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Difficult Mathematical Puzzles - 5

Test your level of expertise in advanced mathematical puzzles.
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Math riddles and puzzles are based on the basic concepts of algebra, arithmetic, ratio & proportion, time & distance, probability, etc. Here is a compilation of different mathematical puzzles grouped into easy, medium and hard maths puzzles with answers. This article contains a set of 10 arithmetic and algebra puzzles with answers. In this set, you will come across hard math puzzles devised to hone your mental aptitude.
Solve these questions and check your level of preparation:
Q.1. Stranded on a deserted island, Harry Puttar is left with only a 40 litres container of milk. To conserve his milk he decides that on the first day he will drink one litre of milk and then refill the container back up with water. On the 2nd day he will drink 2 litres and refill the container. On the 3rd day he will drink 3 litres and so on... By the time all the milk is gone, how much water has he drunk?
It is given that the man has 40 litres container of milk. Also, he will drink 1 litre on the first day and refill the container with water, will drink 2 litres on the second day and refill the container, will drink 3 litres on the third day and refill the container, and so on till 40th day. Thus at the end of 40 days, he must have drunk (1 + 2 + 3 + 4 + ..... +38 + 39 + 40) = 820 litres of liquid.
Out of those 820 litres, 40 litres is the milk which he had initially. Hence, he must have drunk 780 litres of water.
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Q.2. A 3 digit number is such that its tens digit is equal to the product of the other two digits which are prime. Also, the difference between its reverse and itself is 99. What is the sum of the three digits?
It is given that the two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, hundreds digit and units digit must be either 2 or 3 i.e. 2_2, 2_3, 3_2 or 3_3 which means that there are four possible numbers - 242, 263, 362 and 393.
Now, it is also given that - the difference between its reverse and it is 99. So 263 and 362 satisfy this condition. Hence, the sum of the three digits is 11 in each case.
Q.3. One side of the bottom layer of a triangular pyramid has 11 balls. How many are there in the whole pyramid? Note that the pyramid is equilateral and solid.
As there are 11 balls along one side, it means that there are 11 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55 and 66 balls in the remaining layers.
Hence, the total number of balls is = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286 balls
Q.4. Substitute numbers for the letters so that the following mathematical expressions are correct.
ZYX/3 = LQ, PQR/6 = LQ, JKL/9 = LQ
Note that the same number must be used for the same letter whenever it appears.
Let's start with JKL = 9 * LQ. Note that L appear on both the side. Also, after multiplying LQ by 9 the answer should have L at the unit's place. The possible values of LQ are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (As all alphabets should represent different digits)
Now, consider PQR = 6 * LQ. Out of three short-listed values, only 73 satisfies the equation.
Also, ZYX = 3 * LQ is satisfied by 73.
Hence, Z=2, Y=1, X=9, P=4, Q=3, R=8, J=6, K=5, L=7
219/3 = 438/6 = 657/9 = 73
Q.5. When Poonia died, he willed his 17 dogs to the sons, to be divided as follows:
First Son to get 1/2 of the dogs, Second Son to get 1/3rd of the dogs & Third Son to get 1/9th of the dogs. The sons are sitting there trying to figure out how this can possibly be done, when a very old wise man goes riding by. They stop him and ask him to help them solve their problem. Without hesitation he divides the dogs properly and continues riding on his way. How did he do it?
The old man temporarily added his dog to the 17, making a total of 18 dogs.
First son got 1/2 of it = 9
Second son got 1/3 of it = 6
Third son got 1/9 of it = 2 for a total of 17.
He then takes his dog back and goes away......
Q.6. There are 3 colored boxes - Yellow, Black and Orange. Each box contains 2 envelopes. Each envelope contains cash - two of them contain Rs. 250000 each, two of them contain Rs. 150000 each and remaining two contain Rs. 100000 each.
There is one statement written on the cover of each box.
* Yellow Box: Both, a yellow box and a orange box contain Rs. 100000 each.
* Black Box: Both, a black box and a yellow box contain Rs. 250000 each.
* Orange Box: Both, a orange box and a black box contain Rs. 150000 each.
Only one of the above 3 statements is true and the corresponding box contains the maximum amount.
Can you tell which box contains the maximum amount and how much?
Orange box contains the maximum amount Rs. 400000
As it is given that only one of the given 3 statements is true; assume in turn, each statement to be true & the other 2 false and check whether the corresponding box contains the maximum amount.
Let's assume that the statement on the Orange box is true. Thus, the given 3 statements can be interpreted as
* At most one, a yellow box or a orange box contains Rs. 100000.
* At most one, a black box or a yellow box contains Rs. 250000.
* Both, a orange box and a black box contain Rs. 150000 each.
Going through all possible combinations, we can conclude that
Yellow Box: Rs. 100000 + Rs. 250000 = Rs. 350000
Black Box: Rs. 100000 + Rs. 150000 = Rs. 250000
Orange Box: Rs. 150000 + Rs. 250000 = Rs. 400000
You can test out for other two statements i.e. assuming Yellow box statement true and then Black box statement true. In both the cases, other statements will contradict the true statement.
