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(D) Let $X$ be subtracted from the numbers $9, 15$ and $27$ such that the resulting numbers are in continued proportion.This means $(9-X), (15-X)$ and

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $X$ be subtracted from the numbers $9, 15$ and $27$ such that the resulting numbers are in continued proportion.This means $(9-X), (15-X)$ and $(27-X)$ are in continued proportion.For three numbers $a, b, c$ to be in continued proportion,they must satisfy the condition $b^2 = ac$.Here,$a = (9-X)$,$b = (15-X)$,and $c = (27-X)$.Substituting these values into the condition:$(15-X)^2 = (9-X)(27-X)$Expanding both sides:$225 - 30X + X^2 = 243 - 9X - 27X + X^2$$225 - 30X + X^2 = 243 - 36X + X^2$Subtracting $X^2$ from both sides:$225 - 30X = 243 - 36X$Rearranging the terms to solve for $X$:$36X - 30X = 243 - 225$$6X = 18$$X = 3$Thus,the value of $X$ is $3$.

When $X$ is subtracted from the numbers $9, 15$ and $27$,the remainders are in continued proportion. What is the value of $X$?

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