Year 10 - Algebra Expressions Test With ANS

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EVERYTHINGMATHS

End of chapter exercises

Problem 1:

If is an integer, is an integer and is irrational, which of the following are rational

numbers?

Problem 2:

Write each decimal as a simplefraction.

0,12

0,006

1,59

Problem 3:

Show that the decimal is a rational number.

Problem 4:

a b c

b

a

c c

a

c

1

c

12,277

3,2118
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Express as a fraction where (show all working).

Problem 5:

Write the following rational numbers to 2 decimal places.

1

Problem 6:

Round off the following irrational numbers to 3 decimal places.

Problem 7:

Use your calculator and write the following irrational numbers to 3 decimal places.

Problem 8:

Use your calculator (where necessary) and write the following numbers to 5 decimal places.

State whether the numbers are irrational or rational.

0,78 a

b a, b Z

1

2

0,111111

0,999991

3,141592654...

1,618033989...

1,41421356...

2,71828182845904523536...

2

3

5

6

8


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Problem 9:

Write the following irrational numbers to 3 decimal places and then write each one as a

rational number to get an approximation to the irrational number.

3,141 592 654...

1,618 033 989...

1,414 213 56...

2,718 281 828 459 045 235 36...

Problem 10:

Determine between which two consecutive integers the following irrational numbers lie,without using a calculator.

768

0,49

0,0016

0,25

36

1960

0,0036

8 0,04

5 80

5

10

20

30

5

3


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Problem 11:

Find two consecutive integers such that lies between them.

Problem 12:

Find two consecutive integers such that lies between them.

Problem 13:

Factorise:

10

3

20

3

30

3

7

15

9a2

36m2

9 81b2

16 25b6 a2

m2 19

5 5a2b6

16b 81ba4

10a + 25a2

16 + 56b + 49b2

2 12ab + 18a2 b2

4 144 + 48b2 b8 b5

(16 )x4

14x + 7xy 14y7x2

7y 30y2

1 x +x2 x3


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Problem 14:

Simplify the following:

Problem 15:

3(1 ) +p + 1p2

x + yx3 y3

2x + 1 x2 y4

4b( 1) + x(1 )x3 x3

3 p3 19

8 125x6 y9

8(2 +p)3 (p + 1)3

a(a + 4)(a 2)2

(5a 4b)(25 + 20ab + 16 )a2 b2

(2m 3)(4 + 9)(2m + 3)m2

(a + 2b c)(a + 2b + c)

p2 q2

p

p + q

pqp2

+ 2

x

x

2

2x

3

1

a + 7

a + 7

49a2

+ 16x + 2

2x3

1 2a

4 1a2a 1

2 3a + 1a21

1 a

+ 2xx2

+ x + 6x2

+ 2x + 1x2

+ 3x + 2x2


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Show that can be simplified to .

Problem 16:

What must be added to to make it equal to ?

Problem 17:

Evaluate if without using a calculator. Show your work.

Problem 18:

With what expression must be multiplied to get a product of ?

Problem 19:

With what expression must be divided to get a quotient of ?

Practise more questions like this

Answer 1:

(a) rational

(b) irrational

(c) irrational

(d) irrational

Answer 2:

1)

2)

(2x 1)2 (x 3)2 (x +2)(3x 4)

x + 4x2 (x + 2)2

+ 1x3

x + 1x2 x= 7,85

(a 2b) 8a3 b3

27 + 1x3 3x + 1

+ = = =110

2100

12100

610

35

=61000

3500
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3)

4)

Answer 3:

This is a rational number since both the numerator and the denominator are integers.

You can also see that the number is rational because it is a recurring decimal number.

Answer 4:

Answer 5:

1 + + = 1510

9100

59100

10x= 122,77777

100x= 1227, 777777777777

90x= 1105

=110590

22118

x= 3,211.8.

x= 3,2118181818...

100x= 321,18181818...

10000x= 32118,181818...

9900x= 31797

x= 317979900

x= 0, 788888.. .

10x= 7, 88888.. .

100x= 78, 88888.. .

90x= 71

x= 7190


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a) To write to two decimal places we must convert to the decimal form.

b) To write to two decimal places just add a comma and two 0's: 1,00

c) We mark where the cut-off point is, determine if it has to be rounded up or not and then

write the answer. In this case there is a 1 after the cut-off point so we do not round up. The

final answer is:

d) Repeat the steps in c) but this time, round up. The answer is:

Answer 6:

We mark where the cut off point is, determine if it has to be rounded up or not and then

write the answer.

a) 3,142 (round up as there is a 5 after the cut off point)

b) 1,618 (no rounding as there is a 0 after the cut off point)

c) 1,414 (no rounding as there is a 2 after the cut off point)

d) 2,718 (round up as there is a 2 after the cut off point)

Answer 7:

a)

b)

c)

d)

= 0,5012

0,11111 0,111

0,99999 1,001

1,414

1,732

2,236

2,449


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Answer 8:

Irrational Irrational Rational Rational Rational

Rational Irrational Rational Rational

Irrational

Answer 9:

a)

b)

c)

d)

Answer 10:

2 and 3 3 and 4 4 and 5 5 and 6 1 and 2 2 and 3 3 and 4 4 and 5

Answer 11:

2; 3

Answer 12:

3; 4

Answer 13:

(a)

