hey!! i got this hard math q i found online! its reeli challenging--well 4 me >.<

**fring** · July 31st 2006, 11:44 PM

on a Saturday morning Sue felt like getting some exercise. So she left home at 7 am and walked to a shop along a flat road. When she reached the shop, she immediatley started to walk to an oval along a raod which was never flat. After she came to the oval, she did not stop but turned around and walked along the same route back home. She got home at noon. What was the distance she covered during the morning, if her speeds on the flat road, uphill and down hill were 4km/h, 3km/h and 6km/h respectively?

could neone help me? it seems reeli difficult

**topsquark** · August 1st 2006, 03:49 AM

Originally Posted by fring

on a Saturday morning Sue felt like getting some exercise. So she left home at 7 am and walked to a shop along a flat road. When she reached the shop, she immediatley started to walk to an oval along a raod which was never flat. After she came to the oval, she did not stop but turned around and walked along the same route back home. She got home at noon. What was the distance she covered during the morning, if her speeds on the flat road, uphill and down hill were 4km/h, 3km/h and 6km/h respectively?

could neone help me? it seems reeli difficult

According to the problem she never walked along the oval, so the entire route was walked on flat road. Thus, since the trip out and back was the same distance, d/t = 4 km/h. t = 5 h, so d = 20 km.

Somehow I doubt this is what the problem was intending. Presumably she actually DID walk along the oval, but since there is no specific information regarding the oval, and an apparent misprint about it, we can't really solve the problem.

And if you found this problem on the internet how could it be "Urgent Homework Help?"

-Dan

**CaptainBlack** · August 1st 2006, 04:58 AM

Originally Posted by topsquark

According to the problem she never walked along the oval, so the entire route was walked on flat road. Thus, since the trip out and back was the same distance, d/t = 4 km/h. t = 5 h, so d = 20 km.

Somehow I doubt this is what the problem was intending. Presumably she actually DID walk along the oval, but since there is no specific information regarding the oval, and an apparent misprint about it, we can't really solve the problem.

And if you found this problem on the internet how could it be "Urgent Homework Help?"

-Dan

The road from the shop to the oval is never flat so the total distance she
walks up hill is the same as that which she walks down hill and is equal to
the distance from the shop to the oval.

RonL

Note: the question does not depend on it being either uphill or down hill all
the way from the shop to the oval.

**Soroban** · August 1st 2006, 05:20 AM

Hello, fring!]

That cant b the wording of the problem, cn it?
If so, the author is reeli silli . . . it's poorli ritten. . .

On a Saturday morning Sue felt like getting some exercise.
So she left home at 7 am and walked to a shop along a flat road.
When she reached the shop, she immediately started to walk
to an oval along a road which was never flat. reeli dum!
The road is rippled, like a washboard?

After she came to the oval, she did not stop but turned around
and walked along the same route back home. She got home at noon.
Why mention the "oval"? (whatever that is!)
Why not make it a Starbucks or an apple tree?

What was the distance she covered if her speeds on the flat road,
uphill and down hill were 4km/h, 3km/h and 6km/h respectively?

I think I understand what the problem is trying to say . . .

Sue walked $x$ km on the flat road at 4 kph.
Then she walked uphill for $y$ km at 3 kph.

She turned around and walked downhill for $y$ at 6 kph.
Then home on the flat road for $x$ km at 4 kph.

She walked a total distance of $2x + 2y \,=\,\boxed{2(x+y)\text{ km}}$

Let's calculate her time for each leg of the walk.

She walked $x$ miles at 4 kph; this took $\frac{x}{4}$ hours.

She walked $y$ km at 3 kph; this took $\frac{y}{3}$ hours.

She walked $y$ km at 6 kph; this took $\frac{y}{6}$ hours.

She walk $x$ km at 4 kph; this took $\frac{x}{4}$ hours.

Her walk took a total of: . $\frac{x}{4} + \frac{y}{3} + \frac{y}{6} + \frac{x}{4} \:=\:\frac{x + y}{2}$ hours.

Since she walked from 7 AM to noon (5 hours): . $\frac{x + y}{2}\:=\:5\quad\Rightarrow\quad x + y\,=\,10$

Therefore, her total distance is: . $2(x + y) \:=\:2(10)\:=\: \boxed{20\text{ km}}$

**CaptainBlack** · August 1st 2006, 05:28 AM

Originally Posted by Soroban

Hello, fring!]

That cant b the wording of the problem, cn it?
If so, the author is reeli silli . . . it's poorli ritten. . .

I think I understand what the problem is trying to say . . .

Sue walked $x$ km on the flat road at 4 kph.
Then she walked uphill for $y$ km at 3 kph.

She turned around and walked downhill for $y$ at 6 kph.
Then home on the flat road for $x$ km at 4 kph.

Soroban, see my interpretation of the wording, which will give the
same answer but does not depend on the road having constant slope
beteen the shop and oval.

RonL

**Soroban** · August 1st 2006, 04:28 PM

Hello, CaptainBlack!

Soroban, see my interpretation of the wording, which will give the same answer
but does not depend on the road having constant slope beteen the shop and oval.

Absolutely correct, Capn!

Funny, I never considered that concept . . . it's so obvious now.

Reminds of an experience I had only recently.
My wife was giving directions while I drove.
At one point, she said, "Turn right at this intersection and take a quick left."

I complied, but my mind was working on memorizing the return trip.
"So," I said to myself, "on the way back, I must make ... um ...
. . a right and a quick left . . . Hey!"

Evidently, I had never charted my moves both ways on an S-curve . . .

Thread: hey!! i got this hard math q i found online! its reeli challenging--well 4 me >.<

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