Solutions to Grade 8 Math Practice Test

Detailed solutions to the Grade 8 math practice test questions are presented.

    1 - Numbers


  1. Solution to Expressions with Radicals


  2. ? is NOT a rational number.


  3. The value of the digit 5 in the number 34.6597 is
    5 Hundredths

    2 - Sequences


  4. We notice that as we go from one term to the next, we add 3; hence the next term is equal to : 12+3=15

  5. We notice that as we go from one term to the next, we multiply by 3; hence the next term is given by: 27×3=81

  6. a) The first five terms of the sequence starting with n=1 are obtained by substituting n by 1,2,3,4,5 in the expression 2n+1 :
    For n=1   ,   2n+1=2(1)+1=3
    For n=2   ,   2n+1=2(2)+1=5
    For n=3   ,   2n+1=2(3)+1=7
    For n=4   ,   2n+1=2(4)+1=9
    For n=5   ,   2n+1=2(5)+1=11
    b) Going from one term to the next, we add 2 and hence it is an arithmetic sequence with common difference equal to 2.


  7. a) The first five terms of the sequence starting with n=1 are obtained by substituting n by 1,2,3,4,5 in the expression 3×2n1 :
    For n=1   ,   3×2n1=3×211=3×20=3
    For n=2   ,   3×2n1=3×221=3×21=6
    For n=3   ,   3×2n1=3×231=3×22=12
    For n=4   ,   3×2n1=3×241=3×23=24
    For n=5   ,   3×2n1=3×251=3×24=48
    b) Going from one term to the next, we multiply by 2 and hence it is a geometric sequence with common ratio equal to 2.

    3 - Sets


  8. a) The intersection of two sets is the set of all elements common to both sets. Hence
    S1S2={2,9,12}
    b) The union of two sets is the set of all elements in the two sets ( without repetition ). Hence
    S1S2={0,2,9,10,11,12}


  9. Any real number is either rational or irrational but not both, hence
    QP=Empty Set
    The set of all real numbers is the union of the rational and irrational numbers, hence
    QP=R
    b) and d) are true.

    4 - Factors, Multiples and Divisibility of Numbers


  10. a)   345=3×5×23
    b)   150=2×3×5×5
    c)   210=2×3×5×7

  11. The Greatest Common Factor (GCF) of 100 and 180 is equal to 20.

  12. The Least Common Multiple (LCM) of 100 and 15 is equal to 300.


  13. A number is divisible by 3 if the sum of its digits is divisible by 3.

    a)
    add all digits in the given number 101899: 1+0+1+8+9+9=28.
    The result 28 is not divisible by 3 and therefore the given number 101899 is not divisible by 3

    b)
    add all digits in the given number 900234: 9+0+0+2+3+4=18
    The result 18 is divisible by 3 and therefore the given number 900234 is divisible by 3

    c)
    add all digits in the given number 134567280: 1+3+4+5+6+7+2+8+0=36
    36 is divisible by 3 and therefore the given number 134567280 is divisible by 3


  14. A number is divisible by 4 if its two digits on the right form a number that is divisible by 3.

    a)
    The two digits on the right of the given number 189001 are 01 which form a number that is not divisible by 4 and therefore 189001 is not divisible by 4.
    b)
    The two digits on the right of the given number 1005612 are 12 which form a number that is divisible by 4 and therefore 1005612 is divisible by 4.
    c)
    The last two digits in the given number 1003456024 are 24 which form a number that is divisible by 4 and therefore 1003456024 is divisible by 4.


  15. For a number to be divisible by 6, it has to be divisible by 2 and by 3

    a)
    234 is divisible by 2 since its last digit on the right is 4 . It is also divisible by 3 since the sum of its digits 2+3+4=9 is divisible by 3. Therefore 234 is divisible by 6
    b)
    12345 is not divisible by 6 because it is not divisible by 2
    c)
    12114290910 is divisible by 2 since its last digit on the right is 0. The sum of its digits 1+2+1+1+4+2+9+0+9+1+0=30 is divisible by 3 and therefore the given number 12114290910 is also divisible by 3. Since the given number is divisible by 2 and by 3, it is divisible by 6.

