Speed ,Time & Success

·         Prologue

·         Speed(S)

·         Time (T)

·         Distance (D)

·         Time & Work (In a nutshell)

·         Easy Methods

·         Solved Problems

·         Permutations and combinations

·         Prime Number & Number Sequence in Brief

Just an aspirant like you, so do forgive me if you find any mistakes.  Chosen Speed, Time and distance topic because it’s a favorite topic for all public exams

Speed, Time & Distance (STD) are always in a committed triangle relationship

Speed - Speed is how fast something is going. Another way to think of this is as how far you can go in a certain amount of time or how fast I am writing this article (btw it took me 4 hours to complete this article). Measured as distance traveled per unit of time.

Example: The speed of these cars is over 150 kilometers per hour (150 km/h)

Just a passing Reference difference between speed and velocity

Velocity - Velocity is speed with a direction.
So if something is moving at 5 km/h that is a speed.
But if you say it is moving at 5 km/h westwards that is a velocity.
If something moves backwards and forwards very fast it has a high speed, but a low (or zero) velocity

Time – Precious thing (I guess no need to explain time)

Distance is the space between two objects or points (So in our example distance is starting point of car to its finish line)

Unit of Measurement

Unit of measurement of distance = km (kilo meter) , m , cm

Unit of measurement of time = hour (hr), minute (min), second (s)

Unit of measurement of speed = km/hr, m/s

 Conversions

1 km = 1000 m

1 hour = 60 min

1 min = 60 sec

1 hour = 60 * 60 = 3600 sec

1 km/hr = 1000/3600

             = 5/18 m/s

e.g.

 

 For simple calc

1 km/hr = 0.27 m/s

1 m/s = 3.6 km/h

Now let’s turn our attention from more basic concepts to exam oriented questions and ways to solve it

Some important formulas and you can solve any questions related to (STD)

1)      If a TRAIN covers a certain distance at x km/ph and an equal distance at y km/hr ,the average speed of the whole journey = 

For e.g. If a train covers Pune to Mumbai 250 km at x=50 km/hr and Mumbai to pune 250 km at y=60 km/hr

Then average speed    = 54.54 km/hr

2)      Speed and time are inversely proportional (when distance is constant) ⇒ Speed µ  (when distance is constant)

This means as the speed increases the time decreases (or the time taken is less).

3)      If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is  b : a or :

These are the only formulas required for STD type of problems.

Now relating this to train formulas as for train also speed, time and distance logic will be same. The only difference will be when a pole or a man or any “xyz object standing”.

Laymen e.g. The Monorail started in Mumbai (from Chembur to Wadala). Imagine its length is 100 m

1)      Time taken by a train of length d1 meters to pass a pole or standing man or a signal post is equal to the time taken by the train to cover d1 meters.  

Answer = Time taken for the train to cover 100 meters

2)      Time taken by a train of length d1 meters to pass a stationery object (chembur station= 50 m) of length d2 meters is the time taken by the train to cover (d1 + d2) meters.

Answer = Time taken for the train to cover 150 meters

3)      Suppose two trains or two objects bodies are moving in the same direction at s1 m/s and s2 m/s, where s1 > s2, then their relative speed is = (s1 –s2) m/s.

4)      Suppose two trains or two objects bodies are moving in opposite directions at s1  m/s and s2 m/s, then their relative speed is = (s1+s2) m/s.

Easy = opposites attract. And any kind of attraction is a positive thing

      

5)       If two trains of length d1 meters and m2 meters are moving in opposite directions at s1 m/s and s2 m/s, then:

The time taken by the trains to cross each other =		.

Few Problems with super short cut methods

1)      A monorail covers a distance (d1) of 5 km with a speed (s1) of 4 km/hr and next 6 km with a speed of 3 km/hr in travelling from A to B. What will be the train’s average speed during the whole journey?

Solution = If d1=first distance=5 km

And d2=second distance=6 km (given in above example)

And s1=first speed=4 km/hr

And s2=second speed=3 km/hr

Then average speed= =

Putting the value of d1, d2, s1 and s2 from the given example in the above formula

We get

Average speed=

Average speed= (11x4x3)/ (15+24)

                                = (11x4x3)/39

                                = (11x12)/39

                                =3.38 km/hr 

Continuing with the concept of Monorail only

2)      A man  reaches Wadala station(to board mono rail) from home  late by 30 minutes from his scheduled time  if he walks at a speed(s1) of 5 km/h, but if he walks at a speed(speed s2) of 6 km/h, he will reach his home 5 minutes early. What will be the distance from station to home?

Rule- if a person reaches his destination t1 time early by walking at a speed of s1 and t2 time later by walking at a speed of s2, then the distance between both places is

                                                 = =

Now equate the question given above to the rule also given just above.

Now t1=30 minutes,   t2=5 minutes

And   s1=5 km/h, s2= 6 km/h

                                  =       = 17.5 km

3)      If a monorail does a journey in 'H' hrs, the first half at 's1' km/hr and the second half at 's2' km/hr. The total distance covered by the car:

=     

A monorail does a journey in 10 hrs, the first half at 21 km/hr and the second half at 24 km/hr. Find the distance?
Ans: Distance = (2 x 10 x 21 x 24) / (21+24)
= 10080 / 45
= 224 km.

