2015 BECE Mathematics (Maths) Past Questions – Paper Two

1. (a) Find the difference between the product of 2.5 and 7.5 and the sum of 2.75 and 9.55.

(b) Solve $\frac{3x+2}{3}-\frac{3-x}{8}=\frac{1}{6}$

(c) A container is 24 m long, 9 m wide and 8 m high. How many books can it hold if each book is 20 cm long, 16 cm wide and 6 cm thick.

(a) Product of 2.5 and 7.5
$= 2.5 × 7.5\\ = \frac{25}{10}\times\frac{75}{10}\\ = \frac{1875}{100}\\ = 18.75\\$

Sum of 2.75 and 9.55
=
2.75
+ 9.55
__________
12.30
__________

Difference between 18.75 and 12.30
= 18.75
– 12.30
__________
6.45
__________

(b) Solving $\frac{3x+2}{3}-\frac{3-x}{8}=\frac{1}{6}\\ ⇒ 24\left(\frac{3x+2}{3}\right)-24\left(\frac{3-x}{8}\right)=24\left(\frac{1}{6}\right)\\ ⇒ 8\left(3x+2\right)-3\left(3-x\right)=4\left(1\right)\\ ⇒ 24x + 16 - 9 + 3x = 4\\ ⇒ 24x + 3x = 4 - 16 + 9\\ ⇒ 27x = - 3 \\ ⇒ \frac{27x}{27}\ \ =\ \ \frac{-3}{27}\\ ⇒ x =\ -\frac{\mathbf{1}}{\mathbf{9}} \\$

Volume of each book = 20 cm × 16 cm × 6 cm
= 320 cm2 × 6 cm
= 1920 cm3

$\text{No. of books the container can hold } = \frac{Volume\ of\ container}{Volume\ of\ each\ book}\\ = \frac{1728000000\ {cm}^3}{1920\ {cm}^3}\\ = 900,000 books$

2. (a) In a test consisting of 90 questions, Ama answered 75% of the first 40 questions correctly. If she had to get a score of 80% in the test,
(i) how many questions did she answer correctly out of the first 40 questions?
(ii) how many questions should she answer correctly out of the 90 questions ?
(iii) what percentage of the remaining 50 questions should she answer correctly in order to get the 80%?

(b) Three interior angles of a pentagon are 100°, 120° and 108°. Find the size of each of the remaining two interior angles, if one of them is three times the other.

2(a) (i) No.of questions Ama answered correctly out of first 40 questions
$= 75\text{% of first 40 questions} = \frac{75}{100}\times40 = \frac{75\times4}{10} = 30 \ \ questions$

(ii) To score 80% in the test, then she needs to answer

$= 80 \text{ % x 90 questions } = \frac{80}{100}\times90 = 8 × 9 \ = 72 \ questions \ correctly$

(iii) No. of questions she must answer correctly in the remaining 50 questions
= 72 – 30 questions
= 42 questions

Percentage of 42 out of 50 questions

= $= \frac{42}{50}\times100%$
= 42 × 2%
= 84%

(b) Sum of interior angles of a pentagon (5-sided polygon)
= (n – 2) × 180° , where n = no.of sides
= (5 – 2) × 180° [n = 5 sides]
= 3 × 180°
= 540°

Let size of smaller missing angle = x
then, size of bigger missing angle = 3x

Now, if sum of interior angles = 540°,
⇒ 100° + 120° + 108° + x + 3x = 540°
⇒ 328° + 4x = 540°
⇒ 4x = 540° – 328°
⇒ 4x = 212°
⇒ x = 212/4
⇒ x = 53°

Hence, the other missing angle = 3x
= 3 × 53°
= 159°
The sizes of the two remaining interior angles = 53° and 159°

3. (a) Given that vectors $p = \left(\begin{array}{c}2\\ 2\end{array}\right) \ and \ q = \left(\begin{array}{c}x\\ y\end{array}\right), find:$

(i) q if q – p = $\left(\begin{array}{c}12\\ 9\end{array}\right)$ ;

(ii) the magnitude of the vector q – p

(b)

(c) In the diagram |AB| = |AC|, angle ADC = 30° and angle ACD = 7x – 25°. Find
the value of x;
(ii) angle DAC;
(iii) angle BAD.