    5 - Fractions and Mixed Numbers


  16. Start with the fraction in reduced terms and multiply by a factor in order to obtain the second fraction if possible.
    a)   Multiply numerator and denominator of the fraction 73 by 5 and simplify
    7×53×5=3515
    We obtain a fraction with the same denominator but not the same numerator as the given fraction 1015 hence the two fractions are NOT equivalent.

    b)   Multiply numerator and denominator of the fraction 23 by 4 and simplify
    2×43×4=812
    We obtain a fraction with the same denominator and the same numerator as the given fraction 812 hence the two fractions are equivalent.

    c)   Multiply numerator and denominator of the fraction 712 by 3 and simplify
    7×312×3=2136
    We obtain a fraction with the same denominator and the same numerator as the given fraction 2136 hence the two fractions are equivalent.


  17. a)   Rewrite the three fractions to the lowest common denominator which is the lowest common multiple (LCM) of the denominators 5,10 and 15. LCM of 5,10,15 = 30
    Hence
    25+310115 =2×65×6+3×310×31×215×2 =1230+930230 =1930

    b)   Multiply denominators together and numerators together
    716×414=7×416×14
    Factor the terms 16=4×4 and 14=2×7 in the denominator
    =7×4(4×4)×(2×7)
    Cancel common factors and simplify
    =7×4(4×4)×(2×7)=18


    c)   Rewrite the division of the fractions as a multiplication by the reciprocal; hence
    112÷4=112×14
    Simplify
    =118

    d)   Group the whole parts together and the fractions together
    434112+118=(41+1)+(3412+18)
    Simplify
    =438

    e)   Convert the expressions 134   and   3+13 into fractions.
    134=74        and        3+13=103
    Rewrite the given expression using fractions only
    134÷(3+13)=74÷103
    Rewrite the division as a multiplication by the reciprocal
    =74×310
    Simplify
    =2140


  18. a)   0.2÷0.6=26=13
    b)   1÷0.4=104
    It is an improper fraction and may therefore be written as a mixed number
    =8+24=84+24=212


  19. Dalia spends "14 of her salary" on food and beverages
    15 of "14 of her salary" is spent on soft drinks
    16 of "14 of her salary" is spent on cookies
    Total spent on soft drinks and cookies : 15 of "14 of her salary" plus 16 of "14 of her salary"
    which may be written as
    15×14+16×14=14(15+16)=11120
    Dalia spends 11120 of her salary on soft drinks and cookies.

  20. James spends: 5×2=10 hours on homework during weekdays
    Ben spends : 34×10=7.5 hours on homework during weekdays
    Linda spends: 54×10=12.5 hours on homework during weekdays


  21. Using mixed numbers, a liter and a half of juice is written as : 112
    Using fractions, one sixth of a liter is written as : 16
    Number of glasses that can be filled = 112÷16=9

    6 - Exponents and Scientific Notation


  22. a)   (2)353+(3)4 =8125+81 =52

    b)   (1)350+42(2)4 =1(1)31+1616 =111+1=1

    c)   (34)2+(43)2 =3242+4232 =916+3242 =98


  23. a)   10000=104
    b)   0.0000001=107
    c)   1100000=1105=105

  24. Write in Scientific Notation
    a)   12.4×103=1.24×104
    b)   0.0023×102=2.3×105
    c)   12100000=12105=12×105=1.2×104

