Question

find a number whose double is 45 than its half​

Answer 1

Answer :-

• The number is 30.

To find :-

• The number whose double is 45 more than it's half.

Solution :-

Let the number be "x".

Then double the number will be :-

• $\rm2x$

And half of the number will be :-

• $\rm \dfrac{1}{2} x$

45 more than half the number will be :-

• $\rm\dfrac{1}{2} x + 45$

Now, it is given that :-

• Double the number is 45 more than it's half.

Therefore,

$:\longmapsto \sf2x = \dfrac{1}{2} x + 45$

$:\longmapsto\sf2x = \dfrac{x}{2} + 45$

$:\longmapsto\sf2x = \dfrac{x \times 1 + 45 \times 2}{2}$

$:\longmapsto\sf2x = \dfrac{x + 90}{2}$

$:\longmapsto\sf2x \times 2 = x + 90$

$:\longmapsto\sf4x = x + 90$

$:\longmapsto\sf4x - x = 90$

$:\longmapsto\sf3x = 90$

$:\longmapsto\sf x = \dfrac{90}{3}$

$:\longmapsto \underline{\boxed{\sf x = 30}}$

Hence :-

• The number is 30.

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