KUULCHAT
S.H.S MATHEMATICS MOCK

OBJECTIVE TEST

1.

If P = {-2, 0, 2, 4, 6} and Q = {-3, -1, 0, 2, 3, 5}, find the set P ∩ Q.

A.

{}

B.

{-3, -1, 3, 5}

C.

{-2, 4, 6}

D.

{0, 2}

Answer: D

Intersection (∩) are elements that can be found in both sets

P = {-2, 0, 2, 4, 6}

Q = {-3, -1, 0, 2, 3, 5}

P ∩ Q = {0, 2}

2.

Simplify: [( 16 9 ) -3 2 x 16 -3 4 ] 1 3

A.

1 4

B.

3 8

C.

9 16

D.

3 4

Answer: B

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = [( 16 9 ) -3 2 x 16 -3 4 ] 1 3

Notes:

1. 16 = 4 x 4 = 42

Expressed this way in order to cancel the denominator 2 of the -3 2 power

2. 9 = 3 x 3 = 32

Expressed this way in order to cancel the denominator 2 of the -3 2 power

3. 16 = 2 x 2 x 2 x 2 = 24

Expressed this way in order to cancel the denominator 4 of the -3 4 power

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = [( 42 32 ) -3 2 x (24) -3 4 ] 1 3

From the law of indices,

( a b ) c = ac bc

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = [ 42 x -3 2 32 x -3 2 x 24 x -3 4 ] 1 3

Notes:
1. The power 2 cancel the denominator 2 of the power -3 2

2. The power 4 cancel the denominator of 4 of the -3 4

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = [ 4-3 3-3 x 2-3] 1 3

4-3 3-3 = ( 4 3 ) -3

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = [( 4 3 )-3 x 2-3] 1 3

The 1 3 affects each of the powers in the bracket.

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = ( 4 3 )-3 x 1 3 x 2-3 x 1 3

The power 3 cancels the denominator 3 in 1 3

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = ( 4 3 )-1 x 2-1

Notes:

1. The power -1 means inverse which means you reciprocate (the numerator becomes the denominator and the denominator becomes the numerator).

2. ( 4 3 ) -1 = 3 4

3. Every number is being divided by 1 but it is not written.

2 = 2 1

4. 2-1 = 1 2

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = 3 4 x 1 2

[( 16 9 ) -3 2 x 16 -3 4 ] 1 3 = 3 x 1 4 x 2 = 3 8

3.

Find the first quartile of 7, 8, 7, 9, 11, 8, 7, 9, 6 and 8.

A.

8.5

B.

7.0

C.

7.5

D.

8.0

Answer: B

Arrange the numbers in ascending order.

6, 7, 7, 7, 8, 8, 8, 9, 9, 11

First Quartile (ungrouped data) = n 4 th value

Where n is the number of items.

n = 10

First Quartile (ungrouped data) = 10 4 = 2.5th value

First Quartile = 7 + 7 2 = 7.0

4.

A fair die is tossed twice. What is the probability of getting a sum of at least 10?

A.

5 36

B.

2 3

C.

5 18

D.

1 6

Answer: D

1. Calculate the total number of outcomes when a fair die is tossed twice: 6 × 6 = 36

At least 10 means 10 or more.

2. Find the favorable outcomes for a sum of at least 10: (4,6), (6,4) for 10; (5,5) for 10; (5,6), (6,5) for 11, (6,6) for 12

Number of favourable outcome is 6

Probability Formula

Probability = Number of possible selection Number of samples

P(Sum is at least 10) = 6 36 = 1 6

You can also draw a table for all the possible outcomes.

1 2 3 4 5 6
1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
Probability Formula

Probability = Number of possible selection Number of samples

Number of samples = 6 x 6 = 36

Number of possible selection = 6

P(Sum is at least 10) = 6 36 = 1 6

5.

Find the quadratic equation whose roots are 1 2 and - 1 3 .

A.

6x2 - x - 1 = 0

B.

3x2 + x - 1 = 0

C.

6x2 + x - 1 = 0

D.

3x2 + x + 1 = 0

Answer: A

If α and β are the roots, then

(x - α)(x - β) = 0

(x - 1 2 )(x - - 1 3 ) = 0

Note: -- = +

(x - 1 2 )(x + 1 3 ) = 0

x(x + 1 3 ) - 1 2 (x + 1 3 ) = 0

x2 + x 3 - x 2 - 1 x 1 2 x 3 = 0

x2 + x 3 - x 2 - 1 6 = 0

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 3, 2 and 6.

The L.C.M of 3, 2 and 6 is 6.

6 x x2 + 6 x x 3 - 6 x x 2 - 6 x 1 6 = 0 x 6

6x2 + 2x - 3x - 1 = 0

6x2 - x - 1 = 0

6.

From a point T, a man moves 12 km due West and then moves 12 km due South to another point Q.

Calculate the bearing of T from Q.

A.

225°

B.

315°

C.

045°

D.

135°

Answer: C

Since two sides are equal, the right angle is isosceles and the base angles are the same.

Base angle = 180 - 90 2 = 45°

Bearing starts from the north clockwise. If the angle is less than 100°, begin with a 0.

7.

In the diagram PQ is parallel to RS, ∠ QFG = 105° and ∠ FEG = 50°.

Use the diagram to answer the question below

Find the value of m.

A.

55°

B.

75°

C.

105°

D.

130°

Answer: A

QFG + ∠ EFG = 180°

105° + ∠ EFG = 180°

EFG = 180° - 105°

EFG = 75°

FEG + ∠ EGF + ∠ EFG = 180° (Sum of angles of a triangle)

50° + m + 75° = 180°

m + 125° = 180°

m = 180° - 125°

m = 55°

8.

The area of a sector of a circle and the length of its arc are 231 cm2 and 66 cm respectively. Calculate the radius of the circle

[Take π = 22 7 ]

A.

3.5 cm

B.

7.0 cm

C.

10.5 cm

D.

14.0 cm

Answer: B

Area of a sector

Area of a sector = θ 360 x area of the circle

Area of a sector = θ 360 x π x radius x radius

Where θ = angle subtended by the arc at the centre of the circle

Length of an arc

Length of an arc = θ 360 x circumference of the circle

Length of an arc = θ 360 x 2 x π x radius

Where θ = angle subtended by the arc at the centre of the circle

The angle (θ) substended are the same.

