unit: 2 natural numbers and whole numbers natural … · 2020. 9. 3. · 1. the set of natural...

Second Term Notes

For Class 6 St. Francis’ High School Hayatabad, Peshawar

Mathematics

UNIT: 2 NATURAL NUMBERS AND WHOLE NUMBERS

NATURAL NUMBERS

The numbers used for counting objects around us are called natural or counting numbers.

Examples: 1, 2, 3, 4, 5 …………. and so on.

WHOLE NUMBERS

The numbers consisting of (0) zero and all the natural numbers are called whole numbers.

Example: 0, 1, 2, 3, 4, 5……………………… and so on.

EXERCISE 2a

Question # 1 and Question # 2 (In Book)

EXERCISE # 2b

Q1, Q2, Q4, Q5, Q7 all should be done in book.

Q3. Find the difference between the largest odd number of three digits and the smallest even

number of 3 digits.

SOLUTION:

Largest three digit odd number = 999

Smallest three digit even number = 100

Difference = 999 – 100 = 899 Ans.

Q6. Find the value of:

i). 542 × 92 + 8 × 542 iii). 6 × 612 + 4 × 612

Solution Solution

= 542 × 92 + 8 × 542 = 6 × 612 + 4 × 612

= 49864 + 4336 = 3672 + 2448

= 54200 Ans = 6120 Ans

Multiple Choice Questions 2

Q. Select the correct answer from the given options.

1. The set of natural number, with 0 added to it, becomes “set of whole numbers”. The statement is

True.

A. not always true B. True C. False D. None of the above

2. Position of 4 in the number 3, 24, 901 is thousand.

A. Thousand B. Units C. Tens D. Hundreds

3. Position of 8 in the number 859, 637 is lakhs.

A. million B. Ten thousand C. Lakhs D. Crore

4. 5 billion > 5 Arabs. True OR false false

A. false B. 5 billion < 5 Arabs C. True D. Billion = crore

5. Sum of two 3 digit number is always a 3 digit number. Not always true.

A. True B. False C. Not always true D. none of these.

6. 1 million is equal to how many lakhs? 1 million = 10 lakhs.

A. 1 million = 1 lakhs B. 1 million = 10 lakhs C. 1 lakhs = 10 million D. none of these

UNIT# 8 “INTRODUCTION TO ALGEBRA”

Learn definitions from book

EXERCISE 8a

Q1. Are the following statements true, false or open?

i). Zero is a whole number. (True) ii). 5 + 9 = 18 – 5 (False).

iii). 2 is the only even prime. (True) iv).17 is a natural number. (True)

v). 7 x =21, when = 4. (False). vi). 4 (5) = 45 (False)

vii). 5x + 5 = 10, then x = 1 (True) Viii). 0.5 = 1/5 (False)

ix). 12 + 3 = 3 + 12 (True)

Q2. Replace the letters by numbers to make the statements true.

i). x + 2 = 9 ii). 7 – p = 2 iii). 3y = 15 iv). 45 ÷ x = 5

Sol. Sol. Sol. Sol.

X + 2 = 9 7 – p = 2 3y = 15 45 ÷ x = 5

X = 9 – 2 7 – 2 = p 3y / 3 = 15/3 45 / x = 5 (× ing ‘x’)

X = 7 5 = p y = 5 45 = 5x (÷ ing by “5”)

So, 7 + 2 = 9 so, 7 – 5 = 2 so, 3(5) =15 45/5 = 5x/5

9 = x SO, 45/9=5

Q3. Write the following as algebraic expressions.

i). The sum of p and q ii). The difference of a and b, when a is greater than b

Ans. P +q Ans. a - b

iii). Two thirds of x added to 2y. iv). The product of m and n added to their sum

Ans. 𝟐

𝟑 x + 2y Ans. mn + (m + n)

v). The difference of 2p and 3q when 2p is less than 3q

Ans. 3q – 2p

vi). Three times “a” added to four times “b”.

