10th Edition
Math – 10
Chapter Three – Section One (3-1) Least Common Multiples
Least Common Multiples (LCM) Common Denominator
Smallest number that is a multiple of all the numbers.
Method One:
a)
Determine whether the largest is a multiple of the others. If so,
it is the LCM.
b)
Check the multiples of the largest number until you get one that
is a multiple of each of the others.
LCM?
1. Of: 9 and 15
15, 30, 45
2. Of: 8 and 10
10, 20, 30, 40
3. Of: 10 and 15
15, 30
4. Of: 6 and 8
8, 16, 24
5. Of: 5 and 10
10
6. Of: 20, 40, 80
80
Method Two:
a)
Find the prime factorization of each number.
b)
Create a product of each factor, using that factor the greatest
number of times that it occurs in any one factorization.
7. Of: 8 and 10
8 = 2 x 2 x 2 10 = 2 x 5
8. Of: 18 and 40 18 = 2 x 3 x 3
LCM = 2 x 2 x 2 x 3 x 3 x 5 = 360
LCM = 2 x 2 x 2 x 5 = 40
40 = 2 x 2 x 2 x 5
9. Of: 32 and 54 32 = 2 x 2 x 2 x 2 x 2
LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 = 864
54 = 2 x 3 x 3 x 3
10. Of: 24, 35 and 45
24 = 2 x 2 x 2 x 3 35 = 5 x 7
5
LCM = 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520
1
45 = 3 x 3 x
Chapter 3
10th Edition
Math – 10
11. 24 = 23 x 31
12. – 16.
35 = 51 x 71 45 = 32 x 51 LCM = 23 x 32 x 51 x 71 = 2520
skip
Method Three:
a)
Divide all the numbers by prime numbers with no remainders.
Bring down any numbers not divisible by the prime.
b)
Repeat until all are primes.
17.
12, 75, 120
2 12 75 120
2 6 75 60
2 3 75 30
3 3 75 15
5 1 25
5
5 1 5
1
1 1
1
2 x 2 x 2 x 3 x 5 x 5 = 600
18.
27, 90, 84
2 27 90 84
2 27 45 42
3 27 45 21
3 9 15 7
3 3 5 7
2 x 2 x 3 x 3 x 3 x 3 x 5 x 7 = 3780
2
Chapter 3
10th Edition
Math – 10
19.
12, 24, 75, 120
2 12 24 75 120
2 6 12 75 60
2 3 6 75 30
3 3 3 75 15
5 1 1 25
5
5 1 1 5
1
1 1
1
2 x 2 x 3 x 5 x 5 = 600
Chapter 3 – Section 2
Addition and Applications
Addition Using Fraction Notation:
To add when denominators are the same:
a)
Add the numerators
b)
Keep the denominator
c)
Simplify, if possible
1.
ଵ
ହ
+
ଷ
ହ
=
ସ
ହ
2.
1 2 3
+ =
3 3 3
3.
5
1
6 1x6 1
+
=
=
=
12 12 12 2 x 6 2
4.
9
3 12 3 x 4 3
+ =
=
=
16 16 16 4 x 4 4
then simplify
3
=1
3
To add when denominators are different:
a)
Find the lest common multiple of the denominators – The
Least Common Denominator, LCD.
b)
Multiple by 1 (N/N notation) to express each number in
terms of the LCD.
c)
Add the numerators, keeping the same denominator.
d)
Simplify, if possible.
3
Chapter 3
10th Edition
Math – 10
5.
ଶ
ଷ
+
ଵ
଺
LCD of 3 and 6 = 6
ଶ
ଷ
6.
3 5
+
8 6
ଶ
( ∗ )+
ଶ
ଵ
଺
=
ସ
଺
LCD of 6 and 8
+
ଵ
଺
=
ହ
଺
6=2*3 8=2*2*2
8 * [] = 24 (3)
LCD=2*2*2*3=24
6 * [] = 24 (4)
3 3
5 4
9 20 29
( * )+( * ) =
+
=
8 3
6 4
24 24 24
7.
8.
9.