(a) $if \ q- p \ = \left(\begin{array}{c}12\\ 9\end{array}\right) \\ then \ \left(\begin{array}{c}x\\ y\end{array}\right)-\left(\begin{array}{c}2\\ 2\end{array}\right) = \left(\begin{array}{c}12\\ 9\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right) = \left(\begin{array}{c}12\\ 9\end{array}\right) + \left(\begin{array}{c}2\\ 2\end{array}\right) \\ \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}12+2\\ 9+2 \end{array}\right) \\\left(\begin{array}{c}x\\ y\end{array}\right) = \left(\begin{array}{c}14\\ 11\end{array}\right) \\ q = \left(\begin{array}{c}14\\ 11\end{array}\right)$

(ii) Magnitude of vector q – p
$=\text{magnitude of }\left(\begin{array}{c}12\\ 9\end{array}\right)\\ = \sqrt{{12}^2+9^2}\\ = \sqrt{144+81}\\ = \sqrt{225}\\ = 15 \ units\\$

(b) (i) Since |AB| = |AC|
⇒ angle ABC = angle ACB [Base angles of isosceles triangle equal]
Let angle ABC = angle ACB = y

Then, y + y + 50° = 180° [interior angles of a triangle =180°]
⇒ 2y = 180° – 50°
⇒ 2y = 130°
⇒ y = 130/2
⇒ y = 65°

Now, 65° + (7x – 25°) = 180° [angles at a point on a straight line=180°]
⇒ 7x + 65° – 25° = 180°
⇒ 7x + 40 = 180°
⇒ 7x = 180° – 40°
⇒ 7x = 140°
⇒ x = (140°)/7
⇒ x = 20°

(ii) Angle DAC + 7x – 25° + 30° = 180° [interior angles of a triangle =180°]
Let angle DAC = a
⇒ a + 7x – 25° + 30° = 180°
⇒ a + 7(20°) – 25° + 30° = 180°
⇒ a + 140° – 25° + 30° = 180°
⇒ a + 115° + 30° = 180°
⇒ a + 145° = 180°
⇒ a = 180° – 145°
⇒ a = 35°
⇒ angle DAC = 35°

(iii) angle BAD = angle BAC + angle DAC
= 50° + 35°
= 85°

4. (a) The Value Added Tax (VAT) paid by a man on a deep freezer was GHC 90.00. If VAT was charged at 15%,
what was the price of the deep freezer?
How much did the man pay including VAT?
(b) The average of the numbers 5, 7, 2, 6, x, (x+1), 7 and 4 is 5. Find the value of x.
(c) Simplify: $\frac{mn+mp+nq+pq}{n+p}$

(a) (i) If (VAT) 15% → GH¢ 90.00
Then (Original price) 100% → ? (more)

If more, less (15%) divides, hence

= $\frac{100%}{15%}\times GHc\ 90$
= 100 × GHc 6
= GHc 600
Original price = GHc 600.00

(ii) Total amount paid = Original price + VAT
= GHc 600.00 + GHc 90.00
= GHc 690.00

(b) If the average of 8 no.s: 5, 7, 2, 6, x, (x+1), 7 and 4 = 5, then
$⇒ \frac{5+7+2+6+x+x+1+7+4}{8}\ =\ \ 5 \\ ⇒ \frac{32+2x}{8}\ =\ \ 5 \\ ⇒ 8\left(\frac{32+2x}{8}\right)\ =\ 8\ (5) \\ ⇒ 32 + 2x = 40 \\ ⇒ 2x = 40 - 32 \\ ⇒ 2x = 8 \\ ⇒ x = 8/2 \\ ⇒ x = 4 \\$

(c) Simplification of $\frac{mn+mp+nq+pq}{n+p}\\ \frac{m(n+p)+q(n+p)}{n+p}\\ = \frac{\left(n+p\right)(m+q)}{n+p}\\ = \frac{\left(n+p\right)(m+q)}{n+p}\\ = m + q\\$

