Between-10-PM-and-7-45AM-the-water-level-in-a-swimming-pool-decreased-by-13-16-13-16-inch-Assuming-that-the-water-level-decreased-at-a-constant-rate-how-much-did-it-drop-each-hour-The-water-lev

Question Number 170706 by help12345 last updated on 29/May/22

$$ \\ $$Between 10:PM and 7:45AM the water level in a swimming pool decreased by 13÷16 (13/16)inch.
Assuming that the water level decreased at a constant rate, how much did it drop each hour?
The water level decreased by inch each hour.

Commented by Rasheed.Sindhi last updated on 29/May/22

$$\frac{\mathrm{1}}{\mathrm{12}}\:{inch}/{hour} \\ $$

Commented by help12345 last updated on 29/May/22

$${brother}\:{i}\:{need}\:{the}\:{solution}\:{step}\:{by}\:{step}\:{please} \\ $$

Answered by Rasheed.Sindhi last updated on 29/May/22

$$\mathcal{T}{otal}\:{period}\:{between}\:\mathrm{10}\:{pm}\:{and} \\ $$$$\:\mathrm{7}:\mathrm{45}\:{am}:\:\mathrm{9}\:{hours}\:\mathrm{45}\:{minutes} \\ $$$$=\mathrm{9}\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{39}}{\mathrm{4}}\:{hours} \\ $$$${Decrease}\:{during}\:{the}\:{period} \\ $$$$=\mathrm{13}/\mathrm{16}\:{inch} \\ $$$${Decrease}\:{per}\:{hour}=\frac{\mathrm{13}/\mathrm{16}}{\mathrm{39}/\mathrm{4}} \\ $$$$\frac{\mathrm{13}}{\mathrm{16}}×\frac{\mathrm{4}}{\mathrm{39}}=\frac{\mathrm{1}}{\mathrm{12}}\:{inch} \\ $$

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