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3 years ago

Lisa takes 3 hours to mow the lawn, while her cousin, Barb, takes 2 hours. How long will it take them working together?

Mathematics

2 answers:

Colt1911 [192]3 years ago

5 0

Answer:6 hours

Step-by-step explanation:

3x2=6

RSB [31]3 years ago

4 0

Answer:

t=72 minutes

Step-by-step explanation:

If it takes Lisa 3 hours to mow the lawn, then she mows 13 of the lawn per hour. Barb takes 2 hours to do it alone so she completes 12 of the lawn each hour. Working together, they mow the lawn in some time t. This can be written as

$\frac{1}{3}$ + $\frac{1}{2}$ = $\frac{1}{t}$

Multiply each side by the least common denominator, 6t.

6t( $\frac{1}{3}$ + $\frac{1}{2}$ )=6t( $\frac{1}{t}$ )

Simplify and solve.

2t+3t=6

We can multiply the fraction by 60 to convert the fraction of an hour to minutes.

t= $\frac{6}{5}$ (60 minutes)

To mow the lawn, it takes the two working together

t=72 minutes.

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Rina8888 [55]

To solve this we are going to use the formula for speed: $S= \frac{d}{t}$
where
$S$ is the speed
$d$ is the distance
$t$ is the time

Let $S_{l}$ be the speed of the boat in the lake, $S_{a}$ the speed of the boat in the river, $t_{l}$ the time of the boat in the lake, and $t_{a}$ the time of the boat in the river.

We know for our problem that the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
 $S_{a}=S_{l}-2$
We also know that in the lake the boat sailed for 1 hour longer than it sailed in the river, so:
 $t_{l}=t_{a}+1$

Now, we can set up our equations.
Speed of the boat traveling in the river:
 $S_{a}= \frac{6}{t_{a} }$
But we know that $S_{a}=S_{l}-2$ , so:
$S_{l}-2= \frac{6}{t_{a} }$ equation (1)

Speed of the boat traveling in the lake:
$S_{l}= \frac{15}{t_{l} }$
But we know that $t_{l}=t_{a}+1$ , so:
$S_{l}= \frac{15}{t_{a}+1}$ equation (2)

Solving for $t_{a}$ in equation (1):
$S_{l}-2= \frac{6}{t_{a} }$
$t_{a}= \frac{6}{S_{l}-2}$ equation (3)

Solving for $t_{a}$ in equation (2):
$S_{l}= \frac{15}{t_{a}+1}$
$t_{a}+1= \frac{15}{S_{l}}$
$t_{a}=\frac{15}{S_{l}}-1$
$t_{a}= \frac{15-S_{l}}{S_{l}}$ equation (4)

Replacing equation (4) in equation (3):
$t_{a}= \frac{6}{S_{l}-2}$
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$

Solving for $S_{l}$ :
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$
$(15-S_{l})(S_{l}-2)=6S_{l}$
$15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}$
$S_{l}^2-11S_{l}+30=0$
$(S_{l}-6)(S_{l}-5)=0$
$S_{l}=6$ or $S_{l}=5$

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.

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