5. (a) A cylinder which has a height of 90 cm and diameter 14 cm is closed at both ends.
Find:
(i) its total surface area;
(ii) the volume of the cylinder
[Take π = 22/7]
(b) (i) Using a ruler and a pair of compasses only, construct triangle PQR such that
|PQ| = 8cm, angle PQR = 120° and |QR| = 6 cm.
(ii) Measure:
(α) |PR|;
(β) angle QPR

5(a) (i) h = 90cm, d = 14 cm,
⇒ r = 14cm÷2
r = 7cm

Total Surface Area of closed cylinder
= 2πr2 + 2πrh , where r = radius, h = height
$= \left(2\times\frac{22}{7}\times7\times7\right)+\ \left(2\times\frac{22}{7}\times7\times90\right)\\ = (2 × 22 × 7) + (2 × 22 × 90)\\ = 308 + 3960\\ = 4268 cm^2\\$

(ii) Volume of cylinder
$= π r^2 h , \text{ where r = radius, h = height}\\ = \frac{22}{7}\times7\times7\times90\\ = 22 × 7 × 90\\ = 154 × 90\\ = 13860 cm^3\\$

(b) (i)

(ii) Measure:
(α) |PR| = 12.1 cm (± 0.1cm)

(β) angle QPR = 25° (± 1°)

6. The table shows the distribution of grades of candidates in an examination.

Grade	1	2	3	4	5	6
Frequency	2	3	6	5	4	10

(a) Using a graph sheet, draw a bar chart for the distribution
(b) If all candidates who obtained grades above grade 3 were awarded credit, find the probability that a candidate selected at random obtained credit.
(c) Calculate, correct to the nearest whole number, the mean grade of the candidates.

(a) Bar chart for the frequency distribution table below
G

Grade	1	2	3	4	5	6
Frequency	2	3	6	5	4	10

(b) Number of candidates who obtained credit (grades above grade 3 for the distribution)
= Frequencies of Grade 1 and Grade 2
= 2 + 3
= 5

Total number of candidates = 2 + 3 + 6 + 5 + 4 + 10
= 30

Probability of selecting a candidate who obtained credit

= (No.of candidates who obtained credit)/(Total no.of candidates)
= 5/30 = 1/6

(c) $Mean \ grade = \frac{Sum\ of\ all\ grades}{Total\ no.\ \ of\ candidates}\\ = \frac{\left(1\times2\right)+\left(2\times3\right)+\left(3\times6\right)+\left(4\times5\right)+\left(5\times4\right)+(6\times10)}{2+3+6+5+4+10}\ \\ = \frac{2\ +6+18+20+20+60}{2+3+6+5+4+10}\\ = \frac{126}{30} = \frac{42}{10} = 4\frac{2}{10}\\ = 4\frac{1}{5} \ or \ 4.2\\ ≈ 4 \text{(to nearest whole number)} \\$

JUNE 2015 (Second Sitting)
MATHEMATICS 2
ESSAY
1 hour

1. (a) In a class of 70 students, 40 belong to the Red Cross Society, 27 belong to the Girls’ Guide Society and 12 belong to both the Red Cross Society and the Girls’ Guide Society. The remaining students do not belong to any of the two societies.
Illustrate the information on a Venn diagram
How many students belong to the Red Cross Society only?
How many students do not belong to any of the two societies?

(b) A farmer uses 1/3 of his land to plant cassava, 2/5 of the remaining land to plant maize and the rest vegetables. If vegetables cover an area of 10 acres, what is the total area of the farmer’s land.

