GRADE 9 MATHEMATICS EXAM 6 JUNE 2016 PAPER 1 - PNHS

GRADE 9 MATHEMATICS 6 JUNE 2016 PAGE 8 OF 11 QUESTION 5 ( 6 marks ) 5.1 How much money would you have to invest at 10% simple interest p.a. to have a ...

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GRADE 9 MATHEMATICS · EXAM · 6 JUNE 2016 · PAPER 1 MEMO

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GRADE 9 MATHEMATICS

·Q UESTION 1

6 JUNE 2016

PAGE 2 OF 11

( 20 marks )

Answer the following questions on the answer sheet 1.1 Circle the rational numbers in the list of numbers below. √ √ √ 5 ; −16 ; 1, 2˙ ; π ; −2 ; 6 ; 38 ; Solution: 6

√ 3 8

;

;

√ 5

;



−16

1, 2˙

;

(2) 3 4

; π

;

1.2 Circle the prime numbers in the list of numbers below. 2

; 3

; 4

Solution: 2

; 5 ;

; 31 3

; 4

; 12

; 24

Solution: 8

;

3 4

(2)

; 49 ;

;

5

; 49

31

all correct

1.3 Circle the factors of 24 in the list of numbers below. 8

−2

(1)

; 48 ;

12

;

24

; 48

all correct

1.4 Evaluate the following without using a calculator. Show all working! 1.4.1 30 + 2−1

(2)

Solution: =1 =

+

1 2

3 1 or 1 2 2

1.4.2 ( 52 )5 × 45

(2)

Solution: 5 = ( × 4)5 2 20 5 =( ) 2 = 105 = 100000

1.4.3

42015 42016

(1)

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GRADE 9 MATHEMATICS

6 JUNE 2016

PAGE 3 OF 11

Solution: = = 1 = 4

42015−2016 4−1

1.5 Find the values of the variables in the following equations. 1.5.1 5 + a = 21

(1)

Solution: a = 16 1.5.2 6x = 18

(1)

Solution: x = 3 1.5.3 (y + 9)(y + 5) = 0 Solution: y = −9

(2) or y = −5

1.5.4 2b = 4b

(1)

Solution: b = 0 1.5.5 2x = 16

(1)

Solution: x = 4 1.6 How many terms are in the following expression?

(1)

x2 ÷ 3 + (2 + x)y − y 2 × 5x Solution: 3 1.7 Write down a common factor of x2 + x and x2 − 1

(1)

Solution: (x + 1) 1.8 Write down 2 factors of (x + 1)(x)(x − 4) Solution: (x + 1)or(x − 4) or x

one tick for each answer

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(2)

GRADE 9 MATHEMATICS

·Q UESTION 2

6 JUNE 2016

PAGE 4 OF 11

( 10 marks )

No calculator may be used in this question 2.1 Write 300 as a product of its prime factors.

Solution:

3 5 5 2 2

3·5·5·2·2

300 100 20 4 2 1

(2)

for correct working

written as product

2.2 What is the smallest number that 300 must be multiplied by to make a square number?

(1)

Solution: 3 2.3 Write 0, 495 as a fraction.

(2)

495 Solution: 0, 495 = 1000 99 only if working has been shown 200

2.4 Write 0, 2˙ 7˙ as a fraction.

(3)

Solution: x = 0, 2˙ 7˙ 100x = 27, 2˙ 7˙ 100x − x = 27, 2˙ 7˙ − 0, 2˙ 7˙ 99x = 27 27 x= 99 3 = 11

2.5 Evaluate the following expression, and express the final answer as a decimal. r 144 √ 2 + 3 −8) × ( 25 3

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(2)

GRADE 9 MATHEMATICS

6 JUNE 2016

PAGE 5 OF 11

Solution: 12 1 − 2) × 5 2 12 − 10 3 = × 5 4 6 = 20 3 = anything that shows they didn’t use a calculator 10 = 0.3 =(

·Q UESTION 3

( 12 marks )

3.1 Simplify the following, leaving the answers with positive exponents. 3.1.1 3a2 × 4a

(1)

Solution: = 12a3

3.1.2 x3 y −3 × x3 y

(2)

Solution: = x6 y −2 x6 = 2 y 3.1.3 (3x2 y 3 )3

(2)

Solution: = 27x6 y 9

3.1.4

applying power to at least one factor

correct

x4 y 2 z −3 x−1 y 4 z 2

Solution:

(4) x5 y2 z5

one for each power

for positive exponents

3.2 Write in scientific notation: 1, 03 × 10−2 + 13, 8 × 10−2

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(2)

GRADE 9 MATHEMATICS

6 JUNE 2016

PAGE 6 OF 11

Solution: = 14, 83 × 10−2 = 1, 483

× 10−1

3.3 Write in expanded form: 9, 34 × 10−3

(1)

Solution: 0, 00934

·Q UESTION 4

( 12 marks )

4.1 Consider a number pattern that has the rule Tn= 3n−1. 4.1.1 Calculate the value of the 4th term.

(2)

Solution: Tn = 3(4) − 1 = 11

4.1.2 Which term has a value of 68?

(2)

Solution: 68 = 3(n) − 1 n = 23

sub

ans

4.2 Inspect the following number pattern and answer the questions that follow. T1

T2

T3

4.2.1 Write down the number of circles in the next two terms in the sequence.

(2)

Solution: T4 = 9 T5 = 11 4.2.2 Find a formula for the sequence, in the form Tn = .....

