- 1 -

G.C.E. (O.L.) Support Seminar – 2016Mathematics I (Part A)

Answer Guide

Question Number Answer

Marks Other facts

1. 82 = 64 2

By considering the angle in a semicircle and taking 90°

2. x = 50°

x + 90° + 40° = 180° or x + 40° = 90°

1

2

3. Rs. 64 800

Rs. 540 000 × 2—100

× 6

1 2

4. (x + 4) (x + 1)

x2 + 4x + x + 4

1

2

5. The two equal sides are WY and YZ. YWZ^ = 20°

1

2

6. L1 ∩ M

2

7. 60°

2

8. 40

1—5

= 8n(S)

1

2

9. Q3 − Q1 = 6

Q1 = 3, Q3 = 9

1

2

10. Δ PTS and Δ QTR Condition - A.A.S.

1 1

2

11. x2 (x + 1)

2

12. r = 7 cm

2 × 22—7

× r = 44

1

2

13. x + y = 90°

2x + 2y = 180°

1

2

14. 23x 33x

−

13x

1

2

[See page 2

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[3 jeks mssgqj n,kak

Question Number Answer

Marks Other facts

15. 9 2

Marking 6 cm

from BC to DE

16. ADB^ = 40° OAD

^ = 40° or ACB^ = 40°

1 2

17.

( )−5 4 2 1 2 × 2B =

B = ( ) 1 0 0 1 2 × 2( ) 6 −4 −2 0 2 × 2+ 1

2

18. y = 70°Observing that DE CB. (Midpoint theorem) 1

2

19. y = − x + 3m = −1 or c = 3

1

2

20. x = 120°

x + 60° = 180° (opposite angles of a cyclic quadrilateral are supplementary)

1

2

21.−4 −3 −2 −1 1 2 3 4 5 0

x

2

22. r = 224 = 3 × r 4 − 1

1

2

23. 9Number of man days needed for the task = 8 × 9 = 72

1

2

24.

4 cm

3 cm 3 cm

A

C

B

E

D

Drawing the line parallel to BC

and marking 3cm from A Marking the points D and E

1 1

2

25. 60 km h−1

360 km 6 h

or 240 km 4 h

or 120 km 2 h 1

2

[See page 3

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Mathematics I (Part B)Answer Guide

Question Number Answer

Marks Other facts

1. (i) 1—8

+ 2—3

= 19—24

1

1

2

10

The answer can be obtained by finding thenumber of letters as well

(ii) Remaining letters as a fraction = 5—24

Express letters as a fraction = 5—24

× 1—5

= 1—24

1

1

1

3

(iii) Foreign letters as a fraction = 4—24

4—24 of the letters = 520

Total number of letters = 520 4

× 24

= 3120

Registered letters = 3120 × 1—8

= 390

1

1

1

3

(iv) Ordinary letters: Express letters = 2—3

: 1—24

= 16 : 1

1 1 2

2. (i) Length of BE = 1 8

× 2πr

= 1 8

× 2 × 22 7

× 14

= 11 m

Perimeter of ABED = (11 + 15 + 10 + 11) m = 47 m

1

1

1 3

(ii) Area of the section where sand is spread = 1 8 πr2

= 1 8

× 22 7

× 14 × 14

= 77 m2

1

1 2

(iii)

Area of the section without sand = (25 + 15) × 10 − 77 2

= 200 − 77 = 123 m2

2 1 3

15 m

10 m

14 m 6 m

A B

CDE

45°

[See page 4

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Question Number Answer

Marks Other facts

(iv)

(i)

(ii)

(i)

(ii)

B

CE

A

D4.1 m

10 m

Drawing the triangle Marking 4.1 m

1 1

2

5

10

3. (a) Rates for a year = 1500 × 4 = Rs' 6000

Annual rates percentage = 6000 75000

× 100%

= 8%

1

1

1

3

Discount = 6000 × 10100

Amount saved = Rs' 600

1

1 2

(b) Number of shares = 270000 9

= 30 000

Income = Rs' 30 000 × 2 = Rs' 60 000

1

1

2

Gain per share = Rs' 1.50Capital gain = Rs' 1.50 × 30 000 = Rs' 45 000

1 1 1

3 5 10

(30 000 × 10.50) − 270 000= 315 000 − 270 000= 45 000

4. (i)

40 − 50 14 50 − 80 12

1 1 2

(ii)

4

8

12

16

20

24

100−10 20 30 40 50 60 70 80 90×

× ×

×

×

×

× ××

×

Num

ber

of st

uden

ts

MarksCompleting the histogram

2 2

(iii) Marking the endpoints of the frequency polygon Marking the midpointsCompleting the frequency polygon

1

1

1 3

(iv) Percentage = 12 60

× 100%

= 20%

2 1 3 10

[See page 5

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Question Number Answer

Marks Other facts

5. (a) (i)

(ii)

4 6

2 6

Boy

Boy

Boy

First picking (sing)

Second picking (play an instrument)

