Question

2/9 of the people in a restaurant are adults. If there are 95 more children than
adults, how many children are there in the restaurant?
will be? please help me fast in the question​

Answer 1

$\huge \mathbb \red{ANSWER}$

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$\huge \sf \underline{Question}$

2/9 of the people in a restaurant are adults. If there are 95 more children than

adults, how many children are there in the restaurant will be?

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step by step explanation:

$\bold{let \: total \: adults \: and \: children \: be \: x}$

Given,

2/9 of the people in a restaurant are adults

$\bold{adult \:will \: be \: \frac{2}{9}(x)}$

$\bold{children \: will \: be \: \frac{2}{9}(x) + 95}$

Given there are 95 more children than

adults

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Now equation:

$\rm \red{ \frac{2}{9}x + \frac{2}{9}(x) + 95 = x}$

$\rm \blue{ = \frac{4}{9}(x) + 95 = x}$

$\rm \orange{ = 4x + (95)(9) = (x)(9)}$

$\rm \pink{ = 9x - 4x = 855}$

$\rm \yellow{ = \frac{5x}{5} = \frac{855}{5}}$

5x/5 = 855/5

$\tt \blue{x = 171}$

171 is total adult and children

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so now children,

$\tt{ = \frac{2}{9}(x) + 95}$

$\tt{ = \frac{2}{9} (171) + 95}$

$\tt{ = 38 + 95 = 133}$

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Now check the verify adult plus children

$\bf{ \frac{2}{9} (171) + 133 = 171}$

$\bf{ = 38 + 133 = 171}$

$\bf{171 = 171}$

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hence proved