Ans. 3a = 4b

vii). The sum of m and the quotient of n divided by 2

Ans. m + n/2

viii). The quotient of a and b when a is divided by b, added to their product

Ans. 𝒂

𝒃 + ab

ix). One third of x multiplied by the difference of x and y when x is greater than y

Ans. 𝟏

𝟑 x (x – y)

x). Three fourths of x multiplied by the difference of 2p and 5q when 2p is less than 5q

Ans. 𝟑

𝟒 x (5q – 2p)

Q4. Write down separately the terms of the following algebraic expression.

Expression Terms

i). a + b – 2c a, b, - 2c

ii). – 2xyz – 3xy + z -2xyz , - 3xy , z

iii). abc + 2fgh - af2 – bg2 – ch2 abc, 2 fgh , - af2 , -bg2 , -ch2

Q5. Write down the algebraic expression whose terms are given below

Terms Expression

i). a, - 3b, 4c a – 3b + 4c

ii). – 5abc, - 7bcd, 3abd - 5abc – 7bcd + 3abd

iii). 3u, - ½ gt 3u – ½ gt

EXERCISE 8b

Q2. Add the following

i). 2x, - 3x, 5x ii). 2a + 3b , 3a – 4b

Sol Sol

= 2x, - 3x, 5x = 2a + 3b ……. (i

= 2x- 3x + 5x = 3a – 4b ……… (ii

Taking “x” common now adding (i and (ii

= X (2 – 3 + 5) = 2a + 3b + (3a - 4b)

= X (4) = 2a + 3b + 3a – 4b

= 4x = 2a + 3a + 3b – 4b

iv). a+ b – c, 3a + b – 2c = a (2 + 3) - b (-3 + 4)

Sol = a (5) -b (1)

a+b –c …………… (i = 5a – b Ans.

3a + b -2c,………… (ii

Now adding (I and (ii vii). 3x3 – 4x2 + 5x +1, x3 +2x2 – 3x +4,

= a+ b – c + (3a + b – 2c) 4x3 – 3x2 + 4x - 5

= a+ b – c + 3a + b – 2c Sol

= a + 3a + b + b – c – 2c = 3x3 – 4x2 + 5x +1

= a (1 + 3) + b (1 + 1) – c (1 +2) = x3 + 2x2 – 3x +4

= a (4) + b (2) - c (3) = 4x3 – 3x2 + 4x - 5

= 4a + 2b – 3c Ans. = 8x3 – 5x2 +6x -0 Ans.

Q3. Subtract

i). 5a from 8a iii). – 2x from x + 1 iv). 2x + 1 from – x

Sol: Sol: Sol:

8a x + 1 - x

+ 5a -2x +2x + 1

3a 3x + 1 -3x - 1

vii). x – 2y – 3z from 2x + y + 3z

Sol

2x + y + 3z

+X - 2y - 3z (Change the signs)

X + 3y + 6z Ans.

Q5. What must be added to 3a3 – 3a2 + 3a - 1 to get a3 + 3a2 – 3a +1 ?

Sol:

Let, A = 3a3 – 3a2 + 3a – 1

B = a3 + 3a2 – 3a +1

X = ?

Now according to condition

A + X = B (to get x)

X = B – A (putting the values of B and A)

a3 + 3a2 – 3a + 1

+3a3 – 3a2 + 3a – 1 (change signs)

X= -2a3 + 6a2 – 6a + 2 Ans

Q6. What must be subtracted from a4 – 1 to get a4 – 4a3 + 6a2 – 4a + 1?

Sol

Let, A = a4 – 1

B = a4 – 4a3 + 6a2 – 4a + 1

X =?

A – X = B

x = A – B (putting the values of A and B to get x)

a4 – 1

+ a4 – 4a3 + 6a2 – 4a + 1 (change the signs)

+ 4a3 – 6a2 + 4a – 2 Ans

Q9. If A = x – y + z, B = 2x – 3y + 4z and C = 4x – 5y – 6z, find

i). A + B + C

Sol

A = x – y + z

B = 2x – 3y + 4z

C = 4x – 5y – 6z

A + B + C = 7x – 9y – z Ans.

ii). A – B + C

Sol

A = x – y + z

B = 2x – 3y + 4z

C = 4x – 5y – 6z

A - B + C (putting the values)

= (x – y + z) – (2x – 3y + 4z) + (4x – 5y – 6z)

= x – y + z – 2x + 3y – 4z + 4x – 5y – 6z

= x – 2x + 4x – y + 3y – 5y +z – 4z – 6z

= x (1 – 2 +4) – y (1 – 3 +5) + z (1 – 4 – 6)

= x (-1 +4) – y (- 2 + 5) + z (-3 – 6)

= x (3) –y (3) +z (-9)

= 3x – 3y – 9z Ans.

