Higher nov 2008_p1old

Higher Paper 1November 2008

1.(a)The nth term of a sequence is n2+ 5. 

List the first three terms of the sequence.

1.(a)The nth term of a sequence is n2+ 5. List the first three terms of the sequence.

When n = 1, n2+ 5 = 6

When n = 2, n2+ 5 = 9

When n = 3, n2+ 5 = 14

6, 9, 14

Write down the OUTPUT when n is INPUT into the number machine.

Write down the OUTPUT when n is INPUT into the number machine.

(n+7) X 4 or

4(n+7) or

4n + 28

2.A recipe for making 10 fruit oat bars has the following ingredients.10 Fruit Oat Bars80 grams Butter80 grams Brown Sugar2 tablespoons Golden Syrup130 grams Porridge Oats140 grams Dried Fruit2 tablespoons Sunflower Seeds

(a) Gillian is making fruit oat bars for a charity stall. Complete the following table to show thequantity of each ingredient needed to make 150 fruit oat bars.

150 Fruit Oat Bars................grams Butter................grams Brown Sugar................tablespoons Golden Syrup................grams Porridge Oats................grams Dried Fruit................tablespoons Sunflower Seeds

2.A recipe for making 10 fruit oat bars has the following ingredients.10 Fruit Oat Bars80 grams Butter80 grams Brown Sugar2 tablespoons Golden Syrup130 grams Porridge Oats140 grams Dried Fruit2 tablespoons Sunflower Seeds

150 Fruit Oat Bars......1200..........grams Butter......1200........grams Brown Sugar.......30.......tablespoons Golden Syrup.......1950.........grams Porridge Oats.......2100.........grams Dried Fruit.........30.......tablespoons Sunflower Seeds

X 15

(b) When Linda makes 100 fruit oat bars, she buys a 2kg bag of porridge oats. Find the weight of porridge oats left over after making the bars. Give your answer in grams.

HINT:

To make 100

Multiply each ingredient by 10

130 X 10 =

(b) When Linda makes 100 fruit oat bars, she buys a 2kg bag of porridge oats. Find the weight of porridge oats left over after making the bars. Give your answer in grams.

130g X 10 = 1300g

2Kg = 2000g

So 700g left over

(c) A recipe book states that 1 ounce is equivalent to 25 grams. 

Using this information find whether 5 ounces of butter is sufficient to 

make 20 fruit oat bars. 

Show calculations to support your answer.

(c) A recipe book states that 1 ounce is equivalent to 25 grams. Using this information find whether 5 ounces of butter is sufficient to 

make 20 fruit oat bars. Show calculations to support your answer.

5oz = 5 X 25

= 125g

We need 80g for 10

which is 160g for 20

NO 5oz is not enough!

3.(a)On the grid below, draw the enlargement of the given shape using a scale factor of 3 and centre O.

3.(a)On the grid below, draw the enlargement of the given shape using a scale factor of 3 and centre O.

(b) Describe fully the transformation that transforms triangle ABC into triangle XYZ.

(b) Describe fully the transformation that transforms triangle ABC into triangle XYZ.

The above diagram shows three points X, P and Y which lie on a straight line. The bearing of Y from P is 150°. Find the bearing of X from P.

The above diagram shows three points X, P and Y which lie on a straight line. The bearing of Y from P is 150°. Find the bearing of X from P.

30

360 – 30 = 330 °

°

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

(d) Rotate the triangle A through 90° clockwise about the origin.

4.(a) Solve 10x = 30 + 5x.

10x  - 5x = 30 

5x = 30 

x = 6

(b) Solve 6(x –3) = 42

 6x – 18 = 42

 6x  = 42 + 18

 6x  = 60

 x  = 10

(c) Simplify 5(a + 3b) + 3a – 20b

 5a + 15b + 3a – 20b.

 8a - 5b

5.(a) Express 792 as a product of prime numbers in index form.

Prime Numbers

792

5.(a) Express 792 as a product of prime numbers in index form.

Prime Numbers

7922

3962

1982

993

3 3311 11

11

5.(a) Express 792 as a product of prime numbers in index form.

Prime Numbers

7922

3962

1982993

3 33

11 11

1

23 X 32 X 11

(b) Write down the least positive whole number that 792 must be multiplied by to make the result a 

perfect square.

23 X 32 X 11 = 792

We need an extra 2 and an extra 11 so all the indices are EVEN.

