question_answer 20)

Determine, if the following ratios form a proportion. Also, write the middle term and extreme terms, where the ratios form a proportion. (a) 25 cm : 1 m and  Rs. 40 : Rs. 160                                        (b) 39 L : 65 L and 6 bottles : 10 bottles (c) 2 kg : 80 kg and 25 g : 625 g                                     (d) 200 mL : 2.5 L and  Rs. 4 : Rs. 50 TIPS Firstly, convert the terms of a ratio in same unit, then write the given ratios in their lowest form. If lowest form of both ratios are same, then they will be in proportion and their first and fourth terms are known as extreme terms, second and third terms are known as middle terms.

Answer:

                (a) Here, 25 cm : 1 m = 25 cm : 1 \[\times \] 100 cm                 [\[\because \]1 m = 100 cm] \[=25:100=\frac{25}{100}=\frac{25\div 25}{100\div 25}=\frac{1}{4}=1:4\] [dividing numerator and denominator both by 25] and Rs. 40 : Rs. 160 = 40 : 160 \[=\frac{40}{160}=\frac{4}{16}=\frac{1}{4}=1:4\] [dividing numerator and denominator both by 10] Here, 1 : 4 = 1 : 4 i.e. 25 cm : 1 m = Rs. 40 : Rs. 160 So, the ratios of 25 cm : 1 m and Rs. 40 : Rs. 160 are in proportion. i.e. 25 cm : 1 m :: Rs. 40 : Rs. 160 Now, middle terms are 1 m and Rs. 40 and extreme terms are 25 cm and Rs. 160. (b) Here,\[39L:65L=39:65=\frac{39}{65}=\frac{39\div 13}{65\div 13}\]      [\[\because \]HCF of 39 and 65 = 13] \[=\frac{3}{5}=3:5\] and 6 bottles : 10 bottles \[=6:10=\frac{6}{10}=\frac{6\div 2}{10\div 2}=\frac{3}{5}=3:5\] [dividing numerator and denominator both by 2] Here, 3 : 5 = 3 : 5 i.e. 39 L : 65 L = 6 bottles : 10 bottles So, the ratio of 39 L : 65 L and 6 bottles : 10 bottles are in proportion. i.e. 39 L : 65 L : : 6 bottles : 10 bottles. Now, middle terms of ratios are 65 L and 6 bottles and extreme terms of ratios are 39 L and 10 bottles. (c) Here,  \[2kg:80kg=2:80=\frac{2}{80}=\frac{2\div 2}{80\div 2}=\frac{1}{40}=1:40\] [dividing numerator and denominator both by 2] and\[25g:625g=25:625=\frac{25}{625}=\frac{25\div 25}{625\div 25}=\frac{1}{25}=1:25\][\[\because \] HCF of 25 and \[625=5\times 5=25\]] Since, both ratios are not equal. \[\therefore \]  \[\text{2 kg : 80 kg }\ne \text{ 25 g : 625 g }\] Hence, the given ratios are not in proportion. (d) Here, \[200mL:2.5L=200\times \frac{1}{1000}L:2.5L\] \[\left[ \because 1mL=\frac{1}{1000}L \right]\] \[=\frac{200}{1000}L:2.5L=0.200L:2.5L\] \[=0.200:2.5=\frac{0.200}{2.5}=\frac{2}{25}=2:25\] [multiplying numerator and denominator both by 10] and  Rs. 4 : Rs. 50 = 4 : 50 \[=\frac{4}{50}=\frac{4\div 2}{50\div 2}=\frac{2}{25}=2:25\] [dividing numerator and denominator both by 2] Here, 2 : 25 = 2 : 25 i.e. 200 mL : 2.5 L = Rs. 4 : Rs. 50 So, the ratios of 200 mL : 2.5 L and Rs. 4 : Rs. 50 are in proportion. i.e. 200 mL : 2.5 L : : Rs. 4 : Rs. 50 Now, middle terms of ratios are 2.5 L and Rs. 4 and extreme terms of ratios are 200 mL and Rs. 50.