Area of a sector = θ 360 x π x radius x radius

Area = 231 cm2

θ 360 x π x radius x radius = 231

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator 360.

360 x θ 360 x π x radius x radius = 360 x 231

θ x π x radius x radius = 360 x 231

Divide both sides by π x radius x radius

θ = 360 x 231 π x radius x radius --- eqn 1

Length of an arc = θ 360 x 2 x π x radius

Length = 66 cm

θ 360 x 2 x π x radius = 66

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator, 360.

360 x θ 360 x 2 x π x radius = 360 x 66

θ x 2 x π x radius = 360 x 66

Divide both sides by 2 x π x radius

θ = 360 x 66 2 x π x radius --- eqn 2

--- eqn 1 = --- eqn 2

360 x 231 π x radius x radius = 360 x 66 2 x π x radius

Cross multiply.

360 x 231 x 2 x π x radius = 360 x 66 x π x radius x radius

Divide both sides by 360 x π x radius to cancel out.

360 x 231 x 2 x π x radius 360 x π x radius = 360 x 66 x π x radius x radius 360 x π x radius

231 x 2 = 66 x radius

Divide both sides by 66

Radius = 231 x 2 66 = 7 cm

9.

Find the truth set of the equation (x - 2)2 + 3 = (x + 1)2 - 6.

A.

{-2}

B.

{-1}

C.

{1}

D.

{2}

Answer: D

Concept

(a + b)2 = a2 + 2 x a x b + b2 = a2 + 2ab + b2
(a - b)2 = a2-2 x a x b + b2 = a2 -2ab + b2

Alternatively
(a + b)2 = (a + b)x(a + b) [Expand and simplify]
(a -b)2 = (a - b)x(a - b) [Expand and simplify]

(x - 2)2 = x2 - 2 x x x 2 + 22 = x2 -4x + 4
(x + 1)2 = x2 + 2 x x x 1 + 12 = x2 + 2x + 1

(x - 2)2 + 3 = (x + 1)2 - 6
x2 - 4x + 4 + 3 = x2 + 2x + 1 - 6
x2 - 4x + 7 = x2 + 2x - 5
x2 - x2 - 4x - 2x = -5 - 7
-6x = -12

Divide both sides by -6

x = -12 -6 = 2

10.

The sum of the interior angles of a regular polygon with k sides is (3k - 10) right angles.

Find the size of the exterior angle.

A.

60°

B.

40°

C.

90°

D.

120°

Answer: A

Interior angles of a regular polygon

Sum of interior angles = (n - 2) x 180°

Where n = the number of sides of the polygon.

n = k.

Sum of interior of regular polygon with k sides = (k - 2) x 180

But the sum is given as (3k - 10) x 90°

Note: right angle = 90°

(k - 2) x 180 = (3k - 10) x 90

Divide both sides by 90

(k - 2) x 2 = (3k - 10) x 1

2k - 4 = 3k - 10

10 - 4 = 3k - 2k

6 = k

∴ the polygon has 6 sides.

Exterior angle of a regular polygon

Exterior angle of a regular polygon = 360° n

Where n is the number of sides of the polygon.

Number of sides of the polygon = 6

Exterior angle of the polygon = 360° 6 = 60°

11.

If the area of the trapezium MNOP is 300 cm2, find the value of x.

A.

15 cm

B.

12 cm

C.

10 cm

D.

20 cm

Answer: A

The area of the trapezium is the summation of the area of the triangle and the rectangle.

Area of triangle = 1 2 x base x height

Area of a rectangle = Length x Breadth

Base of the triangle = 24 - 16 = 8

Height of the triangle = x
Breadth of rectangle = x

300 = 1 2 x 8 x x + 16 x x

2 divides itself once and 8, 4 times.

300 = 4x + 16x
300 = 20x

Divide both sides by 20

x = 300 20 = 15

12.

The diagram shows a circle centre O. Use it to answer the question below.

Find the value of x.

A.

43°

B.

47°

C.

54°

D.

86°

Answer: B

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

OBC = ∠ OCB

ABD = ∠ OBC = 43°

OBC = ∠ OCB = y = 43°

OBC + ∠ OCB + ∠ BOC = 180° (Sum of angles in a triangle)

OBC = ∠ OCB = 43°

43° + 43° + ∠ BOC = 180°

86° + ∠ BOC = 180°

BOC = 180° - 86°

BOC = 94°

Central Angle

Central angle is twice any inscribed angle subtended by the same arc.

x = BOC 2 = 94° 2 = 47°

13.

In the diagram, O is the centre of the circle. If ∠ NLM = 74°, ∠ LMN = 39° and ∠ LOM = x, find the value of x.

A.

106°

B.

113°

C.

126°

D.

134°

Answer: D

Central angle is twice any inscribed angle subtended by the same arc.

LOM = 2 x ∠ LNM

LNM = ∠ LOM 2 = x 2

LNM + ∠ NLM + ∠ LMN = 180° (Sum of angles in a triangle)

LNM = x 2

NLM = 74°

LMN = 39°

x 2 + 74 + 39 = 180

Get rid of the fraction by multiplying both sides of the equation by the denominator 2.

x + 2 x 74 + 2 x 39 = 180 x 2

x + 148 + 78 = 360

x + 226 = 360

x = 360 - 226

x = 134

14.

When the point (4, 5) is rotated through an angle in the anticlockwise direction about the origin, its image is (-5, 4). What is the angle of rotation?

A.

90o

B.

270o

C.

180o

D.

300o

Answer: A

Rotation through 90o anticlockwise

When rotating a point 90o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-y,x).

In other words, switch x and y and make y negative

Rotation through 270o anticlockwise

When rotating a point 270o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(y,-x).

This means, we switch x and y and make x negative

Rotation through 180o anticlockwise

When rotating a point 180o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-x,-y).

So all we do is make both x and y negative

15.

Calculate the gradient of the line which passes through the points (1, 4) and (-2, 6).

A.

- 3 2

B.

