Chapter 4 Exercise 4 - Biology Leaving Cert

Chapter 4 Exercise 4.1 Q. 1.

(i) Pattern repeats every 5th tile 12 ÷ 5 = 2 R 2 ∴ The 12th tile is a square (ii) 35 ÷ 5 = 7 R 0 ∴ The 35th tile is a circle (iii) 105 ÷ 5 = 21 R 0 ∴ The 105th tile is a circle (iv) 206 ÷ 5 = 41 R 1 ∴ The 206th tile is a triangle

Q. 2.

(i) Pattern repeats every 6th marble 22 ÷ 6 = 3 R 4 ∴ The 22nd marble is green (ii) 61 ÷ 6 = 10 R 1 ∴ The 61st marble is red (iii) 27 ÷ 6 = 4 R 3 Full pattern repeats 4 times giving 4 × 2 = 8 red marbles plus 2 more from the next 3 marbles. ∴ A red marble appears 10 times

Q. 3.

(i) Adding 3 each time ∴ Pattern 4 needs 13 matchsticks (ii) Adding 3 more pattern 5 needs 16 matchsticks (iii) ____________ Pattern 4

______________

Pattern 5

(iv) Total number of matchsticks needed to make the first 7 patterns = 4 + 7 + 10 + 13 + 16 + 19 + 22 = 91 matchsticks Q. 4.

(i) 2, 6, 10, 14 … (a) T1 (start term) = 2 (b) 2

6 +4

10 +4

14

The common difference is +4

+4

(c) T5 (5th term): 14 + 4 = 18 T6 (6th term): 18 + 4 = 22 T7 (7th term): 22 + 4 = 26 The next 3 terms are 18, 22, 26 (d) Start with 2 and then add 4 to every term. Active Maths 2 (Strands 1–5): Ch 4 Solutions

1

(ii) 7, 19, 31, 43 … (a) T1 = 7 (b) 7

19 +12

31

43

+12


+12

(c) T5: 43 + 12 = 55 T6: 55 + 12 = 67 T7: 67 + 12 = 79 The next 3 terms are 55, 67, 79 (d) Start with 7 and then add 12 to every term. (iii) 9, 24, 39, 54 … (a) T1 = 9 (b) 9

24 +15

39

54

+15


+15

(c) T5: 54 + 15 = 69 T6: 69 + 15 = 84 T7: 84 + 15 = 99

The next 3 terms are 69, 84, 99

(d) Start with 9 and then add 15 to every term. (iv) 3, −5, −13, −21 … (a) T1 = 3 (b) 3

–5 –8

–13 –8

–21

The common difference is −8

–8

(c) T5: −21 −8 = −29 T6: −29 − 8 = −37 T7: −37 − 8 = −45 The next 3 terms are −29, −37, −45 (d) Start with 3 and then subtract 8 every term. (v) 1, −10, −21, −32 … (a) T1 = 1 (b) 1

–10 –11

–21 –11

–32


–11

(c) T5: −32 −11 = −43 T6: −43 −11 = −54 T7: −54 −11 = −65 The next 3 terms are −43, −54, −65 (d) Start with 1 and then subtract 11 every term.

2

Active Maths 2 (Strands 1–5): Ch 4 Solutions

Q. 5.

(i) 6, 9, 12, 15 … (a) T1 = 6 (b) 6

9 +3

12 +3

15


+3

(c) T1: 3 × 1 + 3 = 6 T2: 3 × 2 + 3 = 9 ∴ Tn: 3 × n + 3 The nth term (general term): Tn = 3n + 3 (d) T9: 3 × 9 + 3 = 30 The 9th term is 30 (ii) 9, 14, 19, 24 … (a) T1 = 9 (b) 9

14 +5

19 +5

24


+5

(c) T1: 5 × 1 + 4 = 9 T2: 5 × 2 + 4 = 14 ∴ Tn: 5 × n + 4

The nth term Tn = 5n + 4

(d) T13: 5 × 13 + 4 = 69 The 13th term is 69 (iii) 0, −3, −6, −9 … (a) T1 = 0 (b) 0

–3 –3

–6 –3

–9


–3

(c) T1: −3 × 1 + 3 = 0 T2: −3 × 2 + 3 = −3 ∴ Tn: −3 × n + 3

Tn = −3n + 3

(d) T19: −3 × 19 + 3 = −54 The 19th term is −54 (iv) −10, −7, −4, −1 … (a) T1 = −10 (b) –10

–7 +3

–4 +3

–1


+3

(c) T1: 3 × 1 − 13 = −10 T2: 3 × 2 − 13 = −7 ∴ Tn: 3 × n − 13

Tn = 3n − 13

(d) T52: 3 × 52 − 13 = 143 The 52nd term is 143 Active Maths 2 (Strands 1–5): Ch 4 Solutions

3

(v) 1000, 748, 496, 244 … (a) T1 = 1000 (b) 1,000 –252

748

496

–252

244


–252

(c) T1: −252 × 1 + 1252 = 1,000 T2: −252 × 2 + 1252 = 748 ∴ Tn: −252 × n + 1252 Tn = −252n + 1252 (d) T100: −252 × 100 + 1252 = −23,948 The 100th term is −23,948 Q. 6.

(i) Tn = 3n

(iv) Tn = 4 − n

(a) T1: 3 × 1 = 3

(a) T1: 4 − 1 = 3

(b) T3: 3 × 3 = 9

(b) T3: 4 − 3 = 1

(c) T45: 3 × 45 = 135

(c) T45: 4 − 45 = −41

(ii) Tn = 2n + 5

(v) Tn = 9n − 1

(a) T1: 2 × 1 + 5 = 7

(a) T1: 9 × 1 − 1 = 8

(b) T3: 2 × 3 + 5 = 11

(b) T3: 9 × 3 − 1 = 26

(c) T45: 2 × 45 + 5 = 95

(c) T45: 9 × 45 − 1 = 404

(iii) Tn = n − 3

Q. 7.

(vi) Tn = 3 − 2n

(a) T1: 1 − 3 = −2

(a) T1: 3 − 2 × 1 = 1

(b) T3: 3 − 3 = 0

(b) T3: 3 − 2 × 3 = −3

(c) T45: 45 − 3 = 42

(c) T45: 3 − 2 × 45 = −87

(i)

Student A

Student B

Student C

Start amount

€10

€0

€90

End amount

€130

€600

€180

Amount saved over 6 wks

€120

€600

€90

Amount saved per wk

€20

€100

€15

(ii) Using the general term to find the formula for the amount each student saves. Let m = amount of money saved t = time in weeks Student A Common difference (amount saved per week) = (130 − 10) ÷ 6 = 20 Starting amount at zero weeks = 10 ∴ m = 20t + 10 Student B Common difference = (600 − 0) ÷ 6 = 100 Starting amount at zero weeks = 0 ∴ m = 100t + 0 m = 100t

4


Student C Common difference = (180 − 90) ÷ 6 = 15 Starting amount at zero weeks = 90 ∴ m = 15t + 90 (iii) After 12 weeks Student A

Student B

Student C

m = 20t + 10

m = 100t

m = 15t + 90

if t = 12, m = 20 × 12 + 10

if t = 12, m = 100 × 12

if t = 12, m = 15 × 12 + 90

m = 250

m = 1,200

Student A has €250

Student B has €1,200

m = 270 Student C has €270

(iv) The graph for Student B is directly proportional, as the line goes through the origin. (v) Constant of proportionality = 100 In the context of the question, the constant of proportionality means that for every week that goes by, the amount saved increases by €100. Q. 8.

