What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?

Class 10 | Exercise 7B | ICSE | Ratio and Proportion | Selina

Let the number subtracted to be x.

therefore,

$\begin{array}{*{35}{l}} \left( 7\text{ }-\text{ }x \right):\text{ }\left( 17\text{ }-\text{ }x \right)::\text{ }\left( 17\text{ }-\text{ }x \right):\text{ }\left( 47\text{ }-\text{ }x \right) \\ ~ \\ \end{array}$

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(B) - 10

$\begin{array}{*{35}{l}} \left( 7\text{ }-\text{ }x \right)\left( 17\text{ }-\text{ }x \right)\text{ }=\text{ }{{\left( 17\text{ }-\text{ }x \right)}^{2}} \\ 329\text{ }-\text{ }47x\text{ }-\text{ }7x\text{ }+\text{ }{{x}^{2}}~=\text{ }289\text{ }-\text{ }34x\text{ }+\text{ }{{x}^{2}} \\ 329\text{ }-\text{ }289\text{ }=\text{ }-34x\text{ }+\text{ }54x \\ 20x\text{ }=\text{ }40 \\ x\text{ }=\text{ }2 \\ \end{array}$

Therefore, the required number which must be subtracted is 2.

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