Q.7. There are 511 oranges and 9 empty boxes. An orange vendor asks his son to place all the 511 oranges in all the 9 boxes in such a manner that if he asks for any number of oranges from 1 to 511, his son should be able to pick them in terms of boxes. How did the son place all the oranges among the 9 boxes, given that any number of oranges can be put in one box?
1, 2, 4, 8, 16, 32, 64, 128, 256
The orange vendor can ask for only 1 orange, so one box must contain 1 orange.
He can ask for 2 oranges, so one box must contain 2 oranges.
He can ask for 3 oranges, in that case box one and box two will add up to 3.
He can ask for 4 oranges, so one box i.e. third box must contain 4 oranges.
Now using box number one, two and three containing 1, 2 and 4 oranges respectively, his son can give upto 7 oranges. Hence, forth box must contain 8 oranges.
Similarly, using first four boxes containing 1, 2, 4 and 8 oranges, his son can give upto 15 oranges. Hence fifth box must contain 16 oranges.
You must have noticed one thing till now that each box till now contains power of 2 oranges. Hence the answer is 1, 2, 4, 8, 16, 32, 64, 128, 256. This is true for any number of oranges, here in our case only upto 511.
Q.8. 100 Engineers and diploma holders were employed by company XYZ for a website designing task. The company did not allowed any lady engineer to work without her spouse who should also be an engineer. Also, atleast half the men engineers working came with their wives who are also engineers. The company paid five hundred rupees per day to each man engineer, four hundred rupees to each woman engineer and one hundred rupee to each diploma holder. the company gave out 20000 rupees every evening.
How many men engineers, women engineers and diploma holders were working with the company?
Let's assume that there were X men engineers, Y women engineers and Z diploma holders working with the company.
Hence, M(men) + W(women) + D(diploma) = 100
500M + 400W + 100D = 20000, which gives 5M + 4W + D = 200
Eliminating M and W in turn from these equations, we get
M = 3D - 200
W = 300 - 4D
As if woman works, her spouse also works and atleast half the men working came with their wives; the value of W lies between M and M/2. Substituting these limiting values in equations, we get
If W = M,
300 - 4D = 3D – 200       
7D = 500     
D = 500/7 i.e. 71.428
If W = M/2,
300 - 4D = (3D - 200)/2  
600 - 8D = 3D - 200
11D = 800   
D = 800/11  i.e. 72.727
But D must be an integer, hence D=72. Also, M=16 and W=12
There were 16 men engineers, 12 women engineers and 72 diploma holders working with the company.
Q.9. A bag of Apricots was divided between Amitabh and Abhitabh. Abhitabh said, "It's not fair! You have 3 times as many Apricots I have." Amitabh said, "OK, I will give you one Apricot for each year of your age." Abhitab replied, "Still not fair. Now, you have twice as many Apricots as I have." "Dear, that's fair enough as I am twice older than you", said Amitabh. Amitabh went to Kitchen to drink juice. While he was in Kitchen, Abhitabh took apricots from Amitabh's pile equal to Amitabh's age. Who has more apricots now?
Let's assume that initially Abhitabh got N apricots and his age is T years. Hence, initially Amitabh got 3N apricots and his age is 2T years.
Operation
Amitabh's Apricots
Abhitabh's Apricots
Initially
3N
N
Amitabh gave T apricots to Abhitabh
(equals age of Abhitabh)
3N - T
N + T
Abhitabh took 2T apricots from Amitabh's pile (equals age of Amitabh)
3N - 3T
N + 3T
It is given that after Amitabh gave T apricots to Abhitabh, Amitabh had twice as many apricots as Abhitabh had.
3N - T = 2*(N + T)
3N - T = 2N + 2T
N = 3T
From the table, at the end Amitabh has (3N - 3T) apricots and Abhitabh has (N + 3T) apricots.
Substituting N = 3T, we get
Amitabh's apricots = 3N - 3T = 9T - 3T = 6T
Abhitabh's apricots = N + 3T = 3T + 3T = 6T
Thus, at the end Amitabh and Abhitabh, both have the same number of apricots.
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Q.10. Hansa ate a meal at Jugju hotel costing Rs.210. He gave the manager a Rs. 1000 note. He kept the change, came back a few minutes later and had some food packed for his girl friend 'Hansi'. He gave the accountant a Rs. 500 note and received Rs. 120 in change. Later the bank told the accountant that both the Rs. 1000 and the Rs. 500 notes were counterfeit.
How much money did the restaurant lose? Ignore the profit of the food restaurant.
Hotel lost Rs. 1500.
First time restaurant has given food worth Rs. 210 and Rs. 790 change. Similarly second time, food worth Rs. 380 and Rs. 120 change. Here, we are not considering food restaurant profits.
Mathematical Puzzles: Practice Exercises
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