Question 6 Ratio and Proportion Exercise 7.2
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Answer:
Consider x to be added to 5, 11, 19 and 37 to make them in proportion
5 + x: 11 + x :: 19 + x: 37 + x
It can be written as
(5 + x) (37 + x) = (11 + x) (19 + x)
By further calculation
\begin{array}{l} 185+5 x+37 x+x^{2}=209+11 x+19 x+x^{2} \\ 185+42 x+x^{2}=209+30 x+x^{2} \end{array}
So we get
42 x-30 x+x^{2}-x^{2}=209-185
12x = 24
x = 2
Hence, the least number to be added is 2.
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What least number must be added to each of the numbers 5, 11, 19 and 37, so that they are in proportion?
Let x be the number added to 5, 11, 19, 37.`(5 + x)/(11 + x) = (19 + x)/(37 + x)`App. comp. and divi.`(5 + x + 11 + x)/(5 + x - 11 - x) = (19 + x + 37 + x )/(19 + x - 37 - x)``(16 + 2x)/(-6) = (56 + 2x)/(-18)`3(16 + 2x) = 56 + 2x6x - 2x = 56 - 484x = 8
x = 2.
Concept: Concept of Proportion
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