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3 years ago

11

Williiam lost 25% of his marble,gave 1/6 of the remainder to his.Brother and had 120 marbles left.How many marbles did he have a

t first?

1 answer:

bija089 [108]3 years ago

5 0

Answer:

Number of marbles with Williams at the starting = 192

Step-by-step explanation:

Let total number of marbles with Williams = m

He lost marbles = 25% of the total number

= $\frac{25}{100}\times m$

= $\frac{m}{4}$

Remaining marbles with Williams = m - $\frac{m}{4}$

= $\frac{3m}{4}$

He gave marbles to his brother = $\frac{1}{6}$ th of the remaining marbles

= $\frac{1}{6}\times \frac{3m}{4}$

= $\frac{m}{8}$

Remaining marbles after giving marbles to his brother = 120

Therefore, $\frac{3m}{4}-\frac{m}{8}=120$

$\frac{6m}{8}-\frac{m}{8}=120$

$\frac{5m}{8}=120$

m = $\frac{120\times 8}{5}$

m = 192

Number of marbles with Williams at the starting = 192

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