Can't fathom out how MENSA came to this answer.....

[Kos]

I am struggling to work out how MENSA came up with an answer of 472mph to this question.........

An aeroplane covers its outward journey at 590mph. It returns, over exactly the same distance, at 393.33 mph. What is the plane's average speed over the whole journey?

Would anybody care to explain please?

[Norberto]

If the distance is x miles, the outward journey takes x/590 hours. Return journey takes x/393.33 hours.
The total journey is 2x which takes x/590 + x/393.33 hours so average speed is :-
2x/(x/590 +x/393.33) = 472 mph

[R1Geezer]

well perhaps that's why you are not in Mensa!

I imagine that you are missing the time factor what is the elepsed time of the 2 trips?

[Vagrant]

To put it another way, to calculate average speed, you have to take the total distance covered and divide by the total time taken - you can't just take the 2 speeds and take the average.

[Kos]

Thanks Norberto and Vagrant. Understand now

[gazzawazza]

I can't for the life of me understand any of the above answers.
Taking the information given as the average speed each way I think it is a misprint.
Outward = 590mph, inward = 393.33mph. 590+393.33 = 983.33mph. Divide by 2 = average speed of 492mph.

[derrynoose]

On a journey each way of 23206.47miles the outward journey would take 393.33 hours. The return journey would take 590 hours. The total time is 983.33hours. The total of both journeys is 46412,94 miles. Average speed is 46412.94/983.33 which gives 472.
Norberto, Vagrant and Mensa are perfectly correct!

[gazzawazza]

Derry, are you taking the p or what?!
I can't make out if you're being serious or facetious!

[Vagrant]

gazzawazza, derrynoose is both serious, and accurate.
Your calculation is doing exactly what I said you cannot do!

[another-view]

Ok let's simplify it. Say the aeroplane flew out at 100mph and returned at 1 mph. It travelled 100 miles. What's the average speed? 100+1 divided by 2 = 50.5mph? No, as it was travelling for 100 times longer on the return journey.

[gazzawazza]

Ok, I'd just like to say right now that I'm not a thicko, I'm normally brilliant at working out puzzles and stuff, and I was very good at maths at school, but I still can't get my head round this!!!

[gazzawazza]

It has finally sunk in!!!

Using similar figures and Norberto's method;
Distance to travel 12000 miles.
Outward speed, 600mph,
Inward speed, 400mph.

Outward = 12000/600 = 20.
Inward = 12000/400 = 30.
Total journey = 2x, so 12000 x 2 = 24000.

24000 / (20 + 30) = 24000 / 50 = 480.

I just couldn't see the way to solve this one myself! You learn something every day!

[mo183]

Any problem designed to determine IQ is primarily designed to determine the ability to recognise relationships. The application of knowledge (mathematics, physics, geometry, language etc.) to the recognised relationship is secondary to the purpose of measuring intelligence.

The failure to recognise a relationship leads to the inability to resolve the facts.

The relationship ... 590 and 393.33 - mmm! That's about 1200 (2x600,3x400 - knowledge), but I see it is 2x10 and 3x6.66 less. The big clue comes from the glaringly obvious 393.33 borne of the knowledge that it must be the result of dividing by three (or a multiple of 3)

Now then the SIMPLE maths, without even knowing what you are trying to calculate, what else goes into 1180 ... that'll be 5 at a guess. Two lots of 1180 (2360) divided by 5 is 472 whatever-they're-called ... and five is the sum of the factors 2 and 3 (intelloigent assumption not proof).

Double check (for those without confidence). 4 lots of 1180 (2 journeys would have to be divided by 10). That wouldn't be 4720 thingies, would it? Simple maths, again!