The whole questions on Time speed and distance and Train revolve around these basic formulas. But practice is needed to solve it more efficiently

Permutations and combinations

Q1) what is combination?

We always use the word combination in our day to day life.

 e.g "The fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.

e.g. The G20 includes a combination of the largest developed and industrialized countries, which make up nearly 85% of the world’s economy

If the order doesn't matter, it is a Combination

Q2) What is permutation?

The word permutation is though less used but often applied in our everyday life.

e.g.

"The combination to my bank password is 472"(not really just a hypothetical one). Now we do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2. So, in Mathematics we use more precise language: If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation

So let’s focus on permutation first (coz order do matters in life)

Permutation

A Permutation is an ordered Combination.

More easy to remember … permutation = position (p..p)

There are basically 2 types of permutation

1) Permutation with repetition

Above example – I can keep my bank password as 474 or 444 also (note in this the order matters, but the numbers are repeated)

Technically speaking

When we have n things to choose from ... we have n choices each time!
When choosing r of them, the permutations are:

n × n × ... (r times)

(In other words, there are n possibilities for the first choice, AND THEN there are n possibilities for the second choice, and so on, multiplying each time.)
Which is easier to write down using an exponent of r?

n × n × ... (r times) = n^r

Example: in the above bank password, there are 10 numbers to choose from (0, 1,...9) and we choose 3 of them:

10 × 10 × ... (3 times) = 10³ = 1,000 permutations

So, the formula is simply:

nr

where n is the number of things to choose from, and we choose r of them (Repetition allowed, order matters)

2) Permutation without Repetition

In this case, we have to reduce the number of available choices each time.

In the above bank password example.

We have to create a 3 number password out of the 10 numbers (0,1,2,3,4,5,6,7,8,9). But remember repetition is not allowed.

So we choose number 4 (out of ten options we have chosen one).

We choose number 7 (as 4 is already taken and repetition is not allowed so now we can choose a number out of nine)

We choose number 2 (as 4 and 7 is already taken and repetition is not allowed so now we can choose a number out of eight)

what if we wanted to select just 3, then we have to stop the multiplying after 14. How do we do that? There is a neat trick ... we divide by 7! ...

10 × 9 × 8 × 7 × 6 ...		= 10 × 9 × 8 = 720
7× 6 ...

The formula is written:

where n is the number of things to choose from, and we choose r of them (No repetition, order matters)

Some examples (It’s very easy. Just figure if repetition is there or not)

1)

