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Tricky math problems to check your IQ

Top 10 Tricky Math Problems To Check Your IQ

The internet is full of tricky math questions. While many of us quickly solve some, there are many we fail to solve. Here are 10 tricky math problems to check your IQ, making you question how you graduated!

List Of Top 10 Tricky Mathematics Questions:

Question 1

Puzzle: The Monty Hall Problem:

Puzzle here refers to the monty hall problem

Imagine you are a game show contestant. In front of you, there are three doors. Behind each door, there is one thing. While behind one door is a car, the other two have one lion each. Now you are asked to choose. Let’s say you chose Door 1. Before revealing what’s behind Door 1, the host opens another door, say Door 3, showing a lion. The twist is the host knows the correct door. But to intensify the excitement, he offers you a chance. You can either stick to your original choice or make a new choice.

What should you do to maximize your chances of winning the car? Should you stick with your original choice, switch doors, or does it not matter?

Solution

This traditional probability conundrum frequently astounds people with its answer. Think about it, and let me know if you need any further explanation!

The best course of action is to switch doors. It may surprise you to learn that switching increases your chances of winning the car. This is the reason why:

There is a 1/3 chance that the car is behind the door you initially selected and a 2/3 chance that it is behind one of the other two doors. The probability that the automobile is behind the door you initially selected stays at 1/3, but the total probability that it’s behind one of the other two doors increases to 2/3 when the host opens the door to reveal a lion. Consequently, switching doors raises your chances of winning to 2/3.

You might think your intuition plays the key role here. But instead, follow the host’s reactions to get the correct door!

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Question 2

Puzzle: Fermi Problem Puzzle: The Piano Tuners of Chicago

Chicago’s beauty increases 10 fold because of its vibrant culture and music scene. Thus, one usually ends up pondering: How many piano tuners does the city have?

Solution:

Let’s see how the Fermi Problem was solved:

It is estimated that 2.7 million people are living in Chicago. Of them, between 1% and 2% of American homes have a piano. For Chicago, let’s assume a comparable percentage. Therefore, based on our estimate of 2.7 million people, between 27,000 and 54,000 Chicago households own a piano. For best results, pianos usually require tuning once or twice a year. For our estimation, let us say that it occurs once a year on average. Depending on many criteria such as work efficiency, tuning complexity, and travel time, a piano tuner can often tune two to four pianos daily. Assuming a reasonable estimate, each day, there are two pianos.

Consider additional pianos in businesses, schools, churches, and other institutions. Let’s estimate that about 30% of pianos are in such settings.

So, the estimated number of piano tuners:

– Number of households with pianos: 27,000 to 54,000

– Number of pianos needing tuning per year: 27,000 to 54,000

– Number of pianos needing tuning per day (assuming once a year): 27,000 to 54,000 / 365 ≈ 74 to 148

– Number of pianos in businesses and institutions: 30% of total pianos ≈ 22 to 44

– Total number of pianos needing tuning per day: 74 to 148 + 22 to 44 ≈ 96 to 192

Given that a piano tuner can handle around 2 pianos per day, we divide the total number of pianos needing tuning per day by 2 to estimate the number of piano tuners:

– Estimated number of piano tuners: 96 to 192 / 2 ≈ 48 to 96

Thus, approximately 48 to 96 piano tuners are available in Chicago.

Question 3

Puzzle: The Birthday Paradox

 Birthday party puzzle refers to the probability of two people sharing the same birthday in group of 23

In a 23-person group, what is the probability of two people sharing the same birthday?

Solution

The probability is approximately 50%. How?

Well,
1. Compute the Probability of No Shared Birthdays: to determine the likelihood that none of the 23 individuals have the same birthday, we begin with the first person (probability of 1), the second (probability of 364/365, since there are 365 days in a year and one birthday is already taken), the third (probability of 363/365), and so on, up to the 23rd individual. All of these probabilities are multiplied collectively.

2. Use the Complement Rule: The likelihood that no two persons share a birthday is equal to the likelihood that at least two people do. Thus, we deduct the likelihood of no common birthdays from 1 to determine the latter.

Based on these calculations, we determine that among 23 persons, there is a 50% chance that at least two of them have the same birthday. This indicates a more than 50% chance that two or more of the twenty-three randomly selected individuals in the group will have the same birthday.

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Question 4

Puzzle: Where Was Your Car Parked?

This puzzle shows a car being parked and asks the readers to find out the parking lot number

Solution

Just flip your device, and then you will know it is 87! Too many calculations failed at once? Well, you are not alone!

Question 5

Puzzle: How Many Do You See?

This puzzle asks readers about the number of triangles

Solution

Text The answer to the triangle puzzle is 18

Question 6

Puzzle: The Missing Dollar Riddle

A hand holding three dollars

A $30 lunch is ordered by three friends at a restaurant. They each donate $10 since they agree to divide the cost equally. Subsequently, the waiter discovers that the lunch was actually only $25 and that an error occurred. He returns the five dollars to the friends. They choose to return $1 to each person and save $2 for a tip instead. At this point, the waiter has $2, and each customer has paid $9, for a total of $27. How did the missing money end up?