(i) Pattern 1: 7 matchsticks Pattern 2: 12 matchsticks Pattern 3: 17 matchsticks 7

12 +5

22

17 +5

+5

27 +5

32 +5

Adding 5 each time pattern 6 needs 32 matchsticks. (ii) T1: 5 × 1 + 2 = 7 ∴ Tn: 5 × n + 2 = 5n + 2 Let m = number of matchsticks n = pattern number ∴ m = 5n + 2 (iii) The first pattern will take 5 + 2 i.e. 7 matches and thereafter there are an additional 5 matches for each new pattern. (iv) Pattern 25 i.e. n = 25 ∴ m = 5 × 25 + 2 = 127 Pattern 25 needs 127 matches (v) 182 matchsticks used i.e. m = 182 ∴ 182 = 5n + 2 180 = 5n 36 = n Pattern 36 requires 182 matchsticks


5

Q. 9.

(i) Adding 3 extra marbles each time ∴ The 5th pattern needs 18 marbles. (ii) T1 = 6 Common difference = 3 T1: 3 × 1 + 3 = 6 ∴ Tn: 3 × n + 3 = 3n + 3 The nth pattern needs 3n + 3 marbles. (iii) The number of marbles needed for any pattern in the sequence is 3 times the pattern number plus an extra 3 marbles (iv) 93 marbles ⇒ 3n + 3 = 93 3n = 90 n = 30 The 30th pattern requires 93 marbles.

Q. 10.

(i) 4, 7, 10 … Adding 3 each time ∴ The 7th pattern needs 22 coins. (ii)

T1: 3 × 1 + 1 = 4 T2: 3 × 2 + 1 = 7 ∴ Tn: 3 × n + 1 = 3n + 1 The nth pattern needs 3n + 1 coins.

(iii) 4 + 7 + 10 + 13 + 16 = 50 ∴ The student will have made 5 complete patterns. Q. 11.

(i)

Time (mins) 0 2 4 6 8 10 12 14 16

Cost (€) Company A 0.3 0.37 0.44 0.51 0.58 0.65 0.72 0.79 0.86

Cost (€) Company B 0.38 0.43 0.48 0.53 0.58 0.63 0.68 0.73 0.78

(ii) Company A: cost for 2 minutes is €0.44 − €0.37 = €0.07 ∴ Cost for 1 minute = €0.035 (iii) Company B: cost for 2 minutes is €0.48 − €0.43 = €0.05 ∴ Cost for 1 minute = €0.025 (iv) Company B has a higher connection fee of 38 cents (€0.38), which is higher than company B by 8 cents (€0.08). (v) Company A costs an initial 30 cents (€0.30) to make a phone call plus an additional 3.5 cents (€0.035) per minute. Company B costs an initial 38 cents (€0.38) to make a phone call plus an additional 2.5 cents (€0.025) per minute.

6


(vi) c = Cost (€), m = number of minutes Company A: c = 0.035 m + 0.3 Company B: c = 0.025 m + 0.38 (vii) From the graph an 8-minute call costs the same from both companies (i.e. €0.58). (viii) cA = 0.035 m + 0.3 cB = 0.025 m + 0.38 Solve to find when cA = cB i.e. when the cost of calls from companies A and B are the same: 0.035 m + 0.3 = 0.025 m + 0.38 0.01 m = 0.08 0.08 m = ____ 0.01 m = 8 minutes (ix) Company A: c = 0.035 m + 0.3 if m = 23 minutes then c = 0.035 × 23 + 0.3 ∴ c = 1.105 Company A charges €1.105 for a 23-minute call. Company B: c = 0.025 m + 0.38 if m = 23 minutes then c = 0.025 × 23 + 0.38 c = 0.955 Company B changes €0.955 for a 23-minute call. Q. 12.

(i) Company A: from the graph, 10 km costs €50. ∴ 1 km costs €5 ∴ Company A charges €5/km. Company B: from the graph, 50 km costs €150 [i.e. €200 − €50 = €150] ∴ 1 km costs €3 ∴ Company B charges €3/km (ii) Company B has a standing charge of €50. (iii) The trip would cost €5 per km with company A. The trip would cost €50 plus €3 per km with company B. (iv) c = cost (€), d = kilometres travelled Company A

Company B

common difference = 5

common difference = 3

standing charge = 0

standing charge = 50

∴ c = 5d + 0

∴ c = 3d + 50

c = 5d

(v) From the graph at about 25 km both companies charge the same price.


7

(vi) When both companies charge the same price 5d = 3d + 50 2d = 50 ∴ d = 25. Distance equals 25 km as required. (vii) Budget of €500 Company A

Company B

c = 5d

c = 3d + 50

∴ 500 = 5d

500 = 3d + 50

100 = d

450 = 3d 150 = d

The teacher will use Company B, as this gives the greatest distance. The total distance covered on the tour is 150 km. (viii) Company A, because there is no standing charge; charges are only incurred per kilometre travelled. Q. 13.

(i) Time (hours) 0 1 2 3 4 5 6 7 8

(ii)

(iv) The formula means that you start with 50 (litres) of water and subtract 2.5 (litres) every hour. The subtraction describes the leak from the water storage tank.

Capacity (litres) 50 47.5 45 42.5 40 37.5 35 32.5 30

(v) After 6 hours V = −2.5t + 50 ∴ V = −2.5 × 6 + 50 V = 35 litres (vi) Time for tank to empty if V = 0 then −2.5t + 50 = 0

Capacity (litres)

50 45

50 = 2.5t

40

20 = t

35

It takes 20 hours for the tank to empty.

30 25 20

Q. 14.

15

(i) y

10 25

5 1

20 2

3 4 5 6 Time (hours)

7

8

(iii) V = Volume of water left in tank

15 10

t = time (hours)

5

Common difference = −2.5

0

V = −2.5t + 50

8

No. of cars

0


x 0

1

2

3 4 5 Tonne (metal)

6

7

8

(ii) 8 tonnes = 20 cars

(iv) m = 2.5c

1 tonne = 2.5 cars

(v) 2.5c = m m c = ___ or 0.4 m 2.5 ∴ c = 0.4 m

Slope of line = 2.5 ∴ Constant of proportionality = 2.5

(vi) c = 0.4(525) = 210 cars

(iii) For every tonne of metal, 2.5 cars are produced.