    7 - Roots

  25. Simplify
    a)   16=4 because 42=16
    b)   9=3 because 32=9
    b)   38=3 because 33=8

  26. Reduce to simplest form
    a)   3×25=3×25=53
    b)   36×5=36×5=65
    b)   38×7=38×37=237

    8 - Proportionality and Related Problems

  27.  
    a) Use a point on the graph. For example, when t=1 , d=4
    Substitute t by 1and d by 4 in the equation d=k×t to obtain
    4=k×1
    Simplify to obtain
    k=4
    Hence the relationship between the distance d and the time is given by
    d=4×t , with d in km and t in hours.
    b) Find time it takes Leila to walk d = 10 km by solving the equation
    10=4t
    Solve for t
    t=10÷4=2.5 hours
    2.5 hours may also be written as 2:30
    She is 10 km away from her starting point at : 8+2:30=10:30

    Distance Against Time Graph


  28.  
    A column that contains the ratio y/x was added and it shows that y/x is constant and equal to 3. Hence
    y is proportional to x?
    Tables of Proportionality Solution
    a)
    Since y/x=3, we can write y=3x
    Hence
    k=3
    b)
    y=3×10.2=30.6


  29. a) Form the given information, we can write three points of the form (V,t) and they are: (2,10) , (4,20) and (6,30) which are plotted below.

    Volume Versus Time
    b) The three points are located in the same line and therefore there is a proportionality relationship between V and t.
    c) The constant of proportionality k is defined in the equation V=kt. Hence
    k=V÷t
    Use any of the three points above, k is found as follows
    k=V÷t=10÷2=5
    or k=V÷t=20÷4=5
    or k=V÷t=30÷6=5
    Hence
    V=5t
    d) Since we have the relationship between V=kt and V=100, we substitute V by 100 in the equation V=5t.
    100=5t
    Solve the above equation for t
    t=100÷5=20 minutes are needed to fill a tank of 100 Liters.

    9 - Percent and Related Problems


  30. Price after increase = 120 + increase = 120 + 12% of 120
    which is written mathematically as
    Price after increase = 120+12%×120=120+12100×120 =120+14.4 =$134.40


  31. The percent of Jimmy's salary spent on bills = 15% of 50% of his salary
    which is mathematically written as
    15100×50100 =15×50100×100 =75010000 =7.5100 =7.5%


  32. Cost after tax = 40+15% of 40=40+15100×40=$46
    Cost after tip = 46+5100×46=$48.30


  33. Kamelea's spending =$400+$1200+$200+$1200+$600=$3600
    Savings = Salary - spending =$5000$3600=$1400
    Kamelea's savings in percent of salary = 14005000=0.28=28%


  34. Let x be the unknown number. We are given that
    10% of 13 of x = 3
    which is written mathematically as
    10100×13×x=3
    The above equation may be written as
    10x300=3
    Multiply both sides of the equation by 300
    10x300×300=3×300
    Simplify
    10x=900
    Solve for x
    x=900÷10=90


  35. Percentage increase of gas in the US = 433=0.33333=33.33%
    Percentage increase of gas in the France = 21.51.5=0.33333=33.33%
    The US and France saw the same percentage of gas increase in that year.

    10 - Convert Units of Measurement


  36. Divide both sides of the equality 1 m=3.28084 ft by 3.28084 ft
    1 m3.28084 ft=3.28084 ft3.28084 ft
    Simplify to obtain
    1 m3.28084 ft=1
    We now write the given length 10.5 ft as
    10.5 ft=10.5 ft×1
    Substitute 1 by 1 m3.28084 ft. Hence
    10.5 ft=10.5 ft×1 m3.28084 ft
    Cancel  ft
    10.5 ft=10.5 ft×1 m3.28084 ft
    Calculate to obtain
    10.5 ft=10.5 ft×1 m3.28084 ft=3.20039 m


  37. 1.3 km=1.3×1093.61 yd=1421.69 yd


  38. Square both sides of the given equality 1m=1.09361yd to obtain
    (1m)×(1m)=(1.09361yd)×(1.09361yd)
    Simplify
    1m2=1.19598yd2
    1.2m2=1.2×1.19598yd2=1.435176yd2


  39. 1km=1000m and 1hr=3600sec
    Hence
    100km/hr=100km1hr=100×1000m1×3600sec
    Simplify
    100km/hr=27.77777m/sec

