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# Algebra

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Will and Grace are canoeing on a lake.  Will rows at \$50\$ meters per minute and Grace rows at \$30\$ meters per minute. Will starts rowing at \$2\$ p.m. from the west end of the lake, and Grace starts rowing from the east end of the lake at \$2{:}45\$ p.m. If they always row directly towards each other, and the lake is \$2800\$ meters across from the west side of the lake to the east side, at what time will the two meet?

May 31, 2024

#1
+806
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So first off, let's calculate the distance Will fovers before Grace even begins to row.

We have \((45 min)(50 m/min) = 2250 m \)

When Grace starts to row, the two have to cover \((2800 m) – (2250 m) = 550 m \)

Since they are rowing at eachother at the same time, if we let t be the time it takes, we have \( (50)(t) + (30)(t) = 550 \)

Solving this, we get

\( 80t = 550 \\ t = 550/80 = 6.875 \)

This rounds to approximately 6 minutes and 52 seconds.

This means in addition to the 45 minutes Will takes at the beggining, they will take an additional 6 minutes and 52 seconds to meet.

Now, we some math to find that they meet at \(2:51:52\) P.M.

Thanks! :)

May 31, 2024

#1
+806
+1

So first off, let's calculate the distance Will fovers before Grace even begins to row.

We have \((45 min)(50 m/min) = 2250 m \)

When Grace starts to row, the two have to cover \((2800 m) – (2250 m) = 550 m \)

Since they are rowing at eachother at the same time, if we let t be the time it takes, we have \( (50)(t) + (30)(t) = 550 \)

Solving this, we get

\( 80t = 550 \\ t = 550/80 = 6.875 \)

This rounds to approximately 6 minutes and 52 seconds.

This means in addition to the 45 minutes Will takes at the beggining, they will take an additional 6 minutes and 52 seconds to meet.

Now, we some math to find that they meet at \(2:51:52\) P.M.

Thanks! :)

NotThatSmart May 31, 2024