(i)

Time (hours)

Electrician A

Electrician B

300

0 (call-out)

50

0

250

1

90

60

2

130

120

3

170

180

4

210

240

5

250

300

Electrician charges (€)

Q. 15.

(vii) m = 2.5(125) = 312.5 tonnes of metal

Electrician B

200 Electrician A 150 100 50

0

1

2 3 Time (hours)

4

5

(ii) Graph B is directly proportional, as it passes through the origin and the equation of this line would be in the form y = mx, where m the slope is 60. (iii) c = cost (€), t = time worked (hours) Electrician A

Electrician B

common difference = +40


call-out charge = 50

call-out charge = 0

∴ c = 40t + 50

∴ c = 60t + 0 c = 60t

Q. 16.

(i)

Week

Carla (Amount saved €)

Alvaro (Amount saved €)

0

150

60

1

175

100

2

200

140

3

225

180

4

250

220


9

(ii) Amount saved = s Time in weeks = t Carla: common difference = 25

Alvaro: common difference = 40

Start amount = 150

Start amount = 60

∴ s = 25t + 150

∴ s = 40t + 60

Carla: s = 25t + 150; Alvaro: s = 40t + 60 (s = amount saved; t = time in weeks). (iii) 25t + 150 = 40t + 60

If Carla has €500 saved, then:

150 − 60 = 40t − 25t

500 = 25t + 150

90 = 15t

350 = 25t

6=t

14 = t

∴ Answer: 6 weeks

14 − 11 = 3

(iv) After 11 weeks Alvaro has:

It takes Carla an additional 3 weeks to have the same amount as Alvaro in her account.

s = 40(11) + 60 = 500 = €500 saved

(v) s = 25t + 150 + 40t + 60 s = 65t + 210 Q. 17.

(i) Time (minutes)

Máire (distance to school m)

Siobhán (distance to school m)

0

300

500

1

250

400

2

200

300

3

150

200

4

100

100

5

50

0

6

0

(ii)

Distance to school (m)

500

Siobhán

400

300

200

100 Máire 0

1

2

3

4

Time (minutes)

(iii) Siobhán arrives at school first in 5 minutes

10


5

6

distance travelled (iv) Speed = _______________ time taken Máire 300 m Speed = ______ 6 min Speed = 50 m/min

Siobhán 500 m Speed = ______ 5 mins Speed = 100 m/min

(v) Siobhán lives further from school because she had the greater distance to travel of 500 m compared to Máire of only 300 m. (vi) From the graph or table we can see Máire and Siobhán are both the same distance from school (100 m) after 4 minutes. (vii) Let d = distance to school (m) t = time taken (minutes) Máire

Siobhán

common difference = −50

common difference = −100

starting distance (0 mins) = 300

starting distance (0 mins) = 500

∴ d = −50t + 300

∴ d = −100t + 500

1 1 (viii) The students were 25 m apart at 3__ minutes and at 4__ minutes. 2 2 (Answer can be read off a graph with a good scale) OR using algebra

Q. 18.

−50t + 300 + 25 = −100t + 500

−50t + 300 −25= −100t + 500

50t = 175

50t = 225

1 t = 3__ 2

1 t = 4__ 2

Number of letters in pattern = 1 + 2 + 3 + 4 + … + 25 + 26 = 351 letters 1,000 ÷ 351 = 2 R 298 Working backwards from 351 − 26 (letter Z) = 325 325 − 25 (letter Y) = 300 ∴ The 1,000th letter is X

Q. 19.

79

74 –5

–5

59 and 7

64

69 –5

–5

11 +4

T1 = −5 × 1 + 84

T1 = 4 × 1 + 3

Tn = −5n + 84

Tn = 4n + 3

19

15 +4

+4

23 +4

If the mth sequences are the same then −5 m + 84 = 4 m + 3 81 = 9 m 9=m The 9th term is the same. Active Maths 2 (Strands 1–5): Ch 4 Solutions

11

(b) The value of the 9th term is T9 = −5 × 9 + 84

or

T9 = 4 × 9 + 3

T9 = 39

T9 = 39

The value of the 9th term is 39 Q. 20.

(i)

(vi) C = 75A

Area (m2)

Cost (€)

1

75

2

150

3

225

4

300

(vii) €1,170 = 18 m2

5

375

€65 = 1 m2

6

450

7

525

(ii)

C ∴ A = ___ 75 4,500 A = ______ = 60 m2 75

(viii)

Company A

Company B

1 m2 = €75

1 m2 = €65

30 m2 = €2,250

30 m2 = €1,950

y 600 550 500

Company B charges €300 less than Company A to build a shed of 30 m2.

450 Cost (€)

400 350 300 250 200 150 100 50

x

0 0

1

2

3

4 5 Area (m2)

6

7

(iii) Slope of line = 75 ∴ Constant of proportionality = 75 (iv) Cost of shed (C) Area per square metre (A) C = 75A (v) C = 75(12) = €900 Q. 21.

(i) 6 blue counters in pattern 6 (ii) 22 red counters in pattern 7 (iii) 5, 9, 13, 17, 21, 25, 29, 33, 37 Adding 4 each time, pattern 9 needs 37 counters. (iv) T1 = 4 × 1 + 1 T2 = 4 × 2 + 1 Tn = 4n + 1 ∴ There are 4n + 1 counters in the nth pattern.

12


(v) Red counters in pattern 4

10 …

7 +3

+3

common difference = +3 T1 = 3 × 1 + 1 Tn = 3n + 1 ∴ There are 3n + 1 red counters in the nth pattern. (vi) Blue counters in pattern 1, 2, 3 … common difference = +1 T1 = 1 × 1 T2 = 1 × 2 ∴ Tn = 1 × n Tn = n ∴ There are n counters in the nth pattern. (vii) (a) The number of red counters in the pattern is calculated by 3 times the pattern number plus 1 (b) The number of blue counters in the pattern is the same as the pattern number. Q. 22.

(i) Blue tiles 1

5…

3 +2


+2

There will be 21 blue tiles in pattern 11. (ii) Tiles in pattern 1

7…

4 +3


+3

There will be 58 tiles in pattern 20. (iii) (a) Red tiles 0

2…

1 +1

+1

T1 = 1 × 1 − 1 T2 = 1 × 2 − 1 ∴ Tn = 1 × n − 1 Tn = n − 1 (b) Blue tiles 1

5…

3 +2

+2

T1 = 2 × 1 − 1 Tn = 2n − 1 (c) Tiles 1

7…

4 +3

+3

T1 = 3 × 1 − 2 T2 = 3 × 2 − 2 Tn = 3 × n − 2 ∴ Tn = 3n − 2 Active Maths 2 (Strands 1–5): Ch 4 Solutions

13

(iv) 30 blue tiles + 15 red tiles = 45 tiles in total 1 + 4 + 7 + 10 + 13 + 15 = 50 tiles ∴ The decorator can only make 5 complete patterns using 35 of the tiles. Q. 23.