Q10. Simplify

i). a + 2b – 3c – 4a – b + 2c ii). (a2 + 2a + 1) – (b2 + 2a - 1)

Sol Sol

= a + 2b – 3c – 4a – b + 2c = (a2 + 2a + 1) – (b2 + 2a - 1)

= a – 4a + 2b – b – 3c + 2c =a2 + 2a + 1 – b2 - 2a + 1

= a (1 – 4) + b (2 – 1) –c (3 - 2) =a2 + 2a – 2a - b2 + 1 + 1

= a (-3) +b (1) –c (1) =a2 – b2 + 2 Ans.

= -3a + b – c Ans.

Exercise 8c

Q. Simplify

Q2. 8a – {4a – (3a + 5a)} Q4. 3 {a – 2 (b – a – b)}

Sol Sol

= 8a – {4a – (3a + 5a)} = 3 {a – 2 (b – a – b)}

= 8a – {4a – 8a} = 3 {a -2 (b – a + b)}

= 8a – {- 4a} = 3 {a – 2b + 2a – 2b}

= 8a + 4a = 3 {a + 2a – 2b – 2b}

= 12 a Ans = 3 {3a – 4b}

= + 9a – 12b Ans

Q8. 10a – [4 {5a – 3(a – 1)} – 3 (4a – 3a + 1)]

Sol

= 10a – [4 {5a – 3(a – 1)} – 3 (4a – 3a + 1)]

= 10 a – [4 {5a -3(a – 1)} – 3 (4a – 3a – 1)]

= 10 a – [4 {5a – 3a +3} - 12a + 9a + 3]

= 10a – [4 {2a + 3} – 12a + 9a +3]

= 10a – [8a + 12 – 12a + 9a + 3]

= 10a – [8a – 12a + 9a + 12 + 3]

= 10a – [-4a + 9a + 15]

= 10a – [5a + 15]

= 10a – 5a – 15

= 5a – 15 Ans.

Q9. 2(a2 – b2) – 3 [𝒂𝟐 – {b2- a2 + (a2 – b2 – a2)}]

Sol

= 2(a2 – b2) – 3 [a2 – {b-2 a2 + (a 2– b2 – a2)}]

= 2a2 - 2b2 – 3 [a2 – {b2- a2 +a2 – b2 – a2}]

= 2a2 – 2b2 – 3[a2 – b2 + a2 – a2 + b2 + a2]

= 2a2 – 2b2 – 3[a2 + a2]

= 2a2 – 2b2 – 3[2a2]

= 2a2 – 2b2 - 6a2

= 2a2 – 6a2 – 2b2

= -4a2 – 2b2

= - 2 ( 2a2 + b2) Ans.

Exercise 8d

Q1. If x = -3, y = 5 and z = - 2, evaluate the following

iii). x2 – yz / z2 v). 𝑥+𝑦

𝑧 +

𝑦+𝑧

𝑥

Sol: Sol:

= x2 – yz / z2 = 𝑥+𝑦

𝑧 +

𝑦+𝑧

𝑥

= (-3)2 – (5) (-2)/ (-2)2 = −3+5

−2 +

5+(−2)

−3

= 9 + 10 / 4 = 2

− 2 +

5−2

−3

= 19

4 =

2

− 2 +

3

−3

= 4 3

4 Ans. = – 1 – 1 => -2 Ans

viii). X3 + 3xyz – y3 + z3

Sol

= x3 + 3xyz – y3 + z3

= (-3)3 + 3(-3) (5) (-2) – (5)3 + (-2)3

= - 27 + 3 (30) – 125 + (-8)

= - 27 + 90 – 125 – 8

= 63 – 133

= - 70 Ans

Q2. If a = 4, b = 9 and c = 25, evaluate b2 – 4ac

Sol

b2 – 4ac

= (9)2 – 4 (4) (25)

= 81 – 4(100)

= 81 – 400

= - 319 Ans.

