NATIONAL SCIENCE AND MATHS QUIZ PAST QUESTIONS-2
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 125x + 1 = 1/25
ANSWER: x = -5/3
[53(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 Un is defined by Un + 1 = 2Un + 3 and Un = 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
[SP = CP + markup = 1800 + 0.4SP, 0.6SP = 1800, Sp = 3000]
2. An item is sold for GHc780. The markup was 30% of the cost price. Find the cost price.
ANSWER: GHc600
[780 = Cp + 0.3CP, CP = 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 = Cp + 0.4Cp, Cp = 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
[ 6Crxr(1/x2)6 – r = 6Crx3r- 12, 3r – 12 = 0, r = 4, 6C4 = (6 x 5)/(2 x 1) = 15
2. (x2 + 1/x)9
ANSWER: 84
[ 9Crx2r(1/x)9- r = 9Crx3r – 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
[8Crxr(1/x)8 – r = 8Crx2r – 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
[mAB = -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
[mAB = 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
[mAB = -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]
About Author
