NATIONAL SCIENCE AND MATHS QUIZ PAST QUESTIONS-2

Find the volume of the given figure (Leave answer in π)

1. A cylinder of height 10cm and diameter 10 cm.

ANSWER: 250π cm³

[ V = πr²h = π(10)5² = 250π cm³]

2. A right circular cone of base diameter 18 cm and height 20 cm.

ANSWER: 540π cm³

[ V = πr²h/3 = π(81)20/3 = 27(20)π = 540π cm³]

3. A sphere of diameter 12 cm

ANSWER: 288πcm³

[V = 4πr³/3 = 4π6³/3 = 4π(72) = 288π cm³]

Two fair dice are thrown. Find the probability that the positive difference of the scores is

1. 3

ANSWER: 1/6

[ A = {(4, 1), (1, 4), (5, 2),( 2, 5), (6, 3), (3, 6)}, P(A) = 6/36 = 1/6]

2. 4

ANSWER: 1/9

[ B = { (5, 1), (1, 5), (6, 2), (2, 6)} , P(B) = 4/36 = 1/9]

3. 2

ANSWER: 2/9

[ C = {(3, 1), (1, 3), (2, 4), (4, 2), (5, 3), (3, 5), (6, 4), (4, 6)}, P(C ) = 8/36 = 2/9]

1. Solve the quadratic equation 5x² + 3x – 2 = 0

ANSWER: x = -1, 2/5

[ 5x² + 5x – 2x – 2 = 5x(x + 1) – 2(x + 1) = (5x – 2)(x + 1) = 0, x = -1, 2/5]

2. If the angles of a quadrilateral measure (x + 25)°, (2x – 25)°, (x + 30)°, (2x)° in degrees, find the value of x.

ANSWER: x = 55

[ 6x + 30 = 360, 6x = 330, x = 55]

Find two integers such that

1. their mean is 15, and 2 less than the larger number is thrice the smaller number,

ANSWER: 7, 23

[ a + b = 30, a – 2 = 3b, a = 3b + 2, 4b + 2 = 30, b = 7 and a = 23]

2. their mean is 25, and one number plus twice the other number is 79,

ANSWER: 21, 29

[ a + b = 50, a + 2b = 79, a = 100 – 79 = 21, b = 29]

3. their mean is 18, and their difference is 10.

ANSWER: 23 and 13

[a + b = 36, a – b = 10, 2a = 46, a = 23 and b = 13]

Factorise as a difference of two squares

1. x² + 6x + 9 – 4y²

ANSWER: (x + 2y + 3)(x – 2y + 3)

[ (x + 3)² – (2y)² ]= (x + 2y + 3)(x – 2y + 3)

2. 9x² + 6x + 1 – 25y²

ANSWER: (3x + 5y + 1)(3x – 5y + 1)

[ (3x + 1)² – (5y)²= (3x + 5y + 1)(3x – 5y + 1)

3. x² – (25y² + 10y + 1)

ANSWER: (x + 5y + 1)(x – 5y – 1)

[ x² – (5y + 1)² = (x + 5y + 1)(x – 5y – 1)]

Place two numbers between a and L such that the 4 numbers form a linear sequence given

1. a = 12 and L = 33

[d = (33 – 12)/3 = 21/3 = 7, and hence 19, 26]

ANSWER: 19, 26

2. a = 2, and L = 20

[ d = (20 – 2)/3 = 18/3 = 6, and hence 8, 14]

ANSWER: 8, 14

3. a = 3.4 and L = 10.3

[ d = (10.3 – 3.4)/3 = 6.9/3 = 2.3, and hence 5.7, 8.0]

ANSWER: 5.7, 8.0

1. Find the diameter of the circle with equation x² + y² = 16

ANSWER: 8 [radius r = 4 and diameter D = 2r = 8]

2. Given that i² = -1, evaluate (3 – 5i)(5 + 3i)

ANSWER: 30 – 16i

[ 15 – 15i² – 25i + 9i = 30 – 16i]

3. Find the solution set of the linear inequality 8 – 2(2x + 1) < 2x – 3

ANSWER: {x : x > 3/2}

[9 < 6x, 6x > 9, x > 3/2]

Solve the trigonometric equation in the range 0 < x < π. Give answer in radians.

1. tan²x = 1

ANSWER: x = π/4, 3π/4 radians

[ tanx = ± 1, tanx = 1 gives x = π/4, and tanx = -1 gives x = 3π/4]

2. sin²x = 1/4

ANSWER: x = π/6, 5π/6 radians

[sinx = ± 1/2, sinx = ½ gives x = π/6, 5π/6 and sinx = -1/2 gives no solution]

3. cos²x = 1/2

ANSWER: π/4, 3π/4 radians

[cosx = ± 1/√2, cosx = 1/√2 gives x = π/4, cosx = -1/√2 gives x = 3π/4]

Solve the simultaneous equations

1. 3x – 2y = 2 and 9x – 4y = 1

ANSWER: x = -1, y = -5/2

[6x – 4y = 4, 9x – 4y = 1, 3x = – 3, x = -1 and y = -5/2]

2. x – 2y = 5 and 3x + y = 8

ANSWER: x = 3, y = -1

[x – 2y = 5, 6x + 2y = 16, 7x = 21, x = 3 and y = -1]