(a) (i) Let n(R) = Number of Red Cross Society members
n(G) = Number of Girls’ Guide Society members
U = Universal set (total number in class)
x = Remaining students (not belonging any of the two)

(ii) No. of students in Red Cross Society only
= 40 – 12
= 28

(iii) No. of students who do not belong to any of the two societies
= 70 – (28 + 12 + 15)
= 70 – 55
= 15

(b) $Cassava \ Fraction = \frac{1}{3}\\ Remaining\ fraction = 1-\frac{1}{3} = \frac{2}{3}\\ Maize \ fraction = \frac{2}{5} \ of\ remaining \\ = \frac{2}{5}\times\frac{2}{3}\\ = \frac{4}{15}\\ Vegetable \ fraction = Remaining \ fraction - maize \ fraction\\ = \frac{2}{3}-\frac{4}{15}\\ = \frac{5\left(2\right)\ -\ 1(4)}{15}\\ = \frac{10\ -\ 4}{15}\\ = \frac{6}{15} = \frac{\mathbf{2}}{\mathbf{5}}\\ Now, \ by \ simple \ proportion,\\ If \ (vegetable) \ \frac{2}{5} → 10 \ acres,\\ Then, \ (total \ area \ ) 1 \ → \ ? \ (more)\\ If \ more, \ less \ (2/5) \ divides\\ ⇒ \frac{1}{\frac{2}{5}}\times10\ acres\\ ⇒ \frac{5}{2}\times10\ acres\\ = 5 × 5 acres\\ Total \ area = \ 25 acres\\$

2. (a) Solve for x, if $2-\frac{3}{5}x\ \le\ 1{\frac{1}{3}}$

(b) At a rally attended by 520 people, 30% were Fantes, 25% Ewes, 15% Nzemas, 20% Gas and the rest Gonjas
How many Gonjas were at the rally?
How many more Fantes than Nzemas were at the rally ?
Draw a pie chart to illustrate the information.

(a) $\text{Solving for x in } 2 - \frac{3}{5} \le 1 \frac{1}{3}\\ ⇒ 2\ \ -\ \ \frac{3}{5}x\ \ \ \ \le\ \ \ {\frac{4}{3}}\\ ⇒ 15\left(2\right)-15\left(\frac{3}{5}x\right)\ \le\ \ 15\left(\frac{4}{3}\right)\\ ⇒ 15\left(2\right)-3\left(3x\right)\ \le\ \ 5\left(4\right)\\ ⇒ 30 - 9x \le 20\\ ⇒ 30 - 20 \le 9x\\ ⇒ 10 \le 9x\\ ⇒ \frac{10}{9}\ \ \le\ \ \frac{9x}{9}\\ ⇒ \frac{10}{9}\ \ \le\ \ x\\ ⇒ x \ge \frac{10}{9}\\$

(b) (i) Percentage of Gonjas = Total percentage – Sum of other percentages
= 100% – (30%+25%+15%+20%)
= 100% – 90%
= 10%

Number of Gonjas at the rally = 10% × Total number at rally
= 10% × 520
= 10/100 × 520
= 52

(ii)
Number of Fantes = 30 % of 520
= 30/100 × 520
= 3 × 52
= 156

Number of Nzemas = 15% of 520
= 15/100 × 520
= 3/2 × 52
= 78

Number of Fantes more than Nzemas = No. of Fantes – No.of Nzemas
= 156 – 78
= 78

(iii)

3. (a) Mr. Mensah’s farm is 20 km from his house. He uses his car to travel y km of the distance from his house and then walks $1{\frac{1}{2}}$ hours at the rate of 3 km per hour to get to his farm. Find y

(b) The perimeter of a square field is the same as that of a rectangular field. If the length of the rectangular field is 8 km and the width is 5 km, calculate the area of the square field.

(a) Distance travelled by car = y km
Distance walked = one and half hr × 3 km / h
= 4.5 km
Total distance = 20 km

Dist.by car + Dist. walked = Total distance
y + 4.5 = 20
y = 20 – 4.5
= 15.5

(b)

Perimeter of square = Perimeter of rectangle
= 2 × length + 2 × width
= 2 × 8 km + 2 × 5 km
= 16 km + 10 km
= 26 km

Hence perimeter of square = 26 km
Now perimeter of square = 4 ×Length
Length of square = 26 km ÷ 4
= 6.5 km

Now, area of square = Length × length
= 6.5 × 6.5
= 42.25 cm2

(c) Gradient of line = $\frac{y_2-\ y_1}{x_2-{\ x}_1}\\ = \frac{6-(-1)}{-3-2}\\ = \frac{\ \ 6+1}{-3-2}\\ = \frac{\ \ 7}{-5}\\ = -\frac{\ 7}{5}\\$

4. (a) Using a ruler and a pair of compasses only, construct triangle PST with angle PST = 30°, |ST| = 9 cm and |PS| = 12 cm

(b) With T as centre, draw a circle of radius 6 cm

(d) Label the intersections of the circle and the mediator Q1 and Q2.

