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What least number must be subtracted from each of the numbers 23, 30, 57 and 78, so that the remainders are in proportion?

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Question

What least number must be subtracted from each of the numbers 23, 30, 57 and 78, so that the remainders are in proportion?

Sum
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Solution

Let the number to be subtracted be $$x$$.

Then, $$(23 - x), (30 - x), (57 - x), (78 - x)$$ are in proportion.

Therefore, $$(23 - x)(78 - x) = (30 - x)(57 - x)$$

$$1794 - 23x - 78x + x^2 = 1710 - 30x - 57x + x^2$$

$$1794 - 101x + x^2 = 1710 - 87x + x^2$$ 

Subtracting $$x^2$$ from both sides:

$$1794 - 1710 = 101x - 87x$$

$$84 = 14x$$

$$x = \frac{84}{14} = 6$$

Hence, the least number to be subtracted is 6.

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Chapter 7: Ratio and Proportion - EXERCISE 7B [Page 103]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 7 Ratio and Proportion
EXERCISE 7B | Q 8. | Page 103
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