3300u50-1 · 2019. 3. 5. · unit 1: non-calculator higher tier friday, 10 november 2017 –...

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Surname

Other Names

CandidateNumber

0

CentreNumber

GCSE

3300U50-1

MATHEMATICSUNIT 1: NON-CALCULATORHIGHER TIER

FRIDAY, 10 NOVEMBER 2017 – MORNING

1 hour 45 minutes

A17-3300U50-1

ADDITIONAL MATERIALS

The use of a calculator is not permitted in this examination.A ruler, a protractor and a pair of compasses may be required.

INSTRUCTIONS TO CANDIDATES

Use black ink or black ball-point pen. Do not use gel pen or correction fluid.You may use a pencil for graphs and diagrams only.Write your name, centre number and candidate number in the spaces at the top of this page.Answer all the questions in the spaces provided.If you run out of space use the continuation page at the back of the booklet. Question numbers must be given for all work written on the continuation page.Take � as 3·14.

INFORMATION FOR CANDIDATES

You should give details of your method of solution when appropriate.Unless stated, diagrams are not drawn to scale.Scale drawing solutions will not be acceptable where you are asked to calculate.The number of marks is given in brackets at the end of each question or part-question.In question 4, the assessment will take into account the quality of your linguistic and mathematical organisation and communication.

For Examiner’s use only

Question MaximumMarkMark

Awarded

1. 3

2. 5

3. 4

4. 7

5. 4

6. 3

7. 5

8. 3

9. 4

10. 4

11. 3

12. 9

13. 5

14. 2

15. 4

16. 3

17. 4

18. 1

19. 7

Total 80

(3300U50-1)02

2

Volume of prism = area of cross-section × length

Volume of sphere = �r3

Surface area of sphere = 4�r2

Volume of cone = �r2 h

Curved surface area of cone = �rl

In any triangle ABC

Sine rule

Cosine rule a2 = b2 + c2 – 2 bc cos A

Area of triangle = ab sin C

The Quadratic Equation

The solutions of ax2 + bx + c = 0 where a ≠ 0 are given by

Annual Equivalent Rate (AER)

AER, as a decimal, is calculated using the formula , where i is the nominal interest rate

per annum as a decimal and n is the number of compounding periods per annum.

length

cross-section

r

h

r

l

asin A

bsin B

csin C= =

C

BA

a

c

b

xb b ac

a=– ( – )± 2 4

2

Formula List - Higher Tier

Area of trapezium = (a + b)h

b

h

a

13

43

12

12

( ) −1 1in+n

© WJEC CBAC Ltd.

(3300U50-1) Turn over.

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1. Look at the following descriptions of special quadrilateral shapes. Circle the correct name for each one.

(a) Its diagonals intersect at 90°. Only one diagonal is a line of symmetry. [1]

Kite Rhombus Square Trapezium Rectangle

(b) Only one pair of sides are parallel. [1]


(c) All four sides are equal. Its diagonals are not equal in length. [1]


kite Rhombus"

I\ /

-

-

- ..

He. - - \ -

-

1 K¥- ', - -× -; =\: -

t il

,

1

!

Isosceles TrapeziumTrapezium

7 >

>>

4

(3300U50-1)04

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2. (a) Complete the table below. Draw the graph of y = 2x2 – 5 for values of x between –2 and 3. Use the graph paper below. Choose a suitable scale for the y-axis. [4]

x –2 –1 0 1 2 3

y = 2x2 – 5 3 –5 –3 3 13

11 2 3 4– 1– 2– 3 0

y

x

:Ex"*¥t-

- 10


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05 © WJEC CBAC Ltd.

5Examiner

only (b)

y

xO

The sketch above can represent only one of the equations given below. Circle this equation. [1]

y = x2 y = x2 − 3 y = −x2 y = x2 + 3 y = 3x 0

6

(3300U50-1)06

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3. (a) Rotate triangle A through 90° anticlockwise, about the point (–2, 3). [2]

2 6 8 9741 53– 2 0– 4 – 1– 3– 5– 6– 7– 8– 9

– 4

– 6

– 7

– 8

– 9

– 2

– 3

– 5

– 1

2

6

7

9

A

8

4

1

5

3

y

x

•

• •

•


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2 6 8 9741 53– 2 0– 4 – 1– 3– 5– 6– 7– 8– 9

– 4

– 6

– 7

– 8

– 9

– 2

– 3

– 5

– 1

2

6

7

9

8

4

1

5

3

B

y

x

(b) Enlarge triangle B by a scale factor of , using (0, 0) as the centre of enlargement. [2]12

07

•

:

•

(3300U50-1)08 © WJEC CBAC Ltd.

