Question

find the number whose 5th part exceeds its 9th part by 4

Answer 1

let number be x.
x/5-x/9=4
LCM=45
9x-5x/45=4
4x=45*4
therefore, x=45
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Answer 2

Answer:

$Required\:number = 45$

Step-by-step explanation:

$Let \: the \: number \:be \:x$

$Fifth \: part \: of \: the \\ number = \frac{x}{5}$

$Ninth \: part \: of \: the \\ number = \frac{x}{9}$

According to the problem given,

$\frac{x}{5}-\frac{x}{9}=4$

$\implies \frac{9x-5x}{45}=4$

$\implies \frac{4x}{45}=4$

$\implies x=4\times \frac{45}{4}$

$\implies x=45$

Therefore,

$Required\:number = x = 45$

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