MATHEMATICS NOTES CLASS - 5

2ND TERM (PART – A)

TEXT BOOK OXFORD NEW COUNTDOWN BOOK – 5

THIRD EDITION

UNIT 2: HCF and LCM

Exercise # 2a: Q # 1,2 in book

Q # 4 (a,c), Q # 5 (a,d), Q # 6 (b,c), Q # 7 (a,b), Q # 8 (a,c) in notebook

Exercise # 2b: Q # 1,2 in book

Q # 4 (a,d), Q # 6 (a,d), Q # 7 (b,d), Word Problem Q # 1 in notebook

UNIT 3: FRACTIONS

Exercise # 3a: Q # 4 (a,f), Q # 5 (a,g), Word Problem Q # 1 in notebook

Exercise # 3b: Q # 4,5 in book

Q # 6 (a,c,f), Q # 12 (b,e), Q 17 (b,f), Q # 18 (a,d), Q # 19 (a.b.e)

Word Problem Q # 1 in notebook

UNIT 4: DECIMAL AND PERCENTAGES

Exercise # 4a: Q # 1,2,3 in book

Q # 4 (a,d,f,g) in notebook

Exercise # 4b: Q # 2,3 in book

Q # 4 (c,h), Q # 5 (a,d), Q # 6 (c) Q # 9 (b,f,g) Q # 10 (a) Q # 11 (b,d)

Q # 12 (a), Q # 13 (c,f), Q # 14 ( a,c,f) in notebook

UNIT 7: GEOMETRY

Exercise # 7c: in books.

 Unit 2: HCF AND LCM
Exercise 2a
Rules of divisibility HELP Page # 24 – 27
Note: This work will be done in book.
Q1. Fill in the blanks. (Page # 28)
Ans. a. 6 b. 9 c. 6 d. 8 e. 9,10
Q2. True / False. (Page # 28)
Ans. a. True b. False c. False d. True e. True
Note: This work will be done in notebooks.
Q4. Which of the following numbers are divisible by 4? (Page # 29)
a. 1 9 9 6 c. 2 3 6 0 6
Solution:- Divisible by 4:-
Solution:- Divisible by 4:-
23606
1996
TU
TU
01
24
4 06
4 96
- 04
-8
2
16
-1 6
X
As the digits in the tens and units place is not
As the digits in the tens and units place is
divisible by 4, therefore 23606 is not divisible by
divisible by 4 therefore. 1996 is divisible by 4
4.

Q5. Which of the following numbers are divisible by 6. Apply the rule and show the working. (Page # 29)
d. 39582
Solution Divisible by 6:-
39582
i. Yes 3 9 5 8 2 is an even number therefore, 3 9 5 8 2 is divisible by 2.
ii. Sum of the digits = 3+9+5+8+2
Sum of the digits = 27
9
3 27
-27
X
As the sum of the digits is divisible by 3, therefore 3 9 5 8 2 is divisible by 3.
The number 3 9 5 8 2 is divisible by both 2 and 3. Hence it is divisible by 6.
Q6. Which of the following numbers are divisible by 3 but not divisible by 6. Show your working also.
b. 801 (Page # 29)
i. Solution
801
Divisible by 3:-
Sum of the digits = 8+0+1= 9
3
39
-9
X
As the sum of the digits is divisible by 3, therefor 8 0 1 is divisible by 3.
ii. Divisible by 6:-
A) No, 8 0 1 is not an even number therefore, 8 0 1 is not divisible by 2.
B) Sum of the digits = 8+0+1 = 9
3
39
-9
X
As the sum of the digits is divisible by 3, therefore 8 0 1 is divisible by 3.
The number 8 0 1 is not divisible by 2 but it is divisible by 3. Hence the number 8 0 1
is not divisible by 6.

