credit maths

Credit Maths

Revision Chapters 1 - 4

5.02 104 =

50 200

47 000 000 = 4.7 107

3.1 10-4 =

0.00031

0.0000704 = 7.04 10-5

Standard Form/Scientific Notation

Large Numbers

a. Change to standard form

b. Change to normal form

i. 99000 = 9.9 104

i. 6.03 103 = 6.0360.3603.6030.

ii. 610400000 = 6.104 108

iii. 5223000 = 5.223 106

ii. 8.15 105 = 8.1581.5815.815000.8150.81500.iii. 2.7 106 = 2.7270.2700.2700000.27000.270000.27.

Small Numbers

a. Change to standard form

b. Change to normal form

i. 0.0039 = 3.9 10-3

i. 3.07 10-3 = 3.070.3070.03070.00307

ii. 0.000502 = 5.02 10- 4

iii. 0.0000004 = 4 10-7

ii. 5.91 10-5 = 5.910.5910.05910.00005910.005910.000591iii. 8.6 10-6 = 8.60.0860.00860.00000860.000860.0000860.86

5.07 104 =

50 700

457 000 000 =4.57 108

3.1 107 = 31 000 000

9 070 000 = 9.07 106

E.g. Fractions Non calc

7

31

10

12. a

9

22

3

13. b

4

32

3

15. c

7

10

10

21

3

11

1

31

3

9

20

3

10

20

9

3

10

1

21

3

2

11

2

3

4

3

3

13

12

9

12

43

12

9

12

162

12

52

Ex Fractions Non calc

5

22

6

14. a

15

22

3

22. b

4

14

5

43. c

5

12

6

25

5

11

2

101

10

15

32

3

8

32

15

3

8

1

41

5

4

11

4

5

4

1

5

47

20

5

20

167

20

217

20

18

Similar Shapes

5cm

x cm

2cm 8cm

15cm

5 cm

y cm

6cm

6cm

Small

BigEnl s.f. = 4

2

8

x = 4 × 5 = 20cm

big

SmallRed s.f. = ..333.0

15

5

y = 0.333. × 6 = 2cm

Ex 1 Find x

20m

40 m

x m 22 m

big

smallRed s.f. = 5.0

40

20

x = 0.5 × 22 = 11m

Ex 2 Find x

big

smallRed s.f. = 75.0

8

6

x = 0.75 × 20 = 15cm

20cm x cm

6 cm8cm

Find x to 1 d.p.

20 m

2 m

x m

12m

Find x to 1 d.p.

22 m 20 m

x m

12m

small

bigEnl s.f. = 1.1

20

22

x = 1.1 × 12 = 13.2m

Similar Areas and Volumes

8

5.

big

smallfsred

64

25

8

5.

2

fsArea

e.g.1 e.g.2

8m5m

21m2?m2

12m9m

?m3

60m3

22.82164

25? m

9

12.

small

bigfsEnl

27

102

9

12.

3

fsVol

32.1429

214260

27

102? m

Enlargement

Linear Scale Factor = k =

Area Scale factor = k2 =

Missing Length = 2.4 x 10 = 24 cm

4.276.5

? cm

10 cm

25 cm2 144cm2

76.525

144

Find the missing dimension

Enlargement

Linear Scale Factor = k =

Area Scale factor = k2 = 32 = 9

Missing area = 9 x 30 = 270 cm2

35

15

5cm30cm2

? cm2

15cm

1.

Enlargement

Linear Scale Factor = k =

Area Scale factor = k2 =

Missing area = 4 x 80 = 320 mm2

215

30

15 mm

30mm

?mm2

2.

80 mm2

22 = 4

Reduction

Linear Scale Factor = k =

Area Scale factor = k2 =

Missing area =

3

1

18

6

3.

? mm2

540 mm2

6 mm

18 mm

9

1

3

12

2mm605409

1

Reduction

Linear Scale Factor = k =

Area Scale factor = k2 =

Missing Length = 0.3125 x 7.5 = 2.34 cm

3125.009765625.0

09765625.012.5

5.0

4.

0·5 m2

5.12 m2

? m

7·5 m

Enlargement

Linear Scale Factor = k =

Volume Scale factor = k3 =

Missing Volume = 20 x 4.096 = 81.92cm3

6.15

8

5cm 8cm

V= 20cm3 V= x cm3

5.

