- 1 - g.c.e. (o.l.) support seminar – 2016 mathematics i ... · 4 6 2 6 boy boy boy first picking...
TRANSCRIPT
- 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
= 8
n(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
- 2 -
[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 × 2
B =
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
- 3 -
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
- 4 -
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
- 5 -
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 × 18
100 × 1
12
= 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
×
×
×
×
×
×
×
×
- 3 -
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
- 4 -
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
- 5 -
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
- 6 -
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.
QPO
RT
S
45°
- 7 -
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 = 6a3 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
- 8 -
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
- 9 -
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 // FC
BC // 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