- 2 3

C.

2 3

D.

3 2

Answer: B

Gradient = Change in Y Change in X = Y2 - Y1 X2 - X1

Y2 = 6
Y1 = 4
X2 = -2
X1 = 1

Gradient = 6 - 4 -2 - 1 = 2 -3 = - 2 3

16.

If P = {x: 1 ≤ x ≤ 6} and Q = {x: 2 < x < 9} where xR, find PQ.

A.

{x: 2 ≤ x < 6}

B.

{x: 2 ≤ x ≤ 6}

C.

{x: 2 < x ≤ 6}

D.

{x: 2 < x < 6}

Answer: C

P = {x: 1 ≤ x ≤ 6}

Read as x is such that 1 is less than or equal to x and x is less than or equal to 6.

P = {1, 2, 3, 4, 5, 6}

Q = {x: 2 < x < 9}

Read as x is such that 2 is less than x and x is less than 9.

Q = {3, 4, 5, 6, 7, 8}

Intersection (∩) are elements that can be found in both sets

(PQ) = {3, 4, 5, 6}

(PQ) = {x: 2 < x ≤ 6}

Note: ≤ means less than or equal to, hence the number at the right of it is inclusive.

17.

Evaluate : 2 28 - 3 50 + 72

A.

4 7 + 2

B.

4 7 - 9 2

C.

4 7 - 11 2

D.

4 7 - 21 2

Answer: B

2 28 - 3 50 + 72 = 2 28 - 3 50 + 72

Express each surds to be in the form Perfect square x a number

28 = 4 x 7

50 = 25 x 2

72 = 36 x 2

2 28 - 3 50 + 72 = 2 4 x 7 - 3 25 x 2 + 36 x 2

From a x b = a x b

2 28 - 3 50 + 72 = 2 x 4 x 7 - 3 x 25 x 2 + 36 x 2

2 28 - 3 50 + 72 = 2 x 2 x 7 - 3 x 5 x 2 + 6 x 2

2 28 - 3 50 + 72 = 4 7 - 15 2 + 6 2

2 28 - 3 50 + 72 = 4 7 - 9 2

18.

Which of these statements about an acute-angled triangle is true?

A.

It has three equal angles.

B.

It has two equal sides.

C.

It has all its angles less than 90°.

D.

It has one angle less than 90°.

Answer: C

19.

Which of the following about parallelograms is true?

A.

Opposite angles are supplementary

B.

Opposite angles are complementary

C.

Opposite angles are equal

D.

Opposite angles are reflex angles

Answer: C

Angles of a Parallelogram

1. The opposite angles of a parallelogram are equal

A = ∠ C; ∠ D = ∠ B.

2. All the angles of a parallelogram add up to 360°.

A + ∠ B + ∠ C + ∠ D = 360°.

3. All the respective consecutive angles are supplementary.

A + ∠ B = 180°; ∠ B + ∠ C = 180°; ∠ C + ∠ D = 180°; ∠ D + ∠ A = 180°

20.

The gradient of the line passing through the points (3, 6) and (x, 4) is - 2 5 . Find the value of x.

A.

3

B.

8

C.

6

D.

5

Answer: B

Gradient = Change in Y Change in X = Y2 - Y1 X2 - X1

Y2 = 4
Y1 = 6
X2 = x
X1 = 3

Gradient = 4 - 6 x - 3

Gradient = -2 x - 3

But gradient = - 2 5 = -2 5

-2 x - 3 = -2 5

Since the numerator at the left and right are the same, the denominators are also the same.

x - 3 = 5

x = 5 + 3

x = 8

21.

If 9 y + 1 = ( 1 27 ) y - 2, find y.

A.

4 5

B.

- 4 5

C.

1

D.

-4

Answer: A

9 = 3 x 3 = 32
27 = 3 x 3 x 3 = 33

a-b = 1 ab

1 27 = 1 33 = 3-3

32(y + 1) = 3-3(y - 2)

Since the bases are the same, the powers are also the same.

2(y + 1) = -3(y - 2)
2y + 2 = -3y + 6

Note: - x - = +
-3 x -2 = + 6

2y + 2 = -3y + 6
2y + 3y = 6 - 2
5y = 4

Divide both sides by 5

y = 4 5

22.

The interquartile range of a distribution is 7. If the 25th percentile is 16, find the upper quartile.

A.

9

B.

23

C.

30

D.

35

Answer: B

Upper quartile - Lower quartile (25th percentile) = Interquartile range

Lower quartile (25th percentile) = 16

Interquartile range = 7

Upper quartile - 16 = 7
Upper quartile = 7 + 16
Upper quartile = 23

23.

Fati buys milk at ₦ x per tin and sells each at a profit of ₦ y. If she sells 10 tins of milk, how much does she receive from the sales?

A.

₦ 10(x + y)

B.

₦ (10x + y)

C.

₦ (x + 10y)

D.

₦ (xy + 10)

Answer: A

Selling Price = Cost Price + Profit

Selling Price of each = ₦ (x + y)

For the 10 tins, selling price = 10 x ₦ (x + y)

Amount received for the 10 tins = ₦ 10(x + y)

24.

To arrive on schedule, a ferry is to cover a distance of 40 km at 50 km/h. If the ferry delays for 18 minutes before starting the journey, at what speed must it move so as to arrive on schedule?

A.

70 km/h

B.

80 km/h

C.

90 km/h

D.

100 km/h

Answer: B

Speed

Speed = Distance Time

Distance = Speed x Time

Time = Distance Speed

The regular time to reach destination = 40 km 50 km/h = 0.8 hours

18 minutes = 18 60 hours = 0.3 hours

Time left to get to destination = 0.8 hours - 0.3 hours = 0.5 hours.

The ferry must cover the 40 km in 0.5 hours.

Speed = Distance Time = 40 km 0.5 h = 80 km/h

25.

The pie chart represents fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there?

A.

40

B.

80

C.

90

D.

70

Answer: C

Angle of sector of orange = 360° - (100° + 60° + 120°)

Angle of sector of orange = 360° - 280°

Angle of sector of orange = 80°

Angle of sector of apple = 120°

Number of oranges = 60

If 80° = 60

120° = 60 x 120° 80° = 90

There are 90 apples

26.