Q4. Express the following in the simplest form and then evaluate it when a= 7 and b = 6

4a2 – 2 [b + a {3 – a} + 3b2]

Sol

= 4a2 – 2 [b + a {3 – a} + 3b2]

= 4a2 – 2 [b + 3a – a2 + 3b2]

= 4a2– 2b - 6a + 2a2- 6b2

= 4a2 + 2a2 - 6a – 2b – 6b2

= 6a2 – 6a – 2b – 6b2

= 6a (a – 1) – 2b (1 + 3b)

Putting the values

= 6(7) (7 – 1 ) -2(6) {1 + 3(6)}

= 42(6) – 12 (1 + 18)

= 252 – 12(19)

= 252 – 228

= 24 Ans.

UNIT # 12 “ANGLES”

Exercise 12

Q2. Fill in the boxes with the appropriate symbol (>, =, <) between the following pairs of angles.

i) ii). iii)

m<ABC m<DEF m<XYZ m<RST m<PQR m<MNL

Q4. Guess the degree measures of the angles in the following cases.

i) East to south direction ii). East to west direction

O E W O E

S

Right angle Straight angle

Q6. Name the angles marked in each of the given figures and classify them according to their possible

measurement by inspection.

i) ii).

<DAB = 900 (Right angle) <CAB = 400 (Acute angle)

<ABC = 1100 (Obtuse angle) <ABC = 900 (Right angle)

<BCD = 700 (Acute angle) <BCA = 500 (Acute angle)

<CDA = 900 (Right angle)

Q8. Classify the angles whose sizes are given below

i). 300 ii). 10 iii). 900 iv). 1200

Acute angle Acute angle Right angle Obtuse angle

v). 1800 vi). 2700 vii). 890 viii). 3600

Straight angle Reflex angle Acute angle Complete angle

ix). 2100 x). 3000

Reflex angle Reflex angle

= > <

Q8. Calculate how many degrees there are in

i) Four right angles ii). 2/3 right angles

Sol Sol

Right angle = 900 Right angle = 900

Four right angles = 900 x 4 2/3 right angles = 900 x 2/3

= 3600 Ans = 30 x 2

= 600ans

iii). 1 ½ right angles

Sol

Right angle = 900

1 1/2 right angles = 900 x 1 ½

= 900 3/2

= 450 x 3

= 1350Ans

Q9. Classify the angles between the hands of the clock at

ii) 6 p.m. iv). 1 p.m. v). 9: 30 p.m.

1800 400 900

Straight angle Acute angle right angle

Q12. Write the sizes of the complements of the following

i). 700 ii). 230

Sol Sol

Complementary angles = 900 Complementary angles = 900

Complement of 700 = 900 - 700 Complement of 230 = 900 - 230

= 200 Ans. = 670 Ans.

iii). 590 iv). 450

Sol Sol



= 310 Ans = 450 Ans.

v). 00 vi). 900

Sol Sol



= 900 Ans = 00 Ans.

vii). 800 viii). 10 ½ 0

Sol Sol


Complement of 00 = 900 - 800 Complement of 900 = 900 – 10 ½ 0

= 100 Ans = 900 – 21/2

ix). 300 = 900 – 10.5

Sol = 79 .50

Complementary angles = 900

Complement of 300= 900 - 300

= 600

x). ½ 0

Sol

Complementary angles = 900

Complement of 1/20= 900 – 1/20

= 900 – 0.50

= 89.50

Q13. Write the sizes of the supplements of the following angles.

i). 900 ii). 800

Sol Sol

Supplementary angles = 1800 Supplementary angles = 1800

Supplement of 900= 1800 – 900 Supplement of 800 = 1800 - 800

= 900 Ans = 1000 Ans

iii). 00 iv). 1800

Sol Sol



= 1800 Ans = 00 Ans

v). 1620 vi). 850

Sol Sol



= 180 Ans = 950 Ans

vii). 80.50 viii). 1000

Sol: Sol


Supplement of 80.50= 1800 – 80.50 Supplement of 1000 = 1800 -1000

= 99.50 Ans = 800 Ans

Q14. Name each pair of adjacent angles in the following figures A D

i). C A ii).

