(Puzzles) Mostly Asked Important Puzzles (10 Puzzles)

Mostly Asked Important Puzzles (10 Puzzles)


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.

Answer :
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.
 


A contractor had employed 100 labourers for a flyover construction task. He did not allow any woman to work without her husband. Also, atleast half the men working came with their wives. He paid five rupees per day to each man, four ruppes to each woman and one rupee to each child. He gave out 200 rupees every evening. How many men, women and children were working with the constructor?

Answer :


16 men, 12 women and 72 children were working with the constructor.
Let's assume that there were X men, Y women and Z children working with the constructor. Hence,
X + Y + Z = 100
5X + 4Y + Z = 200

Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z

As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get
if Y = X,
300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428
if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727
But Z must be an integer, hence Z=72. Also, X=16 and Y=12
There were 16 men, 12 women and 72 children working with the constructor.


Four friends - Arjan, Bhuvan, Guran and Lakha were comparing the number of sheep that they owned. It was found that Guran had ten more sheep than Lakha. If Arjan gave one-third to Bhuvan, and Bhuvan gave a quarter of what he then held to Guran, who then passed on a fifth of his holding to Lakha, they would all have an equal number of sheep.How many sheep did each of them possess? Give the minimal possible answer.

Answer :

Arjan, Bhuvan, Guran and Lakha had 90, 50, 55 and 45 sheep respectively.
Assume that Arjan, Bhuvan, Guran and Lakha had A, B, G and L sheep respectively. As it is given that at the end each would have an equal number of sheep, comparing the final numbers from the above table.
Arjan's sheep = Bhuvan's sheep
2A/3 = A/4 + 3B/4
8A = 3A + 9B
5A = 9B

Arjan's sheep = Guran's sheep
2A/3 = A/15 + B/5 + 4G/5
2A/3 = A/15 + A/9 + 4G/5 (as B=5A/9)
30A = 3A + 5A + 36G
22A = 36G
11A = 18G

Arjan's sheep = Lakha's sheep
2A/3 = A/60 + B/20 + G/5 + L
2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18)
2A/3 = A/6 + L
A/2 = L
A = 2L

Also, it is given that Guran had ten more sheep than Lakha.
G = L + 10
11A/18 = A/2 + 10
A/9 = 10
A = 90 sheep

Thus, Arjan had 90 sheep, Bhuvan had 5A/9 i.e. 50 sheep, Guran had 11A/18 i.e. 55 sheep and Lakha had A/2 i.e. 45 sheep.


You are locked inside a room with 6 doors - A, B, C, D, E, F. Out of which 3 are Entrances only and 3 are Exits only.

One person came in through door F and two minutes later second person came in through door A. He said, "You will be set free, if you pass through all 6 doors, each door once only and in correct order. Also, door A must be followed by door B or E, door B by C or E, door C by D or F, door D by A or F, door E by B or D and door F by C or D."

After saying that they both left through door B and unlocked all doors. In which order must you pass through the doors?

Answer:

The correct order is CFDABE

It is given that one person came in through door F and second person came in through door A. It means that door A and door F are Entrances. Also, they both left through door B. Hence, door B is Exit.

As Exit and Entrance should alter each other and we know two Entrances, let's assume that the third Entrance is W. Thus, there are 6 possibilities with "_" indicating Exit.
(1) _W_A_F (2) _W_F_A (3) _F_W_A (4) _F_A_W (5) _A_W_F (6) _A_F_W

As door A must be followed by door B or E and none of them lead to the door F, (1) and (6) are not possible.
Also, door D must be the Exit as only door D leads to the door A and door A is the Entrance.
(2) _W_FDA (3) _F_WDA (4) _FDA_W (5) DA_W_F

Only door D and door C lead to the door F. But door D is used. Hence, door C must be the Exit and precede door F. Also, the third Exit is B and the W must be door E.
(2) BECFDA (3) CFBEDA (4) CFDABE (5) DACEBF

But only door B leads to the door C and both are Exits. Hence, (2) and (5) are not possible. Also, door F does not lead to door B - discard (3). Hence, the possible order is (4) i.e. CFDABE.


There is a safe with 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.

Answer:

65292

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 required number is 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.


A person travels on a cycle from home to church on a straight road with wind against him. He took 4 hours to reach there.

On the way back to the home, he took 3 hours to reach as wind was in the same direction. If there is no wind, how much time does he take to travel from home to church?

Answer:

Let distance between home and church is D.

A person took 4 hours to reach church. So speed while travelling towards church is D/4.

Similarly, he took 3 hours to reach home. So speed while coming back is D/3.

