Sonntag, 16. November 2014

Exercises On Fractions


I. Convert Mixed Fractions into Improper Fractions


a)    2
1
8
  = 
 
b)  17
3
5
  = 
c)  12
3
7
  = 
d)  10
2
3
  = 
     
  
     
    
    
e)  28
1
3
  = 
 
f)     4
2
5
  = 
g)  19
1
4
  = 
h)  10
4
5
  = 
     
  
     
     
     
i)     2
1
3
  = 
 
j)     7
1
2
  = 
k)    6
2
3
  = 
l)     1
1
3
  = 
     
  
     
     
     


II. Convert Improper Fractions into Mixed Fractions

a) 
17
8
     b) 
88
5
     c) 
87
7
     d) 
32
3
     e) 
85
3
     f) 
22
3
     g) 
77
4
     h) 
54
5
i) 
3
3
     j) 
15
2
     k) 
20
3
     l) 
2
1
     m) 
28
9
     n) 
17
4
     o) 
226
15
     p) 
13
5


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Samstag, 15. November 2014

Addition and Subtraction of Fractions


ADDITION and SUBSTRACTION of Like FRACTIONS:

In order to add two or more like fractions, we may follow the two steps below:
  • add or substract the numerators of all fractions;
    • and
  • retain the common denominator of all fractions.

Example: Add or Substract the following like fractions
 i)  
1
6
   +  
4
6
   =  
5
6
 ii)  
2
9
   +  
5
9
   +  
7
9
   =  
14
9
 iii)   2
3
5
   +  
4
5
   +   1
2
5
   =  
(2 × 5) + 3
5
   +  
4
5
   +  
(1 × 5) + 2
5
   =  
24
5
 iv)   1
1
4
   +   2
3
4
   +   7
1
4
   =   (1 + 2 + 7)
1 + 3 + 1
4
   =   10
5
4
 v)  
8
10
   −  
3
10
   =  
 5  1
 10  2
 (Dividing the numerator and denominator by their HCF = 5)  = 
1
2
 vi)  
5
12
   −  
7
12
   +  
11
12
   =  
(5 − 7 + 11)
12
  = 
 9  3
 12  4
 (HCF = 3)  = 
3
4
 vii)   4
2
3
   +  
1
3
   −   4
1
3
   =  
(4 × 3) + 2
3
   +  
1
3
   −  
(4 × 3) + 1
3
   =  
2
3


ADDITION and SUBSTRACTION of Unlike FRACTIONS:

To add or substract unlike fractions, we need to:
  • firstly convert them into corresponding equivalent like fractions
    • by Finding the LCM of the denominators;
    and
  • secondly follow the same steps as in Like fractions
    • add or sustract the numerators of all fractions;
      • and
    • retain the common denominator of all fractions.

Example: Add or Substract the following unlike fractions

 i)  
2
3
   +  
3
7
  • To find the LCM of 3 and 7, we proceed to the following division:
  •   3   3, 7
      7   1, 7
      21   1, 1
  • As you can see, 3 × 7 = 21 is the LCM of 3 and 7.
    Since 21 can be divided by both 3 and 7: 21 ÷ 3 = 7 and 21 ÷ 7 = 3.

Now we can convert the given fractions into equivalent fractions with denominator 21 as follow:
2
3
   +  
3
7
   =  
7 × 2
7 × 3
   +  
3 × 3
3 × 7
   =  
14
21
   +  
9
21
   =  
23
21
   =   1
2
21


 ii)  
1
6
   +  
3
8
  • To find the LCM of 6 and 8, we proceed to the following division:
  •   2   6, 8
      2   3, 4
      2   3, 2
      3   3, 1
         1, 1
  • As you can see, 2 × 2 × 2 × 3= 24 is the LCM of 6 and 8.
    Since 24 can be divided by both 6 and 8: 24 ÷ 6 = 4 and 24 ÷ 8 = 3.

