How much faster is my car than yours?

[Divagating the Future]

If it takes twice as long for my BMW m-5 to pass your Corvette after it first overtakes it as it takes the the 2 cars to pass when going in the opposite directions,

 

How many times faster is my car than yours?

{1} guess

Edited February 26, 2012 by Divagating the Future

[md65536]

Good puzzle. I voted it up, but I guess someone disagrees.

[Phi for All]

Good puzzle. I voted it up, but I guess someone disagrees.

I'm guessing it was the "my car's better than your car" attitude.

 

But I agree, taken by itself it's a good puzzle. I'm sure the neg rep came from someone who followed this member's other posts.

[imatfaal]

Seems wrong - but here is my stab

 

 

 

Relative speed overtaking = Speed_Beemer - Speed_{lilredcorvette} = Speed_B - Speed_C

 

Relative speed crossing = Speed_B + Speed_C

 

Distance overtaking and crossing is length of Corvette (ie from when Beemer's front fender is level with Corvettes tail to when Beemer and Corvette fronts are equal) + length of Beemer Distance = B + C

 

Time for overtaking = 2t

Time for crossing = t

 

speed = distance / time

 

1. Speed_B - Speed_{C =} (B+C)/(2t)

 

2. Speed_B + Speed_C = (B+C)/t

 

multiply eq1 by 2

 

3. 2Speed_B - 2Speed_{C =} (B+C)/(t)

 

Two RHSs are equal

 

4. 2Speed_B - 2Speed_{C =} Speed_B + Speed_C

 

5_. Speed_B = 3SpeedC

 

The Beemer is doing 3 times the speed of the Corvette.

 

Strikes me as wrong - but I am not sure where/if I have tripped up

 

 

 

[md65536]

I'm guessing it was the "my car's better than your car" attitude.

 

But I agree, taken by itself it's a good puzzle. I'm sure the neg rep came from someone who followed this member's other posts.

Yes, one must take one's proper place even in the imaginary world of a puzzle. A lesser person cannot be imagined to have a superior car. A respectful puzzle with a proper attitude would display appropriate subordination.

 

Admittedly, I was offended at the notion that I would drive a Corvette, at least in the presence of faster cars. I think it is a better puzzle if I am the one driving the faster imaginary car.

 

 

 

But I think you're right though. I would guess that one might think that if another posted puzzle is incomplete, then this one might be as well. Perhaps an attitude of "my brain's better than your brain" so if I can't solve the puzzle, there must be something wrong with it. It's just a suspicion though... figuring out the post ratings is a puzzle in itself.

 

Seems wrong

 

I get the same answer.

 

 

I assumed the "overtaking distance" was some large initial separation of the vehicles. It doesn't matter though because the answer is independent of that distance.

 

 

 

 

 

[Phi for All]

It's just a suspicion though... figuring out the post ratings is a puzzle in itself.

Are you able to click on the number in the rep points box to see who voted which way?

[hypervalent_iodine]

The neg rep was from me and was a legitimate mistake (the touchpad on my Mac is a bit sensitive sometimes). I would have asked someone to fix it, but frankly I didn't care enough.

[md65536]

Are you able to click on the number in the rep points box to see who voted which way?

No. It's probably best that way; it would just feed my penchant for conspiracy theories!

[mertol]

Here is another problem:

Say the bmw goes at 90mph and the vette goes at 30mph, the bmw is travelling from NY to Washington (204 miles), 1hour after the bmw departed, the vette starts from Washington to NY. What is the distance between the 2 cars 1 hour before they meet?

[raj bhramar]

Hello friends, this is my first posting in this forum. I take very much interest in puzzles, brain teasers. Hope you enjoy my company.

I am replying

 

Here is another problem:

Say the bmw goes at 90mph and the vette goes at 30mph, the bmw is travelling from NY to Washington (204 miles), 1hour after the bmw departed, the vette starts from Washington to NY. What is the distance between the 2 cars 1 hour before they meet?

 

I will denote bmw as car 'M' and Vette as car 'N'; NY as 'A', Washington as 'B', and their meeting place as 'C'.

It is given that M departed one hour before N started. Say after one hour M was at a place D, So at that moment the distance DB - between the two cars was 204 - 90 = 114 miles.

Say after M reached D, the two cars meets at C after 'T' hrs.

So DC - the distance traveled by M = 90T

Also BC - the distance traveled by N = 30T

But DC + BC = DB = 114 miles. Therefore 90T + 30T = 114......[a]

From equation [a], T = 57 minutes.

Now the question is what the distance was between M and N, before one hour they meet.

Clearly before M & N meet, M had traveled for 57 minutes after reaching point D, so M was at 3 min distance before reaching the place D, i.e. 3/60 * 90 = 4.5 miles back from D; and at that moment car N did not even started from B.

Hence distance between M & N at that moment (one hour before the two cars meet) = 114 + 4.5 = 118.5 miles.

 

SORRY I DON'T SEE ANY OPTION TO POST IN SPOILER. Please help me....