1 7 1 3 7
3 7 10 2 x5 5
+ = x + = + =
=
=
6 18 6 3 18 18 18 18 2 x9 9
4
1
3
4 100
1 10
3
400
10
3
413
+
+
= x
+
x +
=
+
+
=
10 100 1000 10 100 100 10 1000 1000 1000 1000 1000
7
2 1
+ +
using prime fractorization.
10 21 7
7
2
1
+
+
LCD = 2*3*5*7
2*5 3*7 7
7
3*7
2
2*5
1 2 * 3* 5
(
*
)+(
*
)+( *
)
2*5 3*7
3* 7 2 * 5
7 2 * 3* 5
7 *3*7
2*2*5
1 * 2 * 3 * 5 147 20
30 197
+
+
=
+
+
=
2 * 3 * 5 * 7 2 * 3 * 5 * 7 2 * 3 * 5 * 7 210 210 210 210
10.
7
5 11 7 4 5 3 11 2 28 15 22 65
+
+
= x +
x + x =
+
+
=
18 24 36 18 4 24 3 36 2 72 72 72 72
Page 161
11.
4 1 4 2 1
8
1
9
+
= x +
= + = mile
5 10 5 2 10 10 10 10
Chapter 3 Section 3
Subtraction, Order, and Applications
To subtract, when denominators are the same:
a)
Subtract the numerators.
b)
Keep the denominators.
c)
Simplify, if possible.
1.
7 3 4 1x 4 1
− = =
=
8 8 8 2x4 2
4
Chapter 3
10th Edition
Math – 10
2.
10 4 10 − 4 6 2 * 3 3
− =
=
=
=
16 16
16
16 2 * 8 8
3.
8
3
5 1x5 1
− =
=
=
10 10 10 2 x5 2
To subtract, when denominators are different:
a)
Find the Least Common Multiple of the Denominators,
LCD.
b)
Multiply by 1 (N/N) to express each number in terms of the
LCD.
c)
Subtract the numerators, keeping the denominators.
d)
Simplify, if possible.
4.
3 2
−
LCD 3 and 4 = 12
4 3
3 3
2 4
9
8 9 −8 1
( * )−( * ) = − =
=
4 3
3 4 12 12
12
12
5.
5 1 5 3 1 2 15 2 13
− = x + x = − =
6 9 6 3 9 2 18 18 18
6.
4 3 4 2 3
8
3
5 1x5 1
− = x − = − =
=
=
5 10 5 2 10 10 10 10 2 x5 2
7.
11 5 11 4 5 7 44 − 35
9
− =
x − x =
=
28 16 28 4 16 7
112
112
ORDER – greater than, less than, or equal
To determine, which of two numbers is greater when there is a common
denominator, compare the numerators.
8.
3
5
[]
8
8
3[]5
9.
7 6
[]
10 10
7
6
>
10 10
3<5
5
Chapter 3
10th Edition
Math – 10
When denominators are different, multiply by I (N/N) to make the
denominators the same.
10.
2
5
[]
3
8
2 8 5 3
* [] *
3 8 8 3
LCD 3, 8 = 24
16 15
[]
24 24
16 15
>
24 24
11.
3 8
[]
4 12
3 3 8
x []
4 3 12
9 8
[]
12 12
9
8
>
12 12
12.
5 7
[]
6 8
5 4 7 3
x [] x
6 4 8 3
20 21
[]
24 24
20 21
<
24 24
Equations
2 5
=
3 6
13.
X+
14.
3
7
+T =
5
8
X+
2 2 5 2 2
− = − x
3 3 6 3 2
X =
5 4 1
− =
6 6 6
3 3
7 3
− +T = −
5 5
8 5
T=
7 5 3 8 35 − 24 11
x − x =
=
8 5 5 8
40
40
2 2
7 2
− +D= −
3 3
8 3
D=
7 3 2 8 21 − 16 5
=
x − x =
8 3 3 8
24
24
Page 168
15.
2
7
+D=
3
8
Chapter Three Section Four
Mixed Numerals
Mixed Numerals – Whole Number and (plus) a Fraction
ଶ
ଶ
1.
1+ =1
ଷ
ଷ
2.
2+ =2
ଷ
ଷ
3.
8+ =8
ଷ
ଷ
4.