Q7. With the help of the rule of divisibility for 8, find out which of the following are divisible by 8.
a. 4 1 8 9 b. 1 1 7 0 0 0
Solution:- Divisible by 8:- Solution:- Divisible by 8:-
4189 117000

HTU HTU
23 The number 1 1 7 0 0 0 formed by the last three
8 189 digits is zero, therefore 1 1 7 0 0 0 is divisible by
8.
- 16
29
-24
5
The number 4 1 8 9 formed by the last three
digits is not divisible by 8, therefore 4 1 8 9 is
not divisible by 8.

Q8. Check with the help of the rule of divisibility which of the following are divisible by 9.
a. 1 8 7 3 c. 6 0 8 3
Solution:- Divisible by 9:- Solution:- Divisible by 9:-
1873 6083
Sum of the digits = 1+8+7+3 = 19 Sum of the digits = 6+0+8+3 = 17
2 1
9 19 9 17
-18 -9
1 8
As the sum of the digits is not divisible by 9, As the sum of the digits is not divisible by 9,
therefore the number 1873 is not divisible by 9. therefore the number 6083 is not divisible by 9.
 Exercise 2b
Note: This work will be done in book.
Q1. Fill in the blanks. (Page # 35)
Ans. a. 1 b. greater c. highest common factor d. product e. 72
Q2. True/False.
Ans. a. True b. True c. False d. True e. False
Note: This work will be done in notebooks.
Q4. Find the HCF by prime factorisation method. (Page # 35)
a. 64, 148, 60
Solution
HCF by prime factorisation method:-
64, 148, 60

2 64 2 148 2 60
2 32 2 74 2 30
2 16 37 3 7 3 15
2 8 1 5 5
2 4 1
2 2
1

64 =2×2×2×2×2×2
148 = 2 × 2 × 37
60 =2×2×3×5
Common factors = 2 , 2
HCF of 64, 148 and 60 = 2 × 2
= 4 Ans
d. 27, 130, 46
Solution
HCF by prime factorisation method
27, 130, 46

3 27 2 130 2 46
3 9 5 65 23 2 3
3 3 13 1 3 1
1 1

27 =3×3×3×1
1 3 0 = 2 × 5 × 13 × 1
4 6 = 2 × 23 × 1
Common factors = 1
HCF of 27, 130 and 46 = 1 Ans
Q6. Using division method find the HCF of the following numbers. (Page # 36)
a. 39, 93, 54
Solution
HCF by Division method.
39, 93, 54
39, 93 3,54
2 18
8 13
39 93 3 54
- 78 2 -54
15 39 X
-30 1
9 15
-9 1
6 9
-6 2
3 6
-6
X
HCF of 39, 93, 54 = 3 Ans

d. 40, 52, 76
Solution
HCF by Division method.
40, 52, 76
40, 52 4,76
1 19

40 52 4 76
-40 3 -76
3
12 4010 X
-36 1
4 12
- 12
X
HCF of 40, 52, 76 = 4 Ans

Q7. Find the LCM of the following using division method. (Page # 36)
b. 14, 15, 21, 28
Solution LCM by Division method: 14, 15, 21, 28
2 14, 15, 21, 28
2 7, 15, 21, 14
3 7, 15, 21, 7
5 7, 5, 7, 7
7 7, 1, 7, 7
1, 1, 1, 1
LCM of 14, 15, 21 and 28 = 2  2  3  5  7
LCM of 14, 15, 21 and 28 = 4  15  7
LCM of 14, 15, 21 and 28 = 60  7
LCM of 14, 15, 21 and 28 = 420 Ans
 d. 12, 26, 16, 39
Solution LCM by Division method: 12, 26, 16, 39
2 12, 26, 16, 39
2 6, 13, 8, 39
2 3, 13, 4, 39
2 3, 13, 2, 39
3 3, 13, 1, 39
13 1, 13, 1, 13
1, 1, 1, 1
LCM of 12, 26, 16 and 39 = 2  2  2  2  3  13
LCM of 12, 26, 16 and 39 = 4  4  39
LCM of 12, 26, 16 and 39 = 16  39
LCM of 12, 26, 16 and 39 = 624 Ans

 Word Problems (Page # 36)