Now we can convert the given fractions into equivalent fractions with denominator 24 as follow:
1
6
   +  
3
8
   =  
4 × 1
4 × 6
   +  
3 × 3
3 × 8
   =  
4
24
   +  
9
24
   =  
13
24



 iii)   4
2
3
   −   3
1
4
   +   2
1
6
   =  
(4 × 3) +2
3
   −  
(3 × 4) + 1
4
   +  
(2 × 6) + 1
6


=  
14
3
  −  
13
4
  +  
13
6


Let's first find the LCM 0f 3, 4 and 6:
  2   3, 4, 6
  2   3, 2, 3
  3   3, 1, 3
     1, 1, 1
  As you can see, 2 × 2 × 3 = 12 is the LCM of 3, 4 and 6.
  For, 12 can be divided by 3, 4 and 6: 12 ÷ 3 = 4, 12 ÷ 4 = 3 and 12 ÷ 6 = 2.

Now let's convert the given fractions into equivalent fractions with denominator 12 as follow:

14
3
   −  
13
4
   +  
13
6
  =  
4 × 14
4 × 3
   −  
3 × 13
3 × 4
   +  
2 × 13
2 × 6
  =  
56
12
   −  
39
12
   +  
26
12
=
43
12
  =    3
7
12



 iv)  
15
16
   −  
17
24
 ,   The LCM of 16 and 24 is: 2 × 2 × 2 × 2 × 3 = 48, (see table)  
  2   8, 12
  2   4, 6
  3   2, 3
  3   1, 3
     1, 1

The fractions are now converted into equivalent fractions with like denominators, 48:
 ↔ 
15
16
   −  
17
24
 = 
3 × 15
3 × 16
   −  
2 × 17
2 × 24
 = 
45
48
   −  
34
48
 = 
11
48







◄◄Fractions Core Lessons

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Donnerstag, 13. November 2014

Fractions
Core Lessons


➪  Fractions:
      

FRACTIONS:

A fraction is a number representing a part of a whole thing. The whole thing might be a single object or a group of objects.

Example:
Assume 
5
7
   is a fraction. This fraction is read as "five-seventh" which means that 5 parts
out of 7 equal parts in which the whole is divided.
In the fraction, 5 is called the numerator and 7 is the denominator.



LIKE FRACTIONS:

Fractions with the same denominators are called Like Fractions.

Example:

                
1
7
 ;       
2
7
 ;       
3
7
 ;       
5
7
 ;       
6
7
 .
As you can see "7" is the bottom number and is the same (like or alike) for the whole fractions above.


UNLIKE FRACTIONS:

These are fractions with different denominators.

Example:

                
1
5
 ;       
2
7
 ;       
3
13
 ;       
4
9
 ;       
6
17
 .
Check this out: The bottom numbers of the fractions above aren't the same instead they are different (unlike), what leads us to name these fractions: unlike fractions.


REPRESENTATIONS of FRACTIONS on a NUMBER LINE

Here we will show fractions on a number line. To do that, we draw a number line, mark points on it and divide the gaps between points in parts to represent fractions.


As you can see, the gap between 0 and 1 is divided into 5 equal parts and the fractions:
1
5
 ,  
2
5
 ,  
3
5
   and  
4
5
 are represented on the numbers Line.   remark: 1 =
5
5


CONVERSION of MIXED FRACTIONS into IMPROPER FRACTIONS
Let  2
3
5
  be the mixed fraction:
As you can see, this mixed fraction has two different parts:
  • a whole number: 2
    • and
  • a proper fraction:
    • 3
    5
    		 , consisting of a numerator (3) and a denominator (5).

The conversion includes 3 steps:
  • multiply the whole number (2) by the denominator (5);
  • add the numerator (3);
  • keep the same denominator (5).

The Formula

Improper fraction =
(Whole Number × Denominator) + Numerator
Denominator


Example: Express each of the following mixed fractions as improper fractions

  		•  2
    3
    5
    		 =
    (2 × 5) + 3
    5
    		 =
    13
    5

  		• 4
    5
    9
    		 =
    (4 × 9) + 5
    9
    		 =
    41
    9

  		• 3
    2
    7
    		 =
    (3 × 7) + 2
    7
    		 =
    23
    7

  		• 7
    1
    4
    		 =
    (7 × 4) + 1
    4
    		 =
    29
    4


CONVERSION of IMPROPER FRACTIONS into MIXED FRACTIONS

Let 
28
9
  be the improper fraction.