12 + = 12
ସ
ସ
ଶ
ଷ
ସ
ସ
ଶ
ଷ
6
Chapter 3
10th Edition
Math – 10
Convert to Fraction Notation
ଶ
ଶ
ସ
ଶ
ସ ହ
ଶ଴
ଶ
5.
4 =4+ = + =( ∗ )+ + =
ହ
6.
6
ହ
ଵ
ଵ଴
଺଴
=
ଵ଴
ଵ
ଵ
+ =
଺
ହ
ଵ
ହ
ହ
ହ
ଶଶ
ହ
଺ଵ
ଵ଴
Convert from a Mixed Numeral to Fraction Notation (Faster Method)
a. Multiply the WHOLE number by the DENOMINATOR.
b. Add result to Numerator.
c. Keep the Denominator.
ହ
(ସ∗଺)ାହ
ଵ
ଷ଺ାଵ
7.
4 =
8.
9 =
9.
20 =
଺
ସ
ଶ
ଷ
଺
ସ
ଷ
଺
=
ଶଽ
଺
ଷ଻
=
଺଴ାଶ
ଶସାହ
=
ସ
=
଺ଶ
ଷ
Convert from Fraction Notation to Mixed Numeral
a. The Divisor = the Denominator
b. The Quotient (answer) = the Whole Number
c. The Remainder = the Numerator
10.
11.
12.
଻
ଷ
ଵ
=2ܴ1=2
ଵଵ
ଵ଴
ଷ
=1ܴ1=1
ଵଵ଴
଺
ଵ
ଵ଴
ଶ
= 18 ܴ 2 = 18 = 18
଺
ଵ
ଷ
ଶ
13.
4846 ÷ 6 = 807 R 4 = 807
14.
6053 ÷ 45 = 134 R 23 = 134
ଷ
ଶଷ
ସହ
7
Chapter 3
10th Edition
Math – 10
Chapter Three Section Five
Addition and Subtraction Using Mixed
Numerals; Applications
Addition
a. First, add the fractions
b. Then, add the whole numbers
1.
2
ଷ
ଵ଴
+ 5
ଵ
ଵ଴
2
5
7
2.
ଶ
8 +3
ହ
଻
8
ଵ଴
3
ଷ
ଵ଴
ଵ
ଵ଴
ଶ
ହ
ଶ
ହ
଻
=
ଵ଴
ଵ
ଵ଴
ଷ
9 +3
ସ
ହ
9
଺
3
ଷ
ସ
ହ
଺
଻
12 + 1
ଷ
ଵ଴
ଶ
=7
11 + 1
3.
=
+
=
ଵ
ଵ଴
ସ
=
ଵ଴
ଶ
=
ହ
ହ
ସ
ଵ଴
ସ
=
ଵ଴
= 12
=
ଵଶ
+
ଷ
ସ
ହ
଺
ଵ
ଵଵ
ଵ଴
ଵ
ଵ଴
ଵ଴
ଷ
∗ =
ଷ
ଶ
∗ =
= 13
=1
ଶ
଻
ଽ
ଵଶ
ଵ଴
ଵଶ
+
ଽ
ଵଶ
=
ଵଽ
ଵଶ
=1
଻
ଵଶ
ଵଶ
Subtraction
a. First, sub fractions
b. Then, sub whole
4.
଻
10 − 9
଼
ଷ
଼
10
9
1
5.
ଶ
8 − 5
ଷ
ଵ
ଶ
8
5
3
଻
଼
ଷ
଼
ଵ
ଶ
ଶ
ଷ
ଵ
ଶ
ଵ
଺
଻
଼
ଵ
=1
=
=
−
ଷ
଼
=
ସ
଼
=
ଵ
ଶ
ଶ
ସ
଺
ଷ
଺
=3
ସ
ଵ
଺
−
ଷ
଺
=
ଵ
଺
଺
8
Chapter 3
10th Edition
Math – 10
6.
ଵ
8 − 4
ଽ
ହ
ଵ
8-1
଺
ଽ
ହ
4
3
଺
ହ
=
=
ଶ
ଵ଼
ଵହ
+
ଵ଼
ଵ଼
=
ଶ଴
ଵ଼
ଵ଼
ଶ଴
−
=
ଶ
ଵ଼
ଵହ
ଵ଼
=
ହ
ଵ଼
ଵ଼
7.
