# A question on an IQ test question (semi-math related) **URL:** \ **Category:** Factual Questions\ **Created:** [June 30, 2015, 1:55am UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830 "2015-06-30T01:55:03Z")\ **Posts on this page:** 15\ **Page:** 2
**Author:** ![pulykamell](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/pulykamell/32/3166_2.png) [@pulykamell](https://boards.straightdope.com/u/pulykamell)\ **Post date:** [June 30, 2015, 2:56pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/21 "2015-06-30T14:56:49Z")
> [@Recusant](#): > > 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? Yes, your average rate time is 3 hours per day. It would be analagous to the IQ question in the OP if it were phrased as “If one hour, Jack walks 2 miles, and the second hour, Jack walks 4 miles, what is the average speed?” In that case, yes, it’s 3 miles per hour. But that’s not the question in the OP. In that question, we need to figure out the total time before we can make a rate calculation. Using your rain example, an admittedly convoluted analogy would be something like “When the storm started, two inches of rain fell at a rate of two inches an hour, and then the storm kicked up, dumping another two inches of rain at a rate of four inches an hour before clearing up. What is the average rate in inches per hour for the storm?” (2 2/3 inches per hour.) ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [June 30, 2015, 3:20pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/22 "2015-06-30T15:20:22Z")
> [@Senegoid](#): > > The only way to solve this is: > [@Senegoid](#): > > You can’t get meaningful results by adding or averaging the MPH figures directly. I may be being overly pedantic but I take issue with these statements. When distance is held constant you can simply calculate the [harmonic mean](https://en.wikipedia.org/wiki/Harmonic_mean) of the speeds to find the average speed. 2/(1/2 + 1/6) = 3 mph From the wikipedia page: > [@](#): > > In certain situations, especially many situations involving rates and ratios, the harmonic mean provides the truest average. For instance, if a vehicle travels a certain distance at a speed x (e.g. 60 kilometres per hour) and then the same distance again at a speed y (e.g. 40 kilometres per hour), then its average speed is the harmonic mean of x and y (48 kilometres per hour), and its total travel time is the same as if it had traveled the whole distance at that average speed. ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [June 30, 2015, 3:21pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/23 "2015-06-30T15:21:25Z")
> [@pulykamell](#): > > But that’s not the question in the OP. In that question, we need to figure out the total time before we can make a rate calculation. Or, alternatively, you could calculate the **harmonic mean** of the two rates. [Here’s a page](http://www.cut-the-knot.org/arithmetic/HarmonicMean.shtml) that explains what that is and how it relates to a problem like this one. (Everyone else’s answer is correct, but it may or may not make the OP feel better to point out that the word “average” can have more than one meaning. “Add the numbers together and divide by how many numbers there are” gives the **arithmetic average** (or “arithmetic mean”). That said, I am dubious whether a “legitimate” IQ test would include a question like this, since it depends so much on knowledge of a specific thing (the meaning of “average speed”) and the results are going to depend so much on whether the test-taker has encountered this kind of question before. ETA: I see **Lance Turbo** has ninja’d me on the harmonic mean. I still think the link I gave is relevant, though. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [June 30, 2015, 4:10pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/24 "2015-06-30T16:10:05Z")
> [@Recusant](#): > > 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? As has been noted above, this isn’t a good analogy because it deals with cumulative quantities rather than rates. A closer analog would be the following question: suppose there was a snowstorm yesterday. First we got two inches of snow at the rate of 2" per hour. Then we got another two inches of snow at a rate of 6" per hour. Over the course of the whole snowstorm, what was the average rate of snowfall in inches per hour? ---
**Author:** ![JohnGalt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johngalt/32/184_2.png) [@JohnGalt](https://boards.straightdope.com/u/JohnGalt)\ **Post date:** [June 30, 2015, 5:26pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/25 "2015-06-30T17:26:57Z")
> [@Bill\_Door](#): > > 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. I use this question when I teach Excel class to show that you CAN’T AVERAGE RATES. Out of perhaps 250 students who have taken the class, not one answered correctly, but after explaining it they seem to all get it. I dare them to do this as a bar bet, but warn them they’ll probably get in an argument or get punched in the nose. The reason I do this in Excel is we do a spreadsheet with a payroll example. Each employee has a different pay rate, and we multiple their hours worked by their pay rate, and then get then total amount paid. In the next step, I have them get the AVERAGE pay rate, and multiply that by the TOTAL hours worked to get another “total amount paid”. Of course the two totals don’t agree, because you can’t average rates. The climbing hill question is another example of trying to average rates. It’s like driving 100 miles using 2 gallons of gas, for 50 MPG, then driving 1000 miles using 100 gallons of gas at 10 MPG. The average MPG is not (50+10)/2 for 30 MPG, but rather (100+1000)/(2+100) or 10.78 MPG. ---
**Author:** ![excavating\_for\_a\_mind](https://avatars.discourse-cdn.com/v4/letter/e/e8c25b/32.png) [@excavating\_for\_a\_mind](https://boards.straightdope.com/u/excavating_for_a_mind)\ **Post date:** [June 30, 2015, 6:02pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/26 "2015-06-30T18:02:03Z")
> [@Bill\_Door](#): > > 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. Well, if your speed for the second mile is 186,000 miles per second, you might just be able to make it (but you’d be just a tad bit younger). ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [June 30, 2015, 6:04pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/27 "2015-06-30T18:04:21Z")
