Logic Problem: The Trainee Technician

A 120 wire cable has been laid firmly underground between two telephone exchanges located 10km apart.Unfortunately after the cable was laid it was discovered to be the wrong type, the problem is the individual wires are not labeled. There is no visual way of knowing which wire is which and thus connections at either end is not immediately possible.

You are a trainee technician and your boss has asked you to identify and label the wires at both ends without ripping it all up. You have no transport and only a battery and light bulb to test continuity. You do have tape and pen for labeling the wires.

What is the shortest distance in kilometers you will need to walk to correctly identify and label each wire?

Know here the efficient way! 

Source 

To Be A Skilled Technician

What was the task to test the skill? 

The shortest distance is 20 km! Surprised? Read further.

Let's name the two exchanges as a 1 & 2 respectively. Now at end 1, let's make a groups of wires having 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 number of wires. Now somebody might ask why not 15 groups having 8 wires in each. After reading the entire process here, we'll get the answer of it.So we 15 groups have total 1 + 2 + 3....+ 15 = 120 wires. Let's name these groups as A, B, C, D ...... O. That means group A has 1, B has 2, C has 3 wires & so on.

Now join together all the wires of the particular group. For example, 2 wires of group B should be joined together, 7 wires of G tied together & so on. The sole wire of A is left as it is.

We will take the battery & bulb to other end traveling 10 km. We will say a wire is paired with the other if the bulb gets illuminated if battery & bulb connected in between.

Now let's take any wire at the other end & find the number of wires that are pairing with that particular wire under test. We will group such wires & label with those exactly how we labeled at end 1.

For example, if we find 2 wires pairing with particular wire then that wire & 2 paired wires together to be grouped in 3 wires & labeled as C.The sole wire not getting paired with any will be labeled as A. And group with wire pairing with 7 other wires together should be labeled as H.

In this way, we will have the exact group structure that we have at end 1. By now, we have identified & labeled correctly wires in groups of 1,2,3,....15 wires at both ends.

Now, we are going to label each wire of group by it's group & count number. For example, the only wire in group A labeled as A1, 2 wires in B are labeled as B1,B2, wires in G are labeled as G1,G2,G3,G4,G5,G6,G7 and so on.

Now, at end 2 itself, what we are going to do is connecting first wire of each group to A1. Second wire of each group to be connected to B2, third of each to be connected to C3 and so on. (refer the diagram above, where labels of wires that are to be connected together are written in same color).

Time Taken For The Journey

A RED ant is sitting on one side of a table (point X) and a BLACK ant is sitting on the opposite side of the table (point Y).

Now both of them decides to exchange their places and starts crawling. On the way, both of the ants meet and after that, it takes 20 seconds for the RED ant to move to point Y and it takes 5 seconds for the BLACK ant to reach point X.

Find out the total time taken by the RED and the BLACK ant to make the journey.

Here are steps to calculate! 

Source 

Calculation of Time For The Journey

What was the journey?

Let the speed of RED ant is R & that of BLACK ant is B

Let time taken by them to meet be T.

Now we will apply the basic formula of distance:

Distance = Speed * Time.

The RED ant travels R T distance before meeting and 20 R after the meeting.

The BLACK ant travels B T distance before meeting and 5 B after the meeting.

Now as per the question,The distance traveled by RED ant before they both meet will be equal to the distance covered by BLACK ant after they meet. We can say the same for the vice versa case as well.

Thus,

RT = 5B and BT = 20R i.e. B = 20R/T, putting in RT = 5B

R T = [20R/T] * 5

RT = 100R/T

T^2 = 100

T = 10.

Thus the RED ant will require 10 + 20 = 30 seconds to travel the distance.

And the BLACK ant will take 10 + 5 = 15 seconds to travel the distance. 

Distance Between Houses?

Four friends built a colony for themselves. They built their own houses at different
distances from each other.

Chris lived 60 km away from Alex.
Darren lived 40 km away from Bill.
Chris lived 10 km nearer to Darren than he lived to Bill.

Can you find out how far was Darren's house from Alex?

Click here to know the distance! 

Locations of 4 Friends

What was the given data & question? 

