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# not so easy Math Problem

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Posted · Report post

A truck travels from A to B.

Going uphill, it goes at 56 mph.

Going downhill, it goes at 72 mph.

On level ground, it goes at 63 mph.

If it takes 4 hours to travel from A to B, and 5 hours to come back,

what is the distance between A and B?

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Posted (edited) · Report post

283.5Mi

252 miles of it is downhill from A to B and 31.5 miles are flat.

It takes 1/2 hour to go the flat part (31.5 mi/63mi/hour)

For the downhill portion , it takes 3.5 hours (252/72)

Going back (uphill) it takes 4.5 hours (252/56)

Edited by kevink2
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Posted (edited) · Report post

Same answer here, is ofc correct.

72*t = 56 * (t+1) .. t=3,5h, the time it takes to drive downhill from A to B, rest is just filling the gaps

Edited by Potok
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Posted · Report post

283.5Mi

252 miles of it is downhill from A to B and 31.5 miles are flat.

It takes 1/2 hour to go the flat part (31.5 mi/63mi/hour)

For the downhill portion , it takes 3.5 hours (252/72)

Going back (uphill) it takes 4.5 hours (252/56)

Same answer here, is ofc correct.

72*t = 56 * (t+1) .. t=3,5h, the time it takes to drive downhill from A to B, rest is just filling the gaps

turned out to be so easy I guess. well done

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Posted (edited) · Report post

Interesting! I got the same answer...,

0 miles flat

15.75 miles downhill (from A toward B)

267.75 miles uphill (from A toward B)

Edited by CaptainEd
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Posted · Report post

So in your solution A is lower (compared to sea level) than B and you travel from A to B mostly uphill ? But the path from A to B is (timewise) shorter than from B to A.

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Posted · Report post

Suppose D is the distance traveled downhill from A to B, L is the distance traveled on level ground from A to B, and U is the distance traveled uphill from A to B. Then the total time to travel from A to B is (D/72)+(L/63)+(U/56)=4. We can multiply through by 504 to get 7D+8L+9U=2016 (call this equation X). On the way back from B to A, D and U switch roles, with D being uphill and U being downhill. So, for the return trip, we get (D/56)+(L/63)+(u/72)=5. Multiply both sides by 504 to get 9D+8L+7U=2520. Adding this to equation X gives 16D+16L+16U=4536. Simplify this to D+L+U=283.5. Since D+U+L is the distance from A to B, we have determined that distance to be 283.5 without specifying what D, L, and U are.

Note that this problem only has a unique solution for some sets of speeds. {D=72,L=63,U=56} happens to be one such set.

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Posted · Report post

So in your solution A is lower (compared to sea level) than B and you travel from A to B mostly uphill ? But the path from A to B is (timewise) shorter than from B to A.

Oops, you're right, I got it backward.

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Posted · Report post

Here is one way to solve this...

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Posted · Report post

x/63 + y/72= 4

x/63 + y/56= 5

x = 31.5 Mi ; net ground level distance.

y = 252 Mi ; net inclined distance

Total x + y = 283.5 Mi.

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Posted · Report post

Suppose D is the distance traveled downhill from A to B, L is the distance traveled on level ground from A to B, and U is the distance traveled uphill from A to B. Then the total time to travel from A to B is (D/72)+(L/63)+(U/56)=4. We can multiply through by 504 to get 7D+8L+9U=2016 (call this equation X). On the way back from B to A, D and U switch roles, with D being uphill and U being downhill. So, for the return trip, we get (D/56)+(L/63)+(u/72)=5. Multiply both sides by 504 to get 9D+8L+7U=2520. Adding this to equation X gives 16D+16L+16U=4536. Simplify this to D+L+U=283.5. Since D+U+L is the distance from A to B, we have determined that distance to be 283.5 without specifying what D, L, and U are.

Note that this problem only has a unique solution for some sets of speeds. {D=72,L=63,U=56} happens to be one such set.

Suppose D is the distance traveled downhill from A to B, L is the distance traveled on level ground from A to B, and U is the distance traveled uphill from A to B. Then the total time to travel from A to B is (D/72)+(L/63)+(U/56)=4. We can multiply through by 504 to get 7D+8L+9U=2016 (call this equation X). On the way back from B to A, D and U switch roles, with D being uphill and U being downhill. So, for the return trip, we get (D/56)+(L/63)+(u/72)=5. Multiply both sides by 504 to get 9D+8L+7U=2520. Adding this to equation X gives 16D+16L+16U=4536. Simplify this to D+L+U=283.5. Since D+U+L is the distance from A to B, we have determined that distance to be 283.5 without specifying what D, L, and U are.

Note that this problem only has a unique solution for some sets of speeds. {D=72,L=63,U=56} happens to be one such set.

Well said....! Another unique sets of speeds is D=56, U=42, L=48.

In this case (If time of travell remains same); Two equations will be:

6D+7L+8U=6*7*8*4

8D+7L+6U=6*7*8*5

14(D+L+U)= 6*7*8*9

Therefore D+L+U=216

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