Dividing 28 by 9 (28 ÷ 9) gives us:
  • a Quotient: 3;
    • and
  • a Remainder: 1.
  • the Denominator (9) is still the same.
This fraction can be written: 3
1
9

The Formula

Mixed fraction = Quotient ×
Remainder
Denominator


Example: Express each of the following improper fractions as mixed fractions

  		•
    17
    4
    		 = 4
    1
    4

  • 226
    15
    		 = 15
    1
    15

  • 28
    9
    		 = 3
    1
    9

  • 13
    5
    		 = 2
    3
    5



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Montag, 10. November 2014

Exercises on Combinations and Repetitions1


A - Nine cards, each of a different colour, are to be arranged in a line.
  1. How many different arrangements of the 9 cards are possible?

    2. The 9 cards include a pink card and a green card.
  2. How many different arrangements do not have the pink card next to the green card?

    3. Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.
  3. How many different arrangements in total of 3 cards are possible?
  4. How many of the arrangements of 3 cards in part (iii) contain the pink card?
  5. How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card?

Answers














A - Answers and explanations
  1. Here we just need to multiply the series of descending natural numbers of 9, in other words using the factorial function.
    2. 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880

  2. This issue can be solved by reversing the question like this:

    1. The total number of arrangements that do have the pink (P) card next to the green card (G), we assume both card are considered as one, attached together indeed!

      2. Therefore there are two possiblities:
      • pink (P) card next to green card (G): PG = 8! = 40,320
      • green (G) card next to pink (P) card: GP = 8! = 40,320
      >> P next to G = 2 × 8! = 80,640

    2. Now working out the arrangements do not have the pink card next to the green card or P away from G

      3. P away from G = total number of arrangments − P next to G
      ==> P away from G = 9! − 2 × 8! = 362,880 − 80,640 = 282,240

    3. Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.
  3. How many different arrangements in total of 3 cards are possible?
    4. 9P3 =
    9!
    (9 − 3)!
    		  = 
    9!
    6!
    		  = 504

  4. How many of the arrangements of 3 cards in part (iii) contain the pink card?
    5. With the restriction, there must be a pink (P) card, the two other cards are randomly selected and arranged from the 8 cards remaining.
    - 1 pink (P) out of 3 cards: ==> 3P1 =
    3!
    1!
    		  =  3! = 6 ways or possibilities
    - 2 cards out of 8: ==> 8P2 =
    8!
    (8 − 2)!
    		  = 
    8!
    6!
    		  =  56 ways

    Then, there are 56 × 6 = 168 arrangements of 3 cards containing the pink card.

  5. How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card?
    6. Again, we can reverse the question:
    • Firstly, we work out "The arrangements of 3 cards that do have the pink card next to the green card and green next to pink": PG or GP
      • - 2 × 2! = 4 ways
      - 1 card randomly select out of the 7 remaining: 7 ways
      >> 4 × 7 = 28 ways
    • P aways G = 504 (see (iii))
      • Therefore:
      The arrangements of 3 cards that do not have the pink card next to the green card is: 504 − 28 = 476






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Dienstag, 4. November 2014

Exercises on Combinaitions and Repetitions

  1. Combinations
    2.
    1. Dave is the Ceo of a committee. In how many ways can a committee of 5 be chosen from 10 people given, so that Dave ist one of them?
      2. a)

      252
      		b)

      126


      c)

      495
      		d)

      3,024



      Mixed Exercises

    2. Determine whether each of the following situations is a Combination or Permutation.
      • Determining how many different ways you can elect a Chairman and Co-Chairman of a committee if you have 10 people to choose from.

        Your choice: or

      • Choosing the batting order on a baseball team with 13 players. How many different ways?