Girl

Girl

Girl

4 6

2 6

4 6

2 6

Extending the tree diagram

Probability of a boy performingon one occasion and a girl performing on the other occasion. = 16

36

2

1

1 + 1

1

2

4 6

For 4 6

and 2 6

(b) (i)

2

2 3 4 5 6

3

4

5

6

Sing

Play

an

inst

rum

ent

× ×

×× ×

×

×

×

×

×

×

×

×

×

×

× × ×

× × ×

× × × ×

Representing on the grid

1

1

4 10

(ii) Marking the event on the grid

Probability = 20 25

or 4 5

1

2 3

= 4 6

× 2 6

+ 2 6

× 4 6

- 1 -G.C.E. (O.L.) Support Seminar – 2016

Mathematics II (A Part) Answer GuideQuestion Number Answer

Marks Other facts

1. Monthly loan amount for Institution A = 300 000 24

= Rs. 12500

Interest for the monthly loan amount = 12 500 ×

18100

× 112

= Rs. 187.50

Number of month units = 24 2

(24 + 1)

= 300Total interest = Rs. 187.50 × 300 = Rs. 56 250

1

1

1 1 1

4

10

Interest for the first = 300 000 × 10100

year for Institution B = Rs. 30 000 Amount for the second year = Rs. 330 000

Interest for the second year = 330 000 × 10100

= Rs. 33 000Total interest = Rs. 63 000

Since 63 000 > 56 250, more interest has to be paid by taking the loan from Institution B.∴ the friend’s statement is false.

1

1 1

1 1

10

2. (a) (i) y = 5 1 1

(ii) Correct axes Marking 6 points Smooth curve

1 1 1

3

(b) (i) Minimum value = (−4) 1 1

3(ii) Decreases negatively 1 + 1

2

(c) y = (x + 1) (x − 3)

= x2 − 2x − 3 1 x = √3 ∴ x2 = 3

∴ 0 = x2 − 3 2 21 − y = − 2x

Drawing y = − 2x √3 = 1.7 ± 0.1

1

1

1 3 3 10

= 300 000 × 110100

× 110100

= 363 000 = 363 000 − 300 000 = 63 000

- 1

- 1- 1

[See page 2

12

- 2 -

2. (a) (ii)

[See page 3

m%ia;drh

4 2 3

−2

−4

1

2

4

6

8

10

12

−1−2 5 x

y

×

×

×

×

×

×

×

×

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Question Number Answer

Marks Other facts

3. (i) 2 1 1

10

For the fdcolumn − 1Σfd − 1A + Σfd

40 − 1

(ii)Class

interval Mid

value (x) Frequency

(f) fx

4 − 8

9 − 13

14 − 18

19 − 23

24− 28

29 − 33

34 − 38

6111621263136

23581552

12338016839015572

Σf = 40 Σfx = 910

For the mid value column For the fx column Σfx = 910Mean number of trips = 910

40 = 22.75 = 23 (to the nearest whole number)

1 1 1 1

1 5

(iii) Cost of the soil = Rs. 23 × 4 × 2000

= Rs. 184 000 1 1

(iv) Expected cost for 2 days = Rs. 184 000 × 2 × 40

= Rs. 14 720 000

Since 14 000 000 < 14 720 000, the engineer’s

statement could be true.

1

1

1 3

4. Number of children = x or any other unknown term

Number of adults = y or any other unknown term

3x + 2y = 186 1

2x + y = 114 2

2 × 2 4x + 2y = 228 3

3 − 1 x = 42

By substituting x = 42 in 2

2 × 42 + y = 114

y = 114 − 84

y = 30

1

1

1

1

1

1

[See page 4

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Question Number Answer

Marks Other facts

Number of children = 42, Number of adults = 30

Total amount for a day = 42 × 100 + 30 × 150

Total amount for a week = (42 × 100 + 30 × 150) × 7

= Rs. 60 900

1 1 1 1

10 10

10

For squaring one of the two expressions from 2x – 1 and x + 3

For the formula -1

For substituting -1

5. (i) Length of the hypotenuse = 2x − 1 cm 1 1

(ii) Length of the remaining side = x + 3 cm

(2x − 1)2 = x 2 + (x + 3)2

1

1 2

(iii) 4x2 − 4x + 1 = x2 + x2 + 6x + 9

2x2 − 10x − 8 = 0 }x2 − 5x = 4 1

1 2

(iv) x2 − 5x + 25 4

= 4 + 25 4

( x − 5 2

)2 = 41 4

x =± √41 + 5

2

x = 2

6.4 + 5 x =

26.4 + 5 −

x = 5.7 x = − 0.7x cannot be negative ∴ x = 5.7 cm Length of the hypotenuse = 2 × 5.7 − 1 = 11.4 − 1 = 10.4 cm

1

1

1

1

1 5

6. (i) 25 m

B

A

D C E

20 m

12 m

40°

1 1

(ii) sin 40° = CE 20

0.6428 = CE 20

CE = 12.856 ∴ CE = 13 m (to the nearest metre)

