Difficult Mathematical Puzzles - 1

Maths number puzzles often feature in a number of competitive exams. These are based on various concepts of algebra, arithmetic, time & distance, etc. Here is a compilation of a variety of puzzles grouped into: easy, medium and difficult maths puzzles. Each article contains a set of 10 mathematical puzzles with answers and explanations. In this article, you will come across difficult puzzles based on the basic concepts of mathematics. 
Practice the given questions and check your level of preparation:
Q.1.What is the minimum number of numbers needed to form every number from 1 to 8,000?
Example: To form 1122, you would need 2 – 1s & 2 – 2s but you would not count the numbers again that you had already counted from making 1122 to form another numbers.
You will need 3 of numbers 0, 8 & 9 ( _ 888 , _999 , _ 000 )and 4 of numbers from 1 to 7.( 1111 , 2222 , 3333 , 4444 , 5555 , 6666 , 7777 )
Total number required is : 4 * 7 + 3*3 = 28+9 = 37
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Q.2.To reach Honululu, in the middle of a barren plataeu, I will have to travel overland by foot from the coast. On a trek like this, each person can only carry enough rations for six days and the farthest we can travel in one day is 25 miles. Also, the city is 125 miles from the starting point. What I am trying to figure out is the fewest number of persons, including myself, that I will need in our group so that I can reach the city, stay overnight, and then return to the coast without running out of supplies. How many persons (including myself) will I need to accomplish this mission?
Total 5 persons are required.
It is given that each person can only carry enough rations for six days. And there are 5 persons. Hence, total of 30 days rations is available.
First Day: 5 days of rations are used up. One person goes back using one day of rations for the return trip. The rations remaining for the further trek is for 24 days.
Second Day: The remaining 4 people use up 4 days of rations. One person goes back using 2 days of rations for the return trip. The rations remaining for the further trek is for 18 days.
Third Day: The remaining 3 people use up 3 days of rations. One person goes back using 3 days of rations for the return trip. The rations remaining for the further trek is for 12 days.
Fourth Day: The remaining 2 persons uses up 2 days of rations. One person goes back using 4 days of rations for the return trip. The rations remaining for the further trek is for 6 days.
Fifth Day: The remaining person uses up one day of rations. He stays overnight. The next day he returns to the coast using 5 days of rations.
Thus, total 5 persons, including you are required.
Q.3. John goes to a furniture auction. All purchases must be paid for in cash. He goes to the ATM and draws out Rs.40,000 (Currency notes are of the denomination of Re.1, Rs.2, Rs.5, Rs.10, Rs.20, Rs.50 and Rs.100). Since John does not want to be seen carrying that much money, he places it in 16 envelopes numbered 1 through 16. Each envelope contains the least number of currency notes possible (i.e. no two tens in place of a twenty).
At the auction he makes a successful bid of Rs.8456 for a table. He hands the auctioneer envelopes number(s) 4, 9, and 14. After opening the envelopes the auctioneer finds exactly the right amount. How many Rs.2 notes did the auctioneer find in the envelopes?
Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of currency notes possible" is only to misguide you. This is always possible for any amount!!! One more thing to notice here is that John must have placed money in envelopes in such a way that if he bids for any amount less than Rs.40000, he should be able to pick them in terms of envelopes.
First envelope contains, 20 = Re.1
Second envelope contains, 21 = Rs.2
Third envelope contains, 22 = Rs.4
Fourth envelope contains, 23 = Rs.8 and so on...
Hence the amount in envelopes is Re.1, Rs.2, Rs.4, Rs.8, Rs.16, Rs.32, Rs.64, Rs.128, Rs.256, Rs.512, Rs.1024, Rs.2048, Rs.4096, Rs.8192, Rs. 16384, Rs. 7233.
Last envelope (No. 16) contains only Rs.7233 as total amount is only Rs.40000.
Now as he bids for Rs.8456 and gives envelope number 4, 9 and 14 which contains Rs.8, Rs.256 and Rs.8192 respectively.
Envelope No 4 contains one Rs 5, one Rs 2, one Re 1.
Envelope No 9 contains two Rs.100 note, one Rs.50 note, one Rs.5 note and one Re.1 note.
Envelope No 14 contains eighty-one Rs.100 note, one Rs.50 note, four Rs.10 note and one Rs.2 note.
Hence the auctioneer will find only two Rs.2 notes in the envelopes.
Q.4. An eagle is flying between two trains, each travelling towards each other on the same track at 80 km/h. The eagle reaches one engine, reverses itself immediately, and flies back to the other engine, repeating the process each time. The eagle is flying at 50 km/h. If the eagle flies 150 km before the trains meet, how far apart were the trains initially?
The eagle is flying at the speed of 50 km/h and covers 150 km. Hence, the eagle flies for 3 hours after trains started. It's obvious that trains met 3 hours after they started travelling towards each other. Also, trains were travelling at the speed of 80 km/h. So, each train traveled 240 km before they met. Hence, the trains were 480 km apart initially.