(i) 2000 1800 1600

Dollars ($)

1400 1200 1000 800 600 400 200 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Euros (€)

(ii) The relationship between dollars and euros is directly proportional because we can see from the graph that the line passes through (0,0), i.e. 0 dollar converts to 0 euros. (iii) y-axis represents dollars = d x-axis represents euros = e ∴ d = 1.25e (iv) Constant of proportionality = 1.25 (v) The constant of proportionality represents the exchange rate, i.e. €1 = $1.25. (vi) Henry will receive $1,000 (vii) Formula: d = 1.25e d = 1.25 (800) = 1,000 Q. 24.

(i)

Charges (€)

200

150 Mechanic A 100

50

Mechanic B

0

1

2

3

4

Time (hours)

14


5

(ii) The call-out fee for mechanic A is €75. (iii) Let c = mechanic’s charge (€) and t = time (hours) Mechanic A

Mechanic B

common difference = (200 − 150) ÷ 2

common difference = (200 − 100) ÷ 2

= 25/hour

= 50/hour

call-out fee = 75

call-out fee = 0

∴ c = 25t + 75

c = 50t + 0 ∴ c = 50t

(iv) 6 hours to fix a car Mechanic A

Mechanic B

c = 25t + 75

c = 50t

∴ c = 25 × 6 + 75

∴ c = 50 × 6

c = 225

c = 300

Mechanic A is cheaper at €225 for 6 hours. Q. 25.

Year

No. of houses

1991

780

1992

710

1993

640

1994

570

1995

500

1996

430

1997

360

1998

290

1999

220

2000

150

(i) From 1995 to 2000, building decreased by 350. common difference = 350 ÷ 5 = 70 (decreasing by 70) The number of houses decreased by 70 each year. (ii) Working back from 1995, adding on 70 each year until 1991 gives 780 houses built in 1991. (iii) The number of houses built over the 10 years: 780 + 710 + 640 + 570 + 500 + 430 + 360 + 290 + 220 + 150 = 4,650


15

Q. 26.

(a)

(b)

Time (s)

Bill’s distance (m)

Jenny’s distance (m)

0

7

2

1

9

5

2

11

8

3

13

11

4

15

14

5

17

17

6

19

20

7

21

23

8

23

26

9

25

29

10

27

32 Jenny

32 Distance (m)

28 24

Bill

20 16 12 8 4 1 2 3 4 5 6 7 8 9 10 Time (seconds)

(c) After 5 seconds both runners are the same distance from Tina. (d) After 9 seconds Jenny is furthest from Tina and 4 m ahead of Bill (e) The distance between Bill and Tina increases by 2 m each second. ∴ Common difference = +2 Let d = distance between runners t = time in seconds Starting distance = 7 (when t = 0) ∴d=2×t+7 d = 2t + 7

(f) The distance between Jenny and Tina increases by 3 m every second. ∴ Common difference = +3 Starting distance = 2 ∴d=3×t+2 d = 3t + 2 (g) Using formulas in (e) and (f) to verify the answer to part (c) that both runners are the same distance from Tina after 5 seconds. 3t + 2 = 2t + 7 t = 5 as required

16


(when t = 0)

(h) Jenny stops after 1 min (60 seconds) Jenny: d = 3 × 60 + 2 d = 182 after 1 minute Jenny is 182 m from Tina. Bill: d = 2 × 60 + 7 d = 127 after 1 minute Bill is 127 m from Tina. Distance for Bill to catch up = 182 − 127 = 55 m 1 At 2 m/s it will take (55 ÷ 2) i.e. 27__ seconds 2 for Bill to be level again. (i) –5

–4 +1

–3 +1

–2

–1

0

1

2 ...

+1

The distance between the two runners changes by 1 m every second. T0: 1 × 0 − 5 = −5 T1: 1 × 1 − 5 = −4 ∴ Tn: 1 × n − 4 = n − 5 Letting d = distance between runners and t = time in seconds gives d = t − 5 thus when d = 100 100 = t − 5 105 = t Runners would be 100 m apart after 105 seconds (a) From the data in the table there is a common difference of €18 for every 100 units used. This shows that the relationship between the number of units used and the cost is linear. e.g. 56 − 38 = 18 74 − 56 = 18 92 − 74 = 18 etc. (b) 160 140 120 Cost (€)

Q. 27.

100 80 60

Plan B

40 20

Plan A 100 200 300 400 500 600 700 800 Units used


17

(c) From the graph the standing charge is approximately €20. (d) Common difference = +18 per 100 units T1: 18 × 1 + 20 = 38 T2: 18 × 2 + 20 = 56 ∴ Standing charge when no units used is €20 (e) C = 0.18u + 20, where C is the cost in euro and u is the units used (f) C = 0.18(650) + 20 = 137 155.50 − 137 = 18.50 18.50 Vat rate% = ______ × 100 137 = 13.5% to one decimal place (g)

Units used

100

200

300

400

500

600

700

800

Plan B cost (€)

51.50

67.00

82.50

98.00

113.50

129.00

144.50

160.00

Standing charge = €36, Rate per unit = 15.5 cent Rate per 100 units = €15.50 (h) If Lisa used 650 units on plan B that would cost 36 + 0.155 × 650 = €136.75, which is only 25 cents less than Plan A. Plan B only works out to be cheaper for higher users of electricity.

Cost (€)

(i)

170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10

Plan B Plan A 100

200

300

400 500 Units used

600

700

(j) From the graph, both plans are equal at approximately 640 units. Algebraically, 18n + 20 = 15.5n + 36 2.5n = 16 n = 6.4 ∴ Number of units = 640

18


n = hundreds of units

800

T7: 50 + 19 = 69

Exercise 4.2

The next 3 terms are 34, 50, 69 Q. 1.

