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Find a number whose double is 45 greater than its half.

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#### Solution

Let the number be x .

According to the question,

\[2x = \frac{1}{2}x + 45\]

\[\text{ or }2x - \frac{1}{2}x = 45\]

\[\text{ or }\frac{4x - x}{2} = 45\]

or 3x = 90 [After cross multiplication]

\[\text{ or }x = \frac{90}{3}\]

or x = 30

Thus, the number is 30.

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