Maths riddles and puzzles are based on various concepts of arithmetic, clocks, permutations, time and speed, etc. Solving these puzzles enhances your brain power as well as memory. Additionally, one of the most common puzzles are number puzzles. Here is a compilation of math brain teasers with answers and explanations. Each article consists of 10 simple maths puzzles.
Solve the given questions and check your mental abilities:
Q.1.Set A = (1,3,5,7------------297,299 -150 odd numbers). How many ways are there to choose exactly 18 numbers from set A such that their sum is 191?
0 number of ways is the answer because the sum of 18 odd numbers can never be odd.
Q.2.Vivek, a chain smoker, one day found it extremely difficult to go out to the market to get some cigarettes . He had some cigarette butts with him and he could pull tobacco out of some butts. 5 butts of tobacco make a cigarette. He collected the tobacco and made the cigarettes from the 121 butts available with him. How many cigarettes would he be able to smoke?
120/5 = 24 cigarettes and from (24 + 1) cigarettes, 25/5= 5 cigarettes, and then 5/5 = 1 cigarette. Thus total number of cigarettes, he can roll = 24+5+1 = 30 cigarettes.
Q.3.Find the missing number:
A. 189
B. 191
C. 3
D. 255
We have: 5x2+1=11, 11x2+1=23, 23x2+1=47, 47x2+1=95. So, missing number = 95x2+1=191.
Q.4.In a faulty watch, the minute and the hour hand of a watch meet every 64 minutes. How much does the watch lose or gain time?
The watch gains [65(5/11) - 64] = 1(5/11) minutes because the minute and hour hand meet after every 65(5/11) minutes.
Q.5.The following set of boxes has to be filled in such a way that all the odd digits are utilized in the boxes on the left side while even digits are utilized on the right side boxes. The digits available are: 1, 2, 3, 4, 5, 6, 7, 8, and 9 to make the equation correct.
The following are the two different solutions to this problem where all odd & even digits are placed in left & right boxes respectively.The sum of the integers on the left side is always ODD as it is the sum of three odd numbers While on the right hand side since the two digit number has to be even so the remaining part should be odd and that can be formed in only one way i.e 6 / 2 = 3
Q.6.In an entrance exam, a student has to attempt 250 questions in 4 hours. Out of these 250 questions, 50 questions are on Quantitative Ability (QA). How many minutes should be spent on QA section if his mentor has suggested him that he must spend twice as much time on each maths problem as spent for every other question?
Since the student has to spend on twice the time on quantitative questions as compared to the other questions, so let us assume that there are 100 quant questions instead of 50 taking as much time as any other question.
Therefore, now the target for the student is attempt 300 questions in 240 minutes (i.e. 4 hours).
Applying unitary method, we can say that
300 questions to be attempted in 240 minutes
Thus, 100 questions of quant section should be attempted in (240/300)x100 = 80 minutes.
Q.7.Find the value of 4.4! + 5.5! + ......+n.n!
4.4! can be written as (5! - 4!), similarly 5.5! = (6! - 5!) , 6.6! = (7! - 6!) and so on.
Thus by 4.4! + 5.5! +......+n.n! = (5! - 4!) + (6! - 5!) + (7! - 6!) +......+ (n+1)! - n! = (n+1)! - 4!.
Q.8.The height of a certain flagpole is 33 feet. Grease is applied to the pole. A monkey attempts to climb the pole. It climbs 5 feet every second but slips down 4 ft in the next second. When will the monkey reach the top of the flagpole?
Since monkey climbs 5 feet in 1st second and slips 4 feet in the next thus the monkey gains 1 ft. net in every two seconds.
However, when it has reached 28ft, in the next jump (5ft upwards) it will reach the top.
It will take 28*2 = 56 seconds to reach 28ft & then 1 sec to reach the top.
Hence, 57 seconds is the correct answer.
Q.9.Mangleshacharya was born in 1357 B.C. He had lived one-third of his life as a boy, one-fifth of his life as a youth, one-fourth of his life as a man and the remaining 39 years as an old man. Which year did he die?
Let's assume that age of Mangleshacharya was 'a' years when he died.
(life as a boy) + (life as a youth) + (life as a man) + (life as an old man) = (total life)
(a/3) + (a/5) + (a/4) + 39 = a
20a + 12a + 15a + 39*60 = 60a
13a = 39*60 a ⇒a = 180 years
Thus, he was 180 years old, when he died. As he was born in 1357 B.C., he died in (1357 - 180) = 1177 B.C.
Q.10.What is wrong with this proof?
2 = 1? ;
x = y
x2 = xy
x2 – y2 = xy – y2
(x + y) (x - y) = y (x - y)
x + y = y
2y = y
2 = 1
Since x = y, x - y = 0. So, for any equation in Mathematics division by zero is not allowed . hence the x-y cannot be cancelled out from both sides