2 X 11 = 22

792 x 22 = 17 424 =

√17 424 = 22 x 3 X 11 CAN YOU SEE WHY?

24 X 32 X 112

(c) Explain why 25×36×132 is not a perfect square.

3(1)2 – 2(1) – 53 – 2 – 5-4

-4

(d) Draw the line x = –1·5 on your graph paper and write down the coordinates of the point where this line intersects the curve y = 3x2– 2x – 5.

(a)Complete the table above. (b)On the graph paper, draw the graph of y = 3x2–

2x – 5 for the values of x between –2 and 4. 2(c) Write down the x-values of the points where

the curve y = 3x2– 2x – 5 intersects the x-axis.

(-1.5,4)

(-1,0)(1.5,0)

When a drawing pin is thrown it either lands pin-up or pin-down. The relative frequency of the drawing pin landing pin-up was calculated after a total of 20 throws, 40 throws, 60 throws, 80 throws and 100 throws. The results are plotted on the graph below.

When a drawing pin is thrown it either lands pin-up or pin-down. The relative frequency of the drawing pin landing pin-up was calculated after a total of 20 throws, 40 throws, 60 throws, 80 throws and 100 throws. The results are plotted on the graph below.

(a) Which one of the readings noted by the letters A, B, C, D and E on the graph is likely to give the best estimate of the probability of this drawing pin landing pin-up? You must give a reason for your answer.

E because the more often an experiment is done the better the estimate for the probability

When a drawing pin is thrown it either lands pin-up or pin-down. The relative frequency of the drawing pin landing pin-up was calculated after a total of 20 throws, 40 throws, 60 throws, 80 throws and 100 throws. The results are plotted on the graph below.

(b) Using the graph, find how many times the drawing pin (i) landed pin-up in the first 40 throws,

(ii) landed pin-down in the 100 throws.

0.75 means 75/100

75 300

100 = 400 =

30 out of 40

so 30 times

64 pin-up so 36 pin down

Solve the following simultaneous equations by an algebraic (not graphical) method. Show all your working.3x + 4y = –7 2x + 6y = –3

Make the x co-efficients equal

by multiplying EVERYTHING in the top equation by 2

And EVERYTHING in the bottom equation by 3

Solve the following simultaneous equations by an algebraic (not graphical) method. Show all your working.3x + 4y = –7 2x + 6y = –3

6x + 8y = -14

6x + 18y = -9

Subtract because the

signs are the same!

-10y = -5

10y = 5

y = 0.5

If y = 0.5

3x + 4(0.5) = -7

3x + 2 = -7

3x = -9

x = -3

11. (a) Evaluate each of the following.(i) 4–2

(ii) 60

(b) Simplify √72

116 (i)4–2 =

(ii) 60 = 1

(b) √72 = 7

(c) Write each of the following numbers in standard form.(i) 45 000

(ii) 0·0023

(c) Write each of the following numbers in standard form.(i) 45 000

(ii) 0·0023

45 000 = 4.5 X 104

0·0023 = 2.3 X 10-3

(d) In a test Jayne was asked to write the answer to 20 multiplied by 490 in standard form. She wrote 9800 as her answer. Explain carefully why this was marked incorrect and give the expected answer.

9800 is the correct answer but it is not in standard form9800 = 9.8 X 103

(e) Find, in standard form, the value of

4 .6 X10 -6 2 X 10 -4

2.3 X 10-2

Remember -6 - -4 = -6 + 4

¾ = 0.75

25

= 0.4

320 = 0.15

These fractions are not recurringso 7/11 is the recurring decimal.

HINT:Convert mixed numbers to top-heavy fractions.

1

13 =

43

234 = 11

4

43 X

114 = 44

12 = 38

12 = 323

(h) Two pieces of string have lengths 23 cm and 14 cm measured correct to the nearest cm.What is the maximum possible total length of the two pieces of string?

23cm ± 0.5cm14cm ± 0.5cm

(h) Two pieces of string have lengths 23 cm and 14 cm measured correct to the nearest cm.What is the maximum possible total length of the two pieces of string?

23cm + 0.5cm = 23.5cm14cm + 0.5cm = 14.5cm Total = 38cm

12. When ten times a number x is added to three, the result is less than the result of subtracting x from seven. Write down and simplify an inequality which is satisfied by x.