2, 82843 27, 71281 0, 70000 0, 040000, 500000 6, 00000 44, 27189 0, 06000 8(0,2) = 4, 00000 44, 72136

3, 142 = 3 = 31421000

71500

1, 618 = 1 = 16181000

309500

1, 414 = 1 = 14141000

207500

2, 718 = 2 =7181000

359500

9 = (a 3)(a+ 3)a2


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(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

Answer 14:

36 = (m+ 6)(m 6)m2

9 81 = (3b 9)(3b+ 9)b2

16 25 = (4b+ 5a)(4b 5a)b2 a2

( )=(m+ )(m )m2 19

13

13

5 5 = 5(1 )= 5(1 a )(1 +a )a2b6 a2b6 b3 b3

16b 81b= b(16 81)= b(4 + 9)(4 9)= b(4 + 9)(2a+ 3)(2a 3)a4 a4 a2 a2 a2

10a+ 25 = (a 5)(a 5)a2

16 + 56b+ 49 = (4b+ 7)(4b+ 7)b2

2 12ab+ 18 = (2a 6b)(a 3b)a2 b2

4 144 + 48 = 4 (36 12 + 1)= 4 (6 1)(6 1)b4 b8 b5 b2 b6 b3 b2 b3 b3


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a)

b)

c)

d)

e)

Answer 15:

Answer 16:

1.

2.

(4 )(4 + )= (2 x)(2 +x)(4 + )x2 x2 x2

(7 14x)+ (7xy 14y) = 7x(x 2) + 7y(x 2)x2

= (x 2)(7x+ 7y)

= 7(x 2)(x+y)

(y 10)(y+ 3)

(1 x) (1 x) = (1 x)(1 )= (1 x)(1 x)(1 +x)x2 x2

3(1 p)(1 +p) + (p+ 1) = (p+ 1)[3(1 p) + 1] = (p 1)(3p 2)

x(1 )+y(1 )= x(1 x)(1 +x) +y(1 y)(1 +y)x2 y2

x(x 2) +(1 )(1 + )= x(x 2) + (1 +y)(1 y)(1 + )y2 y2 y2

( 1)(4b x) = (x 1)( +x+ 1)(4b x)x3 x2 3(p )( + + )13

p2 p

319

8 125 =(2 5 )(4 + 10 + 25 )x6 y9 x2 y3 x4 x2y3 y6

8 = [(p+ 2) 2(p+ 1)][ + 2(p+ 2)(p+ 1) + 4 ](2 +p)3 (p+ 1)3 (p+ 2)2 (p+ 1)2

= [p+ 2 2p 2][ + 4p+ 4 + 2 + 6p+ 4 + 4 + 8p+ 4]= (p)(12 + 18p+p2 p2 p2


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Answer 17:

a)

b)

c)

d)

Answer 18:

e)

f) Write all the terms on a common denominator of and simplify:

Answer 19:

Answer 20:

4a+ 4 4a= 8a+ 4a2 a2

125 64a3 b3

(4 9)(4 + 9)= 16 81m2 m2 m4

= + 4ab+ 4 (a+ 2b)2 c2 a2 b2 c2

= = 2pq+(pq)(p+q)

p

p(pq)p+q (p q)

2 p2 q2

6x

= = 12+3 4x2 x2

6x12x2

6x2x

x6

= = 01a+7

a+7

(a+7)(a7)

(a7)(a+7)

(a+7)(a7) =

(x+2)+16(2 )x3

2x332 +x+2x3

2x3

+ = =1

2a

(2a1)(2a+1)a+4

(2a1)(a1) 1a1(a1)(a+4)+(2a1)

(2a1)(a1) 4(2a1)(a1)

=x(x+2)

(x+2)(x+3)

(x+1)(x+1)

(x+2)(x+1)x(x+1)

(x+2)(x+3)

(2x 1)2 (x 3)2

= (2x 1)(2x 1) (x 3)(x 3)=


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Answer 21:

Suppose must be added to to make it equal to .

Therefore must be added to to make it equal to .

Answer 22:

We can factorise as . Therefore

Hence at the expression evaluates to .

Answer 23:

So the expression is:

Answer 24:

4 2x 2x+ 1 ( 3x 3x+ 9)x2 x2= (4 1) + (4 + 6)x+ (1 9)x2

= 3 + 2x 8x2

= (3x 4)(x+ 2)

A x+ 4x2 (x+ 2)2

( x+ 4)+Ax2 = (x+ 2)2

A = (x+ 2)(x+ 2) ( x+ 4)x2

= + 2x+ 2x+ 4 +x 4x2 x2= (1 1) + (2 + 2 + 1)x+ (4 4)x2

= 5x

5x x+ 4x2 (x+ 2)2

+ 1x3 (x+ 1)( x+ 1)x2

+1x3

x+1x2 = (x+1)( x+1)x2

x+1x2

=x+ 1

x= 7,85 x+ 1 = 8, 85

(a 2b)( + 2ab+ 4 )= 8a2 b2 a3 b3

+ 2ab+ 4a2 b2


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We can factorise the cubic as . Therefore, to get a

quotient of we must divide by .

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27 + 1x3 (3x+ 1)(9 3x+ 1)x2

3x+ 1 27 + 1x3 9 3x+ 1x2
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Year 10 - Algebra Expressions Test With ANS