So today I did something I haven’t done since they forced me to back in school and took an IQ test. I’ve tended to avoid them because 1. what they’re actually supposed to be measuring is vaguely defined at best and 2. they amount to little more than an intellectual penis-waving contest, which I have no interest in. But an organization I belong to demanded it, and going along was less trouble than fighting it would’ve been. One question, however, threw me completely- or, rather, its answer did:
A man walks up a 2-mile hill at 2 MPH. He spends no time at the top, but immediately walks down the other side at 6 MPH. What was his average speed for the trip?
We were given answer sheets after we handed the tests in; apparently the answer they wanted was 3 MPH. This baffled me, but the other test-takers I asked assured me this was a common test question (they were all trying to get into Mensa and taking a lot of practice tests). I don’t know if that’s true, but I don’t like the implication if it is.
My reasoning was: it’s a four-mile trip. Two miles are at 2 MPH, 2 miles are at 6 MPH; 6+2=8, 8/2=4, which was my answer. Their reasoning was as follows: the man spends an hour climbing the hill, but only twenty minutes walking down it; a third as much time. I understand how their answer was right, but I don’t understand how I’m wrong. Am I missing something here? Deriving the answer from distance and deriving it from time give two different answers; check. But how was I to know that the time answer was what they wanted? The only thing I can see is that is that it mentions that he spends no time at the top; that wouldn’t matter if it wasn’t about time. But it also specifically asks his average speed for the trip; the key element of a trip isn’t time; it’s distance. There’s no real evidence that they’re looking for one answer over the others, and, again, the desired answer was 3 MPH, not “3 if by time, 4 if by distance”.
So again, am I missing something here? I can see how their answer is right; I don’t dispute that, but how is my answer wrong? Did I miscalculate somehow, or is this a question not assessing a person’s “intelligence”, but their psychic powers? If so, what earthly business does it have on an IQ test?
If the answer they wanted was speed, the only correct answer must be expressed in distance per time. Or in this case, miles per hour. So take total distance divided by total time.
Total distance = 4mi, total time = 1h20m or 80m or 1+1/3rd hr.
By basic algebra total distance / total time = :
4 / (1+1/3) = 4 / (4/3) = 4 * (3/4) = 3 mph.
Any other calculation you may have done was just grabbing some numbers and applying some operators to them with no actual logic or problem solving attached. What you did was exactly as justified as saying the answer is 10 because 2+2+6=10 and your reasoning was that 2, 2, and 6 are the only numbers mentioned in the problem, and addition is simpler than any other operation, so adding them up must be the correct thing to do. In other words, muddled and illogical guesswork mistakenly labeled as “thinking”.
The formula for velocity is distance/time, total distance was 4 miles, total time was 80min. Plug in the numbers and you get 4/80=1/20 miles per min, convert to mph and you get 3 miles per hour.
Where you added 2 MPH to 6 MPH and did 2+6=8, that was meaningless. Speed is a kind of rate. It doesn’t work to add rates like that. Then, dividing that by 4 was even more meaningless.
The only way to solve this is:
(1) Figure out the total distance of the full trip. (You did this part right: 4 miles.)
(2) Figure out the total time of the full trip. (You came close to doing this right. You figured 20 minutes going down. Did you then compute that the full trip was an hour and twenty minutes, or 80 minutes?)
(3) Divide the total miles (4) by the total time (80 minutes, or 4/3 hour): What LSLGuy said.
You can’t get meaningful results by adding or averaging the MPH figures directly. That trick never works.
ETA: We note that he went down the other side of the hill, rather than turning around and going back down the same side. We just have to assume that its the same distance down the other side. Did it say that? Of course not. Those kinds of stupid questions always have stupid holes in them.
There was a Pogo cartoon once, in which Pogo and Churchy were aloft in a gondola hanging from a big hot-air balloon. Pogo says: “We’re pretty far up!”, and Churchy says “It’s twice a far down!”
Determine the portion of the distance involved that each composed (1/2), and thus realize that they each carry the same averaging weight (I’m sure there’s some technical term for that; I don’t know it).
Divide the total of the added speeds (8) by the number of added speeds (2), getting 4.
That is not “just grabbing some numbers and applying some operators to them with no actual logic or problem solving attached”, that’s taking the speed numbers that have already been provided any using them to solve the problem. It’s not muddled, it’s not illogical, and it’s not guesswork.
My question was “how is my answer wrong?”; you have not answered that. Insulting me (inaccurately, no less) doesn’t help.
But why not? That’s what I’m failing to understand here. If day it rains for two hours and the next day it rains for four hours, the average rain time of the two days is three hours. Why does that not work here?
This is the wrong step. You have to determine the portion of the time that each composed rather than the portion of distance. Basically you computed the following average:
Basically, the reason why it doesn’t work is because you give the two legs of his trip equal weight in your average. Since speed is a rate, you can only do that if he spends the same amount of time on each leg. He doesn’t; he spends 3/4 of his trip going up the hill, and only 1/4 of his trip going down the hill.
The previous posters have already given the easier formula for computer his average speed - total distance traveled, divided by time traveled (which is literally the definition of speed). If you wanted to use your method, you would have to do something to weight the trips correctly. So, for example (3 x uphill speed) + (1 x downhill speed) / 4, which gives you 12/4, which gives you 3 MPH.
So, if you want to figure his speed using averages, you can. But you have to weight them correctly to give the correct figure.
It is no different than any trip where you have traffic jams, hills, fast areas etc. He traveled 4 miles in 1 hour and 20 min. It doesn’t matter what he was doing at any given time.
