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You can help us to improve by giving your valuable suggestions atBy using the service of this site, I agree that I will serve wholeheartedly and will not indulge in any sort of activity that threats the integrity of the organisation I am working for / I work for. Knowledge is the power, Dont miss any paper, Subscribe to usWhat number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion ? Let x be subtracted from each term, then23 – x, 30 – x, 57 – x and 78 – x are proportional23 – x : 30 – x : : 57 – x : 78 – x⇒ `(23 – x)/(30 – x) = (57 – x)/(78 – x)`⇒ (23 – x) (78 – x) = (30 – x) (57 – x) ⇒ 1794 – 23x – 78x + x2 = 1710 – 30x – 57x + x2 ⇒ x2 – 101x + 1794 = x2 – 87x + 1710 ⇒ x2 – 101x + 1794 – x2 + 87x – 1710 = 0⇒ –14x + 84 = 0⇒ 14x = 84∴ x = `(84)/(14)` = 6 Hence 6 is to be subtracted. Concept: Concept of Proportion Is there an error in this question or solution?
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Answer:
Consider x be subtracted from each term 23 – x, 30 – x, 57 – x and 78 – x are proportional It can be written as 23 – x: 30 – x :: 57 – x: 78 – x (23 – x)/ (30 – x) = (57 – x)/ (78 – x) By cross multiplication (23 – x) (78 – x) = (30 – x) (57 – x) By further calculation \begin{array}{l} 1794-23 x-78 x+x^{2}=1710-30 x-57 x+x^{2} \\ x^{2}-101 x+1794-x^{2}+87 x-1710=0 \end{array} So we get -14x + 84 = 0 14x = 84 x = 84/14 = 6 Therefore, 6 is the number to be subtracted from each of the numbers.
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