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# A swimmer loses their cap.

## Question

As a swimmer jumps off a small bridge and begins to swim upstream, her swim cap comes off and floats downstream. Ten minutes later she turns around, swimming downstream with the same effort, past her original bridge. At the next bridge, 1000 meters away from the first, she catches the cap. What was the speed of the current? Of the swimmer?
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I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

If you give me how far upstream the swimmer got, I can get the swimmer's base speed, too! Let distance be a; v=a/10-50

so what is the current's speed?

I got the current way back in the first spoiler; c=50 meters per minute

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• 0

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.
b=x(v+c): distance the swimmer goes downstream for the cap
d=(10+x)c: distance the cap floats; 1000 meters.
a+d=b
10v-10c+10c+xc=xv+xc
10v=xv
10=x
sub in:
a=10v-10c
b=10v+10c
d=20c=1000
20c=1000 yields c=50
a=10v-500
b=10v+500
10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

• 1
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• 0

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

##### Share on other sites

• 0

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

If you give me how far upstream the swimmer got, I can get the swimmer's base speed, too! Let distance be a; v=a/10-50

##### Share on other sites

• 0

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

If you give me how far upstream the swimmer got, I can get the swimmer's base speed, too! Let distance be a; v=a/10-50

so what is the current's speed?

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