PUZZLES
Puzzle No : 1
Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided. Answer
18
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X - 4) bullets each.
But it is given that,after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X
Therefore, the equation is
3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18
2) Man Wrinkle spent one-fourth of his life as a boy, one-eighth as a youth, and one-half as an active man. If Man Wrinkle spent 11 years as an old man, then how many years did he spend as an active man?
Solution: Fraction of life as a boy = 1/4
Fraction of life as a youth = 1/8
Fraction of life as an active man = 1/2
Fraction of life as boy, youth and active man = 1/4 + 1/8 + 1/2 = (2 + 1 + 4)/8 = 7/8
Fraction of life as an old man = 1 − 7/8 = 1/8
Thus, one-eighth of Man Wrinkle's life (as an old man) is 11 years.
So, Man Wrinkle's Age = 88 years.
It may be noted that:
Life as boy = 88/4 = 22 years.
Life as youth = 88/8 = 11 years.
Life as active man = 88/2 = 44 years.
Life as old man = 88/8 = 11 years.
The problem may also be solved by setting up the following equation: a/4 + a/8 + a/2 + 11 = a
where a denotes Man Wrinkle's age in years.
The equation may be solved as shown below.
7a/8 + 11 = a
11 = a − 7a/8 = a/8
a/8 = 11 or a = 88.
Therefore, life spent as active man = 88/2 = 44 years.
3) Haretown and Tortoiseville are 45 miles apart. A hare travels at 8 miles per hour from Haretown to Tortoiseville, while a tortoise travels at 1 miles per hour from Tortoiseville to Haretown.
If both set out at the same time, how many miles will the hare have to travel before meeting the tortoise en route?
Solution:
The hare and the tortoise are together covering the distance at 9 miles per hour (i.e., on adding their speeds).
So, they will cover the distance of 45 miles in 5 hours.
Thus, in 5 hours, they will meet and the hare will have traveled 40 miles.
Alternative Solution through Equations:
Note that : Distance = Speed × Time
Let t be the time before the hare and the tortoise meet.
In t hours, the hare will travel 8 t miles.
In t hours, the tortoise will travel 1 t miles.
Now,
8 t + 1 t = 45
So, t = 45 ⁄ 9 = 5 hours.
Thus, distance traveled by hare before meeting = 8 × 5 = 40 miles.
4) Dad gives you money every day to put in your new piggy bank. He gives money to you in such a way that the money in the piggy bank doubles with each passing day.
If you already have 1 cent in the piggy bank and Dad gives you 1 cent the first day, 2 cents the second day, 4 cents the third day and so on, then your piggy bank gets full on the 14th day.
1. On which day will your piggy bank be half-full?
For example, type 6 if your answer is 6th day.
2. In addition to Dad's contributions, if Mom also gave you 1 cent the first day, 2 cents the second day, 4 cents the third day and so on, then on which day would your piggy bank be about half-full?
For example, type 6 if your answer is 6th day.
Solution:
1. Since the money in the piggy bank doubles with each passing day, the piggy bank will be half-full the day previous to the one on which it gets full. Thus, the piggy bank will be half-full on the 13th day.
2. If both Mom and Dad contribute equal amounts to your piggy bank, then each needs to only make your piggy bank quarter-full. When both Mom and Dad contribute, the piggy bank will be half-full two days prior to the day it would be full when only Dad contributes. Thus, the piggy bank will be full on the 12th day when both Mom and Dad contribute.
PUZZLE:Find sum of digits of D.
Let
A= 19991999
B = sum of digits of A
C = sum of digits of B
D = sum of digits of C
(HINT : A = B = C = D (mod 9))
Answer
The sum of the digits od D is 1.
Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)
Consider
A = 19991999
< 20002000
= 22000 * 10002000
= 1024200 * 106000
< 10800 * 106000
= 106800
i.e. A < 106800
i.e. B < 6800 * 9 = 61200
i.e. C < 5 * 9 = 45
i.e. D < 2 * 9 = 18
i.e. E <= 9
i.e. E is a single digit number.
Also,
1999 = 1 (mod 9)
so 19991999 = 1 (mod 9)
Therefore we conclude that E=1.
Puzzle:There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first 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 platoon has moved ahead by 50m.
The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.
The last person covered 120.71 meters.
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 backword - 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 lenght 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.
Puzzle:
If you take a marker & start from a corner on a cube, what is the maximum number of edges you can trace across if you never trace across the same edge twice, never remove the marker from the cube, & never trace anywhere on the cube, except for the corners & edges?
Answer
9
To verify this, you can make a drawing of a cube, & number each of its 12 edges. Then, always starting from 1 corner & 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube.
There is no need to start from other corners or edges of the cube, as you will only be repeating the same combinations. The process is a little more involved than this, but is useful for solving many types of spatial puzzles.