8Examiner

only4. In this question you will be assessed on the quality of your organisation, communication and accuracy in writing.

PQ and PR are tangents to a circle with centre O. RPQ = 30°.

30°

Q

O

R

P

Diagram not drawn to scale

Find the size of OQR. You must indicate any angles you calculate. You must give a reason for each stage of your working. [5 + 2 OCW]

OQR = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . °

$

$

$

150

Tangents meet a radius at 90°

OQ = OR as they we both radii , thatmeans that triangle QOR is an isosceles triangle

OQPR Is a kite so angles add to 3600

So far : 90 36090

- 210

Iso ←angle Qjr

+ 30

angles in a210

triangle 180 3O÷2 = 15°\J

- 150 f isosceles triangle30

15


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9Examiner

only5. (a) Express 0·000 42 in standard form. [1]

(b) Calculate the value of .

Give your answer in standard form. [1]

(c) Calculate the value of (4·7 × 105) – (6.2 × 104). Give your answer in standard form. [2]

7.2 × 106

2 × 10–2

4°2 × 10-4

3°6 × 108

6

47'

0000

6 2 00 °

4 08

= 4.08 × 105

10

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6. Using only a ruler and a pair of compasses, construct a perpendicular line from the point P to the line AB. [3]

B

P

A

(3300U50-1) Turn over.11 © WJEC CBAC Ltd.

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7. A group of pupils from a school took part in The Urdd National Eisteddfod. All of them competed in at least one of the following competitions: Singing, Dancing or Reciting.

• 2 of them only took part in a Dancing competition. • 5 only took part in a Reciting competition. • No one took part in both a Reciting and a Dancing competition. • 3 took part in both a Singing and a Dancing competition. • 9 took part in a Reciting competition. • 22 took part in a Singing competition.

The Venn diagram below shows some of the above information. The universal set, , contains all of the pupils in the group.

One of the pupils in the group is chosen at random. What is the probability that this person only took part in a Singing competition? [5]

0

2

0

Singing

Reciting

Dancing

315

4

5

Total = IS + 3 + 4 + 5 t 2

= 20 + 9

= zq

PConly singing) = 1529

12

(3300U50-1)12

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8. Factorise x2 – 7x – 18, and hence solve x2 – 7x – 18 = 0 . [3]

( x - 9) ( x +2 ) =0

x=9 or x= - 2


9. Solve the following simultaneous equations using an algebraic (not graphical) method. [4]

4 x – 3y = 26 x – 5y = 1

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0 × 5 S S�2� ×3

ag gig,§

a

30 200C - 15g = 10E

�4� 18 )( - 15g= 3

Sub x= 35 in �1�

�3� - �4� 44.5 ) - 3y =2Zx = 7 14 -3g = 2

x = 3.5 14 . 2 = 3g12 = 3g4 = y

Check in �2�

6C 3.5 ) - 5( 4) = I

21 . 20 =|

I =1 ✓

14

(3300U50-1)14

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10. A cylinder just fits inside a hollow cube with sides of length m cm.

m cm

m cm

Diagram not drawn to scale

The radius of the cylinder is cm.

The height of the cylinder is m cm.

The ratio of the volume of the cube to the volume of the cylinder is given by

volume of cube : volume of cylinder

= k : ,

where k is a number.

Find the value of k. You must show all your working. [4]

m2

cmm2

Volume cube = m3 Volume cylinder - t(m)' ( m )

=

IM3m3: Tim 4

4×4 x4 4 : I4m3 : it on

3

÷ m3 ÷m3


15Examiner

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16Examiner

only11.

2 3– 1– 2– 3

– 1

– 3

1

0

2

3

5

4

– 2

1

y

x

Complete the following table to give the set of inequalities that describes the shaded region shown above. [3]

x X 1

÷<

•

^

1

÷/ 5diy=

2

J

y= 3×+1

y 22

Y 7 3x+ 1


17Examiner

only12. Two different squares are constructed. The side length of the smaller square is x cm. The side length of the larger square is 3 cm longer than the side length of the smaller square. The combined area of the two squares is 22·5 cm2.