1

1

1 3

[See page 5

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Question Number Answer

Marks Other facts

7. (i) Tn = a + (n − 1) d a = 50, d = 25, n = 12T12 = 50 + (12 − 1) × 25 = 50 + 275 = Rs. 325

1

1 2

10

50 + (n-1) 25 + 50 (n + 1 - 1) 25 = 425

(ii) Tn + Tn + 1 = 425 a + (n − 1) d + a + nd = 425 2a + 2nd − d = 425 2 × 50 + 2n × 25 − 25 = 425 50n = 350 n = 7 Question numbers 7 and 8

1

1

1 1 4

(iii) Sn = n2 {2a + (n − 1) d}

1300 × 2 = n2 {2 × 50 + (n − 1) 25}

5200 = n (75 + 25n)

208 = n2 + 3n

n2 + 3n − 208 = 0

(n + 16) (n − 13) = 0 n = −16 n = 13 ∴ Number of questions 13 + 1 = 14

1

1

1

1 4

Question Number Answer

Marks Other facts

(iii) tan EAD ^ = 13 + 2512 + 15

= 3827

= 1.4074

EAD ^ = 54° 36ʹ

sin 54° 36ʹ = 38AD

0.8151 = 38AD

AD = 46.62 m

1 1

1

1

1

1 6 10

Part B

[See page 6

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Question Number Answer

Marks Other facts

8. (i) Constructing PQ = 8 cm 1 1

(ii) Constructing QPR ^ = 45°"

Constructing the perpendicular bisector of PQ,

Marking R

1 1

1 3

(iii) Marking the centre

Constructing the circle

1 1

2

(iv) Obtaining S 1 1

(v) Obtaining T

Since PRQ ^ = 90°" QRS ^ = 90°' ∴ Since

PR = RS = RQ" PQR ^ = RQS ^ = 45°' ∴ OQS ^ = 90°

and hence QS is a tangent.

Since QS = ST, ST is also a tangent.

1

1 1 3 10

[See page 7

8.

QP O

RT

S

45°

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Question Number Answer

Marks Other facts

9. Volume of the prism = 1 2

× 2a × 3a × 8a

Volume of the cone = 1 3

× π(2 r)2 × h

1 2

1 3

× 2a × 3a × 8a = × π × 4r2 × 15

πr 2 × 15 4 3

5 = 24a3

20πr2 = 24a3

} r2 = 24a3 20π r2 = 6a

3 5π

1

1

1

1

lg r2 = lg 6 + 3 lg 4.55 − (lg 5 + lg 3.14) = 0.7782 + 3 × 0.6580 − (0.6990 + 0.4969) = 0.7782 + 1.9740 − 1.1959lg r2 = 1.5563 r2 = antilog 1.5563 r2 = 36.0 r = 6 cm Radius of the cone = 2 × 6 = 12 cm

1

2

1

1

1 10

1 mark for 2 logarithms

10. (i)Houses in the village Ranala

ω

A – Houses with television sets

B – Houses with radios7

3578

113120

Naming A and B Marking 78 and 35

11 + 1 3

(ii) 35 1 1

(iii) 85 120

1 1

[See page 8

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Question Number Answer

Marks Other facts

(iv)

ω

A

B

C

3

4926

66

12

Drawing CObtaining 9, 12, 4, 66, 26 and 3 (1 mark for each pair of correct values)

1

3 4

(v) Shading 1 1

10

11. (i) PQX ^ = 20°The angle subtended by an arc at the centre of a circle is twice the angle subtended on the circumference.

1

1 2

10

(ii) RT = TQ Since the straight line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord,

OTQ ^ = 90°Since a tangent is perpendicular to the radius at the point of contact,

OXY ^ = 90°

OTQ ^ + OXY ^ = 180°OTYX is a cyclic quadrilateral since a pair of opposite angles is supplementary.

1 1 1

1

4(iii) XOT ^ = 140° (the angles on a straight line are

supplementary)

XYZ ^ = 140° (the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle)

1

1 2

(iv) Diameter is OY'

∴ OT Y ^ = 90° (the angle in a semicircle is 90o) 1 1

2

[See page 9

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Question Number Answer

Marks Other facts

12. (i)

A

G

CE

B

D F

For drawing the figure 1 1

(ii) CE = EF (data)BE = EG (data)∴ Quadrilateral BCGF is a parallelogram since its diagonals bisect each other.AF = FG (data)BE = EG (data)According to the midpoint theorem,FE // AB ∴ AB // FCBC // FG (opposite sides of a parallelogram)∴ BC // AF

1

1

1 1

1

∴ Quadrilateral ABCF is a parallelogram since pairs of opposite sides are parallel.BC // AG Since parallelograms on the same base (base BC) and between the same pair of parallel lines (BC//AG) are equal in area, BCGF and ABCF are of equal area.

1

1

1 1 9

10