    11 - Evaluate Expressions


  40. Substitute x by 1 in the given expression
    1x+21x2=1(1)+21(1)2
    Evaluate
    =1311=13+1=113


  41. Substitute x by 5 in the given expression
    |x+16|+x21=|(5)+16|+(5)21
    Evaluate
    =|5+16|+251=|1|+251 =1+251=25


  42. Substitute a and b by 2 and 2 respectively in the given expression
    2ab2=2(2)(2)2
    Evaluate
    =44=42=2

    12 - Algebra

    Review
    The distributive property in algebra may be used to expand as follows a(x+y)=a×x+a×y The distributive property may also be used in reverse to factor as follows a×x+a×y=a(x+y)


  43. a)
    Use the distributive property on the expression 3(x+2)
    3(x+2)+x12=3x+3×2+x12 =3x+6+x12
    Group like terms
    =(3x+x)+(612)
    Simplify
    =4x6

    b)
    Use the distributive property on the expression 15(15x+20)
    15(15x+20)+2x+4=15×15x+15×20+2x+4

    Simplify using 15×15x=155x=3x and 15×20=205=4

    =3x+4+2x+4
    Group like terms
    =(3x+2x)+(4+4)
    Simplify
    =5x+8

    c)
    Use the distributive property on the expression 0.2(5x+10)
    0.2(5x+10)+3x4=0.2×5x+0.2×10+3x4
    Simplify
    =x+2+3x4 =(x+3x)+(24) =4x2


  44. Simplify the expressions
    a)
    2x×3x=(2×3)×(x×x)=6x2

    b)
    12x×45x=(12×45)×(x×x)=25x2

    c)
    3x2×5x3=(3×5)×(x2×x3)=15x2+3=15x5


  45. a)
    The greatest common factor of the coefficients 21 and 7 is equal to 7, hence
    21x+7=7×3x+7×1
    Use the distributive property in reverse to factor 7 out.
    =7(3x+1)

    b)
    The greatest common factor of the coefficients 24 and 20 is equal to 4, hence
    2420x=4×64×5x
    Use the distributive property in reverse to factor 4 out.
    =4(65x)

    c)
    The greatest common factor of the coefficients 8, 4 and 32 is equal to 4, hence.
    8b4a+32=4×2b4×a+4×8
    Use the distributive property in reverse to factor the 4 out
    =4(2ba+8)

    13 - Equation with One Variable and Related Problems

  46.  Solve the equations
    a)
    Given the equation 3(x2)=3
    Expand the expression 3(x2) using the distributive property
    3x6=3
    Add 6 to both sides
    3x6+6=3+6
    Simplify
    3x=9
    Divide both sides by 3
    3x÷3=9÷3
    Simplify and solve for x.
    x=3

    b)
    Given the equation 2(9x)=(x+5)
    Expand the parentheses in both sides of the equation using the distributive property
    182x=x5
    Add 2x to both sides of the equation and simplify
    182x+2x=x5+2x
    18=x5
    Add 5 to both sides and simplify
    x=23

    c)
    Given the equation x+13=6
    Multiply both sides by the denominator 3
    x+13×3=6×3
    Simplify
    x+1=18
    Solve for x
    x=17

    d)
    Given the equation 4(x+14)=15
    Expand the parentheses on the left side of the equation using the distributive property
    4x+4×14=15
    Simplify
    4x+1=15
    Solve for x
    4x+11=151
    4x=16
    x=4

    e)
    Given the equation xx2=3
    Multiply all terms by the denominator 2.
    x×2x2×2=3×2
    Simplify and solve for x.
    2xx=6
    x=6


  47. Given
    Rectangular Garden with Path
    a)
    Length of outer perimeter: L=12+x+x=12+2x
    Width of outer perimeter: W=8+x+x=8+2x
    Outer perimeter =2×L+2×W=2(12+2x)+2(8+2x)
    Expand and group like terms
    Outer perimeter =24+4x+16+4x=40+8x
    Perimeter of the garden (in white) =2×12+2×8=40
    Given that "outer perimeter is equal to twice the perimeter of the garden", we can write the equation
    40+8x=2×40

    b)
    Simplify the right side of the equation obtained in a)
    40+8x=80
    Solve for x
    8x=40
    x=5 m

    c) L=12+2x=12+2×5=22 m
    W=8+2x=8+2×5=18 m

    d)
    Area of garden and path = =L×W=22×18=396m2

    e) Area of garden = 12×8=96m2

    f) The area of the path = Area of garden and path - Area of garden =39696=300m2.