C B

B

<COA , < AOB <AOC, < AOD , < DOB, < BOC

<AOD, < DOB , < BOC , < AOC

Q15. In the figure below a = 500, Find the sizes of the other angles

<a = 500

<b = 1300

<c = 500

<d = 1300

m< a and m < c = 500

As both are vertically opposite angles having same vertex.

m< a + m < c = 500 + 500

= 100o

Complete angle = 3600

m< b + m <d = 3600 - 1000

= 2600

As < b and < d are also vertically opposite angles having same vertex then

2600/ 2 = 1300

m <b = 1300 , m < d = 1300

MULTIPLE CHOICE QUESTIONS 12


i). Which of the following is not an acute angle

A. 300 B. 900 C. 780 D. 650

ii). Which of the following is an obtuse angle?

A. 1300 B. 2400 C. 1980 D. 1800

iii). Interior angle of a triangle add up to

A. 3000 B. 3600 C. 900 D. 1800

iv). Two right angles form a straight angle

A. True B. Not true C. True only when they are adjacent

to each other D. true when they overlap each other

v). Two obtuse angles are always supplementary.

A. True if they are adjacent to each other B. False

C. sometimes D. True

sUnit # 13 “Triangles”

Exercise# 13a

Q2. Can a triangle have three equal sides? If so, name the category it belongs to.

Ans. Yes, a triangle can have three equal sides. It is called equilateral triangle. In ABC, mAB = m BC

= m CA and m< A = m < B = m < C

ii). In PQR, m QR = m PR. What type of triangle is it?

Ans. PQR is an isosceles triangle.

m QR = m PR

iii). In ABC , m <BAC = 1050. Name the category of triangle it belongs to.

Ans. ABC is obtuse angles triangle. m< BAC = 1050

Q3. Classify the triangles shown below as scalene, isosceles or equilateral.

i). 2cm 2cm ii). iii). 3cm 3cm

2cm 2cm

3cm

Scalene 2.5cm 3cm

Isosceles Equilateral

Q6. Classify the following as acute angled, right angled or obtuse angled triangles.

I) Ii). iii). iv).

Right angle triangle Acute angled triangle Right angled triangle Obtuse angled

EXERCISE # 13b

Q1. State whether the following are true or false.

i). A right triangle can also be an isosceles triangle. (True)

ii). A scalene triangle is always an acute angled triangle. (False)

iii). An isosceles triangles can also be an obtuse angled triangle. (True)

iv). The exterior angles of an equilateral triangles are not all equal. (False)

Q5. Write the size of the angles of

i). An equilateral triangle ii). An isosceles right angled triangle

S B

U

T A B

Q6. The internal angles of a triangle are in the ratio 2: 3: 5. Find the size of its greatest angle.

What type of triangle is it?

Sol

Let the 3 angles of the triangles be 2x, 3x and 5x

The sum of the 3 angles must be 1800

2x + 3x + 5x = 1800

10x = 1800

X = 1800/ 10

X = 180

So, 2x = 2 × 18 = 360

3x = 3 × 18 = 540

5x = 5 × 18 = 900

Three angles are of size 360, 540 and 900 and triangle is right angled triangle.

Multiple Choice Questions 13


i). A scalene triangle can also be an equilateral triangle.

A. True B. not always true C. False D. none of the above

ii). A right isosceles triangle has one right angle and two angles 450v each.

A. True B. not always true C. False D. None of the above

iii). A triangle can be drawn with 2 obtuse angles

A. Sometimes B. False C. always true D. none of the above

iv).In a triangle, the sum of the lengths of any two sides is……………..

A. greater than the length of the third side

B. less than the length of the third side

C. equal to the length of the third side

D. none of the above

v). An exterior angle of a triangle is bigger than the sum of the two opposite interior angles

A. true B. In right angled triangles only C. in acute angled triangles only

D. False

unit: 2 natural numbers and whole numbers natural … · 2020. 9. 3. · 1. the set of natural...

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