x=69 \\ \\ \end{array}\begin{array}{l} \text {Q3. The ratio of the income to the expenditure of a family is 7: 6. Find the savings } \\ \text { if the income is Rs. 1400 } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that the ratio of income and expenditure } =7: 6 \\ \text {And total income } = \text {Rs. } 1400 \\ \Rightarrow \quad \text {Income} : \text {Expenditure} = 7 :6 \\\Rightarrow \quad 1400 : \text {Expenditure} = 7 :6 \\\text { We know that } \\ \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \Rightarrow \quad 1400 \times 6 = \text {Expenditure} \times 7 \\ \Rightarrow \quad \text {Expenditure} = \frac {1400 \times 6}{7} \\ \Rightarrow \quad \text {Expenditure} = 1200 \\ \therefore \text {Savings = Income-Expenditure} =1400-1200 = \text {Rs. } 200\end{array}\begin{array}{l} \text {Q4. The scale of a map is 1: 4000000. What is the actual distance between the two towns } \\ \text { if they are 5 cm apart on the map? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {The scale of map } =1: 4000000 \\ \text {Let’s assume the actual distance between towns is x cms.} \\ \Rightarrow \quad 1: 4000000=5: x \\ \text { We know that } \\ \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \Rightarrow \quad x \times 1=5 \times 4000000 \\ \Rightarrow \quad x =20000000 \text { cm} \\ \Rightarrow \quad x =20000000 \times \frac {1}{100000} \text { km} \quad \quad [\because \text {1 cm }= \frac {1}{100000} \text { km}]\\ \Rightarrow \quad x =200 \text { km} \\ \therefore \text {The distance between two town is 200 km.} \end{array}\begin{array}{l} \text {Q5. The ratio of income of a person to his savings is 10:1. If his savings of one year are Rs. 6000, what is his income per month? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that the ratio of income and saving } =10: 1 \\ \text {And total yearly saving } = \text {Rs. } 6000 \\ \therefore \text {Monthly saving } =\frac {6000}{12} = \text {Rs. } 500 \\\text {Lets assume income per month be Rs. x, we get } \\ \text {Income} : \text {Saving} = 10:1 \\ \Rightarrow \quad x: 500= 10 :1 \\\text { We know that } \\ \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \Rightarrow \quad x= 500 \times 10 \\ \Rightarrow \quad x = 5000 \\ \therefore \text {Income per month is Rs. 5000}\end{array}\begin{array}{l} \text {Q6. An electric pole casts a shadow of length 20 metres at a time when a tree 6 metres high casts a shadow of length 8 metres.} \\ \text {Find the height of the pole. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that Shadow of pole } = 20 \text { metres.} \\ \text {Height of tree} = 6 \text { metres.} \\ \text {Shadow of tree} = 8 \text { metres.} \\ \text {Lets assume height of pole } = x\text { metres.} \\\text {Height of the tree : Length of the shadow of tree :: Height of the pole : Length of the shadow of pole } \\\Rightarrow \quad x: 20=6: 8 \\ \text { We know that } \\ \text {Product of the extreme terms } = \text {Product of mean terms.} \\ \Rightarrow \quad x \times 8 =20 \times 6 \\ \text {Divde both sides by 8, we get} \\ \Rightarrow \quad \frac {x \times 8}{8} =\frac{120}{8} \\ \Rightarrow \quad x=15 \\ \therefore \text {Height of pole is 15 metres.}\end{array}