096.46.15

8 33

Distance Speed Time

D

S T

T

DS

S

DT TSD

STARTER QUESTIONS

S

DT

sm

m

sm

kmT

/5

2000

/5

2

400T secs

T

DS

hrs

km

hrs

kmS

25.4

272

smin154

272

64S km/h

a. D = 2km and S = 5m/s T= ?

b. D = 272km and T = 4 hrs 15 mins S = ?

c. T = 3mins and S = 4m/s D = ?

STD

smin3/4 smD

ssmD 180/4

720mD

1. A bus leaves the airport at 19.24 and arrives in the city centre at 20.11. If the journey is 17 miles calculate its average speed in mph to 1 decimal place

T

DS

47mins

17m hrs6047

17m

0.783

17m

21.7mphS

2. Stewart drove 40 km from Laurencekirk to the golf course in Montrose. If he averaged 72km/h calculate the time of his journey to the nearest minute

S

DT

hkm

kmT

/72

40 hrsT 55.0 33minsmins 33.3

3. Mr Sim drove to Dover in the holidays. He left at 06.30 and arrived at 14.15. If he averaged 70mph, how far did he travel?

TSD mins 45 7hrs mph 70 D

hrs 7.75 mph 70 D542.5D miles

Positive and Negative Numbers

With Algebra

1. Basic Rules: Pos and Neg = NegNeg and Neg = Pos

Examples

a. 2 + (5) ( 4) b. 4 × (2) c. 24 ÷ (6)

= 2 5 + 4

= 1

= 8 = 4

Exercise

a. 3 (1) + ( 7) b. 42 ÷ (7) c. 6 +(15) ÷ 3

2. Simplify: Pos and Neg = NegNeg and Neg = Pos

Examples

a. 2a + (5b) ( 4a) b. 2a2 +(3a)(4a2)(5a)

= 2a 5b + 4a

= 6a 5b

= 2a2 3a + 4a2 + 5a

Exercise

a. 3x + (y) (5x) b. 5m2 (3n2)(4m2)

= 6a2 + 2a

3. Simplify: Pos and Neg = NegNeg and Neg = PosMultiplying makes powersExamples

a. 2a × 3a × a b. 3c × (c) × (2c) × 4c × (c)

= 6 a3 = 24

Exercise

a. 2y × 3y × y × y × y b. 2m × (5m) × (7m) × 2

c5

4. Multiply out: Pos and Neg = NegNeg and Neg = Pos

Examples

a. –3(x + 4) b. – 4(2y – 3) c. –p(3 – p)

= –3x – 12 = –8y + 12

Exercise

a. –5(w + 2) b. 3(2m 5) c. 3p(p 4)

= –3p + p2

5. Simplify: Pos and Neg = NegNeg and Neg = Pos

Examples

a. –3(x + 2) + 8 b. 2(y + 2) + 4(2y – 3) c. 2p(4p – 3) – 3p(p – 3)

= –3x – 6 + 8 = 2y + 4 + 8y – 12

Exercise

= 8p2 – 6p – 3p2 + 9p

a. 2(3c – 5) + 4(2 + c) b. 3(p – 4) – 2(p 5)

= –3x + 2 = 10y – 8 = 5p2 + 3p

6. Evaluate: a = 3, b = -2, c = -5

ExamplesA) a2 – b B) abc + 2c C) b(2a – c)

= 32 – (–2) = 3 × (–2) × (–5) + (–10)

Exercise

A) 4a + c B) c2 – b C) ab – 3b

= –2(6 – (–5))= 9 + 2 = 30 – 10 = –2(6 + 5)= 11 = 20 = –2(11)

= –22

7. Solve

Examples

a. 5x + 2 = x + 26 b. 2(3y – 1) = 2y + 10 c. 3m – 4 = –2(m – 4)

5x x = 26 – 2 6y – 2 = 2y + 10

Exercise

a. 7w + 2 = w + 14 b. 3(2m – 1) = m + 16

3m – 4 = –2m + 8

c. 2(3c – 5) = 4(2 + c) d. 3(p – 4) = –2(p 5)

4x = 24 6y – 2y = 10 + 2 3m + 2m = 8 + 4 x = 6 4y = 12

y = 3 5m = 12 m = 12/5