Medium Mathematical Puzzles - 4

Mathematical riddles and puzzles form an essential part of many competitive exams. These are mostly based on the concepts of algebra, arithmetic, ratio & proportion or picture puzzles. Here is a compilation of some of the best math puzzles grouped according to their level of difficulty. Each article contains a set of 10 maths riddles and puzzles with answers. In this article, you will come across medium level puzzles devised to hone your mental aptitude.
Practice these questions and check your mental abilities:
Q.1.There is a big sports room with 100 lockers, numbered sequentially from 1 to 100. Any locker can be either open or closed. Initially all the lockers are closed. The first sportsman who walks down the room opens every locker. The second sportsman closes every second locker. The third sportsman flips every third door, opening some that are closed, and closing others, which were open. The fourth sportsman flips every fourth door, and so on. After 100 sportsmen have passed down the room, how many lockers are still open?
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10 lockers are open in the end. These will be the lockers number 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.
Q.2.Use the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 to satisfy the given equations.
1 + 7 = 8, 9 - 5 = 4, 6 = 3 x 2
Hint : Because in the Last column there is a multiplication operation then it has to be less than 10 and cannot consist 1 because any number multiply by 1 will give the number itself . hence The cases will be 2 x 3 and 2 x 4 . And 3 & 4 cannot be placed in the right hand corner cell , because 2 will be used to drive it . hence the right hand bottom cell has to be 2 .
Q.3.A Pizza delivery boy named Mahesh wanted to give 20 pizzas to the office. He had bags which can carry a maximum of 10 pizzas. There were ten guards standing at different positions near the office. Each guard would take one pizza from each of the bag. (i.e. if the Delivery boy carries 2 bags, and there are say 10 pizzas in each bag, the first guard would take 1 from each bag i.e. total of 2. And all other guards would do this.) In order to give 20 pizzas to the office, how any minimum pizzas the delivery boy should carry?
  1. 20
  2. 132
  3. 66
  4. 99
Explain your answer
Answer is 66
The delivery boy would carry 7 bags at the first guard at which he would be giving 7 pizzas to him.
Now he would be left with 59 pizzas which could fit up in 6 bags. So he would carry 6 bags to the 2nd guard and give him 6 pizzas.
Now he would be having 53 pizzas which would again fit into 6 bags only. So at the end 3rd guard he would be getting 6 pizzas.
After this he would be having 47 pizzas which would fit into 5 bags which means the 5th guard will get 5 pizzas.
Similarly the process goes on till the delivery boy has 23 pizzas and he gets to the last guard. He gives hin 3 pizzas acc. to the three bags which he has and gives the office the rest 20 pizzas.
Q.4.I have a half-cup of tea and a half-cup of coffee. I take one teaspoon of tea and mix it with my coffee, then I take one teaspoon of this mixture and mix it with the tea. Which cup contains more of its original contents?
Each will have the same amount of its original content. How much liquid is being moved does not matter (whether it is a teaspoon, half a cup, or, in extreme case the full cup does not make any difference).
Q.5.When the units and tens digits of a certain two-digit number are reversed, the sum of the two numbers is 121 and the difference is 9. What is the tens digit of the original number?
  1. 1
  2. 3
  3. 4
  4. 6
  5. 7
The correct answer is (D). Taking the number 10t + u, when we reverse the digits, we get 10u + t. The sum is then (10t + u) + (10u + t) = 11t + 11u = 121. Dividing by 11, we have t + u = 11. Taking the difference: (10t + u) – (10u + t) = 9t – 9u = 9, and dividing by 9: t – u = 1.
Finally, adding: t + u = 11, t – u = 1
2t = 12, t = 6.
Q.6.When a student weighing 54 kg left a class, the average weight of the remaining 59 students increased by 100g. What is the average weight of the remaining 59 students?
Let the average weight of the 59 students be A. Therefore, the total weight of the 59 of them will be 59A.
When the weight of the student who is left is added, the total weight of the class = 59A + 54
When this student is also included, the average weight decreases by 0.1 kgs.
(59A + 54) / 60 = A – 0.1, => 59A + 54 = 60A – 6, => 54 + 6 = 60A – 59A, => A = 60.
Q.7.A man is paid Rs. 240 plus one coat for one year service. However he left after 9 months and receives Rs. 150 and a coat. Then find the price of the coat.
The man worked for 9 months instead of 12 months, he should receive 9/12 of his annual payment i.e., ¾ (240 + 1C) which is equal to 150 + 1C where C is the price of the coat. ¾ (240 + C) = 150 + C. Therefore Price of the Coat (C) = Rs. 120.
Q.8.Albert fired 5 shots and scored 120 points. All the shots were on target. Which targets did the shots land on?
Shots hit by Albert
Target Hit by Albert are : 5, 5, 27, 36, 47.
Q.9.Can you plug in either addition (+), subtraction (-), multiplication (x), division (/) and parenthesis among five number 7s to make a target result number 50?
7 7 7 7 7 = 50
((7 x 7 x 7) + 7) / 7= 50 or ((7 / 7) / 7 +7) x 7 = 50
Q.10.Pooja went to Singapore with her husband. It rained for 13 days. But when it rained in the morning, the afternoon was lovely. And when it rained in the afternoon, the day was preceded by clear morning.
Altogether there were 11 very nice mornings and 12 very nice afternoons. How many days did their holiday last?
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Let's assume the number of days as follows:
Rain in the morning and lovely afternoon = X days
Clear morning and rain in the afternoon = Y days
No rain in the morning and in the afternoon = Z days
Number of days with rain = X + Y = 13 days
Number of days with clear mornings = Y + Z = 11 days
Number of days with clear afternoons = X + Z = 12 days
Solving above 3 equations, we get X = 7, Y = 6 and Z = 5
Hence, total number of days on holiday = 18 days
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