Q1. There are some pieces of wood 18m, 30m and 12m long. The carpenter wants to cut them
into equal sizes. What size should cut them to avoid wastage of wood.
Data:
Number of pieces of wood = 18m, 30m and 12m
HCF of sizes of the wood = ?
Solution:-
2 18 2 30 2 12
3 9 3 15 2 6
3 3 5 5 3 3
1 1 1
18 = 2  3  3
30 = 2  3  5
12 = 2  2  3
Common factors = 2,3
HCF = 2  3
HCF = 6
The size of the wood = 6m Ans
…………………………………………………………………………………………………………………..
 Unit 3: FRACTIONS
Exercise 3a
Note: This work will be done in notebooks.
Q4. Add the following. (Page # 41)
1 3
a. 
5 20
Solution Working Column
1 3
 2 5, 20
5 20
( LCM  D1 )  N1  ( LCM  D2 )  N 2
2 5, 10
LCM
(20  5)  1  (20  20)  3
= 5 5, 5
20
4  1  1 3
= 1, 1
20
43
= LCM = 2  2  5
20
7
= Ans LCM = 20
20

1 1 1
f. 9 3 4
10 5 15
Solution Working Column
1 1 1
9 3 4 2 10, 5 15
10 5 15
(9 10)  1 (3  5)  1 (4 15)  1
=   3 5, 5, 15
10 5 15
90  1 15  1 60  1
=   5 5, 5, 5
10 5 15
91 16 61
=   1, 1, 1
10 5 15
( LCM  D1 )  N1  ( LCM  D2 )  N 2  ( LCM  D3 )  N 3
LCM = 2  3  5
LCM
(30  10)  91  (30  5)  16  (30  15)  61
= LCM = 30
30
3  91  6 16  2  61
=
30 16
273  96  122
= 30 491
30
491
= -30
30
191
11
= 16 Ans -180
30
11
Q5. Subtract the following. (Page # 41)
3 1
a. 
8 6
Solution Working Column
3 1
 2 8, 6
8 6
(24  8)  3  (24  6)  1
= 2 4, 3
24
.3  3  4  1
= 2 2, 3
24
94
= 3 1, 3
24
5
= Ans 1, 1
24
LCM = 2  2  2  3
LCM = 24
4 2 1
g. 8  3 1
7 5 10
Solution
4 2 1
8  3 1
7 5 10
(8  7)  4 (3  5)  2 (1 10)  1
=   Working Column
7 5 10
56  4 15  2 10  1
=   2 7, 5, 10
7 5 10
60 17 11
=   5 7, 5, 5
7 5 10
(70  7)  60  (70  5) 17  (70  10) 11
= 7 7, 1, 1
70
10  60  14 17  7  11
= 1, 1, 1
70
600  238  77
= LCM = 2  5  7
70
362  77
= LCM = 70
70 4
285
= (  by 5) 14 57
70
57
= -56
14
1
=4 Ans 1
14
  WORD PROBLEM (Page # 41)

3 1
Q1. Rashid walked kilometers and then took a rest. Then he walked kilometer more. How far did he
4 2
walk altogether?

Data:
3
Rashid walked = km
4
1
Again he walked = km
2
Total distance he covered = ?
Solution Working Column
3 1
Total distance he covered =    km 2 4, 2
4 2
 (4  4)  3  (4  2)  1
=   km 2 2, 1
 4
1 3  2  1
=  km 1, 1
 4
3 2
= km LCM = 2  2
4
5
= km LCM = 4
4
1
Total distance he covered = 1 km Ans
4

Exercise 3b
Note: This work will be done in book.
Q4. Work out these, using repeated addition with the help of the diagram. (Page # 48)
5 4 3 3
Ans. a. 2 b. c. d.
4 5 3
Q5. Solve the following. (Page # 48)
8 14 9 7 5 6
Ans. a. b. c. d. e. f.
3 5 4 3 6 2
Note: This work will be done in notebooks.
Q6. Solve the following. Reduce to the lowest term. (Page # 48)
a. 2 c. 10 f. 2
6 2 7
3 2 7
Solution Solution Solution
2 10 2
6 2 7
3 2 7
2 6 10 2 7 2
=  =  = 
3 1 2 1 1 7
26 10  2 72
= = =
3 1 2 1 1 7
2  62 10  21 71  2
= (  by 3) = (  by 2) = (  by 7)
1 3 1 1 2 1 1  71
2 2 10  1 1 2
= = =
1 1 1 1 1 1
4 10 2
= = =
1 1 1
= 4 Ans = 10 Ans = 2 Ans
Q12. Find the products. (Page # 49)
b. 1 7 e. 1 3
 