YOU said it was by time when you gave an answer in miles per hour. “An average of eight miles per hour” means that for every hour he traveled, he averaged 8 miles, which is incorrect. Your mistake is in thinking of MPH as a unit, but it’s not; it’s a ratio of two separate units.
You could also have said that he walked for an average of 20 minutes (or 1/3 hour) per mile, which would be correct, and would be the average by distance (per mile), but that wouldn’t be a speed.
As Senegoid notes, this is a poorly worded problem. The assumption that the other side of the hill is 2 miles long is in no way justified (in the real world, the percentage of hills that are symmetrical must be very small).
So the accurate answer to the question must be “Not possible to determine from the information supplied.”
Suppose you get an A (4 points) in a 1-unit course and a C (2 points) in a 5-unit course. Can you just average the two numbers (4+2) / 2 = 6/2 = 3 to get a grade point average of 3 (B)? No, because the 5-unit grade counts more heavily towards the average than the 1-unit grade. This is called a weighted average, where some scores count more heavily toward the average than others.
Something similar goes on when you work with miles-per-hour, which itself is an average of distance per unit time. You can’t just add several different miles-per-hour figures, then divide by the total distance, to get the overall miles-per-hour. If the guy spent one hour going uphill at one speed, and then only 20 minutes going downhill at another speed, then just like the course grades, those two speeds don’t weigh equally in computing the overall speed. If the uphill speed continued for three times as long (one hour) as the downhill speed (only 20 minutes, or 1/3 hour), then the uphill speed weighs three times as heavily into the overall speed as the downhill speed does. Your calculation doesn’t deal with this.
Gripe! They didn’t specify that the “other side of the hill” was also two miles.
It is perfectly possible to walk up a two mile hill and down a six mile hill! The hill’s symmetry was assumed, and that’s poor test-question construction!
BZZZT! No they don’t. Average speed is a function of time. They don’t carry the same weight because you spend less **time **going down the hill.
If it’s any consolation I always intuit the answer to this question wrong the same way you do. I have to (mentally) grab myself by the scruff of my brain and say “NO Princhester, you always intuit this one wrong. Just use the formula and forget your intuition.”
The key element of the trip may be distance, but the key element for average speed on a constant distance is time. And your answer is wrong because it isn’t the average speed for the trip, which we see if we check the answer:
Your calculations:
Speed up 2 mph (= 1 hour)
Speed down 6 mph (= 20 min)
Average speed half distance at 2 and half at 6 = 4 mph.
Check answer: 4 miles at 4 mph is 1 hour
Why do I pick that way to check the answer? Because that’s the definition of average speed. In the real world the hiker probably didn’t have a constant speed on either leg, but the total average speed is the same if 2 mph and 6 mph are average speeds as well. (Ignoring of course the nitpick about no mention being made of the distance going down, which is a valid criticism of the question, but not particularly interesting.)
Maybe your error becomes clearer to you if you consider other cases, such as the trip up taking much longer and the trip down being much shorter, say 1 mph up and 40 mph down.
Your method gives an average speed of 20.5 mph, which gives us a time for the four mile hike of 12 minutes, while the actual time is 2 hours 6 minutes.
So no psychic powers are needed, since the distance answer is wrong by the definition of average speed, but I agree that it’s debatable if the problem, viewed in isolation, is a good test of IQ.
The classic trick question using this general form is: “On a two mile trip if you drive one mile at 30 miles per hour, how fast do you need to drive the second mile to average 60 miles per hour over the whole trip?”. You’ll get a lot of people who reason as you did saying 90 mph, but actually it can’t be done. Driving one mile at 30 mph uses the two minutes needed to average 60 mph, so the second mile has to happen in zero time to average 60 mph.
Because hours per day and miles per hour are fundamentally different in information content. If you change the speed of the trip, the speed changes and the time changes. If you change the rate of rain in your example, the only thing that changes is the rate of rain.
Let me put it this way, when you have a derived unit, such as miles divided by hours, it matters if what changes is the dividend or the divisor. The IQ-test problem has a constant distance, which is the dividend in speed, and changing time, the divisor. This means you have to do the average based on how much time is spent at each speed. Change the problem to one where the speed changes because you use constant time on each leg, and you can use the change in distance for weighing your average.
2 mph up a 2 mile mountain (=1 hour)
6 mph down a 6 mile slope (=1 hour)
Average = 4 mph -> 4 mph for a 8 mile trip = 2 hours.
Your example with the rain is equivalent to this problem with the distance changing instead of the time. You have a rate in hours per day, and you’re changing the hours.
Here’s your mischief … both time and distance are involved and distance is given with respect to time. You averaged the two rates assuming time was equal, if you walked an hour at 2 mph then walked an hour at 6 mph, indeed you’d average 4 mph. Graph that (distance vs rate) and you’ll have a straight line something of the form y = x.
Here the problem gives distances as equal, again when you graph this function (time vs. rate) you get a curve in the form of y = 1/x.
Yes, that’s correct. In addition to the asymmetrical hills in the OP the hill is a ‘2 mile hill’. What is 2 miles? The difference in altitude between the top and the bottom? The distance from the base on one side to the base on the other side? The circumference of the hill at it’s base? Just as a rate of speed is a ratio a distance like 2 miles has to be applied to some particular attribute of the hill.
I do realize the OP has summarized the question perhaps leaving out all these details. But I often have to work from specs produced by sub-humanoids requiring this stuff to be clarified.