Brain teaser: This logic problem from a Hong Kong elementary school entrance exam has become a viral sensation, leaving many adults stumped. Children, however, can solve it in 20 seconds
A logic puzzle has baffled the internet, but its solution is apparently so simple that a child can solve it in seconds.
The now-viral puzzle comes from a Hong Kong elementary school admission test for six-year-olds, who are required to solve it within 20 seconds.
The test, as reported by Centauro, features a drawing of a parking lot with a car positioned in one of the six numbered spots, blocking the number from view.
Based on the visible numbers, the students are asked to determine the number of the spot where the car is parked.
If you can't solve it right away, you're not alone: apparently, many adults have been stumped by the first-grade entrance exam question.
At first glance it might seem like logic or algebra is needed to find the right answer.
But, as it turns out, the solution is much easier than it first appears. It is explained below.
Keep it simple: Gronwups tend to overthink the question, mistakenly believing that the solution requires mathematical calculations
The trick is to flip over the page - or the computer screen - with the drawing, making it apparent that the parking spots are numbered in a sequence, from 86 to 91.
The car is therefore in the second-left space, which is parking spot number 87.
Adults tend to overthink the question, erroneously believing that the solution is rooted in complex mathematics, but children, who have a tendency to look at things from various angles, quickly grasp the simplicity of the puzzle and nail the answer right away.
British puzzle inventor David Bodycombe told The Guardian this week that the parking lot logic problem is his brainchild, inspired by a car park he had seen in Portugal 20 years ago.
It follows a number of other seemingly difficult maths questions that have gone viral lately, including an Edexcel GCSE Maths question in the UK that had some students complaining on Twitter.
THE OTHER MATHS QUESTIONS THAT STUMPED THE WORLD
When is Cheryl's birthday?
In April this year, people across the world were left baffled by a maths problem set for 14-year-olds in Singapore.
The test asked:
Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.
Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
Bernard: At first I don't know when Cheryl's birthday is, but I know now.
Albert: Then I also know when Cheryl's birthday is.
So when is Cheryl's birthday?
The solution is in the image below.
In June, GCSE students in the UK complained about what they said was a particularly hard Edexcel Maths paper.
There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow.
Hannah takes a sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet.
The probability that Hannah eats two orange sweets is 1/3.
Show that n²-n-90=0
Some students took to Twitter to mock the Edexcel Maths question, which they said was too hard
The solution was to work out that it was not asking for the value of n, but to do some algebra to show that n²-n-90=0.
When Hannah takes her first sweet from the bag, there is a 6/n chance it is orange.
This is because there are 6 orange sweets and n sweets in total.
With her second sweet, there is a 5/(n-1) chance that it is orange.
This is because there are only 5 orange sweets left out of a total of n-1 sweets.
The chance of getting two orange sweets in a row is the first probability multiplied the second one: 6/n x 5/n–1
The question tells us that the chance of Hannah getting two orange sweets is 1/3.
So: 6/n x 5/n–1 = 1/3
Now rearrange this equation.
(6x5)/n(n-1) = 1/3
30/(n² – n) = 1/3
Then times by 3 on both sides to cancel out the fraction gives:
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