(i) 14, 15, 21, 32 …

(iv) 5, 7, 6, 2

(a) T1 (start term) = 14 (b) 14

15 +1

(a) T1 = 5

21 +6

(b) 5

32 +11

+2

+5

+5

T5: 2 − 7 = –5

T6: 48 + 21 = 69

T6: −5 − 10 = −15

T7: 69 + 26 = 95

T7: −15 − 13 = −28


The next 3 terms are −5, −15, −28

(ii) 16, 26, 42, 64 … (a) T1 = 16 26

42 +16

(v) −8, −12, −10, −2 …

64

(a) T1 = −8

+22

(b) –8

+6

–12 –4

The first differences: 10, 16, 22

+2

–2 +8

+6

The first differences: −4, 2, 8

(c) T4: 42 + 22 = 64

The second difference: +6

T5: 64 + 28 = 92

(c) T4: −10 + 8 = −2

T6: 92 + 34 = 126

T5: −2 + 14 = 12

T7: 126 + 40 = 166

T6: 12 + 20 = 32


T7: 32 + 26 = 58 The next 3 terms are 12, 32, 58

(iii) 0, 4, 11, 21 … (a) T1 = 0 4

11 +7

+3

–10

+6


+4

–3

(c) T4: 6 − 4 = 2

∴ T5: 32 + 16 = 48

(b) 0

–4

The second difference: −3

(c) T4: 21 + 11 = 32

+6

–1

2

The first differences: 2, −1, −4


+10

6

–3

The first differences: 1, 6, 11 …

(b) 16

7

21

T2: 3 × 2 = 6

+3

The first differences: 4, 7, 10 (c) T4: 11 + 10 = 21 T5: 21 + 13 = 34 T6: 34 + 16 = 50

(i) 3, 6, 12, 24 … (a) T1: 3

+10


Q. 2.

T3: 6 × 2 = 12 T4: 12 × 2 = 24 The sequence doubles. (b) T5: 24 × 2 = 48 T6: 48 × 2 = 96


19

T7: 96 × 2 = 192

T3: −7.5 × 3 = −22.5

The next 3 terms : 48, 96, 192

The sequence triples. (b) T4: −22.5 × 3 = −67.5

(ii) 13, 39, 117, 351 …

T5: −67.5 × 3 = −202.5

(a) T1: 13

T6: −202.5 × 3 = −607.5

T2: 13 × 3 = 39

T7: −607.5 × 3 = −1,822.5

T3: 39 × 3 = 117

The next 3 terms are −202.5, −607.5, −1,822.5

The sequence triples. (b) T5: 351 × 3 = 1053 T6: 1053 × 3 = 3159

Q. 3.

(i) 4

11 +7

T7: 3159 × 3 = 9477 The next 3 terms are 1053, 3159, 9477

(ii) 23

+15 +4

26 +3

29 +3

32 ... +3

The sequence is linear, as the first difference is constant. (iii) 1

3 +2

(b) T5: 1 × 2 = 2

8 +5

16 +8

+3

+3

T6: 2 × 2 = 4

The sequence is quadratic, as the second difference is constant.

T7: 4 × 2 = 8 The next 3 terms: 2, 4, 8

(iv) 2, 4, 8, 16 …

1 ___ 1 __ 1 1 __ ___

T1: 2

, , , … 81 27 9 3 1 (a) T1: ___ 81 3 1 1 T2: ___ × __ = ___ 81 1 27 3 1 1 T3: ___ × __ = __ 27 1 9 ∴ The sequence triples.

T2: 2 × 2 = 4 T3: 4 × 2 = 8 T4: 8 × 2 = 16 The sequence is exponential, as each term is double the previous term. (v) –11

1 3 (b) T5: __ × __ = 1 3 1 3 T6: 1 × __ = 3 1 T7: 3 × 3 = 9

–8 +3

–5 +3

–2 +3

The sequence is linear, as the first difference is constant. Q. 4.

(v) −2.5, −7.5, −22.5, −67.5 … (a) T1: −2.5 T2: –2.5 × 3 = –7.5

20

37 ...

The sequence is quadratic, as the second difference is constant.

The sequence doubles.

The next 3 terms: 1, 3, 9

+11 +4

1 1 1 (iii) __, __, __, 1 … 8 4 2 1 (a) T1: __ 8 1 2 1 T2: __ × __ = __ 8 1 4 1 2 1 T3: __ × __ = __ 4 1 2

(iv)

22


1, 2, 4 … (i) Exponential doubling sequence ∴ 4 × 2 = 8 blocks needed for diagram 4.

(b) T2: −(2)2 − 4 × 2 + 1 = −11

(ii) Diagram 5 needs 8 × 2 = 16 blocks

(c) T3: −(3)2 − 4 × 3 + 1 = −20

(iii) Diagram 6: 16 × 2 = 32

Square the number, then make it negative, then subtract 4 times the original number and add 1.

Diagram 7: 32 × 2 = 64 Diagram 8: 64 × 2 = 128 Diagram 9: 128 × 2 = 256

(v) Tn = 3n

Diagram 10: 256 × 2 = 512

(a) T1: 31 = 3

Diagram 10 needs 512 blocks

(b) T2: 32 = 9

(iv) If the pattern was quadratic 1

2

(c) T3: 32 = 27

4 ...

+1

Three to the power of the nth number

+2

Q. 6.

+1

(i) 18

T3: 2 + 2 = 4

21 +3

+7

T4: 4 + 3 = 7

∴a=2

Diagram 5 would need 11 blocks.

b + c = 16

(a) T1: 12 + 1 = 2

T2: 2 × 22 + b × 2 + c = 21

(b) T2: 22 + 1 = 5

2b + c = 13

(c) T3: 32 + 1 = 10 Square the number and add 1. (ii) Tn = 2n2 + n − 2

2b + c = 13 b

(b) T2: 2 × 22 + 2 − 2 = 8

= −3

Using b + c = 16

(c) T3: 2 × 32 + 3 − 2 = 19 Square the number, then double it, then add the original number and subtract 2.

gives −3 + c = 16 ∴ c = 19 Since a = 2, b = −3 and c = 19 ⇒ Tn = 2n2 − 3n + 19

− 3n

(a) T1: −(1)2 − 3 × 1 = –4 (b) T2:

Solving simultaneously: − (b + c = 16)

(a) T1: 2 × 12 + 1 − 2 = 1

−(2)2

Tn = an2 + bn + c T1: 2 × 12 + b × 1 + c = 18

(i) Tn = n2 + 1

(iii) Tn =

+11

2a = 4

(v) T5: 7 + 4 = 11

−n2

39 ...

+4

+4

∴ Diagram 4 would need 7 blocks.