(a) Show that 4 x2 + 12x − 27 = 0 . [4]

(b) Find the dimensions of each of the squares. Do not use a trial and improvement method. You must show all your working and justify any decision that you make. [5]

Side length of smaller square = ……….. . . . . . . . . . . . . . . . . . . . . . . . . . .………………. cm

Side length of larger square = ……….. . . . . . . . . . . . . . . . . . . . . . . . . . .………………. cm

x x2(x+3)2 xt 3

x

x+3×2+ ( xt 3) ( xt 3)

2=

22.5×2+ x2 + 6x +9 = 22.5

2×2 + 6×-13.5 = 0

4×2 + 12×-27=0 as required

@ - 108+0 iz }

+ 18 - 6

( 4x + I8) (4×-6)=0( zx +9 ) ( 2x - 3) = 0

x= - 9- or x= z2 2

a X x =L .5

can't have negative side

1.5

4.5


18Examiner

only13. Given that y is inversely proportional to x3 and that y = 120 when x = 2,

(a) find an expression for y in terms of x, [3]

(b) use the expression you found in part (a) to complete the following table. [2]

x 2 10

y 120 15

y = K:

120 = K Y= 960.

-

23 , (3

960 = K

4

0.96

y = 9601000

IS =

960×3×3=

96015×3= 64

x = 4


19Examiner

only14. The two triangles shown below are not drawn to scale.

84°

37° 37°

59°

Which one of the following statements is correct? Give full reasons for your answer. [2]

The correct statement is ………………….. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .……….

This is because …………….. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .……. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .……….

A: the triangles must be congruent

B: the triangles could be congruent

C: the triangles cannot be congruent

84

sq°

84 180+ 37 - 121⇒ 9

'

B

all the angles are the Same

BUI to be congruent all sidesmust also be the same . We

are not told the byte ofthe sides .

20

(3300U50-1)20

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15. (a) Express 0.642 as a fraction. [2]

(b) Evaluate . [2]

. .

136

⎛⎝⎜

⎞⎠⎟− 12

100 x = 6143•124242 ...x = 0 . 64242. . .99x = 63 . 6

X = 63.6 = 636: 990

in :(E.)= 1 × 536 = 6F

(3300U50-1)21

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© WJEC CBAC Ltd.

Examineronly

16. You are given that p = 40 and q = 10. Circle the correct answer in each of the following:

(a) p is equal to [1]

10 4 4 10 10 2 2 10 20

(b) pq is equal to [1]

10 40 40 10 400 200 20

(c) q5 is equal to [1]

100 10 5 10 50 625 10 100

Turn over.

P = ifp = zfo

0Fi × to = Foo

= 20

(a) 5= (a) to )( riokrkro )÷

0 × To

= ioof

(3300U50-1)

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23Examiner

only18. The following diagram shows a sketch of the curve y = f (x).

0

2

– 4

– 2

4

y

y

x

x

The curve is transformed, as shown below.

Using function notation, complete the following to give the equation of the transformed curve. [1]

The equation of the transformed curve is

y = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

0

2

– 4

– 2

4

Turn over.

•

( 0,2 )

£4

C- 4,2 )

:( x + 4)


24Examiner

only19.

2 3 3 4 4 4

Two of the cards shown above are selected at random, without being replaced.

Find the probability that

(a) the product of the two numbers selected is 12, [3]

(b) the sum of the two numbers selected is even. [4]

END OF PAPER

2- 35LE u

y 2 Is

2%68③ €53 I × Is = Yo- ¥ 40# ④ to z x Is = To

¥ 4Iot So = ¥

@

2- 35 € • 'z x I = I

30

y ⑧ Is 2% 30 €5 ⑤ E x I = ⇐

- ÷. 4I * ÷. ② I × Is = I6 € 3

¥ * z x z -- I=

1430

Questionnumber

Additional page, if required.Write the question number(s) in the left-hand margin.

Examineronly

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© WJEC CBAC Ltd.

(3300U50-1)

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3300u50-1 · 2019. 3. 5. · unit 1: non-calculator higher tier friday, 10 november 2017 –...

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