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(2)

GRADE 9 MATHEMATICS

6 JUNE 2016

Solution: Tn = 2n

PAGE 7 OF 11

+1

4.3 Inspect the following number pattern and answer the questions that follow. T1

T2

T3

4.3.1 Write down the number of circles in the next two terms in the sequence.

(2)

Solution: T4 = 10 T5 = 15 4.3.2 Find a formula for the sequence, in the form Tn = ... Solution: Tn = 12 n2 + 12 n or 21 n(n + 1)

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(2)

GRADE 9 MATHEMATICS

·Q UESTION 5

6 JUNE 2016

PAGE 8 OF 11

( 6 marks )

5.1 How much money would you have to invest at 10% simple interest p.a. to have a final value of R5000 after 7 years?

(3)

Solution: A = P (1 + in) R5000 = P (1 + (0.1)(7)) sub R5000 P = rearrange 1.7 = R2941.18

5.2 Calculate the final amount if you invest R500 at 5% compound interest p.a. for 3 years.

(3)

Solution: A = P (1 + i)n = 500(1 + 0.05)3 = 578.8125

·Q UESTION 6

( 13 marks )

Factorize the following expressions completely: 6.1 4xy + 6x2 + 2x

(2)

Solution: = 2x(2y + 3x + 1)

2x

bracket

6.2 32x4 − 2x2

(3)

Solution: = 2x2 (16x2 − 1)

common factor

2

= 2x (4x + 1)(4x − 1)

a

6.3 x2 + 3x + 2

each bracket

(2)

Solution: = (x + 2)(x + 1)

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GRADE 9 MATHEMATICS

6 JUNE 2016

PAGE 9 OF 11

6.4 3(k − 1) − (1 − k)(3 − k)

(3)

Solution: = 3(k − 1) + (k − 1)(3 − k) = (k − 1)(3 + (3 − k))

switch

common factor

= (k − 1)(6 − k) 6.5 (a + 2)(3a − 1) − 2(a2 − 4)

(3)

Solution: = (a + 2)(3a − 1) − 2(a + 2)(a − 2) = (a + 2)((3a − 1) − 2(a − 2)) = (a + 2)(3a − 1 − 2a + 4) = (a + 2)(a + 3)

·Q UESTION 7

( 13 marks )

Simplify the following completely: 7.1 (2y − 3x)(y + 5x − 2)

(3)

Solution: = 2y 2 + 10xy − 4y − 3xy − 15x2 + 6x = 2y 2 − 4y + 7xy + 6x − 15x2

7.2

1 6 (3x

ca

+ 2) − 34 (2x − 1)

(4)

Solution: = =

7.3

(6x + 4 −12x + 13 12

) − (18x − 9) 12

14p+21 7p

(2)

Solution: 7(2p + 3) 7p 2p + 3 = p =

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GRADE 9 MATHEMATICS 7.4

y 2 −x2 2

×

6 JUNE 2016

6 3x+3y

PAGE 10 OF 11 (4)

Solution: (y 2 − x2 )6 2(3x + 3y) 6(y + x)(y − x) DOTS = 6(x + y) com. fac. = (y − x) =

·Q UESTION 8

( 14 marks )

Solve for x in the following equations: 8.1 3x + 2 = x − 8

(2)

Solution: 3x + 2 − 2 = x − 8 − 2 3x = x − 10 3x − x = x − x − 10 2x = −10 x = −5

8.2 x2 − 2x = 8

(3)

Solution: x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0

x = 4 or x = −2 8.3

x+3 4



x+2 8

=

x 2

for both brackets

ca for both

−1

(5)

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GRADE 9 MATHEMATICS

6 JUNE 2016

PAGE 11 OF 11

Solution: 2(x + 3)

− (x + 2)

=

−8

4x

8 8 2(x + 3) − (x + 2) = 4x − 8

multiplying through by common factor

x + 4 = 4x − 8 3x = 12 x=4

x+2

2

8.4 x2 −2x = x−2 Solution:

(4)

2 x+2 = x(x − 2) (x − 2) x+2 2(x) = x(x − 2) x(x − 2) x + 2 = 2x mult. through with com.denom. x=2

ca

END OF EXAM