Mathematical and Logical Riddles

Mathematical logical puzzles are a combination of mathematical and logical concepts. Such puzzle-based questions are frequently asked in various competitive exams and interviews. Here is a compilation of 10 math and logic puzzles devised to hone your mental aptitude. This set contains common maths, logical and analytical puzzles with answers for interviews and exams. 
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Try this combination of maths' riddles and tough reasoning puzzles:
Q.1. The cricket match between India and England was over.
Sehwag scored more runs than Saurabh.
Tendulkar scored more runs than Gautam but less than Rahul
Badani scored as much runs as Ajit but less than Rahul and more than Tendulkar.
Saurabh scored more runs than either Ajit or Rahul.
Each batsman scored 5 runs more than his immediate batsman. The lowest score was 20 runs.
How much did each one of them score?
Use the given facts and put down all the players in order. The order is as follow with Sehwag, the highest scorer and Gautam, the lowest scorer.
Sehwag - 45
Saurabh - 40
Rahul - 35
Badani, Ajit - 30
Tendulkar - 25
Gautam - 20
Q.2. Because of thermal plant shut down in the area there is all blackout and you were in a hall with a candle, an oil-burning stove, a gas lantern and a single match, which do you light first?
The match.
Q.3. Prashant lives in "Puri's Society" where all the houses are in a row and are numbered sequentially starting from 1. His house number is 29. Reshma lives in the same society. All the house numbers on the left side of her house add up exactly the same as all the house numbers on the right side of her house. What is the number of Reshma's house? Find the minimal possible answer.
Let's assume that in the "Puri's Society" there are total N houses numbered from 1 to N and Reshma's house number is X.
Now it is given that all the house numbers on the left side of Reshma's house add up exactly the same as all the house numbers on the right side of her house.
Hence, 1 + 2 + 3 + ..... + (X-1) = (X+1) + (X+2) + (X+3) + ..... + N
Both the sides of the above equations are in A.P. Hence, using A.P. summation formula,
[(X-1)/2][2*(1) + (X-1-1)] = [(N-X)/2][2*(X+1) + (N-X-1)]
[X-1][(2) + (X-2)] = [N-X][(2X+2) + (N-X-1)]
(X-1)(X) = (N-X)(N+X+1)
X2 - X = N2 + NX + N - NX - X2 - X
X2 = N2 + N - X2
2X2 = N2 + N
X2 = (N2 + N)/2
X2 = N(N+1)/2
Now, using Trial and Error method to find values of N and X such that above equation is satisfied, we get
N = 8, X = 6
N = 49, X = 35
But we require minimal possible answer and it is given that prashant's house number is 29. It means that there are atleast 29 houses. Hence, first value is not possible. And the answer is: there are 49 houses and Reshma's house number is 35.
Q.4. Major Sunder is forming five-person Special work force. The group must contain one Manager, two lawyers and two typists. J, K and L are possible lawyers. M, N and P are possible managers. E, F and G are possible typists. Also, J and L prefer to work with each other in the same team. P prefers to work only if F works. How many different possible Groups, Major Sunder can make?
As 2 lawyers to be selected from the given 3 and also J & L prefer to work together, JL must be there in all the possible Groups. Also, P prefers to work only if F works. It doesn't mean that F won't work without P. Hence, possible groups are:
JL - N - EF
JL - N - FG
JL - N - GE
JL - P - EF
JL - P - FG
JL - M - EF
JL - M - FG
JL - M - EG
Hence, there 8 different groups are possible.
Q.5. If a cow eats 50 pounds of grass every day EXCEPT every 5th day on which it only eats 40 pounds of grass. If the cow continues this, how many pounds of grass will it eat in 222 days?
The cow will eat 10,660 pounds of grass in 222 days.
It is given that on every 5th day cow eats 40 pounds of grass i.e. on day number 5, 10, 15, 20, .... 215, 220 the cow eats 40 pounds of grass.
Total number of 5th days = 222/5 = 44 (the cow eats 40 pounds)
Hence, the normal days are = 222 - 44 = 178 (the cow eats 50 pounds)
Thus, in 222 days, the cow will eat
= (178) * (50) + (44) * (40)
= 8900 + 1760
= 10,660 pounds
Q.6. You have 3 points labeled P, Q and R. You then have another 3 points labeled p, q and r. The aim of the puzzle is to connect point P with point p, q and r, point Q with point P, Q and R and point R with point p, q and r. Points can be arranged in any order but follow one rule that while connecting the points, lines cannot cross over each other.
