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BEST BRAIN EXAMINATIONS KONSORTIUM

SPECIAL PRIVATE MOCK EXAMINATIONS FOR BECE CANDIDATES– OCTOBER 2025
MARKING SCHEME – MATHEMATICS
PAPER 2[60 MARKS@ 15 MARKS EACH]
QUESTION SOLUTION Marks
1 (a) H
H
B1
T
H H
T
H T
H
T T
H
T
T

(i) Outcomes = {HHH, HHT, HTH, HTT, THH, THT, TTH, B1

TTT}, = >n (s) = 8
(ii) Exactly two tails = {HTT, THT, TTH} = 3, M1A1
3
= > P (Exactly two tails) = 8
(iii) Only one head = {HTT, THT, TTH} = 3,
3 M1A1
= > P (only one head) = 8

(b) 𝑎 = (−4) and 𝑏 = (𝑞1)

𝑝
Then 𝑎 + 𝑏 = (−1−8
) M½
−𝟒 𝒒 −𝟏
( )+( )=( )
𝒑 𝟏 −𝟖 M½
−4 + 𝑞 = −1; 𝑞 = −1 + 4; 𝑞 = 3 M½
𝑝 + 1 = −8; 𝑝 = −8 − 1; 𝑝 = −9 M½
:. 𝑝 = −9 , 𝑞 = 3 A1
(c) Points: A (1, n) B (- 3, - 3)
5 𝑦 −𝑦 5 −3−𝑛 5 −3−𝑛 𝟏
Gradient (m) = 2 = > Gradient (m) = 𝑥2 −𝑥1 = > 2
= −3−1
=> 2
= −4 M𝟐
2 1

2 (– 3 – n) = 5 (– 4) Cross Multiply = > M𝟐

𝟏

– 6 – 2n = – 20 = > – 2n = – 20 + 6
𝟏
M
𝟐
−2𝑛 −14
−2
= −2
𝟏
M𝟐
𝟏
n=7 ∴ The value of n is 7 𝑨𝟐

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(d)
4 3
3𝑥
= 7 −𝑥
Multiply through by the L.C.M = 3x M1

4 3
(3x) 3𝑥 = (3x) 7 −(3𝑥) 𝑥 = >4 = 3x × 7 – 3 × 3 = > 4 = 21x – 9 =
M1
> 4 + 9 = 21x
13 21𝑥 13 13
21
=
21
=> 21
=𝑥 ∴𝑥=
21
A1

QUESTION 1 TOTAL = 15 MARKS

2 (a)
6 10 14 18
(i) Constant of proportionality = 3 = = = =2 =2= 2=2 M1
5 7 9

∴ The constant of proportionality = 2 A1

(ii) y = 7x Divide both sides by x

𝑦 7𝑥 𝑦
𝑥
= 𝑥 =>𝑥=7 ∴ The constant of proportionality is 7
M1A1

(b) Area of the Bigger circle = 𝜋𝑟 2 , where r = 14cm

22
= 7 × (14𝑐𝑚2 ) B½
22
= 7 × 14𝑐𝑚 × 14𝑐𝑚 B½
= 22 × 2𝑐𝑚 × 14𝑐𝑚
= 44cm × 14cm
= 616𝑐𝑚2
Area of the smaller circle = 𝜋𝑟 2 , where r = 7cm M½
22
= 7 × 7𝑐𝑚 × 7𝑐𝑚
B½
= 22cm × 7cm
= 154𝑐𝑚2 B½
Area of the shaded region = Area of the Bigger circle – Area of the
smaller circle
= 616𝑐𝑚2 − 154𝑐𝑚2 A½
= 462𝑐𝑚2

(c) Estimated number of drops of rain water in tank = 1.12 × 107

Estimated number of microbes = 5.80 × 109
5.80×109
Estimated number of microbes per drop of rain water = 1.16×107 =
580×1011
= 5.0 × 108 M1A1
116×109

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(d) i. Mean M½
𝑆𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
𝑀𝑒𝑎𝑛 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠
𝟏𝟐+𝟏𝟑+𝟏𝟑++𝟏𝟑+𝟏𝟒+𝟏𝟒+𝟏𝟓+𝟏𝟓
= M½
𝟖
109
= 8
= 13.6 A½
:. Mean is Gh¢ 13.63
Median
Arranging the data in alphabetical order
= > 12, 13, 13, 13, 14, 14, 15, 15
13+14
Median = = 13.5
2
:. Median is Gh¢ 13.5 A½
Mode
Mode = most frequently occurring value M½
= 13
:. Mode is Gh¢ 13 A½

ii. I. Well Paid?

