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What 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?
Question 7 Ratio and Proportion Exercise 7.2
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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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