  48. Let x be the original number.
    "10 is subtracted from twice a number" is written as : 2x10
    "the result is multiplied by half" is written as : 12(2x10)
    "the answer is 5" is written as : 12(2x10)=5
    Multiply both sides of the equation and simplify
    12(2x10)×2=5×2
    2x10=10
    The original number is equal to 10

    14 - Inequality with One Variable


  49. a)
    Given the inequality x+2<4
    Subtract 2 from both sides of the inequality and simplify
    x+22<42
    x<2

    b)
    Given the inequality 2(x+3)2 Expand the parentheses on the left side of the inequality using the distributive property and simplify
    2×x+2×32
    2x+62
    Subtract 6 from both sides of the inequality and simplify
    2x+6626
    2x4
    Divide both sides of the inequality by 2 and simplify
    2x242
    x2

    c) Given the inequality 3x+211
    Subtract 2 from both sides of the inequality and simplify
    3x+22112
    3x9
    Divide both sides of the inequality by 3 and change the symbol of the inequality because 3 is negative.
    3x393
    Simplify
    x3

    d) Given the inequality 4x+12x+3
    Multiply both sides of the inequality by the denominator 2
    4x+12×2(x+3)×2
    Simplify
    4x+12x+6
    Subtract 1 from both sides of the inequality and simplify
    4x+112x+61
    4x2x+5
    Subtract 2x from both sides of the inequality and simplify
    4x2x2x+52x
    2x5
    Divide both sides of the inequality by 2 and simplify
    x5/2

    15 - Functions


  50. A function is a relation between two sets such that to each input there corresponds one output only.
    a)
    The relation {(1,2),(3,4),(5,7),(5,9)} is NOT a function because to the input 5 correspond two outputs: 7 and 9.
    b)
    The relation {(1,2),(3,4),(5,7),(7,9)} is a function because to each input corresponds one output only
    c)
    The relation {(3,3),(9,4),(5,7),(9,0)} is NOT a function because to the input 9 correspond two outputs: 4 and 0.


  51. Graph (3) is a line and is therefore the graph of a linear function.
    Graph of Functions

  52.   Given the function y=2x+1,
    a)
    for x=0 , y=2x+1=2(0)+1=1
    for x=1 , y=2x+1=2(1)+1=3
    b) The results in part a) may be represented by ordered pairs (x,y) as (0,1) and (1,3)
    The given function y=2x+1 is a linear function and its graph is a line and therefore the two ordered pairs obtained above may be used to graph the function as shown below.
    Graph of the Function y = 2x + 1


  53. a) The function corresponding to graph (1) has a higher rate of change because it increases faster as x increases.
    b)
    Graph of Linear Functions with Points
    Points on Graph (1) : (0,1) , (3,7) ; there are many other points
    Points on Graph (2) : (0,3) , (8,8) ; there are many other points
    c)

    Rate of change of Graph (1) : r1=Change in yChange in x=7130=2
    Rate of change of Graph (2) : r2=Change in yChange in x=8380=5/8
    d)
    Calculations show that the rate of change of (1) is higher than the rate of change of graph (2) which confirm the answer to part a) above.