To make the wages seem high, you would use the mean (Gh¢ 13.6) B1
because it is slightly higher than both the median and mode.
II. Badly Paid?
To make the wages seem low, you would use the mode (Gh¢ 13)
because it is the most common pay. B1
QUESTION 2 TOTAL = 15 MARKS
3 (a) Value of the angle marked x
3x + 33° = 4x + 8° [vertically opposite angles are equal] B1
33° − 8° = 4x – 3x
25° = x ∴ The value of x is 25° M1A1

Corresponding angles are equal B1

y = 3x + 33°, whereas x = 25°

y = 3 (25°) + 33°= > y = 75° + 33° 𝟏

y = 108° 𝑴𝟏 𝟐
OR
Alternate angles are equal B1
y = 4x + 8°, where x = 25°
y = 4 (25°) + 8° = >y = 100° + 8°
y = 108° 𝟏
𝑴𝟏
𝟐

(b) 5
Given 𝑐 = (−3 ), 𝑑 = (−2 ), 𝑒 = (12)
5 6
𝟖𝒄 + 𝒅 + 𝟐𝒆 = 𝟖(−𝟑) + ( 𝟓 ) + 𝟐(𝟏𝟐
𝟓 −𝟐
)
𝟒𝟎 −𝟐 𝟐𝟒
𝟔 A½
= (−𝟐𝟒) + ( 𝟓 ) + (𝟏𝟐) M½
𝟔𝟐
= (−𝟕 )
A1

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(c) From the diagram

Opposite (O) = 3 B½
Adjacent (A) = 4
Hypotenus (H) = 5

𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 (𝑂) 3
A½
Sin x = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠 (𝐻) = Sin x = 5
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 (𝐴)
Cos x = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠 (𝐻) = Cos x = 5
4 A½
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 3
Tan x = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = Tan x = 4 A½
(d)i. Expressing 31 days in minutes, 60 minutes = 1 hour
1 day = 24 hours = >Number of minutes in a day = 24hrs ×
60 minutes = 1440 mins
In 31 days = 31 × 1440 mins= 44640 minutes M1A
In standard form = 4.4640 × 𝟏𝟎𝟒 𝟏
A
𝟐
893 147 893 ×147 131271
ii. 0.893 × 14.7 = >1000 = 10 =1000 ×10 = 10000
13.1271 (4 decimal places) M1A1

QUESTION 3 TOTAL = 15 MARKS

4 (a)
M1
A1

B1

B2 B2

B2

P (- 2, - 2) Q (- 8, - 8) R (- 2, - 6)
(iii)

(iv)
Reflection on the line y = 0
(x, y) → (x, - y)/ P (- 2, - 2) → 𝑃1 (−2, 2)/

Q (- 8, - 8) → 𝑄1 (−8, 8)

P (- 2, - 6) → 𝑅1 (−2, 6)

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(v)
Reflection in the line x =1
(x, y) → (2k – x, y) or (𝑦𝑥 ) → (2𝑘−𝑥
𝑦
)
𝑃1 (−2
2
) → 𝑃2 (2(1)−(−2)
2
)
−2 4
𝑃1 ( 2 ) → 𝑃2 (2)
𝑄1 (−8
8
) → 𝑄2 (2(1)−(−8)
8
)
−8 10
𝑄1 ( 8 ) → 𝑄2 ( 8 )
𝑅1 (−2
6
) → 𝑅2 (2(1)−(−2)
6
)
−2 4
𝑅1 ( 6 ) → 𝑅2 (6)

𝑃1 (−2, 2) → 𝑃2 (4,2)
B½ B½
𝑄1 (−8,8) → 𝑄2 (10,8)
B½ B½
𝑅1 (−2,6) → 𝑅2 (4,6)
B½ B½

(vi) An anticlockwise rotation of 180° about the origin

(x, y) → (– x, – y)
𝑃1 (−2, 2) → 𝑃3 (2, −2) B½

𝑄1 (−8,8) → 𝑄3 (8, −8) B½

𝑅1 (−2,6) → 𝑅3 (2, −6) B½

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QUESTION 4 TOTAL = 15 MARKS

5 (a)i Number of Frequency No. of vehicles ×
vehicles per day Frequency
0 1 0
1 1 1
2 10 20
3 7 21 B2
4 5 20
5 2 10
6 0 0
TOTALS 26 72

ii.
Vertical axis – B2 (– ½ ee)
Horizontal axis- B2 (– ½ ee)
Graph of mean number of vehicles that stop for routine checks
Drawing each bar correctly
= ½ each x 6= 3 marks