3. 2x + 3y = 0 and 3x – 2y = 13

ANSWER: x = 3 and y = -2

[ 4x + 6y = 0, 9x – 6y = 39, 13x = 39, x = 3 and y = -2]

1. Solve for x from the equation (ax – b)/(cx + d) = 3

ANSWER: x = (b + 3d)/(a – 3c)

[ax – b = 3cx + 3d, x(a – 3c) = b + 3d, x = (b + 3d)/(a – 3c)]

2. Find the next prime after 151

ANSWER: 157

[152, 153, 154, 155, 156, (157) ]

3. A fair die is tossed. Find the probability of obtaining an odd number given that the number rolled is 3.

ANSWER: 1

Factorise

1. 9x² – 12x + 4 – 16y²

ANSWER: (3x + 4y – 2)(3x – 4y – 2)

[ (3x – 2)² – 16y² = (3x + 4y – 2)(3x – 4y – 2)

2. 25x² – 9y² + 6y – 1

ANSWER: (5x + 3y – 1)(5x – 3y + 1)

[ 25x² – (9y² – 6y + 1) = 25x² – (3y – 1)² = (5x + 3y – 1)(5x – 3y + 1)

3. 4x² – 12x + 9 – 36y²

ANSWER: (2x + 6y – 3)(2x – 6y – 3)

[ (2x – 3)² – 36y² = (2x + 6y – 3)(2x – 6y – 3)

Express as the logarithm of a single term

1. 5logx + 2logy – 3logz

ANSWER: log(x⁵y²/z³)

2. 2logx + 3logy – 4logz

ANSWER: log(x²y³/z⁴)

3. 3loga – 2logb – logc

ANSWER: log(a³/b²c)

Express as a single trigonometric ratio

1. 2sin2Acos2A

ANSWER: sin4A

[sin(2A + 2A) = sin4A]

2. cos²3A – sin²3A

ANSWER: cos6A

[ cos(3A + 3A) = cos6A]

3. 2tan2A/(1 – tan²2A)

ANSWER: tan4A

[tan(2A + 2A) = tan4A]

Find the coordinates of the point on the given curve at which its gradient m has the given value

1. y = x² – x + 3, and m = 1

ANSWER: (1, 3)

[ dy/dx = 2x – 1 = 1, x = 1 and y = 1 – 1 + 3 = 3, hence (1, 3)]

2. y = x² + 2/x, and m = 0

ANSWER: (1, 3)

[ dy/dx = 2x – 2/x² = 0, x³ = 1, x = 1, y = 1 + 2 = 3, hence (1, 3)]

3. y = x² – 2x – 15, and m = 2

ANSWER: (2, – 15)

[ dy/dx = 2x – 2 = 2, x = 2, y = 4 – 4 – 15 = -15, hence (2, -15)]

1. If x = 2 is a root of the equation x² – 3x + a = 0, find a and find the other root.

ANSWER: a = 2, and x = 1

[ 4 – 6 + a = 0, a = 2, x² – 3x + 2 = 0, (x – 2)(x – 1) = 0, x = 1]

2. If x – 2 is a factor of ax² – 12x + 4, find a and find the other factor.

ANSWER: a = 5, (5x – 2)

[ 4a – 24 + 4 = 0, 4a = 20, a = 5, 5x² – 12x + 4 = (x – 2)(5x – 2)]

3. If one root of the equation x² + ax – 5 = 0 is x = 5, find a and find the other root.

ANSWER: a = -4, and x = -1

[ 25 + 5a – 5 = 0, 5a = -20, a = -4, x² – 4x – 5 = (x – 5)(x + 1) = 0, x = -1]

1. Solve for x given 125^{x + 1} = 1/25

ANSWER: x = -5/3

[5^{3(x + 1)} = 5^-2, 3x + 3 = -2, 3x = -5, x = -5/3]

2. Find the solution set for the quadratic inequality x² + 3x – 4 < 0.

ANSWER: {x : -4 < x < 1}

[(x + 4)(x – 1) < 0, -4 < x < 1]

3. If the vectors a = 2i + 3j and b = 3xi + 10j are perpendicular, find the value of x.

ANSWER: x = -5

[ a.b = 6x + 30 = 0, x = -5]

Find the value of x if the following are consecutive terms of an exponential sequence.

1. 3, 3x, 12x

ANSWER: x = 4

[ 3x/3 = 12x/3x, x = 12/3 = 4]

2. 60x, 10x, 5

ANSWER: x = 3

[ 10x/60x = 5/10x, 1/6 = 1/2x, 2x = 6, x = 3]

3. 10x, 2x, 2

ANSWER: x = 5

[2x/10x = 2/2x, 1/5 = 1/x, x = 5]

Find the inverse of the given function

1. y = (x + 2)/(x – 1)

ANSWER: y = (x + 2)/(x – 1)

[ x = (y + 2)/(y – 1), x(y – 1) = (y + 2), y(x – 1) = (x + 2), y = (x + 2)/(x – 1]

2. y = (x – 1)/(x – 2)

ANSWER: y = (2x – 1)/(x – 1)

[ x = (y – 1)/(y – 2), x(y – 2) = (y – 1), y(x – 1) = (2x – 1), y = (2x – 1)/(x – 1)]

3. y = x/(x + 3)

ANSWER: y = -3x/(x – 1)

[ x = y/(y + 3), x(y + 3) = y, y(x – 1) = -3x, y = -3x/(x – 1)]

A car travelled from A to B at 100 km/hr for 3 hrs and from B to A at 60 km/hr for 5 hrs. Assume there is no stopping. Find

1. the distance of the round trip from A to B and back to A,

ANSWER: 600 km

[100 x 3 + 60 x 5 = 600 km]

2. average speed for the round trip,

ANSWER: 75 km/hr

[ distance/time = 600/8 = 75 km/hr]

3. the times when the car is midway between A and B.