Puzzle:The distance between Station Atena and Station Barcena is 84 miles. A train starts from Atena towards Barcena. A bird starts at the same time from Barcena straight towards the moving train. On reaching the train, it instantaneously turns back and returns to Barcena. The bird makes these journeys from Barcena to the train and back to Barcena continuously till the train reaches Barcena. The bird finally returns to Barcena and rests. Calculate the total distance in miles the bird travels in the following two cases:
1. the bird flies at 80 miles per hour and the speed of the train is 60 miles per hour
2. the bird flies at 60 miles per hour and the speed of the train is 80 miles per hour
Case 1: Bird flies at a speed greater than that of the train
The train (at a speed of 60 miles per hour) travels 60 miles in 60 minutes.
Therefore, the train travels from Atena to Barcena (84 miles) in 84 minutes.
Importantly, the bird makes the journeys continuously back and forth for this same amount of time (namely, 84 minutes).
Thus, the total distance traveled by the bird
= 80 miles per hour × 84 minutes = 80 × 84 / 60 miles = 112 miles.
Case 2: Bird flies at a speed less than that of the train
In 36 minutes, the bird travels 36 miles, the train travels 48 miles, and the two meet.
Now, the train (which is traveling at a speed greater than that of the bird) will reach Barcena before the bird.
So, the bird simply returns to Barcena (a return journey of 36 miles).
Thus, the total distance traveled by the bird is 72 miles.
Puzzle:One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and another is a Doctor.1)If the Doctor is a male, then the Engineer is a male.2)If the Engineer is younger than the Doctor, then the Engineer and the Doctor are not blood relatives.3)If the Engineer is a female, then she and the Doctor are blood relatives. Can you tell who is the Doctor and the Engineer?
Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.
Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.
Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr. Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.
Now from 2, Mr. Bajaj's mother can not be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further.
PUZZLE:Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not necessarily in the same order.
Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge. No wife partners her husband and Cam does not play bridge.
Who is married to Cam?
Carrie is married to Cam.
"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."
It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina.
As Cam does not play bridge, Billy's husband must be Laurie.
Hence, Carrie is married to Cam.
PUZZLE: There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.
Find the tractors each originally had?
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Back tracing.
It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)
Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)
Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)
Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.
PUZZLE: There is a pole in a lake. One-half of the pole is in the ground, another one-third of it is covered by water, and 9 ft is out of the water. What is the total length of the pole in ft?
PUZZLE: A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?
The sign-maker will need 192 zeroes.
Divide 1000 building numbers into groups of 100 each as follow:
(1..100), (101..200), (201..300), ....... (901..1000)
For the first group, sign-maker will need 11 zeroes.
For group numbers 2 to 9, he will require 20 zeroes each.
And for group number 10, he will require 21 zeroes.
The total numbers of zeroes required are
= 11 + 8*20 + 21
= 11 + 160 + 21
= 192
PUZZLE: There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?
Answer
It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.
1. Take 8 coins and weigh 4 against 4.
o If both are not equal, goto step 2
o If both are equal, goto step 3
2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
If both are equal, L4 is the odd coin and is lighter.
If L2 is light, L2 is the odd coin and is lighter.
If L3 is light, L3 is the odd coin and is lighter.
o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
If both are equal, there is some error.
If H1 is heavy, H1 is the odd coin and is heavier.
If H2 is heavy, H2 is the odd coin and is heavier.
o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
If both are equal, L1 is the odd coin and is lighter.
If H3 is heavy, H3 is the odd coin and is heavier.
If H4 is heavy, H4 is the odd coin and is heavier.
The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.
o If both are equal, there is some error.
o If X is heavy, X is the odd coin and is heavier.
o If X is light, X is the odd coin and is lighter.
PUZZLE: In a sports contest there were m medals awarded on n successive days (n > 1).On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.2)On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.3)On the nth and last day, the remaining n medals were awarded.4)How many days did the contest last, and how many medals were awarded altogether?
Total 36 medals were awarded and the contest was for 6 days.
On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6
PUZZLE: A number of 9 digits has 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, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
• 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 1-9 in some order.Find the number.
The answer is 381654729
One way to solve it 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 as 0 does not occur.
PUZZLE: 1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day.
What part of the contents of the container is left at the end of the second day?
Assume that contents of the container is X
On the first day 1/3rd is evaporated.
(1 - 1/3) of X is remaining i.e. (2/3)X
On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X
Hence 1/6th of the contents of the container is remaining
PUZVipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.
For how long did Vipul study in candle light?
Answer
Vipul studied for 3 hours in candle light.
Assume that the initial lenght of both the candle was L and Vipul studied for X hours.
In X hours, total thick candle burnt = XL/6
In X hours, total thin candle burnt = XL/4
After X hours, total thick candle remaining = L - XL/6
After X hours, total thin candle remaining = L - XL/4
Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.
(L - XL/6) = 2(L - XL/4)
(6 - X)/6 = (4 - X)/2
(6 - X) = 3*(4 - X)
6 - X = 12 - 3X
2X = 6
X = 3
Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.
ZLE:
