Sum
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.
Is there an error in this question or solution?
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