Simplify: 1 2 (x-1)- 1 3 ( 1 2 x-1).

A.

1 6 (x-1)

B.

1 6 (2x+1)

C.

1 6 (2x-1)

D.

1 6 (1-2x)

Answer: C

1 2 (x-1)- 1 3 ( 1 2 x-1)

Expand the brackets

1 2 (x-1)- 1 3 ( 1 2 x-1) = x 2 - 1 2 - x 6 + 1 3

Note: - x - = +

- 1 3 x - 1 = 1 3

The Least Common Multiples (L.C.M) of the denominators (2,3 and 6) is 6

1 2 (x-1)- 1 3 ( 1 2 x-1) = 3 x x - 3 x 1 - 1 x x + 2 x 1 6

1 2 (x-1)- 1 3 ( 1 2 x-1) = 3x - 3 - x + 2 6

1 2 (x-1)- 1 3 ( 1 2 x-1) = 3x - x - 3 + 2 6

1 2 (x-1)- 1 3 ( 1 2 x-1) = 2x - 1 6

1 2 (x-1)- 1 3 ( 1 2 x-1) = 1 6 (2x - 1)

27.

Calculate the distance between points (3, -2) and (8, 10).

A.

12 units

B.

13 units

C.

14 units

D.

15 units

Answer: B

Distance between two points

Distance = (x2 - x1)2 + (y2 - y1)2

Distance = (8 - 3)2 + (10 -- 2)2

Note: - - = +

Distance = (5)2 + (10 + 2)2

Distance = 52 + 122

Distance = 25 + 144

Distance = 169 = 13 units

28.

Find the value of ∠ RPT in the diagram.

A.

20°

B.

40°

C.

50°

D.

60°

Answer: B

QRS = ∠ PQT = 40° (Corresponding angles)

PQT is an isosceles triangle (Two sides equal).

Isosceles triangle has base angles equal.

RPT = ∠ PQT = 40° (Base angles equal)

29.

If cos(3x + 28°) = sin (2x + 48°), 0° ≤ x ≤ 90°, find the value of x.

A.

2.8°

B.

15.2°

C.

20.0°

D.

22.0°

Answer: A

cos (3x + 28°) = sin (2x + 48°)

Trigonometric Values

sin (90° - θ) = cos θ

cos (90° - θ) = sin θ

sin (90° + θ) = cos θ

cos (90° + θ) = -sin θ

Use the trigonometric values to change either cos to sin or sin to cos and equate the angles to solve for x.

Method I

Changing the cos to sin.

cos (3x + 28°) = sin (90° - [3x + 28°])

cos (3x + 28°) = sin (2x + 48°)

sin (90° - [3x + 28°]) = sin (2x + 48°)

Since both sides are sin, the angles are the same.

90° - (3x + 28°) = 2x + 48°

90° - 3x - 28° = 2x + 48°

90° - 28° - 48° = 2x + 3x

90° - 76° = 5x

14° = 5x

Divide both sides by 5.

x = 14° 5 = 2.8°

Method II

Changing the sin to cos.

sin (2x + 48°) = cos(90° - [2x + 48°])

cos (3x + 28°) = sin (2x + 48°)

cos (3x + 28°) = cos(90° - [2x + 48°])

Since both sides are cos, the angles are the same.

3x + 28° = 90° - (2x + 48°)

3x + 28° = 90° - 2x - 48°

3x + 2x = 90° - 48° - 28°

5x = 14°

Divide both sides by 5.

x = 14° 5 = 2.8°

30.

Make b the subject of the relation lb = 1 2 (a + b)h.

A.

al 2 - h

B.

al 2l - h

C.

2l - h al

D.

ah 2l - h

Answer: A

lb = 1 2 (a + b)h

Get rid of the fraction by multiplying both sides by the denominator 2

lb x 2 = h(a + b)

2lb = ah + bh

2lb - bh = ah

Factorize b out.

b(2l - h) = ah

Divide both sides by (2l - h)

b = ah 2l - h

31.

In the diagram, EFGH are points on a circle centre P. If ∠ EFG = 142o and ∠ FGP = 63o, find ∠ FEP.

A.

63o

B.

76o

C.

79o

D.

89o

Answer: C

Central angle is twice any inscribed angle subtended by the same arc.

EPG = 2 x ∠ EFG

EPG = 2 x 142o

EPG = 284o

EPG + x = 360o (Sum of angles of a circle)

284o + x = 360o

x = 360o - 284o

x = 76o

Sum of interior angles of a quadrilateral is 360o

Note: sum of interior angles = (n - 2) x 180o, where n is the number of sides.

63o + 142o + ∠ FEP + x = 360o

63o + 142o + ∠ FEP + 76o = 360o

FEP + 281o = 360o

FEP = 360o - 281o

FEP = 79o

32.

The sum of the interior angles of a polygon is 1260°. Find the number of sides.

A.

9

B.

6

C.

7

D.

8

Answer: A

Interior angles of a regular polygon

Sum of interior angles = (n - 2) x 180°

Where n = number of sides of the regular polygon.

Sum of interior angles = 1260°

(n - 2) x 180° = 1260°

(n - 2) x 180 = 1260

Divide both sides by 180

n - 2 = 1260 180

n - 2 = 7

n = 7 + 2

n = 9

33.

Find the value of x in the diagram.

A.

16

B.

44

C.

36

D.

28

Answer: D

Sum of exterior angles of a polygon = 360°

3x + 2x + x + 10 + 4x + 50 + 20 = 360

10x + 80 = 360

10x = 360 - 80

10x = 280

Divide both sides by 10

x = 280 10 = 28

34.

A die is rolled once. Find the probability of obtaining a number less than 3.

A.

1 6

B.

5 6

C.

2 3

D.

1 3

Answer: D

Probability = Number of possible selection Number of samples

Numbers in a die rolled once = {1,2,3,4,5,6}

Numbers less than 3 = {1,2}

Number of possible selection = 2
Number of samples = 6

Probability of less than 3 = 2 6 = 1 3

35.

What is the coefficient of x in the expansion of (4x2 + 3x - 1)(3x + 1)?