Q. 5.

28

(ii) 2

− 3 × 2 = −10

7 +5

(c) T3: −(3)2 − 3 × 3 = −18 Square the number, then make it negative, then subtract 3 times the original number. (iv) Tn = −n2 − 4n + 1

17 +10

+5

32 ... +15

+5

2nd difference constant ∴ Sequence is quadratic. Tn = an2 + bn + c

(a) T1: −(1)2 − 4 × 1 + 1 = −4 Active Maths 2 (Strands 1–5): Ch 4 Solutions

21

Second difference = 2a

Using b + c = −3.5

∴ 2a = 5

−8.5 + c = −3.5

a = 2.5

c=5

T1 = 2 and a = 2.5 i.e. T1 :

2.5(1)2

+ b(1) + c = 2

2.5 + b + c = 2

In summary a = 3.5, b = −8.5 and c = 5 gives Tn = 3.5n2 − 8.5n + 5 (iv) 1

∴ b + c = −0.5 Similarly T2 = 7

–3 –4

∴ T2: 2.5(2)2 + b(2) + c = 7 10 + 2b + c = 7 Solve the simultaneous equations: 2b + c = −3

2b + c = −9

c=2 In summary a = 2.5, b = −2.5 and c = 2 Tn = 2.5n2 − 2.5n + 2 11

+7

and a = 1.5

T2: 1.5(2)2 + b(2) + c = −3

⇒ −2.5 + c = −0.5

+9

∴ 2a = 3

b + c = −0.5

Using b + c = −0.5

+2

27... +16

+7

Solving these simultaneous equations: 2b + c = −9 − (b + c = −0.5) b

∴ 2a = 7 giving a = 3.5 Tn = an2 + bn + c T1: 3.5(1)2 + b(1) + c = 0 3.5 + b + c = 0

−8.5 + c = −0.5 c=8 In summary a = 1.5, b = −8.5 and c = 8 gives Tn = 1.5n2 − 8.5n + 8 (v) –10

–9 +1

b + c = −3.5 T2: 3.5(2)2 + b(2) + c = 2 2b + c = −12 Solving these simultaneous equations: 2b + c = −12 − (b + c = −3.5) b = −8.5 Active Maths 2 (Strands 1–5): Ch 4 Solutions

= −8.5

using b + c = −0.5

2nd difference = 7

22

+3

+3

T1: 1.5(1)2 + b(1) + c = 1

= −2.5

2

+2

Tn = an2 + bn + c

− (b + c = −0.5)

(iii) 0

–1

–2 ...

The 2nd difference = +3

2b + c = −3

b

–4

–14 –5

–6

–25 ... –11

–6

∴ 2a = −6 a = −3 Tn = an2 + bn + c T1: −3(1)2 + b(1) + c = −10 b + c = −7 T2: −3(2)2 + b(2) + c = −9 2b + c = 3

Solving simultaneous equations:

1+c=1

2b + c = 3

c=0 ∴ Tn = 1 × n2 + 1 × n + 0

− (b + c = −7)

Tn = n2 + n

= 10

b

There are n2 + n blocks in the nth term.

b + c = −7 and b = 10 10 + c = −7 ∴ c = −17

(iv) T10: 102 + 10 = 110

In summary a = −3, b = 10 and c = −17 gives

(v) The number of blocks needed for the nth pattern is obtained by squaring the number and then adding the number to this.

Tn = −3n2 + 10n − 17 Q. 7.

(i) Pattern

Number of blocks

1

2

2

6

3

12

Q. 8.

(i)

Pattern

1

2

3

4

5

Number of tiles 9 12 21 36 57

(ii) 9

(ii)

12 +3

21 +9

+6

36 +15

+6

2a = 6 ∴a=3 Tn = an2 + bn + c ∴ T1: 3(1)2 + b(1) + c = 9 b+c=6 (iii) 2

6 +4

12 +6

and T2: 3(2)2 + b(2) + c = 12

20 ...

2b + c = 0

+8

Solving simultaneous equations: 2b + c = 0

+2

+2

2a = 2

− (b + c = 6)

∴a=1 Tn =

an2

∴ T1:

b + bn + c

1(1)2

using b + c = 6

+ b(1) + c = 2

−6 + c = 6

b+c=1 and T2:

1(2)2

= −6

c = 12 + b(2) + c = 6

∴ Tn = 3 × n2 − 6 × n + 12

2b + c = 2

Tn = 3n2 − 6n + 12


3n2 − 6n + 12 tiles needed to make nth pattern

2b + c = 2 − (b + c = 1) b

=1

using b + c = 1

(iii) T12: 3(12)2 − 6(12) + 12 = 372 372 tiles in 12th pattern


23

(iv) 3n2 − 6n + 12 = 201


3n2 − 6n − 189 = 0 n2

2b + c = −3

− 2n − 63 = 0

− (b + c = −1)

(n + 7)(n − 9) = 0 ∴ n = −7 OR

9

Using b + c = −1

Since n is a positive number the biggest pattern in the sequence is pattern 9. Q. 9.

= −2

b

−2 + c = −1 c=1 ∴ T n = 2 × n2 − 2 × n + 1

(i) Pattern 4

Tn = 2n2 − 2n + 1 (v) T20: 2(20)2 − 2(20) + 1 = 761 There are 761 discs in the 20th pattern. Q. 10.

(i) 20

30 +10

30 +0

–10

1st 4, 8, 12 This is a quadratic sequence starting at 1, with a first difference increasing at a linear rate and the second difference increasing at a constant rate of 4. (iii) T7 = 85 red discs 5

+4

25 ... +12

+4

2nd difference constant means the sequence is a quadratic; also 2a = 4 ∴a=2 Tn =

+ bn + c

T1: 2(1)2 + b(1) + c = 1 b + c = −1 T2:

2(2)2

+ b(2) + c = 5

2b + c = −3

24

∴ a = −5 Tn = an2 + bn + c T1: −5(1)2 + b(1) + c = 20 b + c = 25 T2: −5(2)2 + b(2) + c = 30 2b + c = 50

13 +8

an2

–10

2a = −10

2nd 4, 4

+4

–10

This is a quadratic sequence, as the 2nd difference is constant.

(ii) 1, 5, 13, 25

(iv) 1

20


Solving simultaneous equations: 2b + c = 50 − (b + c = 25) b

= 25

Using b + c = 25 25 + c = 25 c=0 ∴ Tn = −5 × n2 + 25 × n + 0 Tn = −5n2 + 25n Assuming h = height and t = time gives h = −5t2 + 25t

(ii) Height (m)

y 30 20 10 0

1 2 3 4 x Time (seconds)

(iii) After 3.5 seconds h = −5(3.5)2 + 25(3.5) h = 26.25 Height of the rocket after 3.5 seconds is 26.25 m The height from the graph is close at approximately 25 m and will depend on the scale for further accuracy to be achieved. (iv) Solve for h = 0 (Rocket at ground level) ∴ −5t2 + 25t = 0 −5t(t − 5) = 0 ∴ t = 0 and t = 5 Q. 11.

(i)

Area covered (m2)

The rocket was at ground level at zero seconds and 5 seconds. 45 40 35 30 25 20 15 10 5 0

0

1

2

3 Days Area covered

4

5

(ii) The data follows an exponential pattern, as it is tripling the previous terms. e.g. T1: 1.5 T2: 1.5 × 3 = 4.5 T3: 4.5 × 3 = 13.5 etc. (iii) After 11 days T4 : 13.5 × 3 = 40.5 keep multiplying by 3, 7 more times to get T11 T11 = 40.5 × 3 × 3 × 3 × 3 × 3 × 3 × 3 ∴ T11 = 88,573.5 After 11 days 88,573.5 m2 will be covered. Q.12.