ANSWER: After 1½ hrs and 5½ hrs

[3/2, 3 + 5/2, 1½ hrs and 5½ hrs]

1. Translate the exponential equation 10x = 725 into an equivalent logarithmic equation.

ANSWER: x = log725

2. If f(x + 2) = x2, evaluate f(5)

ANSWER: 9

[x + 2 = 5, x = 3, f(5) = 32 = 9]

3. Simplify (x2 – 5x – 24)/(x – 8)

ANSWER: (x + 3)

[(x – 8)(x + 3)/(x – 8) = (x + 3)]

Place 3 numbers between the numbers a and L such that the five numbers form a linear sequence.

1. a = 3, and L = 23

ANSWER: 8, 13, 18

[ d = (23 – 3)/4 = 5, hence 3 + 5 = 8, 8 + 5 = 13 and 13 + 5 = 18]

2. a = 7 and L = 67

ANSWER: 22, 37, 52

[ d = (67 – 7)/4 = 15, hence 7 + 15 = 22, 22 + 15 = 37, 37 + 15 = 52]

3. a = 28 and L = 4

ANSWER: 22, 16, 10

[ d = (4 – 28)/4 = -6, hence 28 – 6 = 22, 22 – 6 = 16, 16 – 6 = 10]

Solve for x from the given equation

1. 3/x = 2/a + 5/b

[3/x = (2b + 5a)/ab, x = 3ab/(5a + 2b)]

ANSWER: x = 3ab/(5a + 2b)

2. 5/a = 4/x – 3/b

[4/x = 5/a + 3/b = (5b + 3a)/ab, x = 4ab/(5b + 3a)]

ANSWER: x = 4ab/(5b + 3a)

3. a/b = x/c + d

[x = c(a/b – d) = c(a – bd)/b]

ANSWER: x = c(a – bd)/b

Express in partial fractions

1. (2x + 1)/(x + 1)

ANSWER: 2 – 1/(x + 1)

[ (2(x + 1) – 1)/(x + 1) = 2 – 1/(x + 1)]

2. (3x – 2)/(x – 2)

ANSWER: 3 + 4/(x – 2)

[ (3(x – 2) + 4)/(x – 2) = 3 + 4/(x – 2)]

3. (2x + 5)/ (2x – 1)

ANSWER: 1 + 6/(2x – 1)

[ ((2x – 1) + 6)/(2x – 1) = 1 + 6/(2x – 1)]

1. Find the coordinates of the point on the curve y = 1/x2 at which the gradient

is ¼.

ANSWER: (-2, ¼)

[dy/dx = -2/x3 = ¼, x3 = -8, x = -2, y = ¼, hence (-2, ¼)]

2. Evaluate cos81֯cos21֯ + sin81֯sin21֯

ANSWER: ½ or 0.5

[cos(81 – 21) = cos60 = ½]

3. Evaluate 10P3 [10 permutation 3]

ANSWER: 720

[10!/7! = 10 x 9 x 8 = 720]

If α and β are the roots of x² + 5x + 3 = 0, evaluate

1. α² + β²

ANSWER: 19

[ α² + β² = (α + β)² – 2αβ = (-5)² -2(3) = 25 – 6 = 19]

2. 1/α + 1/β

ANSWER: -5/3

[ (α + β)/αβ= -5/3]

3. α²β + β²α

ANSWER: -15

[ αβ(α + β) = 3(-5) = -15

Determine if the pair of lines are perpendicular, or parallel or neither.

1. x + 3y = 5, x – 3y = 5

ANSWER: NEITHER PARALLEL NOR PERPENDICULAR

[ y = -x/3 + 5/3, y = x/3 – 5/3]

2. x – 2y = 2, 2x + y = 3

ANSWER: PERPENDICULAR

[ y = ½ x – 1 and y = -2x – 3, perpendicular since ½ ( -2) = -1]

3. 2x + 3y = 5, 6y = – 4x + 20

ANSWER: PARALLEL

[ y = –(2/3)x + 5/3 and y = -(2/3)x + 10/3, hence parallel]

Given that sinA = 5/13 and cosB = 3/5, and A and B are acute angles, evaluate

1. sin2B

ANSWER: 24/25

[ 2sinBcosB = 2(4/5)(3/5) = 24/25]

2. cos2A

ANSWER: 119/169

[ cos²A – sin²A = 144/169 – 25/169 = 119/169]

3. sin(A + B)

ANSWER: 63/65

[ sinAcosB + cosAsinB = (5/13)(3/5) + (12/13)(4/5) = 15/65 + 48/65 = 63/65]