A.

-1

B.

0

C.

1

D.

2

Answer: B

(4x2 + 3x - 1)(3x + 1) = (4x2 + 3x - 1)(3x + 1)

(4x2 + 3x - 1)(3x + 1) = 3x(4x2 + 3x - 1) + 1(4x2 + 3x - 1)

(4x2 + 3x - 1)(3x + 1) = 12x3 + 9x2 - 3x + 4x2 + 3x - 1

(4x2 + 3x - 1)(3x + 1) = 12x3 + 9x2 + 4x2 - 3x + 3x - 1

(4x2 + 3x - 1)(3x + 1) = 12x3 + 13x2 + 0x - 1

36.

Make U the subject of the relation: x = 2U - 3 3U + 2 .

A.

U = 2x + 3 3x + 2

B.

U = 2x + 3 3x - 2

C.

U = 2x - 3 3x - 2

D.

U = 2x + 3 2 - 3x

Answer: D

x = 2U - 3 3U + 2

Get rid of the fraction by multiplying each sides by the denominator (3U + 2).

x x (3U + 2) = 2U - 3

3xU + 2x = 2U - 3

3xU - 2U = -2x - 3

Factorize U out.

U(3x - 2) = -2x - 3

Divide both sides by (3x - 2).

U = -2x - 3 3x - 2

Factorize (-1) out.

U = -1 -1 x 2x + 3 -3x + 2

-1 cancels -1 and you can rearrange -3x + 2 as 2 - 3x

U = 2x + 3 2 - 3x

Alternatively,

3xU + 2x = 2U - 3

2x + 3 = 2U - 3xU

Factorize U out.

2x + 3 = U(2 - 3x)

Divide both sides by (2 - 3x)

2x + 3 2 - 3x = U

U = 2x + 3 2 - 3x

37.

Simplify: (11two)2

A.

1001two

B.

1101two

C.

101two

D.

10001two

Answer: A

Method I

A number raised to the power 2 (square) means the number multiplied by itself.

1 1
1 1
x 1 1
1 1
+ 1 1
1 0 0 1

Note: 2 in base two is 10.

Method II

Change the binary (base two) to base 10 and simplify and change the result back to base two.

11two = 1 x 21 + 1 x 20

11two = 1 x 2 + 1 x 1

Note: any number raised to the power 0 is 1.

11two = 2 + 1 = 3

(11two)2 = 32 = 3 x 3 = 9

Change the 9 base 10 to base 2.

9 base 10 to base 2

9 2 4 r 1 2 2 r 0 2 1 r 0 2 0 r 1

You list the remainders from the bottom to the top.

9ten = 1001two

38.

If 2a = 64 and b a = 3, evaluate a2 + b2.

A.

48

B.

90

C.

160

D.

250

Answer: B

a = a 1 2

64 = 64 1 2

64 = 26

64 = 26 x 1 2 = 23

2a = 23

Since the bases are the same, the powers are also the same.

a = 3

b a = 3

b = 3 x a

b = 3 x 3

b = 9

a2 + b2 = 32 + 92

a2 + b2 = 9 + 81

a2 + b2 = 90

39.

In the diagram, O is the centre of the circle. SOQ is a diameter and ∠ SRP = 37°.

Find ∠ PSQ.

A.

37°

B.

53°

C.

65°

D.

127°

Answer: B

Notes:

Angles subtended by the diameter (or semi-circle) is 90°.

QPS = 90°

Angles subtended by the same arc are the same.

From arc SP,

PRS = ∠ PQS = 37° (Angles subtended by the same arc)

PQS + ∠ QPS + ∠ PSQ = 180° (Sum of angles in a triangle)

37° + 90° + ∠ PSQ = 180°

127° + ∠ PSQ = 180°

PSQ = 180° - 127°

PSQ = 53°

40.

Abudu can do a piece of work in 6 days and Efah can do the same work in 3 days. What fraction of the work can both do together in a day?

A.

1 3

B.

1 8

C.

2 3

D.

1 2

Answer: D

Abudu takes 6 days to complete work.

Fraction of work done by Abudu in a day = 1 6

Efah takes 3 days to complete the same work.

Fraction of work completed by Efah in a day = 1 3

Fraction of work done by both in a day is the sum of fractions of work done by each.

Fraction of work done by both = 1 6 + 1 3

Fraction of work done by both = 1 + 2 6

Fraction of work done by both = 3 6

Fraction of work done by both = 1 2

Alternatively

Let y the work to be done

Abudu takes 6 days to complete work.

Work completed by Abudu in a day = 1 6 x y

Work completed by Abudu in a day = y 6

Efah takes 3 days to complete the same work.

Work completed by Efah in a day = 1 3 x y

Work completed by Efah in a day = y 3

Work completed by both Abudu and Efah in a day is the sum of work done by each in a day.

Work done by both = y 6 + y 3

Work done by both = y + 2y 6

Work done by both = 3y 6

Work done by both = y 2

Fraction of work done in a day = Work done in a day Total work to be done

Total work to be done is y

Work done by both in a day = y 2

Fraction of work done in a day = y 2 ÷ y

Every number is divisible by 1

Fraction of work done in a day = y 2 ÷ y 1

Reciprocate the fraction at the right of ÷ and change the ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

Fraction of work done in a day = y 2 x 1 y

The y cancels each other.

Fraction of work done in a day = 1 2

41.

Make x the subject of the relation:

E = kx2 2y + z.

A.

x = [ 2y(E - z) k ]2

B.

x = 2y(E + z) k2

C.

x = 2y(E - z) k

D.

x = 2yk(E + z)

Answer: C

E = kx2 2y + z

Multiply both sides the denominator 2y

E x 2y = kx2 2y x 2y + z x 2y

2Ey = kx2 + 2yz

kx2 = 2Ey - 2yz

Divide both sides by k.

x2 = 2Ey - 2yz k

Factorize 2y out.

x2 = 2y(E - z) k

Take square root of both sides.

x = 2y(E - z) k

42.

Factorize: p2q2 - 6pqr + 9r2

A.

(pq - 3r)2

B.

(pq - 3r)(pq + 3r)

C.

