Question 14/20: Set 5 - Try Out - General Reasoning (SNBT)

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A full cylindrical water tank has a height of $4.2\text{ m}$ and a diameter of $3\text{ m}$ . Its base leaks, releasing water at a rate of $12\text{ dm}^3$ per minute. The water will run out in ...

$41\text{ hours }15\text{ minutes}$

$41\text{ hours }25\text{ minutes}$

$42\text{ hours }15\text{ minutes}$

$42\text{ hours }25\text{ minutes}$

$42\text{ hours }45\text{ minutes}$

Explanation

To determine how long it will take for the water to run out, we first need to calculate the volume of water in the tank and then divide it by the leakage rate.

Known Data

Here are the dimensions of the water tank (cylinder) converted to decimeters (dm) to match the flow rate unit ( $\text{dm}^3$ ):

Variable	Initial Value	Conversion (dm)
Height ( $h$ )	$4.2\text{ m}$	$42\text{ dm}$
Diameter ( $d$ )	$3\text{ m}$	$30\text{ dm}$
Radius ( $r$ )	$1.5\text{ m}$	$15\text{ dm}$
Rate ( $Q$ )	$12\text{ dm}^3/\text{minute}$	-

Volume Calculation

The volume of a cylinder is calculated using the formula $V = \pi r^2 h$ .

V = \frac{22}{7} \times 15^2 \times 42

V = \frac{22}{7} \times 225 \times 42

V = 22 \times 225 \times 6

V = 29,700\text{ dm}^3

Time Calculation

The time required for the water to run out is the volume divided by the leakage rate.

t = \frac{V}{Q} = \frac{29,700}{12} = 2,475\text{ minutes}

Next, we convert the time from minutes to hours and minutes:

2,475\text{ minutes} = \frac{2,475}{60}\text{ hours}

2,475\text{ minutes} = 41.25\text{ hours}

2,475\text{ minutes} = 41\text{ hours} + (0.25 \times 60)\text{ minutes}

2,475\text{ minutes} = 41\text{ hours } 15\text{ minutes}

So, the water will run out in $41\text{ hours } 15\text{ minutes}$ .

Command Palette

Number 14

Explanation

Known Data

Volume Calculation

Time Calculation