        Your choice: or

      • Choosing 3 toppings for a pizza if there are 9 choices.

        Your choice: or

      • Voting to allow 10 new members to join a club when there are 25 that would like to join.

        Your choice: or

      • Creating an access code for a computer site using any 8 alphabet letters.

        Your choice: or




  2. Permutations
    3.
  1. A special type of password consists of four different letters of the alphabet, where each letter is used only once. How many different possible passwords are there?


    2. a)

    14,950
    		b)

    358,800


    c)

    426
    		d)

    456,976


  2. How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?


    3. a)

    84
    		b)

    120


    c)

    720
    		d)

    504










Dave is already chosen prior to the selection, so we need to choose another 4 from 9 remaining (no repetitions allow!).
In choosing a committee, order doesn't matter; so we need the number of combinations of 4 people chosen from 9.
9C4 =
9!
4!(9 − 4)!
  = 
9!
4!·5!
  = 
9·8·7·6·5!
4!·5!
  = 
9·8·7·6
4·3·2·1
  =  126



The number of permutations of 4 letters chosen from 26 is
26P4 =
26!
(26 − 4)!
  = 
26!
22!
  = 
26·25·24·23·22!
22!
  =  26·25·24·23 = 358,800



The number of permutations of 3 digits chosen from 10 is
10P3 =
10!
(10 − 3)!
  = 
10!
7!
  = 
10·9·8·7!
7!
  =  10·9·8 = 720




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Montag, 3. November 2014

Permutations

  1. Permutations with repetition 
    2.  Choosing from n things offers n choices each time!
    While choosing r of them, the permutations are:

    n · n · n ... (r times)

    Explanation:
    There are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multplying each time.

    We then use the exponent of r to write it down as follow:

    n · n · n ... (r times) = nr (n exponent of r)


    The Formula

    nr

    where:
    1. n is the number of things to choose from;
      2. and
    2. r the chosen things out of n
      3. (Here Repetition is allowed and order matters)

    Example:
    3 numbers are chosen from 10 available (numbered 0, 1, 2, ..., 9)! What could be the permutations?

    The answer:

    10 · 10 · ... (3 times) = 103 = 1.000 permutations



  2. Permutations without Repetition
    3. In this case, the number of available choices is reduced each time we pick a number.

    Example: Assuming we want to know what order 15 pool balls could be in!

    There're numbered
    1, 2, ...., 15: see Img ➦

    It goes like this: if we choose one number, let's say 5, we won't be able to choose aigain! Our next possibily of choosing will be amoung 14 poolballs:
    So, our first choice has 15 possibilities, the next choice 14 possibilities, then comes 13, 12, etc.
    And the total permutations will be:
    15 · 14 · 13 · ... = 15! = 1.307.674.368.000
    Now, we just want to choose 3 of them.
    So the permutation will be:
    15 · 14 · 13 = 2.730
    (Which means, there are 2.730 different ways to arrange 3 pool balls out of 15 balls)

Without repetition the choices get reduced each time.

To better express this mathematically, we use the Factorial function: symbolized !, which means multipying a series of descending natural numbers.
Examples:

  • 5! = 5 · 4 · 3 · 2 · 1 = 120
  • 4! = 4 · 3 · 2 · 1 = 24
  • 3! = 3 · 2 · 1 = 6
  • 2! = 2 · 1 = 2
  • 1! = 1

    • Note: 0! = 1 is an Axiom

To recap:
  1. If we want to select all of the 15 pool balls, the permutations will be:

    2. 15! = 1.307.674.368.000

    and
  2. If we want to select just 3, the permutations will be:

    3. 15!
    (15 − 3)!
    		  = 
    15!
    12!
    		  = 
    15·14·13·12!
    12!
    		  =  15·14·13  =  2.730

Formula and Notations

P(n, r)  =  nPr  =  nPr  = 
n!
(n − r)!
With:
  1. n is the number of things to choose from;

    2. and

  2. r the chosen things out of n
    3. (No repetition, order matters)






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