(pq + 3r)2

D.

(pr + 3q)(pr - 3q)

Answer: A

Since 3 can go into 6 and 9, split -6pqr into -3pqr and -3pqr

p2q2 - 6pqr + 9r2 = p2q2 - 3pqr - 3pqr + 9r2

p2q2 - 6pqr + 9r2 = pq(pq - 3r) - 3r(pq - 3r)

p2q2 - 6pqr + 9r2 = (pq - 3r)(pq - 3r)

Since pq - 3r = pq - 3r, the multiplication makes it square

p2q2 - 6pqr + 9r2 = (pq - 3r)2

The equation above is similar to the general factorization rule of (a - b)2

(a - b)2 = a2 - 2ab + b2

Thus the first term square then the sign (either + or -) and 2 times the first and second terms + the second term square

Similarly (a + b)2 = a2 + 2ab + b2

The first term for the equation, p2q2 - 6pqr + 9r2 is pq and the second term is 3r and the sign is -

43.

The height of a square base pyramid is thrice the length of a side of its base. If the base area is 324 cm2, find the volume of the pyramid.

A.

17,496 cm3

B.

5,832 cm3

C.

324 cm3

D.

972

Answer: B

Volume of a square base pyramid = Area of base x height 3

Area of a square = Length x Length = L2

Area of the base = 324 cm2

L2 = 324

Take square root of both sides.

L = 324 = 18 cm

The height of a square base pyramid is thrice the length of a side of its base

Height = 3 x Length of its base

Length of base = 18 cm

Height = 3 x 18 cm = 54 cm

Volume of the square base pyramid = 324 x 54 3 = 5,832 cm3

44.

The graph is for the relation: y = d - x.

Find the value of d.

A.

2

B.

1

C.

0

D.

-1

Answer: B

Pick a point on the line of the relation: y = d - x and substitute into the relation to find d.

y = d - x

Using the point (1, 0)

x = 1 and y = 0

0 = d - 1

1 = d

d = 1

OR

Using the point (0, 1)

x = 0, y = 1

1 = d - 0

1 = d

d = 1

OR

Using the point (-2, 3)

x = -2 and y = 3

3 = d - (-2)

Note: -- or -(-) = +

3 = d + 2

3 - 2 = d

1 = d

d = 1

45.

Find the value of x for which 2x - 1 x2 + 2x + 1 is not defined.

A.

1 2

B.

1

C.

2

D.

-1

Answer: D

For a relation to be not defined/undefined, the denominator must be equal to 0

x2 + 2x + 1 = 0

Express 2x as two numbers when you multiply you will get c (1) and when you add you will get b (2)

1 x 1 = 1 and 1 + 1 = 2 hence the number is 1

x2 + x + x + 1 = 0

x(x + 1) + 1(x + 1) = 0

(x + 1)(x + 1) = 0

When x + 1 = 0, x = -1

46.

A farmer cleared 40% of a piece of land the first day and 60% of the remainder the next day. What percentage of the land was remaining at the end of the second day?

A.

30%

B.

24%

C.

20%

D.

15%

Answer: B

Let y = area of the field

First day area cleared = 40% of y

First day area cleared = 40 100 x y = 0.4y

Remaining uncleared area = y - 0.4y

Remaining uncleared area = 0.6y

Second day area cleared = 60% of remaining

Second day area cleared = 60 100 x 0.6y

Second day area cleared = 0.36y

Remaining area after the second day = 0.6y - 0.36y

Remaining area after the second day = 0.24y

Percentage = Number Total x 100%

Percentage of the land remaining at the end of the second day = 0.24y y x 100%

Note: the ys cancel each other.

Percentage of the land remaining at the end of the second day = 0.24 x 1 1 x 100% = 24%

47.

The mean of the numbers 2, 5, 2x and 7 is not greater than 5. Find the range of values of x.

A.

x ≤ 3

B.

x ≥ 3

C.

x < 3

D.

x > 3

Answer: A

Mean = Sum of items Number of items

Mean = 2 + 5 + 2x + 7 4

The mean is not greater than 5 means the mean can be less than 5 or equal to 5.

2 + 5 + 2x + 7 4 ≤ 5

14 + 2x 4 ≤ 5

Get rid of the fraction by multiplying both sides by the denominator, 4

4 x 14 + 2x 4 ≤ 5 x 4

14 + 2x ≤ 20

2x ≤ 20 - 14

2x ≤ 6

Divide both sides by 2.

x 6 2

x ≤ 3

48.

Solve 4x2 - 16x + 15 = 0

A.

x = -1 1 2 or -2 1 2

B.

x = 1 1 2 or -1 1 2

C.

x = 1 1 2 or 2 1 2

D.

x = 1 1 2 or -2 1 2

Answer: C

4x2 - 16x + 15 = 0

Notes:

1. a x c = 4 x 15 = 60

2. Split b (-16) such that two numbers when you sum you will get b (-16) and when you multiply you will get ac (60). The two numbers are -10 and -6

3. -16x = -10x - 6x

4x2 -10x - 6x + 15 = 0

2x(2x - 5) - 3(2x - 5) = 0

(2x - 5)(2x - 3) = 0

When 2x - 5 = 0, x = 5 2 = 2 1 2

When 2x - 3 = 0, x = 3 2 = 1 1 2

49.

x 6.20 6.85 7.5
y 3.9 5.2 6.5

The points on a linear graph are as shown in the table. Find the gradient (slope) of the line.

A.

1 2

B.

1

C.

2

D.

2 1 2

Answer: C

Gradient = Change in Y Change in X = Y2 - Y1 X2 - X1

Choose any two points from the table.

P1(6.20, 3.90) and P2(6.85, 5.20)

Gradient = 5.2-3.9 6.85-6.20 = 1.3 0.65 = 2

50.

The foot of a ladder is 6 m from the base of an electric pole. The top of the ladder rests against the pole at a point 8 m above the ground. How long is the ladder?

A.

7 m

B.

10 m

C.

12 m

D.

14 m

Answer: B

Let x = the length of the ladder.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

x2 = 62 + 82

x2 = 36 + 64

x2 = 100

Take square root of both sides.

x = 100 = 10

The ladder is 10 m long.