For example, what order could 16 pool balls be in? After choosing, say, number "14" we can't choose it again. So, our first choice would have 16 possibilities, and our next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be: 16 × 15 × 14 × 13 × ... = 20,922,789,888,000 But maybe we don't want to choose them all, just 3 of them, so that would be only: 16 × 15 × 14 = 3,360 In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls. 2) How many ways can first and second place be awarded to 10 people? 10! = 10! = 3,628,800 = 90 (10-2)! 8! 40,320 (which is just the same as: 10 × 9 = 90) Note Instead of writing the whole formula, people use different notations such as these: Example: P(10,2) = 90 Combinations There are also two types of combinations (remember the order does not matter now = fruit salad ) Repetition is Allowed: such as coins in your pocket (5,5,5,10,10) No Repetition: such as lottery numbers (2,14,15,27,30,33) Combinations without Repetition This is how Housie game works. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win! It does not matter if 9 is coming first or last. As we complete striking all the numbers we are winners (Just an hypothetical example) The easiest way to explain it is to: assume that the order does matter (ie permutations), then alter it so the order does not matter. Going back to our pool ball example, let's say we just want to know which 3 pool balls were chosen, not the order. We already know that 3 out of 16 gave us 3,360 permutations. But many of those will be the same to us now, because we don't care what order! For example, let us say balls 1, 2 and 3 were chosen. These are the possibilites: 1) Order does matter 2) Order doesn't matter 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 1 2 3 So, the permutations will have 6 times as many possibilities. In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is: 3! = 3 × 2 × 1 = 6 (Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!) So we adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in their order any more): That formula is so important it is often just written in big parentheses like this: where n is the number of things to choose from, and we choose r of them (No repetition, order doesn't matter) It is often called "n choose r" (such as "16 choose 3") Note As well as the "big parentheses", people also use these notations: Example So, our pool ball example (now without order) is: 16! = 16! = 20,922,789,888,000 = 560 3!(16-3)! 3!×13! 6×6,227,020,800 Or we could do it this way: 16×15×14 = 3360 = 560 3×2×1 6 So remember, do the permutation, then reduce by a further "r!" 2) We will do it for the bank password example We have to choose 3 number out of 10 (order does matter) = 10 X 9 X 8 = 720 Out of the 3 numbers chosen order does not matter = 3! = 6 =      =  120 (combinations) 2) Combinations with repetition (It’s a bit tough. You can follow Mrunal Sir article on Permutation and combination) Let us say there are five flavors of ice-cream: banana, chocolate, lemon, strawberry and vanilla. We can have three scoops. How many variations will there be? Let's use letters for the flavors: {b, c, l, s, v}. Example selections would be {c, c, c} (3 scoops of chocolate) {b, l, v} (one each of banana, lemon and vanilla) {b, v, v} (one of banana, two of vanilla) (And just to be clear: There are n=5 things to choose from and we choose r=3 of them. Order does not matter, and we can repeat!) Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out. Think about the ice cream being in boxes, we could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and we will have 3 scoops of chocolate! So it is like we are ordering a robot to get our ice cream, but it doesn't change anything, we still get what we want. We could write this down as (arrow means move, circle means scoop). In fact the three examples above would be written like this: {c, c, c} (3 scoops of chocolate): {b, l, v} (one each of banana, lemon and vanilla): {b, v, v} (one of banana, two of vanilla): OK, so instead of worrying about different flavors, we have a simpler question: "how many different ways can we arrange arrows and circles?" Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container). So (being general here) there are r + (n-1) positions, and we want to choose r of them to have circles. This is like saying "we have r + (n-1) pool balls and want to choose r of them". In other words it is now like the pool balls question, but with slightly changed numbers. And we would write it like this: where n is the number of things to choose from, and we choose r of them (Repetition allowed, order doesn't matter) Interestingly, we could have looked at the arrows instead of the circles, and we would have then been saying "we have r + (n-1) positions and want to choose (n-1) of them to have arrows", and the answer would be the same ... So, what about our example, what is the answer? (5+3-1)! = 7! = 5040 = 35 3!(5-1)! 3!×4! 6×24 Prime number and Number Sequences in short Prime Numbers A prime number can be divided, without a remainder, only by itself and by 1. For example, 17 can be divided only by 17 and by 1. Some facts: The only even prime number is 2. All other even numbers can be divided by 2. If the sum of a number's digits is a multiple of 3, that number can be divided by 3. No prime number greater than 5 ends in a 5. Any number greater than 5 that ends in a 5 can be divided by 5. Zero and 1 are not considered prime numbers. Except for 0 and 1, a number is either a prime number or a composite number. A composite number is defined as any number, greater than 1, that is not prime. To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can't be a prime number. If you don't get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below). Here is a table of all prime numbers up to 1,000: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997 Common Number Patterns Numbers can have interesting patterns. Here we list the most common patterns and how they are made. Arithmetic Sequences An Arithmetic Sequence is made by adding some value each time. Example: 1, 4, 7, 10, 13, 16, 19, 22, 25, ... This sequence has a difference of 3 between each number. The pattern is continued by adding 3 to the last number each time, like this: Example: 3, 8, 13, 18, 23, 28, 33, 38, ... This sequence has a difference of 5 between each number. The pattern is continued by adding 5 to the last number each time, like this: The value added each time is called the "common difference" What is the common difference in this example? 19, 27, 35, 43, ... Answer: The common difference is 8 The common difference could also be negative: Example: 25, 23, 21, 19, 17, 15, ... This common difference is −2 The pattern is continued by subtracting 2 each time, like this: Geometric Sequences A Geometric Sequence is made by multiplying by some value each time. Example: 2, 4, 8, 16, 32, 64, 128, 256, ... This sequence has a factor of 2 between each number. The pattern is continued by multiplying by 2 each time, like this: Example: 3, 9, 27, 81, 243, 729, 2187, ... This sequence has a factor of 3 between each number. The pattern is continued by multiplying by 3 each time, like this: Special Sequences Triangular Numbers 1, 3, 6, 10, 15, 21, 28, 36, 45, ... This Triangular Number Sequence is generated from a pattern of dots which form a triangle. By adding another row of dots and counting all the dots we can find the next number of the sequence: Square Numbers 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, ... They are the squares of whole numbers: 0 (=0×0) 1 (=1×1) 4 (=2×2) 9 (=3×3) 16 (=4×4) etc... Cube Numbers 1, 8, 27, 64, 125, 216, 343, 512, 729, ... They are the cubes of the counting numbers (they start at 1): 1 (=1×1×1) 8 (=2×2×2) 27 (=3×3×3) 64 (=4×4×4) etc... Fibonacci Numbers 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... The Fibonacci Sequence is found by adding the two numbers before it together. The 2 is found by adding the two numbers before it (1+1) The 21 is found by adding the two numbers before it (8+13) The next number in the sequence above would be 55 (21+34) Tricks of Coding Sequence 1)      E, J, O, T, Y =   5, 10, 15, 20, 25 (alphabet and its equivalent numbers. Just remember EJO is in Third year )  The normal English alphabet contains 26 letters in all, as shown above (Usually, questions on alphabet are accompanied by this normal alphabet). From A to M, the alphabet completes its first half, while the other half starts from N and ends at Z. 1)      A-M - 1-13  (First Alphabetical Half )  Easy to remember 1am in New Zealand N-Z –14-26 (Second Alphabetical Half)   Note - A detailed Article on Coding , decoding and pipes & tank  and other topics next month References – www.Mrunal.org www.google.com R S Agarwal Book www.IndiaBix.com