Solution

Actually, there isn’t a dollar missing. Because the $2 tip is added to the $27 that the companions paid, the computation is deceptive. But the $27 already includes the gratuity, so you don’t need to worry about an additional $2.

You think all algebra classes are gone to waste? Not always! Your college algebra classes aren’t wasted. So concentrate and let’s move on!

Question 7

Puzzle: Two Train Problem

Two trains are facing each other while a bee is in the middle

Train X is traveling at 60 mph. It leaves Station A at 9:00 AM. Train Y travels at 40 mph in the opposite direction of Train X. Train Y leaves Station B at 10:00 AM. The distance between Station A and Station B is 300 miles. When do the two trains meet?”

Solution

From the problem, we know that Train X started its journey 1 hour before Train Y. So, within that 1 hour, Train X traveled 60 miles. Since both trains travel opposite directions, we can imagine them moving toward each other.

So, the combined speed of the two trains is 60 mph + 40 mph = 100 mph. To find out how long it will take for the trains to meet, we can use the formula distance = rate x time.

The remaining distance from 300 miles is (300-60) = 240 miles.

So the formula will be 240 miles = 100 mph x time,

where time = 240/100 = 2.4 hours.

We calculated the distance when Train Y started its journey at 10:00 AM, and they will take 2.4 hours or 2 hours and 24 minutes to meet.

Therefore, the two trains will meet at 12:24 PM.

Question 8

Puzzle: The 100 Prisoners and a Light Bulb Problem

One 100 inmates are housed in isolation cells with no means of communication. The living room is in the middle and has just one lightbulb. From their cell, none of the prisoners can see the lightbulb. The warden randomly selects one prisoner to enter the living room each day. The prisoner has two options: either do nothing or flip the lightbulb, turning it on when it’s off or off while it’s on. Before the process starts, the convicts are free to devise a plan. How can they ensure that, at some point, every prisoner has made it into the living room?

Solution ,

The inmates can decide that they must turn on the light exactly once when another prisoner enters the room when it is dark, but only if they have never done it before. They do nothing if a prisoner comes in and the light is on or they have previously toggled it. This plan ensures that all of the prisoners will have entered the room at some point. Sometimes, you don’t need to perform calculus to get the correct answer.

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Question 9

Puzzle: So What is the Value of 6?

If 1=3

2=3

3=5

4=4

5=4

Then, 6=?

Solution

If you have started calculating, I’m sorry to let you know that sometimes it is not about mathematical formulas. It’s all about what’s in front of our eyes. As ‘six’ has three letters, the answer is three! Get it now!?!

Question 10

Puzzle: Cheryl’s Birthday Puzzle

Cheryl just became friends with Albert and Bernard. They both want to know Cheryl’s birthday. When they asked Cheryl, she gave them a list of 10 days.

15th May, 16th May, 19th May, 17th June, 18th June, 14th July, 16th July, 14th August, 15th August, and 17th August.

To give some hint, Cheryl told the month to Albert and the day to Bernard separately.

Albert said that he doesn’t know Cheryl’s birthday, and he also knows that Bernard does not know her birthday too. Then Bernard replied, earlier, he did not know her birthday, but now he knows. Alber then confirms that he also knows Cheryl’s birthday.

Now, you have to determine which date from Cheryl’s list is correct.

Solution

Only 19th May and 18th June are unique on the list of 10 dates. The other dates, 14, 15, 16, and 17, come twice. So, May and June are removed from the probable list.

14th July and August cannot be the option as both dates are the same. So any dates between 16th July, 15th August, and 17th August are the correct option. As August has two dates, it is not the birth month. Hence, on 16th July, you will answer this challenging math question correctly.

BONUS: Question 11

Puzzle: Three Gods Puzzle

Three gods have unique characteristics. One always speaks true, one speaks false, and the other speaks randomly, be it true or false. Determine which god is which. You can ask one yes-no question to each god.

Three god puzzle is being shown here by representing three Buddhas

Solution

Ask any god – “If I were to ask you if you were the truth-teller, would you say yes?” If yes, you can determine the god who speaks the truth. The god who speaks randomly can’t answer the question, and the god who always speaks falsely will answer no.

Later, a computer scientist, John McCarthy, added another difficulty level to this puzzle. Such as,

The gods can understand your question in English, and they will answer in their language, ‘Da’ and ‘Ja,’ for yes and no. But you need to know which means yes and which means no. So, you must find the meanings of ‘Da’ and ‘Ja.’

Answer to the three gods puzzle is given here

The solution to this tricky math question is to ask each god the same question. The true god will always say yes, and the false god will always answer no. This way, you can determine the nature of the gods.

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Conclusion

We believe these 10 tricky math problems and their solutions have given you an idea of how or what logic you should apply when facing such mathematical equations. So, if you can’t solve it by yourself, focus on practicing hard. Also, now that you know all the challenging maths questions and answers, let us know how many you got correct.

Mark Twain