THEORY QUESTIONS

1.

(a)

Solve, correct to one decimal place, tan(θ + 25°) = 5.145, where ° ≤ θ ≤ 90°.

(b)

In the relation t = m n2 + 4r :

(i)

make n the subject of the relation;

(ii)

find the positive value of n when t = 25, m = 5 and r = 4.

(a)

tan(θ + 25°) = 5.145

Take tan inverse (tan-1) of both sides.

θ + 25° = tan-1(5.145)

θ + 25° = 79°

θ = 79° - 25°

θ = 54°

(b)

(i)

t = m n2 + 4r

Get rid of the square root by squaring both sides of the equation.

t2 = m2 x (n2 + 4r)

Divide both sides by m2

t2 m2 = n2 + 4r

t2 m2 - 4r = n2

Every number is being divided by 1

4r = 4r 1

t2 m2 - 4r 1 = n2

1 x  t2 - m2 x 4r m2 = n2

t2 - m24r m2 = n2

Get rid of the square by taking square root of both sides.

n = t2 - m24r m2

(ii)

n = t2 - m24r m2

When t = 25, m = 5 and r = 4

n = 252 - 52  x  4  x  4 52

n = 625 - 400 25

n = 225 25 = 9 = 3

2.

(a)

Copy and complete the following table for the relation: y = 2(x + 2)2 - 3 for -5 ≤ x ≤ 2.

x -5 -4 -3 -2 -1 0 1 2
y -1 -3 5

(b)

Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of the relation y = 2(x + 2)2 - 3 for -5 ≤ x ≤ 2.

(c)

Use the graph to find the solution of:

(i)

2(x + 2)2 = 3;

(ii)

2(x + 2)2 = 5.

(d)

For what values of x, from the graph is y increasing in the interval?

(a)

y = 2(x + 2)2 - 3

When x = -5

y = 2(-5 + 2)2 - 3

y = 2(-3)2 - 3

y = 2 x 9 - 3

y = 18 - 3

y = 15

When x = -4

y = 2(-4 + 2)2 - 3

y = 2(-2)2 - 3

y = 2 x 4 - 3

y = 8 - 3

y = 5

When x = -3

y = 2(-3 + 2)2 - 3

y = 2(-1)2 - 3

y = 2 x 1 - 3

y = 2 - 3

y = -1

When x = -2

y = 2(-2 + 2)2 - 3

y = 2(0)2 - 3

y = 2 x 0 - 3

y = 0 - 3

y = -3

When x = -1

y = 2(-1 + 2)2 - 3

y = 2(1)2 - 3

y = 2 x 1 - 3

y = 2 - 3

y = -1

When x = 0

y = 2(0 + 2)2 - 3

y = 2(2)2 - 3

y = 2 x 4 - 3

y = 8 - 3

y = 5

When x = 1

y = 2(1 + 2)2 - 3

y = 2(3)2 - 3

y = 2 x 9 - 3

y = 18 - 3

y = 15

When x = 2

y = 2(2 + 2)2 - 3

y = 2(4)2 - 3

y = 2 x 16 - 3

y = 32 - 3

y = 29

x -5 -4 -3 -2 -1 0 1 2
y 15 5 -1 -3 -1 5 15 29

(b)

Notes:

1. Each grid value = Interval on the axis Number of grids per interval

2. Each grid value on the y-axis = 5 10 = 0.5

3. Number of grids = Point Grid value

4. Number of grids on the y-axis for 1 = 1 0.5 = 2

5. Number of grids on the y-axis for 3 = 3 0.5 = 6

6. Number of grids on the y-axis for 29 = 29 0.5 = 58

(c)

Notes:

1. Each grid value = Interval on the axis Number of grids per interval

2. Each grid value on the x-axis = 1 10 = 0.1

(i)

Express the relation in such a way that one side equals the given relation.

2(x + 2)2 = 3

2(x + 2)2 - 3 = 0

The solution are the values of x where the line y = 0 cuts the curve.

x = -3.2 ± 0.1 or -0.8 ± 0.1

(ii)

Express the relation in such a way that one side equals the given relation.

2(x + 2)2 = 5

Subtract 3 from each sides.

2(x + 2)2 - 3 = 5 -3

2(x + 2)2 - 3 = 2

The solution are the values of x where the line y = 2 cuts the curve.

x = -3.6 ± 0.1 or -0.4 ± 0.1

(d)

The values of x for which y is increasing from the graph is −2 < x ≤ 2

3.

In a military camp, 50 officers had a choice of beans, plantain and rice. Of these officers, 21 chose beans, 24 plantains and 18 rice. Also 3 chose beans only, 9 plantain only, 2 rice only and 5 chose all three kinds of food.

(a)

Illustrate the information on a Venn diagram.

(b)

Use the Venn diagram to find the number of officers who chose:

(i)

plantain and beans only;

(ii)

exactly two kinds of food;

(iii)

none of the three kinds of food.

(a)

Let U = {Officers}

n(U) = 50

Let B = {Officers who chose beans}

n(B) = 21

Let R = {Officers who choose rice}

n(R) = 18

x = number of officers who chose beans and plantain only

y = number of officers who chose beans and rice only

z = number of officers who chose rice and plantain only

a = number of officers who chose none of the three kinds of food

Covering the rice (R) region

18 + 3 + x + 9 + a = 50

x + a + 30 = 50

x + a = 50 - 30

x + a = 20 ------ eqn 1

Covering the beans (B) region

21 + a + 2 + z + 9 = 50

a + z + 32 = 50

a + z = 50 - 32

a + z = 18 ------ eqn 2

Covering the plantain (P) region

24 + 3 + y + 2 + a = 50

y + a + 29 = 50

y + a = 50 - 29

y + a = 21 -------- eqn 3

Sum of the beans (B) region should give 21.

3 + y + 5 + x = 21

y + x + 8 = 21

y + x = 21 - 8

y + x = 13 ------- eqn 4

Get rid of a by subtracting --- eqn 1 from ---- eqn 3

y - x = 1 -------- eqn 5

Solve ---- eqn 4 and ---- eqn 5 simultaneously.