5 8 11 4
Solution Solution
1 7 1 3
 
5 8 11 4
1 7 1 3
= =
58 11 4
7 3
= Ans = Ans
40 44
Q17. Find the products of the following fractions. (Page # 50)
b. 3 7 f. 7 2
 
4 8 10 3
Solution Solution
3 7 7 2
 
4 8 10 3
3 7 7 2
= = 
48 10 3
21 7 21
= Ans =  (  by 2)
32 5 10 3
7 1
=
5 3
7
= Ans
15
Q18. Verify distributive law by using brackets to help you. (Page # 50)

a. 1 1 1 d. 1 1 1
   
2 4 5 4 5 3
Solution Solution
1 1 1 1 1 1
  Distributive law:-   Distributive law:-
2 4 5 4 5 3
1 1 1 1 1 1 1 1 1 1 1 1
             
 2 4 5 2  4 5  4 5 3 4  5 3
1 1 1 1 1 1
L. H. S =     L. H. S =    
2 4 5  4 5 3
 1 1  1  1 1  1
L. H. S =   L. H. S =  
 2 4  5  45  3
1 1 1 1
L. H. S =  L. H. S = 
8 5 20 3
1 1
L. H. S = L. H. S =
40 60
1 1 1 1 1 1
R. H. S =     R. H. S =    
2  4 5 4  5 3
1  1 1  1  1 1 
R. H S =    R. H S =   
2  45  4  5 3 
1 1 1 1
R. H. S =  R. H. S = 
2 20 4 15
1 1 1 1
R. H. S = R. H. S =
2  20 4  15
1 1
R. H. S = R. H. S =
40 60
L. H. S = R. H. S L. H. S = R. H. S
Hence the distributive law is verified. Hence the distributive law is verified.

Q19. Solve these multiplications making sure each product is in its lowest term. (Page # 50)

2
a. 2 3
5
Solution Working Column
2
2 3
5 7
(2  5)  2
= 3 5 36
5
10  2
= 3 -35
5
12
= 3 1
5
12 3
= 
5 1
 12  3
=
5 1
36
=
5
1
=7 Ans
5

1 7
b. 4 
2 8
Solution Working Column
1 7
4 
2 8 3
(4  2)  1 7
=  16 63
2 8
8 1 7
=  -48
2 8
9 7
=  15
2 8
9 7
=
28
63
=
16
15
=3
16

1 1 1
e. 3  
9 3 6
Solution
1 1 1
3  
9 3 6
(3  9)  1 1 1
=  
9 3 6
28 1 1
=  
9 3 6
14
28  1 1
= (  by 2)
9  3  63
14  1 1
=
9  3 3
14
= Ans
81
  WORD PROBLEM (Page # 50)

3 2
Q1. Ayesha’s grandmother buys m of lace for her handkerchief. She uses only m of the lace.
4 3
What length of lace is left behind?

Data:
3
Total length of lace for handkerchief = m
4
2
Length of the lace used by her = m
3
Length of lace is left behind = ?