Penalties
Wrong/non-labeling of
𝟏
vertices − ee
𝟐
Non-joining or non-use
𝟏
of straight edge− 𝟐 ee
Non-calibration of
𝟏
axes− ee
𝟐
𝟏
Non-labelling of axes− 𝟐 ee

Note: Not drawn to scale

Scale
x-axis – 2cm : 1 unit
y-axis – 2cm: 1 unit

(b) Let the son’s age = x,

1 1 M1
The man’s age = 3 (x) = > 2
3(𝑥) = 21 = > (2) 2 3(𝑥) = 21(2)

3𝑥 21×2
3
= 3 = > x = 7 × 2 = > x = 14
∴ The son is 14 years M1A1

(c) 55 ÷ 11 + 7 × 6 – 24 + 31 = 5 + 42 – 24 + 31 = 47 – 24 + 31 = 23 + 31= M1A1

54

OCTOBER 2025– MATHS MARKING SCHEME

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(d) 𝑎 ÷ 102 = 150

𝒂 = 𝟏𝟓𝟎 × 𝟏𝟎𝟐
𝒂 = 𝟏𝟓𝟎𝟎𝟎 M1
A1

6 (a)i. Monthly salary = GH¢ 3, 600.00

(i) The value of y = 1300 + 500 + 300 + y + 350 = 3600 M1A1
1300 + 500 + 300 = 350 + y = 3600 = > 2450 + y = 3600 = > y =
3600 – 2450 = > y = 1150

Expenditure Amount (GH¢) Angle

Food 1300 1300
3600
×360° = 130°
B½
Education 500 500
×360° = 50°
10
B½
Aged Parents 300 300
×360° = 30°
10
B½
Building Projects 1150 1150
×360° = 115°
10
B½
Miscellaneous 350 350
×360° = 35°
10
B½
3600 360°

Food

130° 50°
35° 30°
115° Aged Parents
B½
(ii)
Building Project

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(b) 1
Wage = Gh¢8 4 per hour
𝟏 𝟏 𝟑 𝟑𝟑 𝟗𝟗
Overtime rate = 𝟏 𝟐 × 𝟖 𝟒 = 𝟐 × = 𝐆𝐡¢ per hour M1
𝟒 𝟖
Amount Michael earns on Tuesday =(8 4 × 8) + ( 8 × 4)
1 99 M1
33 99
= ( 4 × 8) + ( 8 × 4)
𝟗𝟗
= (𝟑𝟑 × 𝟐) + ( 𝟐 )
= 66 + 49.5 M1
= 115.5
:. Michael earns Gh¢115.5 on Tuesday A1
(c) A ∪ B = {1, 2, 3, 4, 5, 6} A1

(d) Araba’s ratio = 3, Akua’s ratio = 5, Araba’s share = x, Akua’s share = x + 600
For x, 3:5 = x : x + 600 M1
3 𝑥
5
= 𝑥+600 = > 5× 𝑥 = 3 × 𝑥 + 600 = > 5x = 3x + 1800 = > 5x –
2𝑥 1800
3x =1800 = > 2
= 2
x = 900

Araba’s share = GH¢ 900.00

Akua’s share = GH¢ 900.00 + GH¢ 600 = GH¢ 1500.00 M1

Total Amount shared = GH¢ 900.00 + GH¢ 1500.00 = 2400.00

∴ They shared GH¢ 2,400.00 A1
QUESTION 6 TOTAL = 15 MARKS

OBJECTIVES
PAPER I
[40 MARKS]
1. C 11. C 21. C 31. B
2. C 12. B 22. C 32. C
3. C 13. A 23. B 33. C
4. D 14. A 24. A 34. A
5. C 15. B 25. A 35. B
6. D 16. C 26. A 36. D
7. A 17. D 27. D 37. A
8. C 18. A 28. A 38. D
9. B 19. C 29. C 39. C
10. B 20. D 30. A 40. B

OCTOBER 2025– MATHS MARKING SCHEME