# These math riddles are sure to confuse you a little!

Math can be confusing at times, and when presented in the form of twisted riddles, it can be really tricky to handle. Want to give these math riddles a try?
Math riddles with answers!

Raise your hands if math was your favorite subject in school! Ask this question in a hall with over 100 people, and you may find only 7-8 raising their hands.

Why? Are mathematicians like Euclid and Pythagoras to be blamed? Well, while many people do blame them, they are innocent. Rather, such mathematicians and others all around the globe have made significant transformations in the ways we look at numbers.

While numbers themselves are tricky to handle, the situation becomes worse in cases when alphabets are added to numbers. Was it a busy day? Do you want to relax and give your mind a fresh break? Well, this blog might be the best piece of content you will read today.

## Math Riddle 1:

What do you get when you combine Einstein and Pythagoras discoveries?

## Math Riddle 2:

The numbers a, b, and c are positive integers.

An apple costs \$a, a banana costs \$b, and a cherry costs \$c.

The cost of b apples, b bananas, and a + b cherries is \$77.

What would the cost be for one apple, two bananas, and one cherry?

## Math Riddle 3:

Prove that 2 = 1?

Eager to check the answers? Scroll down!

Math Riddle 1:

What do you get when you combine Einstein and Pythagoras discoveries?

E m c2 = m ( a2 +  b2 )

Math Riddle 2:

The numbers a, b, and c are positive integers.

An apple costs \$a, a banana costs \$b, and a cherry costs \$c.

The cost of b apples, b bananas, and a + b cherries is \$77.

What would the cost be for one apple, two bananas, and one cherry?

Let's try to solve this…

b*a +b*b + (a+b)*c = 77

b(a+b) + c(a+b) = 77

(a+b)(b+c) = 77

Now see the question - The numbers a, b, and c are positive integers.

and we know that it is a whole number that needs to be multiplied to get a value of 77. So just get the factors of 77. You will realize that this is a simple composite number to find the factors. - 1, 7, 11, and 77.

Basically, the number 77 can be obtained by 1 x 77 or 7 x 11. This realization is key.

Problem solved. A+B or B+C can’t be 1 so it has to be 7 and 11.

So a + 2b + c = (a+b) + (b+c) = 7 + 11 = 18.

Math Riddle 3:

Prove that 2 = 1?

If a = b (let’s suppose)                                        [a = b]

Next, we multiply both sides by a

Next we’ll see that a2                                  [a2 = ab]

When compared with ab

They are the same. Remove b2.                  [a2− b2 = ab − b2]

Both sides we will factorize.

Now each side contains a − b.                     [(a+b)(a − b) = b(a − b)]

We’ll divide through by a

Minus b and olé

a + b = b.                                 [a + b = b]

But since I said a = b

b + b = b you’ll agree?                                 [b + b = b]

So if b = 1

Then this sum I have done                          [1 + 1 = 1]

Proves that 2 = 1

We know that you frowned the moment you saw the alphabet in between the numbers. But hey, you can’t deny the fact that you enjoyed the last 5-10 minutes you spent reading this blog.

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