1. Find the range of values of k if the equation x2 – 3x + k = 0 has no real roots.

ANSWER: {k : k > 9/4}

[9 – 4k < 0, 4k > 9, k > 9/4

2. Find the equation of the tangent to the curve y = x2 – 3x + 2 at the point on the curve at which x = 2

ANSWER: y = x – 2

[dy/dx = 2x – 3, m = 4 – 3 = 1, y = 4 – 6 + 2 = 0, y = 1(x – 2), y = x – 2]

3. Given that a sequence is defined by Un + 2 = Un + 1 + Un, U1 = 2, U2 = 3, find U3.

ANSWER: 5

[ U3 = U2 + U1 = 3 + 2 = 5]

In a rhombus, find the length of a side if the diagonals have lengths respectively

1, 18 cm and 24 cm

[d² = 81 + 144 =225, d = 15]

ANSWER: 15 cm

2. 12 cm and 12√3 cm

[d² = 36 + 108 = 144, d = 12]

ANSWER: 12 cm

3. 10 cm and 20 cm

[d² = 25 + 100 = 125, d = 5√5]

ANSWER: 5√5 cm

Find the value(s) of x if the vectors

1. a = 3i + xj and b = xi + 9j are parallel,

ANSWER: x = ± 3√3

[x/3 = 9/x, x² = 27, x = ±3√3]

2. a = 4xi – 2j and b = 3i + 6j are perpendicular,

ANSWER: x = 1

[12x – 12 = 0, x = 1]

3. a = 2i + j and b = xi – 3j are parallel.

ANSWER: x = – 6

[1/2 = -3/x, x = -6]

Solve the logarithmic equation

1. 2logx – log4 = log(x + 8)

ANSWER: x = 8

[ x² = 4(x + 8), x² – 4x – 32 = 0, (x – 8)(x + 4) = 0, x = 8]

2. 3logx = logx + log(5x – 6)

ANSWER: x = 2, 3

[2logx = log(5x – 6), x² = 5x – 6, x² – 5x + 6 = 0, (x – 3)(x – 2) = 0, x = 3, 2]

3. log4x + logx = log64

ANSWER: x = 4

[4x² = 64, x² = 16, x = 4]

1. Find x given √(12x + 3) – 3√(2x) = 0

ANSWER: x = ½

[ 12x + 3 = 18x, 6x = 3, x = ½]

2. Find the sum of the interior angles of a polygon of 15 sides. Give answer in radians.

ANSWER: 13π radians

3. Solve for x given 5sinx + 3 = 0 for 0 < x < π

ANSWER: NO SOLUTION

[sinx = -3/5 is not possible for 0 < x < π]

Find two numbers between the given pair of numbers such that the four numbers form a linear sequence.

1. 3√7 and 15√7

ANSWER: 7√7, 11√7

[ d = (15√7 – 3√7)/3 = 4√7, hence 7√7 and 11√7]

2. 5 and (20 + 6√3)

ANSWER: 10 + 2√3, 15 + 4√3

[ d = (20 + 6√3 – 5)/3 = 5 + 2√3, hence 10 + 2√3 and 15 + 4√3]

3. (7 – 3√2) and (13 + 3√2)

ANSWER: 9 – √2 and 11+√2

[ d = (13 + 3√2 – (7 – 3√2))/3 = 2 + 2√2, hence 9 – √2 and 11+ √2]

Find the measure of the interior angle in radians of a regular polygon of

1. 12 sides

ANSWER : 5π/6 radians

[ π – 2π/12 = 5π/6]

2. 18 sides

ANSWER: 8π/9 radians

[π – 2π/18 = 8π/9]

3. 24 sides

[π – 2π/24 = 11π/12]

ANSWER: 11π/12 radians

Express the given expression in terms of a and b where a = logx and b = logy.

1. log(x²y³)

ANSWER: 2a + 3b

[ 2logx + 3logy = 2a + 3b]

2. log(y³√x)

ANSWER: 3b + ½ a

[ 3logy + (1/2)logx = 3b + ½a]

3. log(x/√y)

ANSWER: a – ½ b

[logx – ½ logy = a – ½ b]

1. Simplify (2×2 + 17x + 35)/(2x + 7)

ANSWER: (x + 5)

[(2x + 7)(x + 5)/(2x + 7) = (x + 5)]

2. Find the value of x if the terms 4x, (2x – 3), and x are consecutive terms of an exponential sequence.

ANSWER: x = 3/4

[(2x – 3)2 = 4×2, 4×2 – 12x + 9 = 4×2, 12x = 9, x = 9/12 = ¾]

3. Simplify (4√x + 5√y)(4√x – 5√y)

ANSWER: 16x – 25y

List the set of integers n for which

1. 2n² – 13n + 15 < 0

[(2n – 3)(n – 5) < 0, 3/2 < n < 5, n = 2, 3, 4]

ANSWER: {2, 3, 4}

2. 3n² – 4n – 4 < 0

[ (3n + 2)(n – 2) < 0, -2/3 < n < 2, n = 0, 1]

ANSWER: {0, 1}

3. 3n² – 4n – 15 < 0

[(3n + 5)(n – 3) < 0, -5/3 < n < 3, n = -1, 0, 1, 2]

ANSWER: {-1, 0, 1, 2}

1. Kofi can do a job in 10 hours and Kwame can do the same job in 15 hours. If they work together how long will it take them to complete the job?

