Difficult Mathematical Puzzles - 4

Mathematical games 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 with solutions, grouped according to the level of difficulty. This article contains a set of 10 algebra based and arithmetic puzzles with answers. In this set, you will come across difficult mathematical puzzles devised to hone your mental aptitude.
Solve these questions and check your level of preparation:
Q.1. A number of 5 digits have the following properties:
The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, . Each digit in the number is different i.e. no digits are repeated. The digit 0 does not occur in the number i.e. it is comprised only of the digits 5-9 in some order. How many such numbers are possible and find their value/values?.
Two numbers are possible and are 78965 or 98765
The way to solve this question is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5.
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Q.2. Power went off when Vijay was studying. It was around 2:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last four hours and the thin one one hours less. When he finally went to sleep, the thick candle was twice as long as the thin one. For how long did Vinay study in candle light?
Assume that the initial length of both the candle was L and Vijay studied for X hours.
In X hours, total thick candle burnt = XL/4
In X hours, total thin candle burnt = XL/3
After X hours, total thick candle remaining = L - XL/4
After X hours, total thin candle remaining = L - XL/3
Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.
(L - XL/4) = 2(L - XL/3 )   ⇒ (4 - X)/4 =2 (3- X)/3
12 - 3X = 24-8X
5X = 12,    ⇒ X =12/5
Hence, Vijay studied for 2hours and 24 minutes.
Q.3. Rohit kanwar went to hitbullseye bank to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither Rohit nor the cashier noticed that. After spending 23 paise, Rohit counts the money. And to his surprise, he has double the amount he wanted to withdraw. Find X and Y. (1 Rupee = 100 Paise)
As given, the person wanted to withdraw 100X + Y paise.
But he got 100Y + X paise.
After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is
2 * (100X + Y) = 100Y + X - 23
200X + 2Y = 100Y +X - 20
199X - 98Y = -23⇒  98Y - 199X = 23
Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1
Case I: Y=2X
Solving two equations simultaneously
98Y - 199X = 20 Þ Y - 2X = 0
We get X = - 23/3 & Y = - 46/3
Case II: Y=2X+1
Solving two equations simultaneously
98Y - 199X = 20  Þ Y - 2X = 1
We get X = 25 & Y = 51
Now, it’s obvious that he wanted to withdraw Rs. 25.51.
Q.4. Find all sets of consecutive integers that add up to 1000.
There are total 7 such series:
Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to 1000.
(-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000
Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202.
(-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000
Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70.
(-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000
Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52.
(-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000
Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52.
28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000
Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70.
55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000
Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202.
198 + 199 + 200 +201 + 202 = 1000
Q.5. Lala Kirorimal's locker has a 5 digit number as the key. The 4th digit is 4 greater than the second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11. Find the number.
As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9). It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now the required number is a 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9). Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.
Q.6. Consider a number 268, where last digit is the sum of first two digits i.e. 2 + 6 = 8. How many such 3-digit numbers are there?
The last digit can not be 0.
If the last digit is 1, the only possible number is 101. (Note that 011 is not a 3-digit number)
If the last digit is 2, the possible numbers are 202 and 112.
If the last digit is 3, the possible numbers are 303, 213 and 123.
If the last digit is 4, the possible numbers are 404, 314, 224 and 134.
If the last digit is 5, the possible numbers are 505, 415, 325, 235 and 145.
Note the pattern here - If the last digit is 1, there is only one number. If the last digit is 2, there are two numbers. If the last digit is 3, there are three numbers. If the last digit is 4, there are four numbers. If the last digit is 5, there are five numbers. And so on.....
Thus, total numbers are
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Altogether then, there are 45 different 3-digit numbers, where last digit is the sum of first two digits.
Q.7. There is a 50m long soldier’s unit advancing forward. The last individual in the unit desires to give a letter to the foremost person leading the unit. So while the unit is marching he runs to the front, reaches the foremost person and hands over the letter to him and without stopping, he runs and comes back to his original position. In the mean time the whole unit has moved ahead by 50m. How much distance did the last person cover in that time? Assume that he ran the whole distance with uniform speed.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered – while person moving forward and backward - are equal.
Let's assume that when the last person reached the first person, the platoon moved X meters forward. Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters. Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total length of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.
Q.8. Ram says to Sham, "Can you figure out how many Eggs I have in my bucket?" He gives 3 clues to Sham: If the number of Eggs I have
I) is a multiple of 5, it is a number between 1 and 19
II) is not a multiple of 8, it is a number between 20 and 29
III) is not a multiple of 10, it is a number between 30 and 39
How many Eggs does Ram have in his bucket?
Let's apply all 3 conditions separately and put all possible numbers together.
First condition says that if multiple of 5, then the number is between 1 and 19. Hence, the possible numbers are (5, 10, 15, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39)
Second condition says that if not a multiple of 8, then the number is between 20 and 29. Hence, the possible numbers are (8, 16, 20, 21, 22, 23, 25, 26, 27, 28, 29, 32)
Third condition says that if not a multiple of 10, then the number is between 30 and 39. Hence, the possible numbers are (10, 20, 31, 32, 33, 34, 35, 36, 37, 38, 39)
Only number 32 is there in all 3 result sets. That means that only number 32 satisfies all three conditions. Hence, Ram has 32 eggs in his bucket.
Q.9. There are four clustors of Oranges, Apricots and Bananas as follows:
Clustor I : 1 Orange, 1 Apricot and 1 Banana
Clustor II : 1 Orange, 5 Apricot and 7 Bananas
Clustor III : 1 Orange, 7 Apricots and 10 Bananas
Clustor IV : 9 Oranges, 23 Apricots and 30 Bananas
Clustor II costs Rs 300 and Clustor III costs Rs 390.
Can you tell how much does Clustor I and Clustor IV cost?
Assume that the values of one orange, one apricot and one banana are O, A and B respectively.
From Clustor II : O + 5A + 7B = 300
From Clustor III : O + 7A + 10B = 390
Subtracting above two equations: 2A + 3B = 90
For Clustor I :
= O + A + B
= (O + 5A + 7B) - (4A + 6B)
= (O + 5A + 7B) - 2(2A + 3B)
= 300 - 2(90) = 300 – 180 = 120
Similarly, for Clustor IV :
= 9O + 23A + 30B
= 9(O + 5A + 7B) - (22A + 33B)
= 9(O + 5A + 7B) - 11(2A + 3B)
= 9(300) - 11(90) = 2700 – 990 = 1710
Thus, Clustor I costs Rs 120 and Clustor IV costs Rs 1710.
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Q.10. 1000 cats were kept under observation to check their growth rates. After six months, there were 1000R cats. At the beginning of the 3rd year, there were roughly 2828R cats, which was 4 times what the scientists placed in there at the beginning of the 1st year. If R is a positive variable, how many cats would be there at the beginning of the 11th year?
At the beginning, there were 1000 cats. Also, there were 4000 cats at the beginning of third year which is equal to 2828R. Thus, R = 4000/2828 i.e. 1.414 (the square root of 2). Note that 2828R can be represented as 2000*R*R (R=1.414), which can be further simplified as 1000*R*R*R*R
Also, it is given that at the end of 6 months, there were 1000R cats. It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(R^(2N)) i.e. 1000*(2^N). Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 cats.
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