# Some Interesting Math Problems

Updated on August 7, 2018

SOME INTERESTING MATH PROBLEMS

I am searching through my files finding some good Mathematics materials when I come across these Math Excercises. I choose some interesting problems that are of some topics in Algebra and Probability. I present here the problems with their solutions.

Problem Number One :

An apple, an orange, a banana and a pear are laid out in a straight line . The orange is not at either end and is somewhere to the right of the banana. In how many ways can the fruit be laid out ?

Solution :

The orange (O) must be on the second or third place from the left. The banana (B) must be somewhere to the left of the orange. Hence the placement of the banana and the orange may take any of three forms namely BO_ _ , B_O_, or _ BO_. In each case two ways remain to fill in the open positions with an apple (A) and a pear (P). The total number of ways equals 3*2 or 6. The ways can be listed as follows:

BOAP BPOA PBOA BOPA BAOP ABOP

Problem Number Two :

Near the end of a party , everyone shakes hands with everybody else. A straggler arrives and shakes hands with only those people whom the straggler knows . Altogether sixty-eight handshakes occurred. How many other people at the party did the straggler know ?

Solution :

If all n people at a party shake hands with all others present then n (n- 1)/2 handshakes will take place altogether . Hence, the number of handshakes before the straggler’s arrival must have been sixty-six because that is the largest plausible value less than sixty-eigh . The straggler must have known two other people at the party. Constructing a table of possible values of n(n-1/2 clarifies that sixty-six is the only plausible number.

N n(n-1)/2

7 21

8 28

9 36

10 45

11 55

12 66

13 78

Problem Number Three :

Note that 1647/8235 = 1/5, start with 1647/8235, and delete one digit from both the numerator and the denominator to create an equivalent fraction. Then delete another pair to create another equivalent fraction.

Solution :

The successive equivalent fractions are 167/835 and 17/85 . The author Barry R. Clarke notes that this fraction is the only “sequential digital deletion fraction” with four digits in both the numerator and denominator that includes eight different digits.

Problem Number Four :

The supplement of an angle is 78 degrees less than twice the supplement of the complement of the angle . Find the measure of the angle,

Solution :

Let A = be the measure in degrees of the angle.

180 – A = be the supplement of this angle.

90 - A = be the complement of the angle .

Working equation :

(180 - A ) + 78 = 2 ( 180 - (90 – A) )

258 - A = 2 (90 + A )

258 - A = 180 + 2A

78 = 3A

A = 26

12

9

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• JITENDRA RANKHAMBE

5 years ago

I LIKE IT

• ssss

5 years ago

awesome

• Michael

6 years ago

Really helpful for my mathematical science work

• Mrigank Mongia

7 years ago

Interesting collection :)

• priyanka pokharel

7 years ago

awesome!!

• ritik

7 years ago

very nice

• kamal

7 years ago

send me more/////..

kamal.nicecool@hotmail.com

• Mayank kejriwal

7 years ago

That was really wonderful...stil looking for more harder qustions

• indranil

7 years ago

those were really good.can you please send me some more in indranilbhattacharya9874273176@gmail.com?

• BHARATH

8 years ago

NICE

• Shahid Bukhari

8 years ago from My Awareness in Being.

I would play the game, if it were not forbidden in my Belief ... Islam.

• pemekwulu

8 years ago from USA

Problem No. 2 sounds like application of triangular numbers. Sounds interesting!

• jayb23

8 years ago from India

Awesome Maths problems, made me remember my chilhood days.

• Iliveinthailand

8 years ago

Thank you sooo much cause I had to make my own interesting maths problems and I used yours. My name is not IliveinThailand of course but I'm not going to tell you my real name other wise if my maths teacher saw this and there was my name, he will detension me. Hahahaha! well Thanks again!!!! xxx

• Chris

8 years ago

Thanks for the problems.

For the handhake problem, it must have been 12 people at the party becaue more than that, and there'd be more than 68 handhakes. If it were 11 people, then that would mean the straggler knew 68-55 = 13 people - but there are only 11, so that's impossible. The same applies for 10 or less people.

• Coolmon2009

9 years ago from Texas, USA

Interesting hub thanks for sharing

• Bearman

9 years ago

It is interesting.Now I want to ask you one:

Now I have a watermelon ,I just want to cut it 3 times to make it 7 parts;and when I eat all parts ,the corver of it will be 8 parts.Please help me solve this problem.Thanks.

• brainstorming

9 years ago

really good one,can you write few more please.

• creation75

9 years ago from India

Very interesting. Want more from you.

• Research Analyst

9 years ago

Math has always been a fascinating thing to me but the part I like most is once you learn the correct formulas, it pretty much is the same across the board, unlike english which is always changing. Math problems really makes us become solution orientated.

• danny

9 years ago

easy

• scorpian

9 years ago

thank you

• priya shetty

9 years ago

excellent

• Anirudh

10 years ago

Thanks for the questions

Did them all

Really interesting.

• wilson

10 years ago

i finished all the questions.

if u have further interesting questions can u send them to my email? w1ls0n_C@hotmail.com

• AUTHOR

Maria Cristina Aquino Santander

11 years ago from Manila

Your welcome Kenny. God bless you.

• Ashok Rajagopalan

11 years ago from Chennai

Thank you, though my little mind boggles!

working