ANSWER: 6 HOURS

[1/T = 1/10 + 1/15 = 25/150 = 1/6, hence T = 6]

2. A painter can paint a house in 6 hours. If his assistant helps him, they will take 4 hours to paint the same house. How long will the assistant working alone take to paint the same house?

ANSWER: 12HOURS

[1/T + 1/6 = ¼, 1/T = ¼ – 1/6 = 2/24 = 1/12, T = 12]

3. Kwame can weed a farm in 30 hours. If Owusu helps him, they take 12 hours to weed the farm. How long will Owusu working alone take to weed the same farm?

ANSWER: 20 HOURS

[1/T + 1/30 = 1/12, 1/T = 1/12 – 1/30 = (30 – 12)/360 = 18/360 = 1/20, T = 20]

A bag contains 6 black balls and 4 white balls. Two balls are drawn from the bag without replacement. Find the probability that the balls are

1. both black,

ANSWER: 1/3

[ P(BB) = (6/10)(5/9) = 1/3]

2. both white,

ANSWER: 2/15

[P(WW) = (4/10)(3/9) = 2/15]

3. the balls are of different colours.

ANSWER: 8/15

[P(BW, WB) = (6/10)(4/9) + (4/10)(6/9) = 8/15]

1. Given that a sequence U_n is defined by U_{n + 1}= 2U_n + 3 and U_n = U₁ for all positive integers n, find U₁.

ANSWER: U₁ = -3

[U₁ = 2U₁+ 3, U₁ = -3]

2. Evaluate cos120֯

ANSWER: -1/2 or -0.5

[cos120 = cos(180 – 60) = cos180cos60 + sin180sin60 = -cos60 = -1/2]

3. Evaluate antilog(log4.57)

ANSWER: 4.57

Find the value of a given that

1. f(x) = 2x³ + ax² – 5x + 2 is exactly divisible by (x + 1),

ANSWER: a = – 5

[ f(-1) = -2 + a + 5 + 2 = 0, a + 5 = 0, a = -5]

2. f(x) = x³ – ax² + 5x + 3 leaves a remainder of 5 on division by (x + 1),

ANSWER: a = -8

[ f(-1) = -1 – a – 5 + 3 = 5, a = -8]

3. f(x) = 4x³ + 3x² – ax + 4 leaves a remainder of – 2 on division by (x + 1).

ANSWER: a = -5

[ f(-1) = -4 + 3 + a + 4 = – 2, a = – 5]

The sum of two integers is S and their product is P. Find the two integers if

1. S = 17 and P = 72

ANSWER: 8 and 9

[ 72 = 2 x 36 = 4 x 18 = 8 x 9, hence 8 and 9]

2. S = 33 and P = 230

ANSWER: 10 and 23

[230 = 23 x 10, hence 23 and 10]

3. S = 22 and P = 105

ANSWER: 15 and 7

[105 = 3 x 35 = 5 x 21 = 15 x 7, hence 15 and 7]

1. How many liters of a 70% alcohol solution must be added to 30 liters of a 20% alcohol solution to obtain a 40% alcohol solution?

ANSWER: 20 LITERS

[0.4(x + 30) = 0.7x + 0.2(30), 0.3x = 12 – 6 = 6, x = 60/3 = 20]

2. How many liters of water must be added to 50L of a 20% salt solution to obtain a 10% salt solution?

ANSWER: 50 LITERS

[ 0.10(x + 50) = 0.2(50), x = 100 – 50 = 50]

3. How many liters of water must be evaporated from 40L of a 15% salt solution in order to obtain a 30% salt solution?

ANSWER: 20 LITERS

[0.15(40) = (40 – x)0.3, 6 = 12 – 0.3x, x = 6/0.3 = 20]

1. Find the area of a triangle with sides of lengths 15 cm and 12 cm and an included angle of measure 30֯.

ANSWER: 45 cm2

[ ½ x 15 x 12 x ½ = 15 x 3 = 45 cm2]

2. Solve for x from the equation 2/(x + 3) = 3/(2x + 2)

ANSWER: x = 5

[2(2x + 2) = 3(x + 3), x + 4 = 9, x = 5]

3. Find the set of values of k for which the equation x2 + 10x + k2 = 0 has two distinct real roots.

ANSWER: {k : – 5 < k < 5}

[ 100 – 4k2 > 0, k2 < 25, -5 < k < 5]

1. The cost of an item is GHc1800. The markup was 40% of the selling price. Find the selling price.

ANSWER: GHc3000

[S_P = C_P + markup = 1800 + 0.4S_P,0.6S_P = 1800, S_p = 3000]

2. An item is sold for GHc780. The markup was 30% of the cost price. Find the cost price.

ANSWER: GHc600

[780 = C_p+ 0.3C_P, C_P = 780/1.3 = 600]

3. An item is sold for GHc1680 with a markup of 40% of the cost price. Find the cost price.

ANSWER: GHc1200

[1680 = C_p + 0.4C_p, C_p = 1680/1.4 = 1200]

1. Find the radius of a sphere (in cm) if its volume is numerically 4 times its surface area.

ANSWER: 12 cm

[4πr³/3 = 4(4πr²), r = 3(4) = 12]

2. Find the radius of a solid right circular cylinder (in cm) if its volume is numerically twice its curved surface area.