There is a speed difference of 7*D/12, which is the wind helping person in 1 direction, & slowing him in the other direction. Average the 2 speeds, & you have the speed that person can travel in no wind, which is 7*D/24.

Hence, person will take D / (7*D/24) hours to travel distance D which is 24/7 hours.

Answer is 3 hours 25 minutes 42 seconds


If a bear eats 65 pounds in fish every day EXCEPT every 6th day which it only eats 45 pounds of fish. If the bear continues this, how many pounds of fish will it eat in 200 days?

Answer:

The bear will eat 12,340 pounds of fish in 200 days.

It is given that on every 6th day beareats 45 pounds of fish i.e. on day number 6, 12, 18, 24, .... 192, 198 the bear eats 45 pounds of fish.
Total number of 6th days = 200/6 = 33 (the bear eats 45 pounds)
Hence, the normal days are = 200 - 33 = 167 (the bear eats 65 pounds)
Thus, in 200 days, the bear will eat
= (167) * (65) + (33) * (45)
= 10855 + 1485
= 12,340 pounds


Ankit and Tejas divided a bag of Apples between them.

Tejas said, "It's not fair! You have 3 times as many Apples I have." Ankit said, "OK, I will give you one Apple for each year of your age." Tejas replied, "Still not fair. Now, you have twice as many Apples as I have." "Dear, that's fair enough as I am twice older than you.", said Ankit.

Ankit went to Kitchen to drink water. While Ankit was in Kitchen, Tejas took apples from Ankit's pile equal to Ankit's age.

Who have more apples now?

Answer:

At the end, Ankit and Tejas, both have the same number of apples.

Let's assume that initially Tejas got N apples and his age is T years. Hence, initially Ankit got 3N apples and his age is 2T years.
Operation Ankit's Apples Tejas's Apples
Initially 3N N
Ankit gave T apples to Tejas
(equals age of Tejas) 3N - T N + T
Tejas took 2T apples from Ankit's pile
(equals age of Ankit) 3N - 3T N + 3T
It is given that after Ankit gave T apples to Tejas, Ankit had twice as many apples as Tejas had.
3N - T = 2*(N + T)
3N - T = 2N + 2T
N = 3T
From the table, at the end Ankit have (3N - 3T) apples and Tejas have (N + 3T) apples. Substituting N = 3T, we get
Ankit's apples = 3N - 3T = 9T - 3T = 6T
Tejas's apples = N + 3T = 3T + 3T = 6T
Thus, at the end Ankit and Tejas, both have the same number of apples.


The sum of their (father, mother and son) ages is 70. The father is 6 times as old as the son. When the sum of their ages is twice 70, the father will be twice as old as the son. How old is the mother?

Answer:

The mother is 29 years and 2 months old.

Let's assume that son is X years old. Hence, father is 6X years old and mother is (70-7X) years old.
It is given that the sum of their ages is 70, which will total 140 after 70/3 years.
After 70/3 years, son will be (X + 70/3) years old and father will be (6X + 70/3) years old. Also, it is given that after 70/3 years, the father will be twice as old as the son. Thus,

(6X + 70/3) = 2 * (X + 70/3)
6X + 70/3 = 2X + 140/3
4X = 70/3
X = 35/6

Hence, their ages are
Son = X = 35/6 = 5 years and 10 months
Father = 6X = 6(35/6) = 35 years
Mother = (70 - 7X) = 70 - 7(35/6) = 29 years and 2 months


Assume for a moment that the earth is a perfectly uniform sphere of radius 6400 km. Suppose a thread equal to the length of the circumference of the earth was placed along the equator, and drawn to a tight fit.

Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions. By how many cm. will the thread be separated from the earth's surface?

Answer:

The cicumference of the earth is

= 2 * PI * r
= 2 * PI * 6400 km
= 2 * PI * 6400 * 1000 m
= 2 * PI * 6400 * 1000 * 100 cm
= 1280000000 * PI cm

where r = radius of the earth, PI = 3.141592654
Hence, the length of the thread is = 1280000000 * PI cm
Now length of the thread is increasd by 12 cm. So the new length is = (1280000000 * PI) + 12 cm

This thread will make one concentric circle with the earth which is slightly away from the earth. The circumfernce of that circle is nothing but (1280000000 * PI) + 12 cm
Assume that radius of the outer circle is R cm
Therefore,

2 * PI * R = (1280000000 * PI) + 12 cm
Solving above equation, R = 640000001.908 cm
Radius of the earth is r = 640000000 cm
Hence, the thread will be separatedfrom the earth by
= R - r cm
= 640000001.908 - 640000000
= 1.908 cm