Q.5. Shantanu had three sons, to whom he wants to divide his 19 camels, in the following way: The oldest son was to receive one half the property, the next a quarter, the third one fifth. The three brothers were at a loss as how to divide the inheritance among them-selves without cutting up a camel, until a stranger appeared upon the scene.
Dismounting from his camel, he asked if he might help, for he knew just what to do. The brothers gratefully accepted his offer. Adding his own camel to Shantnu's 19, he divided the 20 as per the will. The oldest son received 10, the next 5, the third 4 . One camel remained: this was his, which he mounted and rode away.
Scratching their heads in amazement, they started calculating. The oldest thought: Are not 10 greater than the half of 19? Someone must have received less than his proper share! But each brother discovered that he had received more than his due. How is it possible?
They took their percentages from 20 and not from 19, so they got more than their share.
The oldest son got 1/2 of 20 = 10 which is 0.5 more
The second son got 1/4 of 20 = 5 which is 0.25 more
The third son got 1/5 of 20 = 4 which is 0.2 more
And the stranger got 1/20 of 20 = 1 which is 0.05 more (As he is not supposed to get anything)
All these fractions add to = 0.5 + 0.25 + 0.2 + 0.05 = 1 which stranger took away.
Q.6. There is a magazine made up of 8 large sheets of paper folded in half. The magazine has 32 pages altogether. The second sheet contains pages 3, 4, 29 and 30. If I pick up a sheet containing page 19, what are the other pages that this sheet contains?
First of all, you had to work out that page 19 was on the back of page 20. Then you had to notice that the numbers were arranged in pairs, with the first and last number adding up to 33 and similarly the second and second last number again adding up to 33.
Then the other pages with 19 and 20 are:
33 -19 =14
33 - 20 = 13
So the four pages on this sheet are 13, 14, 19 and 20.
Q.7. How many steps are required by a carpenter to cut a piece of plywood of dimensions a x b into 1 x 1 pieces?
You can cut an existing piece horizontally or vertically.
You cannot cut two or more pieces at once (so no cutting through stacks).
You need (ab – 1) steps. Take an example of 2×3 and it can be verified that 1 step is required to break the larger piece into two pieces of size 1×3 each. Then, 2 steps are further required for each piece to be made of the size 1×1. Thus, total number of steps are 5 i.e., 2×3 – 1.
Q.8. Ruchira went out for shopping. She had in her Da Milano handbag approximately Rs. 15/- in one rupee notes and 20 paise coins.When she returned, she had as many one rupee notes as she originally had 20 paise coins. And as many 20 paise coins as she originally had one rupee notes. She actually came back with exactly one-third of what she had started out with. How much did she spend and exactly how much did she have with her when she started out?
Let us assume that originally Ruchira had X one rupees and Y 20-paise coins. Going shopping she had (100 X + 20 Y) paise as total money. She returned with money equal to (100 Y + 20X) paise.
This last sum, as we know, is one-third of the original and therefore:
3(100 Y + 20 X) = 100 X + 20 Y
Simplifying we get X = 7Y
Now X=7 if Y = 1. Assuming this, Ruchira had Rs.7.20 when she set out for shopping. This is wrong because Ruchira actually had about 15 rupees.
Now if Y = 2, then X = 14. Assuming this, Ruchira had 14.40 rupees, which accords with the condition of the problem.
If we assume that Y = 3 then the sum will be too big- 21.60 rupees. Therefore the only suitable answer is 14.40 rupees.After shopping Ruchira had 2 one rupee notes and 14 twenty Paise coins. This is actually 1/3rd of the original sum 1,440/3 = 480. Ruchira’s purchases, therefore, cost 14.40 - 4.80 = Rs. 9.60
Q.9. A thief wants to open a lock with 5 digits as the key to open it. He knows that the 3rd digit is three less than 2nd digit, while 2nd digit is four smaller than 4th digit. The first digit is three times the fifth digit. There are three pairs whose sum is 11 and third and fifth digits are equal. Find the key to open the lock.
Let the number is abcde. a = 3e, b = d - 4, c+3 = b, c = e.
So we get the number as 3c, c+3, c, c+7, c.
By putting the value of c=1, the sum cannot be 11. By putting c=2, we get 3 pairs whose sum is 11. So the number is 65292.
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Q.10. What is the 504th digit of the series in which natural numbers are written in order i.e., 1234567891011121314…?
From 1- 9 there are 9 digits, 10-99 there are 90x2=180 digits. So in total we have counted 189 digits. Remaining digits= 504 - 189 = 315.
31digits will come in the category of 3 digit numbers. 315/3= 105. So we have to reach at 105 numbers. So 504 digit is 100 + 104 = 204.
So the 504th digit is the last digit of 204 i.e. 4.
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