(e) (i) Measure |Q1Q2|
(ii) Measure angle Q1TQ2

4. (a)

(e) (i) |Q1Q2| = 11.3cm (± 1cm)

(ii) angle Q1TQ2 = 145° (± 1°)

5. (a) Anita bought 51 tubers of yam at 3 for GHC 10.00. If she sold them and made a loss of 40%, how much did she sell each tuber of yam?

(b) The volume of a cylinder closed at one end is 1056 cm3. If its height is 21 cm, find its
(i) diameter
(ii) total surface area.

(a) Cost Price of the 51 tubers = $\frac{51}{3}\times GHc\ 10$
= 17 × GHc 10
= GHc 170

Selling Price percentage = 100% – 40% = 60%

If (Cost Price) 100% → GHc 170
Then (Selling Price) 60% → ? (less)

If less, then more (100) divides
⇒ $\frac{69}{100}\times GHc\ 170$
⇒ 6 × GHc 17
⇒ GHc 102

Hence selling price of all 51 tubers = GHc 102
⇒ Selling price of each tuber = GHc 102 ÷ 51
= GHc 2.00

(b) (i) $\text{The volume of cylinder closed at one end = }1056 cm^3\\ ⇒ π r^2 h = 1056 \\ ⇒ \frac{22}{7}\ \times{\ r}^2\ \times\ 21 = 1056\\ ⇒ 22 × r^2 × 3 = 1056 \\ ⇒ 66 × r^2 = 1056 \\ ⇒ \frac{{\ 66\ r}^2}{66}\ = \frac{1056}{66}\\ ⇒ r2 = 16\\ ⇒ r = \sqrt{16}\\ ⇒ r = 4cm\\ \text{Hence, diameter = 2 × radius}\\ = 2 × 4cm\\ = 8cm\\$

(ii) $\text{Total surface area of cylinder closed at one end}\\ \text{= Circular Bottom + Curved surface}\\ = π r 2 + 2 π r h\\ = \left(\frac{22}{7}\ \times{\ 4}^2\right)+\ \ \ \left(2\times\ \frac{22}{7}\times4\times21\right)\\ = \left(\frac{22}{7}\ \times4\times4\right)+\ \ \ \left(2\times\ \frac{22}{7}\times4\times21\right)\\ = \left(\frac{22\ \times\ 4\ \times\ 4}{7}\ \right)+\ \ \ \left(\ 2\times22\times4\times3\right)\\ = \frac{352}{7} + 528\\ = 50.29 + 528.00\\ = 578.29 cm^2\\$

6. (a) (i) Copy and complete the following mapping:

x 1 2 3 4 5 6 7
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
y 5 7 9 – – – 17

(ii) Determine the rule for the mapping

(b) Draw two perpendicular axes Ox and Oy on a graph sheet.

(c) Using a scale of 2cm to 1 unit on the x-axis and 2cm to 2 units on the y-axis, mark the x-axis from 0 to 8 and the y-axis from 0 to 20.

(d) Plot the point for each ordered pair (x, y) and join them with a straight line.

(e) Find:
(i) y, when x is 0;
(ii) x, when y is 14

6. (a) (i)
x 1 2 3 4 5 6 7
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
y 5 7 9 11 13 15 17

(ii) $Gradient = \frac{7-5}{2-1} = \frac{2}{1} = 2$

Constant = 3 (The image of 0)

Hence from general equation y = mx + c, where m = gradient, c = constant
Equation = y = 2x+3
Therefore Rule = x → 2x + 3

(b), (c), (d)

(e) (i) When x is 0, y = 3

(ii) When y is 14, x = 5.5

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2015 BECE Mathematics (Maths) Past Questions – Paper Two

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