Solution Working Column

3 2
Length of lace is left behind = m m 2 4, 3
4 3
2 2. 3
 (12  4)  3  (12  3)  2 
=   m 3 1, 3
 12
1, 1
3 3  4  2 
=   m LCM = 2  2  3
 12
LCM = 12
9  8
=  m
 12 

1
= m
2

1
Length of lace is left behind = m Ans
2

………………………………………………………………………………………………………………….
 Unit 4: DECIMALS AND PERCENTAGES
Exercise 4a
Note: This work will be done in book.
Q1. Fill in the blanks. (Page # 64)
Ans. a. 175.19 b. 800.101 c. 132.999 d. 1153.881 e. 764.505
Q2. True / False. (Page # 64)
Ans. a. False b. True c. False d. True e. True
Q3. Choose the correct answer. (Page # 65)
Ans. a. 25.395 b. 36.67 c. three decimal places d. 10.99 e. 0.011
Note: This work will be done in notebooks.
Q4. Solve the following. (Page # 65)
a. 302.48–208.24 d. 860.241+564.237
Solution Solution
9
2 10 12 1
302.48 860.241
–208.24 + 564.237
094.24 Ans 14 2 4 . 4 7 8 Ans

f. 3.19+27.974+8.8 g. 49.34–8.7
Solution Solution
1 1 1 8 13
03.190 49.34
27.974 – 08.70
–08.800 40.64 Ans
39.964 Ans

Exercise 4b
Note: This work will be done in book.
Q2. True / False. (Page # 77)
Ans. a. True b. False c. False d. True e. True
Q3. Choose the correct answer. (Page # 77)
Ans. a. Thousandth b. 0.00256 c. 1.69 d. 1.7 e. 3
Note: This work will be done in notebooks.
Q4. Find the product of the following. (Page # 77)
c. 1 0 . 5 × 2 h. 0 . 9 9 8 × 15
Solution Solution
1 4 4 4

10.5 0.998
×2 ×15
1 1
21.0 Ans
1 4 9 90
+0 9 98×
14.970 Ans

Q5. Solve the following sums. (Page # 77)

a. 1 3 . 6  4 d. 2 1 . 9 5  5
Solution Solution
3 . 4 4 . 3 9

4 13.6 5 21.95
-12 -20
16
19
-16
-15
X
45
Quotient = 3.4 Ans -45
X
Quotient = 4.39 Ans

Q6. Multiply the following decimals by 10 and 100. (Page # 77)

c. 364.8
Solution
i. Multiply by 10
364.8×10
=364.8×10
10
= 3 6 4 8 × 10 1 (÷ by 10)
10 1
=3648
1
=3648
3648×10=3648 Ans
 ii. Multiply by 100
364.8×100
=364.8×100
10
= 3 6 4 8 × 100 10 (÷ by 10)
10 1
=3648×10
=3648
3648×100=36480 Ans
Q9. Multiply the following decimal numbers. (Page # 78)
b. 7 . 1 × 1 . 9 f. 3 . 7 8 × 2 . 4
Solution Solution
1 1
7.1 3 3

×1 . 9 3.78
× 2.4
1
63 . 9
1512
+71 ×
13 . 49 Ans +7 5 6 ×
9.072 Ans

g. 9 . 0 8 × 3 . 5
Solution
2
4

9.08
× 3.5
4540
1
+ 2724×
3 1. 7 8 0 Ans

Q10. Solve the following by changing the decimal fractions into common fractions. (Page # 78)
a. 1.732×0.5
Solution
1.732×0.5
1.732 0.5
= 
1000 10
1732 51
= (÷ by 5)
1000 102
866
 1732 1
= (÷ by 2)
1000 21

866  1
=
1000  1
866
=
1000
= 0.866 Ans
Q11. Work out the following divisions. (Page # 78)
b. 0 . 4 9 ÷ 0 . 7 d. 9 . 5 ÷ 1 . 9
Solution
Solution
9.5÷1.9
0.49÷0.7 9.5 1.9
= 
0.49 0.7 10 10
=  95 10 19
100 10 =  (reciprocal of )
49 10 7 10 19 10
=  (reciprocal of ) 95  101
100 7 10 = (÷ by 10)
49  101 101  19
= (÷ by 10)
10010  7 955  1
= (÷ by 19)
49  1
7 1 191
= (÷ by 7) 5 1
10  71 =
7 1 Working Column
1 1
= 0.7
10  1 =5 Ans
10 7.0
7
= -70
10
X
= 0.7 Ans