    16 - Two-Dimensional Figures


  54. Let h be the hypotenuse of the triangle and apply the Pythagorean theorem to write
    h2=62+82
    Solve for h by taking the square root of both sides of the above equation.
    h=62+82 =36+64 =100 =10 cm


  55. Note that AOC=AOB+BOC
    Substitute the known angles by their sizes.
    79=31+BOC
    Hence
    BOC=7931=48
    Note that angles BOC and EOF are vertical and therefore have equal sizes. Hence
    EOF=48
    Vertical Angles


  56. A square has 4 lines of symmetry as shown below.
    Lines of Symmetry of a Square



  57. Angles m1 and m2 are supplementary and therefore their sum is equal to 180. Hence
    40+m2=180
    Solve for m2
    m2=18040=140
    m1 and m3 are vertical and therefore have equal measure, hence
    m3=m1=40
    m2 and m4 are vertical and therefore have equal measure, hence
    m4=m2=140
    m1 and m5 are corresponding angles and therefore have equal measure, hence
    m5=m1=40
    m2 and m6 are corresponding angles and therefore have equal measure, hence
    m6=m2=140
    m4 and m8 are corresponding angles and therefore have equal measure, hence
    m8=m4=140
    m3 and m7 are corresponding angles and therefore have equal measure, hence
    m7=m3=40
    Parallel and Intersecting Lines

    17 - Perimeter and Area of Planar Figures


  58. Radius: r=Diameter÷2=20÷2=10 cm
    Area=π×r2=3.14×102=3.14×100=314cm2


  59. The area A of a right triangle with legs a and b is given by
    A=12×a×b
    We are given the size of one leg a=16 and we need to find the size of the second leg b.
    Use the Pythagorean theorem to find the second leg b of the right triangle
    b2+162=202
    Hence
    b2=202162=144
    b=144=12cm
    Area of the right triangle is equal to: 12×16×12=96cm2

  60. Because of the symmetry, we calculate the area of the lower part of the arrow which is a trapezoid whose A area is given by
    Area of Half Arrow
    A=12(¯FG+¯ED)ׯHE
    ¯FG=12+164=24
    ¯ED=16
    Since ABDE is a square, we have ¯AE=¯AB=16 .
    ¯HE=12¯AE=1216=8
    Hence
    A=12(24+16)×8=160
    The area of the arrow is twice the area of the trapezoid. Hence
    The area of the arrow is equal to 2×160=320unit2

  61. We decompose the given shape into basic shapes whose areas are easily calculated using formulas.
    Area of Compound Shape Decomposed
    Area of isosceles triangle ABG =12×4×4=8
    Area of trapezoid BCFG =12×2×(4+1)=5
    Area of trapezoid CDEF =12×3×(1+3)=6
    Area of semicircle of diameter DE =12×π×1.52=3.14×1.52=3.53
    Total area of the shaded region =8+5+6+3.53=22.53mm2 , rounded to two decimal places.

    18 - Volumes and Surface Area



  62. Cylinder and half a sphere
    The volume of half the sphere =12×43πr3=46×3.14×63=452.16m3
    The volume of the cylinder =π×r2×h=3.14×62×10=1130.4m3
    The surface area of half the sphere =12×4×π×r2=2×3.14×62=226.08m2
    The surface area of the cylinder (without the bottom) =2×π×r×h=2×3.14×6×10=377.00m2
    Total volume of silo =452.16+1130.4=1582.56m3
    Total surface area of silo =226.08+377.00=603.08m2


  63. Because of the symmetry of the rectangular prism, the volume of the triangular prism is equal to half the volume of the rectangular prism
    Triangular Prism
    Volume of the given rectangular prism =6×3×4=72unit3
    Volume of the triangular prism 12×72=36unit3

    The surface area of the triangular prism is equal to half the surface area of the rectangular prism to which we add the area of the rectangle ABCE made by the red diagonals and the edges AE and BC of the rectangular prism.
    Surface area of the rectangular prism =2×(6×3+3×4+6×4)=108unit2
    Use the Pythagorean theorem to find the length d of the diagonal which is the hypotenuse (red) of the right triangle CDE.
    d2=32+42=25
    Use square root to find
    d=5
    Area of the rectangle made by the diagonals and the edges =5×6=30unit2
    Surface area of the triangular prism =12×108+30=84unit2
    Note that there are other ways to find the volume and surface area of the rectangular prism.