Let's represent houses of friends by their initial letter of the name i.e. C is the house of Chris, D is the house of Darren, B is Bill's house & A is name of Alex's house.

Now C & A are located at 60 km away from each other while D & B are 40 km apart i.e. CA = 60 & DB = 40.

As per given data, D is 10 km nearer than B from C. Hence, CB = CD + 10

There are 5 possibilities of the locations of these 4 friends.

Case 1. D & B are in between C & A.

In this case, 

CD + DB + BA = 60

but DB = 40,

CD + BA = 20,

CB - 10 + BA = 20

CB + BA = 30 i.e. CA = 30

But CA = 60, hence this combination is not possible. Other way, CB >= 40, since CD = CB-10, CD >= 30. That means CD + DB >= 70 for which B needs to be beyond A as CA = 60.

Case 2 : A is between D & B.

In this case, CD + DA  = 60, DA + AB = 40 . If we subtract second one from first, we get,

CD - AB = 20

CB - 10 - AB = 20

CB - AB = 30  

CA = 30 but CA = 60. Hence, this combination too is invalid.

Journey of The Dog

Jessica, Warner decided to meet & left their home. And a puppy starts walking down a road. They started at the same time. Their homes are located at 33 KM away from each other.

• Warner walks at 5 miles/hour.
• Jessica walks at 6 miles/hour.

  

The puppy runs from Warner to Jessica and back again with a constant speed of 10 miles/hour.

 
The puppy does not slow down on the turn. How far does the puppy travel in till Jessica and Warner meet?

Know here the distance traveled by puppy! 

Distance in The Dog's Journey!

What was the puzzle? 

First thing on which we need to focus on in how much time Jessica and Warner would meet. Since they are moving towards each other the distance of 33 KM is being covered at 5 + 6 = 11 KM/h.  So they are going to meet each other in 33/11 = 3 hours. Now everything else here can deceive you to find distance covered by puppy. All you need to do is stick to the basics.

Speed = Distance / Time

Distance = Speed * Time 

Distance covered by Puppy = Speed of Puppy * Time for which it traveled.

Distance covered by Puppy = 10 * 3 = 30 KM 

So Puppy travels 30 KM to & fro until Jessica and Warner meets. 

 

Unlock The Distance

Distances from you to certain cities are written below.

BERLIN = 200 miles
PARIS = 300 miles
ROME = 400 miles
AMSTERDAM = 300 miles
CARDIFF = ?? miles

How far should it be to Cardiff ?

 How far? Find Here! 

Source 

The Distance Unlocked

What was the question?

Just count Vowels V & Consonants C in any 2 spelling to get how much they value.


From BERLIN,

2V + 4C = 200

V + 2C = 100             ........(1)

From ROME,

2V + 2C = 400

V + C = 200               .......(2)

Solving (1) & (2), we get,

V = 300 & C = -100

For CARDIFF, we have,


2V + 6C = 100. 

So CARDIFF = 100 miles

Complex Time, Speed & Distance Maths

There is a circular race-track of diameter 1 km. Two cars A and B are standing on the track diametrically opposite to each other. They are both facing in the clockwise direction. At t=0, both cars start moving at a constant acceleration of 0.1 m/s/s (initial velocity zero). Since both of them are moving at same speed and acceleration and clockwise direction, they will always remain diametrically opposite to each other throughout their motion.

At the center of the race-track there is a bug. At t=0, the bug starts to fly towards car A. When it reaches car A, it turn around and starts moving towards car B. When it reaches B, it again turns back and starts moving towards car A. It keeps repeating the entire cycle. The speed of the bug is 1 m/s throughout.

After 1 hour, all 3 bodies stop moving. What is the total distance traveled by the bug?

Find out complex calculations here!

Simple Solution of Complex Problem

Here is that complex looking problem! 

Everything built, written or designed in the given problem is to distract you from basic physics formula.

Speed = Distance / Time

Hence,

Distance = Speed x Time 

All the details given except speed of bug & time for which it traveled are there to confuse you. Speed of bug is 1 m/s & it traveled for 1 hour = 3600 seconds.

Distance = 1 x 3600 = 3600 m

So the total distance traveled by the bug is 3600 m.