(i) 6

10 +4

12 +2

–2

12 +0

–2

As the second difference is constant, this implies a quadratic sequence. Active Maths 2 (Strands 1–5): Ch 4 Solutions

25

Number (′000)

(ii)

14 12 10 8 6 4 2 0 0

1

2 3 Time (weeks)

4

5

Number (thousands)

(iii) From the graph the maximum number of seagulls is just over 12,000; approximately 12,200.

In summary a = −1, b = 7 and c = 0 then Tn: −1 × n2 + 7 × n + 0 = −n2 + 7n

(iv) 2a = −2, a = −1 and Tn = an2 + bn + c

The general term Tn = −n2 + 7n

T1: −1(1)2 + b(1) + c = 6

(v) Assuming b = no. of seagulls and t = time in weeks.

b+c=7 T2: −1(2)2 + b(2) + c = 10

b = −t2 + 7t

2b + c = 14

If no seagulls then b = 0 and −t2 + 7t = 0

Solving these equations simultaneously:

Solving −t2 + 7t = 0 −t(t − 7) = 0

2b + c = 14

∴ t = 0 OR

− (b + c = 7)

At 0 weeks and 7 weeks, there will be no seagulls left in the colony.

=7

b

Using b + c = 7 and b = 7 gives c = 0

(i) Speed (m/s)

Q. 13.

20 18 16 14 12 10 8 6 4 2 0 0

1

2 3 Time (seconds) Speed

26

t=7


4

5

(ii) 11

17 +6

19 +2

–4

Tn = −2n2 + 12n + 1

17

Since v = speed of car and t = any given time

–2

then v = −2t2 + 12t + 1

–4

As the 2nd difference is constant, the graph is a quadratic.

(iv) From the graph, the speed of the car is approximately 18.5 m/s.

(Using the graph, symmetry between 2 and 4 seconds also implies a quadratic.)

Using the formula when t = 2.5 v = −2(2.5)2 + 12(2.5) + 1 v = 18.5 m/s as required

(iii) 2a = −4 ∴ a = −2 and Tn = an2 + bn + c

(v) When v = 1 m/s then −2t2 + 12t + 1 = 1

T1: −2(1)2 + b(1) + c = 11

giving −2t2 + 12t = 0

b + c = 13

−2t(t − 6) = 0

T2: −2(2)2 + b(2) + c = 17

∴ t = 0 OR

2b + c = 25

The speed of the car is 1 m/s at 0 seconds and 6 seconds.

Solving equations simultaneously: 2b + c = 25

t=6

Q. 14.

(i)

Time

Number

Hour 0

20 = 1

Hour 1

21 = 2

Using b + c = 13 and b = 12 gives

Hour 2

22 = 4

Hour 3

23 = 8

c=1

Hour 4

24 = 16

In summary a = −2, b = 12 and c = 1 gives

Hour 5

25 = 32

Hour 6

26 = 64

− (b + c = 13) b = 12

(ii)

70 60

Number (N)

50 40 30 20 10 0 0

1

2

3 4 Time (hours)

5

6

7

Number

(iii) From the graph approx 22–23 amoebas will be present after 4.5 hours


27

(iv) N = 2t When t = 4.5

Guess and check can be used to find the value of t, e.g.

N = 24.5

210 = 1024 too low

N = 22.6 to 1 d.p.

220 = 1,048,576 too high

(v) 2t = 65,536 216 = 65,536 ∴ After 16 hours Q. 15.

(i)

Pattern 1 2 3

Number of green tiles 8 12 16

Number of red tiles 1 4 9

Total number of tiles 9 16 25

(ii) Green tiles increasing by +4 ∴ Pattern 6 needs 28 green tiles. (iii) Red tiles increasing by squaring the pattern number 12, 22, 32 ∴ Pattern 7 needs 72 red tiles, i.e. Pattern 7 needs 49 red tiles. (iv) Pattern 8 needs 100 tiles in total. (v) Green tiles: 8

12 +4

16 ... +4

This is a linear pattern, as the 1st difference is constant. T1: 4 × 1 + 4 = 8 T2: 4 × 2 + 4 = 12 ∴ Tn = 4n + 4 There are 4n + 4 tiles in the nth green pattern. (vi) Red tiles: 1

4 +3

9 +5

+2

16 ... +7

+2

2a = 2 ∴ a = 1 and Tn = an2 + bn + c T1: 1(1)2 + b(1) + c = 1

b+c=0 T2: 1(2)2 + b(2) + c = 4 2b + c = 0 Solving simultaneous equations: 2b + c = 0 − (b + c = 0) b=0 Using b + c = 0 and b = 0 ⇒ c = 0 also. ∴ Tn = 1(n)2 + 0(n) + 0 Tn = n2

There are n2 tiles in the nth red pattern.

28


This is a quadratic pattern, as the second difference is constant.

(vii) 9

16 +7

36 ... This is a quadratic pattern, as the second

25 +9

difference is constant.

+11 +2

+2

2a = 2 ∴ a = 1 Tn = an2 + bn + c T1: 1(1)2 + b(1) + c = 9 T2: 1(2)2 + b(2) + c = 16 b+c=8 2b + c = 12 Solving simultaneous equations: 2b + c = 12 − (b +c = 8) b =4 Using b + c = 8 and b = 4 gives 4+c=8∴c=4 In summary a = 1, b = 4 and c = 4 gives Tn = 1(n)2 + 4(n) + 4 Tn = n2 + 4n + 4 This is a perfect square and can be written also as (n + 2)2. There are n2 + 4n + 4 OR (n + 2)2 tiles in the nth pattern. Q. 16.

(i) Blue tiles 2

6 +4

12 +6

This is a quadratic pattern, as the 2nd difference is constant.

+8 +2

+2

2a = 2 ⇒ a = 1

20 ...

Tn = an2 + bn + c

and

T1: 1(1)2 + b(1) + c = 2

T2: 1(2)2 + b(2) + c = 6

∴b+c=1

2b + c = 2

Solving simultaneous equations: 2b + c = 2 − (b + c = 1) b

=1

Since b + c = 1 and b = 1 this gives c = 0 In summary a = 1, b = 1 and c = 0 ∴ Tn: 1(n)2 + 1(n) + 0 = n2 + n Tn = n2 + n (ii) When n = 10 number of blue tiles T10: 102 + 10 = 110 ∴ 110 blue tiles needed (iii) Red tiles 2

3 +1

4 +1

5 ...

This is a linear pattern, as the 1st difference is constant.

+1

T1: 1 × 1 + 1 = 2 T2: 1 × 2 + 1 = 3 ∴ Tn: 1 × n + 1 Tn = n + 1 Active Maths 2 (Strands 1–5): Ch 4 Solutions

29

(iv) When n = 17, T17: 17 + 1 = 18 ∴ 18 red tiles needed. (v) Total number of tiles = total blue + total red ∴ Tn = n2 + n + n + 1 Tn = n2 + 2n + 1 OR (n + 1)2 (vi) 650 tiles in total 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 = 649 ∴ 11 patterns in total can be made. Q. 17.

