8th Class Mathematics Direct and Inverse Proportions

  • question_answer 7)                 Suppose 2 kg of sugar contains \[9\times {{10}^{6}}\] crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?

    Answer:

                    Suppose the amount of sugar is \[x\,\,kg\] and the number of crystals is \[y\]. As the amount of sugar increases, the number of crystals also increases in the same ratio. It is a case of direct proportion. We make use of the relation of the type \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}\]. (i) Here, \[{{x}_{1}}=2\] \[{{y}_{1}}=9\times {{10}^{6}}\] \[{{x}_{2}}=5\] Therefore, \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}\] gives \[\frac{2}{9\times {{10}^{6}}}\,=\frac{5}{{{y}_{2}}}\] \[\Rightarrow \]               \[2{{y}_{2}}=5\times 9\times {{10}^{6}}\] \[\Rightarrow \]               \[{{y}_{2}}=\frac{5\times 9\times {{10}^{6}}}{2}\] \[\Rightarrow \]               \[{{y}_{2}}=22.5\,\times {{10}^{6}}\] \[\Rightarrow \]               \[{{y}_{2}}=2.25\,\times {{10}^{7}}\] Hence, there are \[225\times {{10}^{5}}\] crystals. (ii) Here, \[{{x}_{1}}=2\] \[{{y}_{1}}=9\times {{10}^{6}}\] \[{{x}_{3}}=1.2\]                 Therefore, \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{3}}}{{{y}_{3}}}\] gives \[\frac{2}{9\times {{10}^{6}}}=\frac{1.2}{{{y}_{3}}}\] \[\Rightarrow \]               \[2{{y}_{3}}=1.2\times 9\times {{10}^{6}}\] \[\Rightarrow \]               \[2{{y}_{3}}=10.8\times {{10}^{6}}\] \[\Rightarrow \]               \[{{y}_{3}}=\frac{10.8\times {{10}^{6}}}{2}\] \[\Rightarrow \]               \[{{y}_{3}}=5.4\times {{10}^{6}}\] Hence, there are \[5.4\times {{10}^{6}}\] crystals.

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