What should be subtracted from 15, 28, 20, an | vedclass

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Question

(A) Let the number to be subtracted be $x$.According to the problem,the numbers $(15-x), (28-x), (20-x),$ and $(38-x)$ are in proportion.Therefore,$\f

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the number to be subtracted be $x$.According to the problem,the numbers $(15-x), (28-x), (20-x),$ and $(38-x)$ are in proportion.Therefore,$\frac{15-x}{28-x} = \frac{20-x}{38-x}$.Cross-multiplying,we get: $(15-x)(38-x) = (20-x)(28-x)$.Expanding both sides: $570 - 15x - 38x + x^2 = 560 - 20x - 28x + x^2$.Simplifying: $570 - 53x = 560 - 48x$.Rearranging terms: $570 - 560 = 53x - 48x$.$10 = 5x$.$x = 2$.Thus,the required number is $2$.

What should be subtracted from $15, 28, 20,$ and $38$ so that the remaining numbers are in proportion?

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