We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

3.) A large snowplow can clear a parking lot in 4 hours. A small snowplow needs more time to clear the lot.

Working together, the large and small snowplow can clear the lot in 3 hours. How long would it take the

small snow plow to clear the lot by itself?

GAMEMASTERX40 Feb 3, 2019

#1**+2 **

Large snow plow rate = 1 lot/4

Small snow plow rate = 1lot/x

(1/4 + 1/x) * 3 = 1 lot

3/4 +3/x =1

3/x=1/4

x = 3/(1/4) = 12 hours

ElectricPavlov Feb 3, 2019

#2**+2 **

For problems like these, I like to use a table. I like to fill in the table with the information given to me and then create an equation that I can solve.

For this table, I will use the equation d=rt as the structure for the table.

D | R | T | |

Large Snowplow | |||

Small Snowplow | |||

Large Snowplow and Small Snowplow |

We will use the equation d=rt as a metaphor for this particular question.

d is metaphorical for the total amount done; for example, in this problem, the large snowplow is clearing one parking lot of its snow. Therefore, d=1 for the large snowplow.

Although we do not know how long the small snowplow takes to clear the parking lot yet, we do know that it has to clear one parking lot by itself, so d=1 for the small snowplow, as well.

When the snowplows work together, they both only clear one parking lot in this problem, so d=1 for the large and small snowplow. Let's fill this information in now.

D | R | T | |

Large Snowplow | 1p | ||

Small Snowplow | 1p | ||

Large Snowplow and Small Snowplow | 1p |

Next, I will fill in the "T" column, which represents time.

The given information states that the large snowplow leaves the parking lot devoid of snow after 4 hours. Therefore, t=4.

We do not know how much time it took the small snowplow to clear this parking lot; after all, that is what we are solving for. I will assign a variable, *x*, for that slot.

Collectively, the snowplows unload all that snow in a matter of 3 hours, so t=3 in the final row.

This is what the table looks like now:

D | R | T | |

Large Snowplow | 1p | 4hr | |

Small Snowplow | 1p | xhr | |

Large Snowplow and Small Snowplow | 1p | 3hr |

To fill in the rates, I will refer back to that metaphorical formula. I can rearrange this formula to solve for *r.*

\(d=rt\) | divide by t to isolate r. |

\(r=\frac{d}{t}\) |

This rearrangement shows that all I have to do is divide the "D" values by the "T" values in order to fill in the elements of the "R" column. Here is an update of what the table looks like.

D | R | T | |

Large Snowplow | \(1p\) | \(\frac{1p}{4\text{hr}}\) | \(4\text{hr}\) |

Small Snowplow | \(1p\) | \(\frac{1p}{x\text{hr}}\) | \(x\text{hr}\) |

Large Snowplow and Small Snowplow | \(1p\) | \(\frac{1p}{3\text{hr}}\) | \(3\text{hr}\) |

Great! We have the table filled out. In order to generate an equation, I want you to focus on the "R" column, the one that represents the rate of the snowplows. When the snowplows are in cahoots, their combined rates equal that of the sum of the individual rates. This allows us to generate an equation.

\(\frac{1}{4}+\frac{1}{x}=\frac{1}{3}\)

\(\frac{1}{4}+\frac{1}{x}=\frac{1}{3}\) | There are many ways to approach this, but probably the simplest way is to multiply both sides by the LCD, 12x. |

\(3x+12=4x\) | Subtract 3x on both sides. |

\(x=12\text{hr.}\) | This is how long the small snowplow will take on its own. |

TheXSquaredFactor Feb 3, 2019