Q12. Work out the following sums by the direct division method. (Page # 78)
b. 9 . 6 ÷ 0 . 0 4 8 Working Column
200
Solution
48 9600
9.6÷0.048
-9600
9.6 0.048 X
= 
10 1000
96 1000
= 
10 48
96  1000100
= (÷ by 10)
101  48
96 100
=
1 48
9600
=
48
9.6÷0.048 = 200 Ans
Q13. Change into decimal fractions. (Page # 78)
c. 7 f. 13
8 25
Solution Solution
7 13
8 25
0.875 0.52

8 7.000 25 13.00
-96 -12 5
60 50
-56 -50
40 X
-40 13
X =0.52 Ans
25
7
=0.875 Ans
8

Q14. Simplify the following expressions using the BODMAS rule. (Page # 78)
a. 6 . 8 × (0 . 1 4 + 1 4 . 6 3) c. 1 0 . 1 × (6 . 8 8 – 3 . 5) f. 9.5 + (2.03 ×1.6) – 4 . 5
Solution Solution Solution
6.8×(0.14+14.63) 10.1 × (6.88–3.5) 9.5 + (2.03 ×1.6) – 4 . 5
= 6.8 × 14.77 = 10.1 × 3.38 = 9.5 + 3.248 – 4.5
6.8 14.77 = 34.138 Ans = 12.748 – 4.5
= 
10 100 = 8.248 Ans
68  1477
=
10  100
100436
=
1000
= 100.436 Ans

…………………………………………………………………………………………………………………..
 Unit 7: GOEMETRY
Exercise 7c
Note: This work will be done in book.
Q1. Fill in the blanks. (Page # 133)
Ans. a. 50o b. equilateral c. Acute angled Triangle d. hypotenuse
e. unequal
Q2. True/ False. (Page # 133)
Ans. a. True b. False c. True d. False e. True
Q3. Choose the correct answer. (Page # 133)
Ans. a. all 3sides are given b. 30o c. an equilateral triangle d. 180o
e. 90o
Q4. Measure the sides of these triangle and write their name equilateral, isosceles or scalene. (Page#134)
Ans. a. equilateral triangle b. equilateral triangle c. isosceles d. scalene
e. scalene f. scalene
Q5. Look at the figures carefully and write the letters in the correct column. (Page # 134)
Ans.
Right angled triangle Acute angled triangle Obtuse angled triangle
C and D A, B and E F and G

Q6. Measure the angles of these triangles and record your findings. (Page # 134/135)
Ans.
Triangle Angles Total of all 3 angles
X Y Z
a. 50o 90o 40o
b. 130o 30o 20o
c. 82o 70o 28o
d. 55o 100o 25o
e. 70o 90o 20o
o
f. 30 120o 30o

Q7. Measure the sides of the given triangle with a ruler. Record the findings in the table. Then, with a
Protractor, measure the angles of the triangle and record your findings. (Page # 135)
Ans.
Length of side (cm) Size of angle (o)
AB 2.7cm CAB 60o
BC 2.7cm ABC 60o
AC 2.7cm BCA 60o
 Equilateral triangle. All three angles are equal
Q8. Measure and record the sides and angles of this triangle. (Page # 135)
Ans.
Length of side (cm) Size of angle (o)
PQ 3.8cm RPQ 65o
QR 3.8cm PQR 65o
PR 3cm QRP 50o
 Isosceles triangle. Two angles are equal
Q9. Measure and record the sides and angles of this triangle. (Page # 136)
Ans.
Length of side (cm) Size of angle (o)
ED 2.9cm FED 50o
DF 2.4cm EDF 40o
EF 3.8cm DFE 90o
 Scalene triangle.
 All three angles are different.
Q10. Measure each triangle’s sides and angles, then name the triangle family to which it belongs.
Present the findings in the form of a table. (Page # 136)
Ans. a. Equilateral triangle b. Scalene triangle3

Q11. Without using a protractor, calculate the size of the angles marked with letters. (Page # 136)
Ans. a. x 45o b. p 74o c. y 47o d. t 90o
e. z 37o