ANSWER: 4 cm

[πr²h = 2(2πrh), r = 4]

3. The area of a circle is numerically 4 times the circumference of the circle. Find its radius (in cm).

ANSW2ER: 8 cm

[πr² = 4(2πr), r = 8 ]

Solve the trigonometric equation for the interval 0 < x < π. Give answer in radians.

1. sin²x = 3/4

ANSWER: x = π/3, 2π/3

[ sinx = ± √3/2 , sinx = √3/2 gives x = π/3, 2π/3; sinx = -√3/2 gives no solution]

2. cos²x = 1/4

ANSWER: x = π/3, 2π/3

[ cosx = ± 1/2, cosx = 1/2 gives x = π/3, cosx = -1/2 gives x = 2π/3]

3. tan²x = 1/3

ANSWER: x = π/6, 5π/6

[tanx = ± 1/√3, tanx = 1/√3 gives x = π/6, tanx = -1/√3 gives x = 5π/6]

1. Find x if (x – 1), (x + 1) and (x + 5) are consecutive terms of an exponential sequence.

ANSWER: x = 3

[(x + 1)2 = (x – 1)(x + 5), 2x + 1 = 4x – 5, 2x = 6, x = 3]

2. Differentiate y = 5×2 – 4/x2

ANSWER: dy/dx = 10x + 8/x3

3. Solve for a from the equation log(3 + a) = log(4b – 1)

ANSWER: a = 4b – 4

[ 3 + a = 4b – 1, a = 4b – 4]

Find the term without x in the expansion of

1. (x + 1/x²)⁶

ANSWER: 15

[ 6C_rx^r(1/x²)^{6 – r} = 6C_rx^{3r- 12}, 3r – 12 = 0, r = 4, 6C₄ = (6 x 5)/(2 x 1) = 15

2. (x² + 1/x)⁹

ANSWER: 84

[ 9C_rx^2r(1/x)^{9- r}= 9C_rx^{3r – 9}, 3r – 9 = 0, r = 3, 9C₃ = (9 x 8 x 7)/(3 x 2 x 1) = 84]

3. (x + 1/x)⁸

ANSWER: 70

[8C_rx^r(1/x)^{8 – r} = 8C_rx^{2r – 8}, 2r – 8 = 0, r = 4, 8C₄ = (8 x 7 x 6 x 5)/(4 x 3 x 2 x 1) = 70]

1. A man is 18 times as old as his son. In 15 years’ time, he will be 3 times as old as his son. Find their present ages.

ANSWER: man 36 years, son 2 years

[ x = 18y, (x + 15) = 3(y + 15), 18y + 15 = 3y + 45, 15y = 30, y = 2, x = 36]

2. The difference in ages of two men is 5. In 2 years’ time, the sum of their ages will be 55. Find their current ages.

ANSWER: 28 years and 23 years

[ x – y = 5, x + y + 4 = 55, 2x = 56, x = 28 and y = 23]

3. A boy is 4 years older than his sister. Five years ago, the sum of their ages was 20. Find their present ages.

ANSWER: 17 years and 13 years

[ x – y = 4, x + y – 10 = 20, x + y = 30, 2x = 34, x = 17, y = 13]

Find an equation of the perpendicular bisector of the line segment joining the points A and B given

1. A(1, 3) and B(3, 1)

ANSWER: y = x

[m_AB = -2/2 = -1, midpoint C(2, 2), y – 2 = 1(x – 2), y = x]

2. A(-1, 1) and B(1, 3)

ANSWER: y = -x + 2

[m_AB = 2/2 = 1, midpoint C(0, 2), y – 2 = -1x, y = -x + 2]

3. A(2, 6) and B(4, 4)

ANSWER: y = x + 2

[m_AB = -2/2 = -1, midpoint C(3, 5), y – 5 = 1(x – 3), y = x + 2]

1. If f(x – 3) = 2x + 5, evaluate f(- 4)

ANSWER: 3

[ x – 3 = – 4, x = -1, f(-4) = 2(-1) + 5 = 3]

2. Solve the trigonometric equation sinx = ½ for 0 < x < π. Leave answer in radians.

ANSWER: x = π/6, 5π/6 radians

3. Solve the equation x + 3/x = 4

ANSWER: x = 1, 3

[x² – 4x + 3 = 0, (x – 3)(x – 1) = 0, x = 3, 1]

Find dy/dx from the implicit equation

1. x³ + 3x²y – y³ = 10

ANSWER: dy/dx = (x² + 2xy)/(y² – x²)

[3x² + 6xy + 3x²dy/dx – 3y²dy/dx = 0, dy/dx = (3x² + 6xy)/(3y² – 3x²)]

2. x³ – 2xy + 2y³ = 15

ANSWER: dy/dx = (3x² – 2y)/(2x – 6y²)

[ 3x² – 2y – 2xdy/dx + 6y²dy/dx = 0, dy/dx = (3x² – 2y)/(2x – 6y²)]

3. 2x³ + xy – 3y² = 18

ANSWER: dy/dx = (6x² + y)/(6y – x)

[ 6x² + y + xdy/dx – 6ydy/dx = 0, dy/dx = (6x² + y)/(6y – x)]

