6th Class Mathematics Ratio and Proportion

  • question_answer 7) Find the ratio of the following. (a) 30 min to 1.5 h          (b) 40 cm to 1.5 m              (c) 55 paise to ` 1           (d) 500 mL to 2 L TIPS Firstly, convert both the quantities in same unit and then find required ratio.

    Answer:

                    (a) Here, we have to convert 1.5 h into min. We know that,  1 h = 60 min \[\therefore \] 1.5 h = 1.5 \[\times \] 60 min = 90 min \[\therefore \]Required ratio \[=\frac{30\min }{90\min }=\frac{30}{90}\] \[\because \]\[30=2\times 3\times 5\] and  \[90=2\times 3\times 5\times 3\] \[\therefore \] HCF of 30 and \[90=2\times 3\times 5=30\] Ratio \[=\frac{30\div 30}{90\div 30}=\frac{1}{3}=1:3\] (b) Here, we have to convert 1.5 m into cm. We know that, m = 100 cm \[\therefore \] \[1.5m=1.5\times 100cm=150cm\] \[\therefore \] Required ratio \[=\frac{40cm}{150cm}=\frac{40}{150}\] \[\because \] \[40=2\times 2\times 2\times 5\] and \[150=5\times 3\times 2\times 5\] \[\therefore \] HCF of 40 and \[150=2\times 5=10\] Ratio \[=\frac{40\div 10}{150\div 10}=\frac{4}{15}=4:15\] (c) Here, we have to convert ` 1 into paise. We know that, ` 1 = 100 paise \[\therefore \] Required ratio \[=\frac{55paise}{100paise}=\frac{55}{100}\] \[\because \] \[55=5\times 11\] and \[100=5\times 2\times 2\times 5\] \[\therefore \]HCF of 55 and \[100=5=\frac{55\div 5}{100\div 5}=\frac{11}{20}=11:20\] (d) Here, we have to convert Litre into mL. We know that, 1 L = 1000 mL \[\therefore \] \[2L=2\times 1000\,mL=2000\,mL\] \[\therefore \] Required ratio \[=\frac{500\,mL}{2000mL}=\frac{500}{2000}=\frac{5}{20}\] [dividing numerator and denominator by 100] Ratio = 1/4 = 1 : 4


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