ଵ
5-1
5-1
ଷ
0+
ଵ
1
ଷ
ଷ
ଷ
ଷ
ଶ
ଷ
−
ଵ
ଷ
ଷ
3
ଷ
Page 182
ଽ
ଵ
ଵ଼
8.
144 + 87 = 144 + 87
ସ
ଵ଴
ଵ
ଶ଴
ଵ
ଶ
ହ
ଶ଴
= 231
ଷ
ଶଷ
ଶ଴
= 231 + 1
଼
ଷ
9.
23 − 18 = 23 − 18 = 22 − 18 = 4
10.
283 − 178 + 250
ଷ
ଶ
଺
ହ
଼
ଵହ
଺
ଷ
ଵ଺
ଷଽ
- 178
104
+ 250 = 354
ଶସ
଺
ଶ଴
= 232
ଷ
ଶ଴
ହ
଺
ଶ
283
ଶସ
ଶଷ
଺
ଷ
ଶସ
− 178
= 282
ଶସ
ଶଷ
ଵ଺
ଶସ
= 104
ଶଷ
ଶସ
ଶସ
Chapter Three Section Six
Multiplication and Division Using
Mixed Numerals; Applications
Multiplication and Division
a.
First, convert to Fraction Notation
b.
Multiply or Divide the Fractions
c.
Then, convert back to Mixed
Multiplication
ଵ
1.
6*3 =
ଷ
ଵ
∗
ଵ଴
ଵ
ଷ
=
2.
2 ∗
=
ହ
∗
ଷ
3.
2*6 =
ଶ
∗
ଷଶ
=
4.
3 ∗2 =
ଵ଴
∗
ହ
ଶ
ଷ
଺
ସ
ଶ
ହ
ଵ
ଷ
ଵ
ଶ
ଶ
ଵ
ଷ
ସ
ହ
ଷ
ଵହ
=
ଶ
଺଴
=1
଼
଺ସ
=
= 20
ହ
଻
଼
= 12
ହ଴
଺
ସ
ହ
ଶ
=8 =8
଺
9
ଵ
ଷ
Chapter 3
10th Edition
Math – 10
Division
ଵ
5.
84 ÷ 5 =
ସ
଼ସ
÷
ଶ଺
÷ =
ଵ
ଵ
6.
26 ÷ 3 =
7.
2 ÷1 =
8.
1 ÷2 =
ଶ
ଶଵ
଻
ଵ
଼ସ
∗
ଶ଺
∗
ଶ
=
ହଶ
ଷ∗ଷ∗ହ
=
ସ
ଶ
ଵ
ଵ
଻
ଵ
ଽ
÷ =
ଽ
∗
ହ
=
ଷ
ଵ
଻
÷
ହ
଻
∗
ଶ
=
ହ
ସ
ସ
ଶ
ହ
ସ
ଶ
ସ
=
ସ
= 16
଺
ହ
ଷ
=7
଻
ଵ
ସ
଺
ସ
ଶଵ
ସ∗ଶ∗ଷ
଻
଻
଼
=1
଻
଼
଻
=
ଶ∗ହ
ଵହ
=
ଵ଴
Page 192
9.
65 + 65 + 65 + ½(65)
ଵ
଺ହ
଻
ଶହହ
ଵ
65 * 3 =
∗ =
= 227
ଶ
10.
11.
௠௜௟௘௦
௚௔௟௟௢௡௦
ଵ
ଶ
ଶ
ଷ଴ଶ ௠௜௟௘௦
=
ଶ
ଷ଴ଶ ௠௜௟௘௦
=
భ
ଵହభబ ௚௔௟௟௢௡௦
ଵ
ଵ
ସହ
ଷଵ
ଷ଴ଶ
=
భఱభ
௚௔௟௟௢௡௦
భబ
ଵ
∗
ଵ଴
௠௜௟௘௦
ଵହଵ ௚௔௟௟௢௡௦
ଵଷଽହ
Chapter Three Section Seven
2.
3.