Solve for x and y from the given proportions

1. (3x – y)/15 = (y – 6)/10 = 3/5

ANSWER: x = 7, y = 12

[ (3x – 6)/25 = 3/5, 3x = 21, x = 7, y = 6 + 6 = 12

2. (x + y)/8 = (x – y + 5)/4 = ¼

ANSWER: x = -1, y = 3

(2x + 5)/12 = ¼, 2x + 5 = 3, 2x = -2, x = -1, -1 + y = 2, y = 3]

3. (x + y – 3)/6 = (x – y + 3)/3 = 2/3

ANSWER: x = 3, y = 4

[ 2x/9 = 2/3, 2x = 6, x = 3. (2y – 6)/3 = 2/3, 2y – 6 = 2, 2y = 8, y = 4]

A and B are events in a sample space S.

1. Find P(AUB) given P(A) = 0.9 and P(B) = 0.6 and A and B are independent.

ANSWER: 0.96

[P(AUB) = P(A) + P(B) – P(A∩B) = 0.9 + 0.6 – 0.54 = 1.5 – 0.54 = 0.96]

2. Find P(AUB) given P(A) = 0.9, P(B) = 0.6 and P(A∩B) = 0.5.

ANSWER: 1

[P(AUB) = 0.9 + 0.6 – 0.5 = 1.5 – 0.5 = 1]

3. Find P(A∩B) given P(A) = 0.8, P(B) = 0.6, and P(AUB) = 0.9

ANSWER: 0.5

[ P(A∩B) = P(A) + P(B) – P(AUB) = 0.8 + 0.6 – 0.9 = 0.5]

1. Find the number of sides of a regular polygon if an exterior angle measure π/15 radians.

ANSWER: 30

[ n = 2π/(π/15) = 2 x 15 = 30]

2. A fair coin is tossed once. Find the probability of obtaining an even number of Heads.

ANSWER: ½ or 0.5

[An even no. of heads is 0 head ie P(T) = ½]

3. Evaluate 10C5 (10 combination 5)

ANSWER: 252

[ 10!/5!5! = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2 x 1)= 28 x 9 = 252]

Solve the given equation

1. x4 – 14×2 + 45 = 0

ANSWER: x = ±3, ±√5

[ x4 – 9×2 – 5×2 + 45 = x2(x2 – 9) – 5(x2 – 9) = (x2 – 9)(x2 – 5) = 0, x = ±3, ±√5]

2. x4 – 13×2 + 36 = 0

ANSWER: x = ± 2, ± 3

[ x4 – 4×2 – 9×2 + 36 = x2(x2 – 4) – 9(x2 – 4) = (x2 – 4)(x2 – 9) = 0, x = ± 2, ± 3]

3. x4 – 29×2 + 100 = 0

ANSWER: x = ± 2, ± 5

[x4 – 25×2 – 4×2 + 100 = x2(x2 – 25) – 4(x2 – 25) = (x2 – 4)(x2 – 25) = 0, x = ± 2, ± 5]

For the point A(3, -5), find the coordinates of the corresponding point symmetric with respect to

1. the x-axis,

ANSWER: (3, 5)

[reflect in the x-axis]

2. the y-axis,

ANSWER: (-3, -5)

[reflect in the y-axis]

3. the origin.

ANSWER: (-3, 5)

[reflect in the origin]

Express the repeating decimal as an infinite exponential series and hence express it as a rational number.

1. 0.5555. . . .

ANSWER: 0.5 + 0.05 + 0.005 + 0.0005 + . . . = 5/9

[1 + 2]

[0.5 + 0.05 + 0.005 + 0.0005 + . . . GP, a = 0.5, r = 0.1 hence = 0.5/(1 – 0.1) = 5/9]

2. 3.3333 . . .

ANSWER: 3 + 0.3 + 0.03 + 0.003 + 0.0003 + . . . = 10/3

[1 + 2]

[ 3 + 0.3 + 0.03 + 0.003 + 0.0003 + . . .GP, a = 3, r = 0.1 hence = 3/(1 – 0.1) = 30/9=10/3 ]

3. 77.7777 . . .

ANSWER: 70 + 7 + 0.7 + 0.07 + 0.007 . . . = 700/9

[1 + 2]

[ 70 + 7 + 0.7 + 0.07 + 0.007 + . . . GP, a = 70, r = 0.1, hence = 70/(1 – 0.1) = 70/0.9 = 700/9]

1. Find two natural numbers with a mean of 25 and such that one number increased by 1 is twice the other number.

ANSWER: 17 and 33

[x + y = 50, x + 1 = 2y, 3y – 1 = 50, 3y = 51, y = 17 and x = 33]

2. Solve for x from the linear equation 5(2x + 4) – 3(4x – 6) = 2(20 – x).

ANSWER: NO SOLUTION

[10x + 20 – 12x + 18 = 40 – 2x, 38 = 40, hence no solution]

3. Determine if the pair of lines 6x – 4y = 5 and 8y – 12x = 5 are parallel, perpendicular or neither.

ANSWER: PARALLEL

[ y = 6x/4 – 5/4, y = 12x/8 + 5/8, same slope 3/2]

Simplify the rational expression

1. (8a³ + 27b³)/(2a + 3b)