    19 - Data and Graphs


  64. a)
    January has the lowest average temperature of 5 and is therefore the coldest month.
    b)
    July has the highest average temperature of 26 and is therefore the hottest month of the year.
    c)
    Difference in temperature between the coldest and hottest months =26(5)=31
    d)
    The smallest increases is from January to February and from June to July
    e)
    The smallest decrease is from July to August

    Average High Temperature in Ottawa

  65. a) The given data in order from smallest to the largest values is as follows
    31,44,45,54,55,56,60,64,67,67,69,70,76,76,77,78,79,84,85,86,88,89,91,92,97
    b)
    Range = Largest value - smallest value =9731=66
    c)
    Start with the class 3039 and add the class width to obtain the remaining classes and cover all data values with a maximum value of 97.
    To obtain the next class, we add 10 to the lower and upper limits of a given class.
    Hence the next class after the class 3039 is given by
    (30+10)(39+10) = 4049
    and continue until all data values are covered as shown in the frequency table below
    Frequency Table

    d)
    A histogram is made using the number of student in the vertical axis and the classes on the horizontal axis as shown below.

    Histogram
    e)
    The scores in the three classes 339 , 4049 and 5059 are below 60 and the number of students in these classes may be found in the frequency table and the histogram.
    1 students in the class 339
    2 students in the class 4049
    3 students in the class 5059
    The total number of students who failed is equal to
    1+2+3=6
    The percentage of students who failed is given by
    Number of Students who failedTotal Number of Students=625=0.24=24%

    20 - Statistics


  66. First order the data values from the smallest to the largest.
    {0,1,2,2,3,3,3,4,9,9,10}
    The median is the value in the middle (red) which is 3
    The lower quartile is the median of the data values below the median 3 which in the above data is {0,1,2,2,3}
    lower quartile = 2
    The upper quartile is the median of the data values above the median 3 which in the above data is {3,4,9,9,10}
    upper quartile = 9



  67. Let x be the score of the fifth quiz. The average of the 5 quizzes is given by
    83+94+97+93+x5
    The average "is at least 90" is mathematically written as
    83+94+97+93+x590
    Multiply both sides of the above inequality by 5
    83+94+97+93+x5×590×5
    Simplify
    83+94+97+93+x450
    Solve for x
    x450(83+94+97+93)
    x83
    Mark needs to score at least 83 in the fifth quiz in order to have an average which is at least 90

    21 - Probabilities


  68. Sample space = all possible outcomes = {1,2,3,4,5,6}
    Set of even numbers among the outcomes = {2,4,6}
    There are 6 possible outcomes from which 3 are even; hence
    probability of getting an even number =Number of elements in the set of even numbersNumber of elements in the sample space=36=12


  69. a)
    Probability of getting a tail is P1=12
    Probability of getting a "4" is P2=16
    The events are independent and therefore
    The probability of getting a tail (coin) and a 4 (die) = P1×P2=12×16=112
    b)
    Probability of getting a head is P3=12
    Probability of getting an odd number is P4=36=12
    The events are independent and therefore
    The probability of getting head (coin) and an odd number (die) P3×P3=12×12=14


  70. a)
    Using the counting principle, we have 3×3=9 possible outcomes written as: (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3) There is one outcome with 3 in both random selections and that is (3,3)
    Hence
    The probability of selecting 3 in both random selections =19
    b)
    Three of the 9 possible outcomes have the same number in both random selections and these are: (1,1) , (2,2) and (3,3)
    The probability of selecting the same number in both random selections =39=13


  71. If 5 said blue was their favorite color and 6 said brown was their favorite number, then
    2056=11 picked a color that is neither blue nor brown
    Hence the probability that the next student surveyed will pick a color that is neither blue nor brown is given by
    1120

More References and links

  1. Grade 8 Math Topics
  2. Fractions Questions and Problems with Solutions