ଶ
∗
ହ
ଵ
∗
ଷ
ଷ
∗ 16 + 8 = ∗
ହ
ଷ
ସ
Average
4.
௠௜௟௘௦
௚௔௟௟௢௡
ଷ
∗ =
= 348
Arearoom = 22 ∗ 15 =
ଶ
ଶ
ଶ
ସ
ସ
ଶ
Arearug = 12 * 9 = 108
ଷ
ଷ
Areafloor = Arearoom – Arearug = 348 - 108 = 240
ସ
1.
= 20
భ
మ
଼
ସ
+
÷
ଵ
=
ଶ∗ହ
ହ
−
ଵ
ସ
଼
ହ∗଼
ଵ଴
ଶ
య
ర
భ
ర
ସ
=
ଷ
ଷ
ଽ ାଵ଴ ାଵ଴ ାଽ
+
ସ
=
మ
ర
ଵ
ସ
=
ସ
Order of Operations; Estimation
ଶ
଼
ଵ
÷
ହ
ଵ଺
+
ଶ଺
+ =
ଵ
ସ
ଵ
ଵ
ସ
ଵ
ଶ
଼
ଵ
= ∗ −
=
ସ
଼
ଵ଴
ସ
ହ
ଵ଴
2
1
4
1
3
−
=
−
=
5 10
10 10
10
ଵ
య
ర
−
଼
=
ଷ
భ
ర
=
ଽ ାଵ଴ ାଵ଴ ାଽ
ସ
ଵଶ
ଵ
=
+
ଶ଺
ଷ
భ
మ
ଷଽ
10
ସ
=
=
ଷ଺
ଷ
ళవ
మ
ସ
+
ଶ଺
=
଻ଽ
ଷ
ଶ
=
଺ଶ
ଷ
ଵ
‫ ݎ݋‬20
∗ =
ସ
଻ଽ
଼
ଶ
ଷ
=9
଻
଼
Chapter 3
10th Edition
Math – 10
5.
6.
7.
భ భ ఱ
ା ା
మ య ల
=
ଷ
య ర
ା
ర ఱ
ଶ
ଷ
భఱ భల
ା
మబ మబ
=
ଶ
య మ ఱ
ା ା
ల ల ల
ଶ
ଷ
=
=
భబ
ల
ଷ
ଷଵ
ଶ଴
ଵ
=
ଵ଴
଺
ଵ
ଶ
−
ହ଺
ସ
଻
ହ଺
=
ଷ
ଶ଻
ଶ
଼
ଵଶ
9.
10.
11.
ଷ
ହ
ଽ
+
ଽ
଻
ቁ÷ −
ଵଶ
ଷ
ଵ
଼
=
ଵ଻
ଵଶ
ଷ
ଵ
∗ − =
଻
଼
ଵ଻
ଶ଼
−
ଵ
଼
=
ହ଺
Estimation with Fractions ( 0 or
8.
ଵ଼
=
ସ଴
ቀ + ቁ ÷ 2 − ( )ଷ = ቀ
ଷ
ଷସ
ଷ
ଵ଴
ଷଵ
∗ =
ଵ
ଵ
∗ =
ଵ
ଶ
‫ ݎ݋‬1)
0
ହଽ
଺ଵ
1
ହଽ
ଶଽ
ହଽ
ହହ଻
ଵ
ଶ
1
ହଽ
12 – 17 Omit
ଽ
ଵ
18.
5
19.
10 ∗ ቀ25
20.
(10 + 7 ) ÷
ଵ଴
+ 26 − 10
ଶ
଻
଼
ସ
ହ
ହ
ଽ
ଵଵ
ଵଷ
ଷ
ଶଽ
ଵ
ଵ
= 6 + 26 − 10 = 32 − 10 = 22
ଶ
ଶ
ଵ
ଶ
ଵ
− 14 ቁ = 11 ∗ (26 − 14) = 11(12) = 132
ଽ
ଵ଻
ଷ଴
ଵ
ଵ
ଵ
ଵ
= ቀ11 + 7 ቁ ÷ = 18 ÷ =
ଶ
ଶ
11
ଶ
ଶ
ଷ଻
ଶ
ଶ
∗ = 37
ଵ
Chapter 3