12. When ten times a number x is added to three, the result is less than the result of subtracting x from seven. Write down and simplify an inequality which is satisfied by x.

10x + 3 < 7- x11x < 7 – 311x < 4x < 4

11

13. The histogram below represents the results of gathering and measuring the lengths of pieces of driftwood.

(a) Use the histogram to find the total number of pieces of driftwood gathered and measured.

13. The histogram below represents the results of gathering and measuring the lengths of pieces of driftwood.

(a) Use the histogram to find the total number of pieces of driftwood gathered and measured.

40 X 4 =

160

20 X 5 =

100

20 X 6 =

120

20 X 3 =

60

40 X 2 = 80

Total = 160 + 100 + 120 + 60 + 80

= 520

13. The histogram below represents the results of gathering and measuring the lengths of pieces of driftwood.

(b) Use the histogram to find an estimate for the median.

160100

120

Half of 520 is 260 so looking at 260th – which is exactly 60

14. The diagram shows a trapezium OABC with OA parallel to CB. The point Q is the mid-point of AB. Given that OA = x, OB = y and CB = 2OA

express, in their simplest form, each of the following in terms of x and y.(a) OC (b) CQ

14. The diagram shows a trapezium OABC with OA parallel to CB. The point Q is the mid-point of AB. Given that OA = x, OB = y and CB = 2OA

express, in their simplest form, each of the following in terms of x and y.(a) OC (b) CQ

xy

2xy-2x

AB = -x + y

CQ = CB + BQ

= 2x – 0.5(-x+y)

= 2x +0.5x – 0.5y

= 2.5x - 0.5y

OC = OB + BC= y – 2x

15. A bag contains 7 yellow beads, 3 white beads and 1 black bead. Two beads are drawn at random without replacement from the bag.

(a) Calculate the probability that the two beads are both yellow.

(b) Calculate the probability that at least one white bead is drawn.

15. A bag contains 7 yellow beads, 3 white beads and 1 black bead. Two beads are drawn at random without replacement from the bag.(a) Calculate the probability that the two beads are both yellow.

(b) Calculate the probability that at least one white bead is drawn.

y

y

y

y

w

w

w

w

b

b

b

b

7

11 3

11

1

11

2

10

3

10

6

10

1

107

10

1

107

10 3

10

0

15. A bag contains 7 yellow beads, 3 white beads and 1 black bead. Two beads are drawn at random without replacement from the bag.(a) Calculate the probability that the two beads are both yellow.(b) Calculate the probability that at least one white bead is drawn.

y

y

y

y

w

w

w

w

b

b

b

b

7

11 3

11

1

11

2

10

3

10

6

10

1

107

10

1

107

10 3

10

0

P(y and y) = 7/11 X 6/10 = 42/110

P(y and w) or (w and y) or (w and b) or (w and w) or (b and w)

= (7/11 x 3/10) + (3/11 X 7/10) +

(3/11 X 1/10) + (3/11 X 2/10) + (1/11 X 3/10)

= 21/110 + 21/110 + 3/110 + 6/110 + 3/110

=54/110

16. (a) The diagram shows a sketch of y = f(x). On the same diagram, sketch the curve y = f(x) + 5. Mark clearly the value of y at the point where this curve crosses the y-axis.

16. (a) The diagram shows a sketch of y = f(x). On the same diagram, sketch the curve y = f(x) + 5. Mark clearly the value of y at the point where this curve crosses the y-axis.

16. (a) The diagram shows a sketch of y = f(x). On the same diagram, sketch the curve y = f(x) + 5. Mark clearly the value of y at the point where this curve crosses the y-axis.

3

x

y

4

-4

g(x)

- g(x)

HINT: Make the denominator (5x-2)(3x+1)

HINT: Make the denominator (5x-2)(3x+1)

4x(3x+1) 3(5x-2)

(5x-2)(3x+1) + (5x-2)(3x+1) = 3

4x(3x+1) + 3(5x-2) = 3(5x-2)(3x+1)

4x(3x+1) + 3(5x-2) = 3 (5x-2)(3x+1)

12x2 + 4x + 15x - 6 = 3 (15x2 – x - 2)

12x2 + 4x + 15x - 6 = 45x2 - 3x - 6

0 = 33x2 - 22x

0 = 11x(3x – 2)

11x = 0x = 0

3x-2 = 0x = 2/3