Graph A is a linear pattern, as the slope of the line equals 2, that is the 1st difference is constant. Graph B is an exponential pattern, as the pattern is doubling, for example, when x = 1, y = 2. when x = 2, y = 4 when x = 3, y = 8 and when x = 4, y = 16 Graph C is a quadratic pattern because of the symmetry in the shape of the graph either side of the minimum at x = 4. Looking at the change in the y-values for every 1 unit change in the x-value, the second difference is constant (+2), indicating a quadratic pattern. 16

9 –7

4 –5

(i) 3

–3 +2

+2

Q. 18.

1

0 –3

–1 +2

–1 –1

+2

0 ...

0 +1

+2

3 +3

+2

Yes, this graph represents a quadratic pattern, as the second difference is constant. (ii) Between 2 and 4 seconds (below x-axis). (iii) 2nd difference = 2 ∴a=1 Tn = an2 + bn + c T1 = 3 T2 = 0 2 1(1) + b(1) + c = 3 1+b+c=3 b+c=2 Solving simultaneous equations: b+c=2 −(2b + c = −4) −b = 6 b = −6

30


1(2)2 + b(2) + c = 0 4 + 2b + c = 0 2b + c = −4

−6 + c = 2 c=8 ∴ Tn = 1n2 − 6n + 8 h = t2 − 6t + 8 (iv) The aeroplane was released at time t = 0. h = 0 − 6(0) + 8 h=8m (v) h = (20)2 − 6(20) + 8 h = 288 m (vi) Barry has assumed that the paper aeroplane will continue to fly in a quadratic pattern. (vii) No, as the paper aeroplane cannot possibly fly that high when launched from 8 m. Q. 19.

(a)

(i) T1

T2

T3

T4

T5

4

–2

–4

–2

4

–6

–2

2

6

4

4

4

2nd difference is constant. ∴ Pattern is quadratic. (ii) T1

T2

T3

T4

T5

T6

5

4.8

4.6

4

3

0

–0.2

–0.2

–0.6 –0.4

0

–1

–3 –2

–0.4

2nd difference is not constant. ∴ Pattern is not quadratic. (iii) T1

T2

T3

T4

T5

T6

–2

2

4

4

2

–2

4

2

0 –2

–2

–2 –2

–4 –2

2nd difference is constant. ∴ Pattern is quadratic. (b) T1

T2

T3

T4

4

2

1

1

–2

–1 +1

0 +1


31

2nd difference = 1

2 + 2b + c = 2

∴ 2a = 1

2b + c = 0

1 a = __ 2

Solving simultaneous equations: b + c = 3.5

Tn = an2 + bn + c

−(2b + c = 0)

T1 = 4

−b = 3.5

1 __

(1)2

+ b(1) + c = 4 2 1 __ +b+c=4 2 b + c = 3.5

Because b = −3.5 b + c = 3.5 −3.5 + c = 3.5 c=7

T2 = 2 1 __ 2

Q. 20.

(2)2

(a)

1 ∴ Tn = __n2 – 3.5n + 7 2

+ b(2) + c = 2

x

−3

−2

−1

0

1

2

3

y

1.5

2.5

2.9

3.0

2.9

2.5

1.5

Student answers may vary slightly. (b) Test to see if the second difference is constant: 1.5

2.5 +1

2.9 +0.4

3.0 +0.1

–0.3

–0.6

No. As the 2nd difference is not constant, this section of the Spanish Arch cannot be represented by a quadratic function. Q. 21.

(i) 800

860 +60

912 +52

–8

956 +44

–8

Shares will be worth €992 after 5 months. (ii) v = value of shares, m = any give month From (i) the 2nd difference is constant ∴ 2a = −8, a = −4 Tn = an2 + bn + c T1: –4(1)2 + b(1) + c = 800

T2: −4(2)2 + b(2) + c = 860

b + c = 804

2b + c = 876

Solving simultaneous equations: 2b + c = 876 −(b + c = 804) b = 72 ∴ c = 804 − 72 c = 732

32


In summary a = −4, b = 72 and c = 732 gives Tn: −4(n)2 + 72(n) + 732 = −4n2 + 72n + 732 ∴ v = −4m2 + 72m + 732 (iii) v = −4(0)2 + 72(0) + 732 = 732 €732 was invested. (iv) v = €1,056 ∴ −4m2 + 72m + 732 = 1,056 −4m2 + 72m − 324 = 0 m2 − 18m + 81 = 0

÷ –4

(m − 9)(m − 9) = 0 (m − 9)2 = 0 ∴m=9 After 9 months v = €1,056. (v) After 10 months: v: −4(10)2 + 72(10) + 732 = 1,052. Up until the 9th month the shares had been increasing in value; however, in the 10th month they start to fall, i.e. start to lose money. That is why the broker advised that the shares are sold after the 9th month when they reached their peak. Q. 22.

(i) 5

20 +15

45 +25

+10

80 ... +35

+10

Yes. The data represents a quadratic pattern, as the 2nd difference of +10 is constant. (ii) After 4 seconds the coin has fallen 80 m. (iii) f = distance coin falls in m, t = time in seconds 2a = 10 ∴ a = 5 and Tn = an2 + bn + c T1: 5(1)2 + b(1) + c = 5

T2: 5(2)2 + b(2) + c = 20

b+c=0

2b + c = 0

Solving simultaneous equations: 2b + c = 0 −(b + c = 0) b=0 ∴ Tn :

∴ c = 0 also 5n2

+ 0 + 0 = 5n2

and f = 5t2 (iv) If f = 200 metres then 5t2 = 200 t2 = 40

___

t = √ 40

t = 6.3 seconds to 1 d.p.


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(v)

Time (s)

Height above ground (m)

0

200

0.5

198.75

1 1.5

195 [= 200 – 5] 188.75 [= 200 – 11.25]

fall = 5 × time2 5(0.5)2 = 1.25 5(1)2 = 5 5(1.5)2 = 11.25 Subtract the amount fallen at each time interval to find the height. (vi) h = height of coin above ground (in m) t = time in seconds ∴ h = 200 − 5t2 OR

h = −5t2 + 200

(vii) When h = 50 m, calculate t: 50 = 200 − 5t2 5t2 = 150 t2 = 30

___

t = √ 30 , t = 5.4772... At 5.5 seconds (to 1 d.p.) (viii) If t = 20 seconds, how far did the coin fall? using f = 5t2 f = 5 × 202 f = 2,000 The coin would have been dropped 2,000 m. (ix) If h = 10 km (10,000 m) 5t2 = 10,000 t2 = 2,000

______

t = √ 2,000

t = 44.721… The coin would have taken 45 seconds (to the nearest second).

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Chapter 4 Exercise 4 - Biology Leaving Cert

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