---- eqn 4 + ----- eqn 5

2y = 14

Divide both sides by 2

y = 14 2 = 7

Substitute y = 7 into ---- eqn 5

7 - x = 1

7 - 1 = x

6 = x

x = 6

Substitute x = 6 into ------ eqn 1

6 + a = 20

a = 20 - 6

a = 14

Substituting a = 14 into --- eqn 2

14 + z = 18

z = 18 - 14

z = 4

Redraw the Venn diagram and substitute the values of the variables

(b)

(i)

plantain and beans only = 6

(ii)

exactly two kinds of food = 6 + 7 + 4 = 17

(iii)

none of the three kinds of food = 14

4.

(a)

Given that P = ( y 8 3 2 ) , Q = ( -3 -5 -4 x ) , R = ( -56 -93 z -27 ) and PQ = R, find the values of x, y and z

(b)

(i)

Draw on a graph paper, using a scale of 2 cm to 1 unit on both axes, the lines x = 1; y = 2; and x + y = 5.

(ii)

Shade the region which satisfies simultaneously the inequalities x + y ≤ 5; y ≥ 2 and x ≥ 1

(a)

2 x 2 Matrix Multiplication

Multiply the elements of each row of the first matrix by the elements of each column in the second matrix (element by element) and add the products

( a b c d ) x ( e f g h ) = ( ae + bg af + bh ce + dg cf + dh )

PQ = ( y 8 3 2 ) ( -3 -5 -4 x ) = ( -3y - 32 -5y + 8x -9 - 8 -15 + 2x )

PQ = ( -3y - 32 -5y + 8x -17 -15 + 2x )

PQ = R

( -3y - 32 -5y + 8x -17 -15 + 2x ) = ( -56 -93 z -27 )

-3y - 32 = -59

-3y = -59 + 32

-3y = -27

Divide both sides by -3

y = -27 -3 = 9

-15 + 2x = -27

2x = -27 + 15

2x = -12

Divide both sides by 2

x = -12 2 = -6

z = -17

(b)

x + y = 5

When x = 0

y = 5

Coordinates = (0, 5)

When y = 0

x = 5

Coordinates = (5, 0)

5.

(a)

Abiola starts a business with $1,250.00. Frances joins the business later with a capital of $1,875.00. At the end of the first year, profits are shared equally between Abiola and Frances. When did Frances join the business?

(b)

In the diagram, CE||AD and ED = EB = AB, find the value of:

(i)

n;

(ii)

m.

(a)

Let's take it that Frances joined the business n months after Abiola started it.

Abiola invested $1,250 for 12 months for the same profit as Frances who invested $1875 for n months.

Since the profit are shared equally, then Abiola's contribution is the same as Frances' contribution.

Contribution = Amount invested x Number of months in the business

Abiola's contribution = $1,250 x 12

Number of months that Frances was in the business = 12 - n

Frances' contribution = $1875 x (12 - n)

1250 x 12 = 1875 x (12 - n)

Divide both sides by 1875

12 - n = 1250 x 12 1875

12 - n = 8

12 - 8 = n

n = 4 months

∴ Frances joined the business 4 months after Abiola started.

(b)

ABE and ∆ BED are isosceles (Two sides equal)

Isosceles triangles have the base angles equal.

(i)

BAE = ∠ CEA = 40° (Alternate angles)

AEB = ∠ BAE = 40° (Base angles equal)

AEB + ∠ BAE + ∠ ABE = 180°(Sum of angles in a triangle)

40° + 40° + n° = 180°

80 + n = 180

n = 180 - 80

n = 100

(ii)

ABE + ∠ DBE = 180° (Sum of angles on a straight line)

n° + ∠ DBE = 180°

But n = 100

100° + ∠ DBE = 180°

DBE = 180° - 100°

DBE = 80°

DBE = ∠ BDE = 80°(Base angles equal)

DBE + ∠ BDE + ∠ BED = 180°(Sum of angles in a triangle)

80° + 80° + m° = 180°

160° + m° = 180°

m° = 180° - 160°

m° = 20°

m = 20

6.

(a)

A cone and a pyramid have equal heights and volumes. If the base area of the pyramid is 154 cm2, find the radius of the cone.

[Take π = 22 7 ]

(b)

A spherical bowl of radius r cm is a quarter full when 6 litres of water is poured into it.

Calculate, correct to three significant figures, the diameter of the bowl.

[Take π = 22 7 ]

(a)

Volume of a cone

Volume = π x r2 x h 3

Where r = the radius and h = the height of the cone.

Volume of a pyramid

Volume = 1 3 x Base area x Height

Let h = height of the pyramid and the cone.

Since the volumes are the same, the two formula can be equated.

π x r2 x h 3 = 1 3 x Base area x h

Base area of the pyramid = 154

π x r2 x h 3 = 1 3 x 154 x h

Divide both sides by h to get rid of the h

π x r2 x 1 3 = 1 3 x 154

Get rid of the fractions by multiplying both sides by the denominator 3

π x r2 = 154

π = 22 7

22 7 x r2 = 154

Get rid of the fraction by multiplying both sides by the denominator 7

22 x r2 = 154 x 7

Divide both sides by 22

r2 = 154 x 7 22

r2 = 49

Take square roots of both sides.

r = 49 = 7

The radius of the cone is 7 cm.

(b)

Volume of a sphere

Volume = 4 3 x π x r3

Where r = the radius of the sphere.

Volume of water poured = 6 litres = 6 x 1000 cm3 = 6000 cm3

The water is a quarter full. Hence the volume of water is 1 4 of the full volume of the sphere

π = 22 7

1 4 x 4 3 x 22 7 x r3 = 6000

4 x 22 4 x 3 x 7 x r3 = 6000

88 84 x r3 = 6000

Get rid of the fraction by multiplying both sides by the denominator 84.

88 x r3 = 6000 x 84

Divide both sides by 88

r3 = 6000 x 84 88

r3 = 5727.27

Take cube root of both sides.

r = 5727.27 1 3

r = 17.8916044472 cm

Diameter is 2 times the radius

Diameter = 2 x 17.8916044472 cm

Diameter = 35.7832088944 cm ≈ 35.8 cm(three significant figures)