ANSWER : 4a² – 6ab + 9b²

[Sum of 2 cubes]

[ (2a)³ + (3b)³)/(2a + 3b) = (2a + 3b)(4a² – 6ab + 9b²)/(2a + 3b) = 4a² – 6ab + 9b²]

2. (125x³ – 8y³)/(5x – 2y)

ANSWER: 25x² + 10xy + 4y²

[difference of 2 cubes]

[((5x)³ – (2y)³)/(5x – 2y) = (5x – 2y)(25x² + 10xy + 4y²)/(5x – 2y) = 25x² + 10xy + 4y²]

3. (64a³ + 27b³)/(4a + 3b)

ANSWER: 16a² – 12ab + 9b²

[sum of 2 cubes]

[(4a)³ + (3b)³)/(4a + 3b) = (4a + 3b)(16a² – 12ab + 9b²)/(4a + 3b)/(4a + 3b) = 16a² – 12ab + 9b²]

A bag contains 7 white balls, 5 black balls and 8 red balls. Three balls are randomly drawn from the bag without replacement. Find the probability that

1. two balls are black and one ball white,

ANSWER: 7/114

[ A = {bbw, bwb, wbb}, P(A) = 3 x (5/20) x (4/19) x 7/18 = (1/19) x (7/6) = 7/114]

2. two balls are white and one ball red,

ANSWER: 14/95

[ B = {wwr, wrw, rww}, P(B) = 3 x (7/20) x (6/19) x (8/18) = 14/95]

3. two balls are red and one black.

ANSWER: 7/57

[ C = {rrb, rbr, brr}, P(C) = 3 x (8/20) x (7/19) x (5/18) = 14/114 = 7/57]

Find the domain of the given function

1. f(x) = (x + 2)/(x² – 2x – 15)

ANSWER: {x : x ≠ 5, -3}

[x² – 2x – 15 = (x – 5)(x + 3) ≠ 0, x ≠ 5, -3]

2. f(x) = (2x² + 1)/(3x² – 2x – 5)

ANSWER: {x : x ≠ -1, 5/3}

[3x² – 2x – 5 = 3x² – 5x + 3x – 5 = (3x – 5)(x + 1) ≠ 0, x ≠ -1, 5/3]

3. f(x) = (x³ + 8)/(2x² – 3x – 9)

ANSWER: {x : x ≠ 3, -3/2}

[2x² – 6x + 3x – 9 = 2x(x – 3) + 3(x – 3) = (2x + 3)(x – 3) ≠ 0, x ≠ 3, -3/2]

1. When a number is decreased by 30%, the result is 280. Find the number.

ANSWER: 400

[0.7x = 280, x = 280/0.7 = 2800/7 = 400]

2. Given the relation R = {(1, 2), (2, 2), (2, 4), (3, 3), (3, 5)}, find the domain A and the range B of R. Determine if R is a function.

ANSWER: A = {1, 2, 3}, B = {2, 3, 4, 5}, R is not a function.

3. Factorise (x + 1)³ + 1

ANSWER: (x + 2)(x² + x + 1)

[((x + 1) + 1)((x + 1)²– (x + 1) + 1) = (x + 2)(x² + 2x + 1 – x – 1 + 1) = (x + 2)(x² + x + 1)]

Solve the trigonometric equation for the interval 0 < x < π. Give answer in radians.

1. sin²x = 1/4

ANSWER: x = π/6, 5π/6

[ sinx = ± ½ , sinx = ½ gives x = π/6, 5π/6; sinx = -1/2 gives no solution]

2. cos²x = 1/2

ANSWER: x = π/4, 3π/4

[ cosx = ±1/√2, cosx = 1/√2 gives x = π/4, cosx = -1/√2 gives 3π/4]

3. tan²x = 3

ANSWER: x = π/3, 2π/3

[tanx = ± √3, tanx = √3 gives x = π/3, tanx = -√3 gives 2π/3]

A bag contains 7 white balls, 5 black balls and 8 red balls. Three balls are randomly drawn from the bag without replacement. Find the probability that

1. two balls are black and one ball white,

ANSWER: 7/114

[ A = {bbw, bwb, wbb}, P(A) = 3 x (5/20) x (4/19) x 7/18 = (1/19) x (7/6) = 7/114]

2. two balls are white and one ball red,

ANSWER: 14/95

[ B = {wwr, wrw, rww}, P(B) = 3 x (7/20) x (6/19) x (8/18) = 14/95]

3. two balls are red and one ball black.

ANSWER: 14/114 = 7/57

[ C = {rrb, rbr, brr}, P(C ) = 3 x (8/20) x (7/19) x (5/18) = 14/114]

A box contains 8 red balls and 10 black balls. Two balls are drawn without replacement. Find the probability

1. both balls are red,

ANSWER: 28/153

[P(RR) = (8/18) x (7/17) = 4/9 x 7/17 = 28/153]

2. both balls are black,

ANSWER: 5/17

[P(BB) = (10/18) x (9/17) = (10/2) x (1/17) = 5/17]

3. the balls are of different colours.

ANSWER: 80/153

[P(RB, BR) = (8/18) x (10/17) + (10/18) x (8/17) = (40 x 2)/(9 x 17) = 80/153]

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