There is no solution to it, if you consider 2 dimensions. It is impossible to join each of points P, Q and R with points p, q and r without lines crossing each other.
There is solution, if you consider 3 dimensions. Consider a circular base and a line perpendicular to it passing from the center. Now take any 3 points along the perimeter of the circular base as points p, q and r. Similarly take any 3 points along the perpendicular line as points P, Q and R. Now it is quite simple to join each of points P, Q and R with points p, q and r without any of the lines crossing each other.
The other possible 3D structure is Pyramid. Take points p, q and r as vertices of the triangular base and points P, Q and R along the height of the Pyramid which is perpendicular to the triangular base and passing through the apex.
Q.7. There are five containers of oil of different weight. They are weighed in pairs of two with all possibilities. The weights in kgs are 165, 168, 169.5, 171, 172.5, 174, 175.5, 177, 180, and 181.5 . How much does each container weigh?
Lets assume that the weights of five containers are B1, B2, B3, B4 and B5 pounds respectively. Also, B1 ≤ B2 ≤ B3 ≤ B4 ≤ B5. It is given that five containers of oil are weighed two at a time in all possible ways. It means that each of the container is weighted four times. Thus,
4*(B1 + B2 + B3 + B4 + B5) = (165 + 168 + 169.5 + 171 + 172.5 + 174 + 175.5 + 177 + 180 + 181.5)
4*(B1 + B2 + B3 + B4 + B5) = 1734
(B1 + B2 + B3 + B4 + B5) = 433.5 pounds..........(1)
Now, B1 and B2 must add to 165 as they are the lightest one.
B1 + B2 = 165...............(2)
Similarly, B4 and B5 must add to 181.5 as they are the heaviest one.
B4 + B5 = 181.5.............(3)
From above three equation, we get B3 = 87 pounds
Also, it is obvious that B1 and B3 will add to 168 - the next possible higher value. Similarly, B3 and B5 will add to 180 - the next possible lower value.
B1 + B3 = 168.............(4)
B3 + B5 = 180.............(5)
Substituting B3 = 87, we get B1 = 81 and B5 = 93
From 2 & 3 equations, we get B2 = 84 and B4 = 88.5
Hence, the weights of five containers are 81, 84, 87, 88.5 and 93 pounds.
Q.8. A token operated toy machine offers three selections - Teddy, Doll or Random (Either Teddy or Doll) but the machine has been strung out wrongly so that each switch does not give what it claims. If each token costs Rs. 25, how many minimum coupons have to be put into the machine to work out which switch gives which selection?
Put 1 token and push the switch named Random. There are only 2 possibilities. It will give either Teddy or Doll.
If it gives Teddy, then the switch named Random is for Teddy. The switch named Doll is for Random selection. And the switch named Teddy is for Doll.
If it gives Doll, then the switch named Random is for Doll. The switch named Teddy is for Random selection. And the switch named Doll is for Teddy. Thus, you can make out which switch is for what by putting just 1 token of Rs. 25 and pressing Random selection first.
Q.9. Five friends ran a 200m race. No friend ended at same position. Vijay did not come first.
Rohit was neither first nor last. Vishal came in one place after Vijay.
Rahul was not second. Rishabh was two place below Rahul. In what order did they finish?
Lets find the possible places 5 friends can finish. Possibilities are:
Vijay - 2, 3, 4 (not 5th as Vishal came one place after him)
Rohit - 2, 3, 4
Vishal - 3, 4, 5
Rahul - 1, 3 (not 4th & 5th as Rishabh is two place after him)
Rishabh - 3, 5
So the result is:
1 Rahul; 2 Rohit; 3 Rishabh; 4 Vijay; 5 Vishal
Q.10. At the entrance to a members club stands a stranger seeking admission. A friend told him that it's easy to get in. You just have to answer a question corrcetly! Answering wrong, however, will result in being shot! To live a little longer, the man waits in a back alley near the entrance for people to go in. After a while a man comes to the entrance. The door warden asks him: "Twelve?" to which he replies "Six!" and goes in. "That's easy." our friend thinks, but he waits a little longer. Another man comes to the door. "Six?" the door warden asks, to which he replies "Three!" and goes in. "That's too good to be true" our friend thinks, and he was not right. Because, when asked "Four?", he answered "Two!" and was found dead in the alley. What was the correct answer?
The answer is the number of letters in the word spoken by the door warden.
"Twelve" contains "Six" letters i.e. T, W, E, L, V, E
"Six" contains "Three" letters i.e. S, I, X Similarly,
"Four" contains "Four" letters i.e. F, O, U, R.
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