If you get paid $20/hour for an hour, and $60/hour for 20 minutes, what’s your average rate of pay for that time? Lessee, you made $20 plus $20 or $40. You worked 1.3333333 hours or 4/3 hour. So in total, you were paid an average of $30/hour. The downhill may have been faster, but it counts for less toward the average because it was shorter. ---
**Author:** ![Knowed\_Out](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/knowed_out/32/7380_2.png) [@Knowed\_Out](https://boards.straightdope.com/u/Knowed_Out)\ **Post date:** [June 30, 2015, 8:07pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/28 "2015-06-30T20:07:55Z")
Remember d = rt. Distance = rate X time. Distance is miles, rate is miles per hour, and time is hours. How long does it take to travel up the hill? We have distance and rate. 2 mi = 2 mi/hr x t hrs. t = 2 mi/2 mph = 1 hr. Assuming the hill is 2 miles from top to bottom in any direction, how long does it take to get back down? This time, t = 2 mi/6 mph = 1/3 hr. Total time to go up and down the hill is 1 hr + 1/3 hr = 1 1/3 hr = 4/3 hr. We know total distance is 2 miles up and 2 miles down, for a total of 4 miles. Go back to d = rt. We know distance and time, and are solving for rate. 4 mi = r x 4/3 hr r = 4 mi divided by 4/3 hr = 3 mph ---
**Author:** ![Hampshire](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hampshire/32/4887_2.png) [@Hampshire](https://boards.straightdope.com/u/Hampshire)\ **Post date:** [June 30, 2015, 9:41pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/29 "2015-06-30T21:41:59Z")
The only question that I can come up with where 4mph is the answer would be if they said “The speed limit going up the hill is posted as 2mph, the speed limit going down the hill is posted as 6mph. What is the average posted speed limit?” (2 + 6) / 2 = 4 But as others have posted, they want the average speed of his entire trip. An 80 minute trip. First 20 minutes: 2mph Second 20 minutes: 2mph Third 20 minutes: 2mph Last 20 minutes: 6mph (2 + 2 +2 + 6) / 4 = 3mph ---
**Author:** ![Cartoonacy](https://avatars.discourse-cdn.com/v4/letter/c/848f3c/32.png) [@Cartoonacy](https://boards.straightdope.com/u/Cartoonacy)\ **Post date:** [July 1, 2015, 6:34pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/30 "2015-07-01T18:34:47Z")
Speed is already an average. It’s the average distance traveled per unit of time. By calculating (2 + 6) / 2, you’re averaging two averages, which means nothing mathematically. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [July 3, 2015, 9:59am UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/31 "2015-07-03T09:59:03Z")
Just adding to this because I came up with, possibly, an even better example to illustrate why some people were a bit quick to ridicule the question. Let’s replace the distance with chocolates and speed with chocolates per person. To avoid fractional people I increase from 2 miles to 6 chocolates. So 6 chocolates are eaten at a rate of 2 per person, and another 6 are eaten at a rate of 6 per person. What was the average rate of chocolate consumption for all twelve chocolates? Averaging the rates with a focus on the number of chocolates, as the OP suggests, give us half weight to either rate and an average of (6 + 2)/2 = 4 chocolates per person. Going the route of figuring out how many people (the equivalent of time in the original) there were shows us there were 3 people sharing the first 6, and 1 person sharing the last 6 for a total of 4. 12 chocolates for 4 persons = 3 chocolates per person. ---
**Author:** ![Wendell\_Wagner](https://avatars.discourse-cdn.com/v4/letter/w/8491ac/32.png) [@Wendell\_Wagner](https://boards.straightdope.com/u/Wendell_Wagner)\ **Post date:** [July 3, 2015, 3:01pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/32 "2015-07-03T15:01:12Z")
Let’s go back to the OP. What I.Q. test was this? Was it this one? > **[Mensa\_miniquiz.pdf](https://www.cbsnews.com/htdocs/pdf/Mensa_miniquiz.pdf)** > > 21.69 KB That’s the one that comes up when I do a search on words in the problem given in the OP. My reaction to this is that it’s a terrible I.Q. test. Some of these questions are clearly culturally dependent. Some of them depend on one’s ability to do simple arithmetic quickly and accurately. Even to the extent that any of this is really a test of intelligence, it’s only a test using a few of the common sorts of questions used in I.Q. tests. Recusant, what organization asked you to take this test? What purpose was served by asking you to do it? ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [July 3, 2015, 3:56pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/33 "2015-07-03T15:56:01Z")
> [@Wendell\_Wagner](#): > > Let’s go back to the OP. What I.Q. test was this? Was it this one? > > [http://www.cbsnews.com/htdocs/pdf/Mensa\_miniquiz.pdf](http://www.cbsnews.com/htdocs/pdf/Mensa_miniquiz.pdf) > > That’s the one that comes up when I do a search on words in the problem given in the OP. My reaction to this is that it’s a terrible I.Q. test. I notice that it doesn’t actually claim to be an I.Q. test. (And no one who’s seriously familiar with the concept of I.Q. would claim that it is one.) ---
**Author:** ![Wendell\_Wagner](https://avatars.discourse-cdn.com/v4/letter/w/8491ac/32.png) [@Wendell\_Wagner](https://boards.straightdope.com/u/Wendell_Wagner)\ **Post date:** [July 3, 2015, 5:44pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/34 "2015-07-03T17:44:55Z")
True, but the claim in the OP was that it was an I.Q. test. Also, note that it claims to be similar to the Mensa Admission Test. Mensa claims to be a high-I.Q. society, so presumably they think that something like it is an I.Q. test. I still want to know what organization gives this test to its members, as Recusant claims. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [July 3, 2015, 5:57pm UTC](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830/35 "2015-07-03T17:57:36Z")
If it helps to visualize the answer, just get rid of the hill and the distances. A man walks for 80 minutes. He walks at 2 mph for sixty of those minutes and 6 mph for the other twenty minutes. What was his average speed for the trip? [Previous page](https://boards.straightdope.com/t/a-question-on-an-iq-test-question-semi-math-related/723830.md?page=1)