Edited March 3, 2012 by raj bhramar

[Spyman]

This other problem is more a trick question than a math question...

 

 

The distance between NY & Washington and the headstart of the bmw is irrelevant. If the bmw goes at 90mph and the vette goes at 30mph, then the bmw is 90 miles from their meeting point one hour before it gets there and the vette is 30 miles from the meeting point one hour before it gets there, therefore they are separated by 90+30=120 miles one hour before they meet.

 

 

The spoiler tags are:

 

 
[spoiler] hidden message [/spoiler]
 

 

LOL - Edited since I failed to spell "spoiler" correct.

Edited March 3, 2012 by Spyman

[raj bhramar]

Thank you spyman, I can now post in spoiler......!

This other problem is more a trick question than a math question...

 

 

The distance between NY & Washington and the headstart of the bmw is irrelevant. If the bmw goes at 90mph and the vette goes at 30mph, then the bmw is 90 miles from their meeting point one hour before it gets there and the vette is 30 miles from the meeting point one hour before it gets there, therefore they are separated by 90+30=120 miles one hour before they meet.

 

 

The spoiler tags are:

 

 
[spoiler] hidden message [/spoiler]
 

 

LOL - Edited since I failed to spell "spoiler" correct.

 

You are right.... but only if the distance between NY and Washington & the head start of the BMW and Corvette is irrelevant. But as seen from the puzzle, it clearly states the relevance....or I did not understand it ....?

 

 

Say the cars are traveling between the two stations A & B. Let the speed of BMW be x miles per hour and that of Corvette is y miles per hour.

Say the meeting place of two cars is C, when they travel in opposite directions.

The time taken by BMW to reach from A to C is say t hr. So the distance AC = xt ………………………..…. [1]

Same time t hr. is taken by Corvette to reach from B to C. So the distance BC = yt …………………….….. [2]

 

Now BMW reaches B and returns to overtake Corvette, say at place D, after 2t hr.

So BMW travels from C to B, B to C and then C to D, in 2t hr. i.e. the distance CB + BC + CD = 2xt ... [3]

And Corvette travels from C to D in 2t hr. i.e. the distance CD = 2yt ………………………………….…………… [4]

 

From [2], [3], and [4], yt + yt + 2yt = 2xt

Hence x = 2y

 

Seems wrong - but here is my stab

 

 

 

Relative speed overtaking = Speed_Beemer - Speed_{lilredcorvette} = Speed_B - Speed_C

 

Relative speed crossing = Speed_B + Speed_C

 

Distance overtaking and crossing is length of Corvette (ie from when Beemer's front fender is level with Corvettes tail to when Beemer and Corvette fronts are equal) + length of Beemer Distance = B + C

 

Time for overtaking = 2t

Time for crossing = t

 

speed = distance / time

 

1. Speed_B - Speed_{C =} (B+C)/(2t)

 

2. Speed_B + Speed_C = (B+C)/t

 

multiply eq1 by 2

 

3. 2Speed_B - 2Speed_{C =} (B+C)/(t)

 

Two RHSs are equal

 

4. 2Speed_B - 2Speed_{C =} Speed_B + Speed_C

 

5_. Speed_B = 3SpeedC

 

The Beemer is doing 3 times the speed of the Corvette.

 

Strikes me as wrong - but I am not sure where/if I have tripped up

 

 

 

Edited March 3, 2012 by raj bhramar

[Spyman]

Post #1 is one puzzle and post #9 is another totally different problem, try to solve them separately.

[Delta1212]

This other problem is more a trick question than a math question...

 

 

The distance between NY & Washington and the headstart of the bmw is irrelevant. If the bmw goes at 90mph and the vette goes at 30mph, then the bmw is 90 miles from their meeting point one hour before it gets there and the vette is 30 miles from the meeting point one hour before it gets there, therefore they are separated by 90+30=120 miles one hour before they meet.

 

 

The spoiler tags are:

 

 
[spoiler] hidden message [/spoiler]
 

 

LOL - Edited since I failed to spell "spoiler" correct.

Your answer is incorrect because they are already less than 120 miles apart when the BMW leaves. This means it tales less than an hour from the time the BMW leaves for the cars to meet. You have to factor in part of the last hour that only has the vette traveling at 90 mph while the BMW is stationary.

[Spyman]

Yes, you and raj bhramar are of course correct, I must have had a brain fart.

 

 

Distance between the cars when the vette starts is 204-90=114 miles, which is 6 miles short just as said. The two cars covered that distance together in 114/(90+30)=0.95 hours, so the bmw was moving alone for 0.05 hours which is 90*0.05=4.5 miles. That means the two cars was 114+4.5=118.5 miles apart one hour before they meet.

 

 

EDIT: And still can't spell...

Edited March 5, 2012 by Spyman

[mertol]

I knew about the common mathematical teaser with the excessive information trying to confuse you so I made this one up to see how many people will asume it is a teaser .

[Spyman]

Mission successful.