Solutions to Grade 8 Math Practice Test
Detailed solutions to the Grade 8 math practice test questions are presented.
1 - Numbers
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? is NOT a rational number.
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The value of the digit 5 in the number 34.6597 is
2 - Sequences
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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
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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
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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.
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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 3×2n−1 :
For n=1 , 3×2n−1=3×21−1=3×20=3
For n=2 , 3×2n−1=3×22−1=3×21=6
For n=3 , 3×2n−1=3×23−1=3×22=12
For n=4 , 3×2n−1=3×24−1=3×23=24
For n=5 , 3×2n−1=3×25−1=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
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a) The intersection of two sets is the set of all elements common to both sets. Hence
S1∩S2={2,9,12}
b) The union of two sets is the set of all elements in the two sets ( without repetition ). Hence
S1∪S2={0,2,9,10,11,12}
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Any real number is either rational or irrational but not both, hence
Q∩P=Empty Set
The set of all real numbers is the union of the rational and irrational numbers, hence
Q∪P=R
b) and d) are true.
4 - Factors, Multiples and Divisibility of Numbers
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a) 345=3×5×23
b) 150=2×3×5×5
c) 210=2×3×5×7
- The Greatest Common Factor (GCF) of 100 and 180 is equal to 20.
- The Least Common Multiple (LCM) of 100 and 15 is equal to 300.
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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
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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.
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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
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.
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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+310−115 =2×65×6+3×310×3−1×215×2 =1230+930−230 =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
434−112+118=(4−1+1)+(34−12+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
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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
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.
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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
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
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a) (−2)3−53+(−3)4 =−8−125+81 =−52
b) (−1)−3−50+42(−2)4 =1(−1)3−1+1616 =1−1−1+1=−1
c) (34)2+(43)−2 =3242+4−23−2 =916+3242 =98
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a) 10000=104
b) 0.0000001=10−7
c) 1100000=1105=10−5
- Write in Scientific Notation
a) 12.4×103=1.24×104
b) 0.0023×10−2=2.3×10−5
c) 12100000=12105=12×10−5=1.2×10−4
7 - Roots
- Simplify
a) √16=4 because 42=16
b) √9=3 because 32=9
b) 3√8=3 because 33=8
- Reduce to simplest form
a) √3×25=√3×√25=5√3
b) √36×5=√36×√5=6√5
b) 3√8×7=3√8×3√7=23√7
8 - Proportionality and Related Problems
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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
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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?
a)
Since y/x=3, we can write y=3x
Hence
k=3
b)
y=3×10.2=30.6
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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.
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
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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
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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%
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Cost after tax = 40+15% of 40=40+15100×40=$46
Cost after tip = 46+5100×46=$48.30
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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%
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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
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Percentage increase of gas in the US = 4−33=0.33333=33.33%
Percentage increase of gas in the France = 2−1.51.5=0.33333=33.33%
The US and France saw the same percentage of gas increase in that year.
10 - Convert Units of Measurement
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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
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1.3 km=1.3×1093.61 yd=1421.69 yd
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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
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1km=1000m and 1hr=3600sec
Hence
100km/hr=100km1hr=100×1000m1×3600sec
Simplify
100km/hr=27.77777m/sec
11 - Evaluate Expressions
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Substitute x by 1 in the given expression
1x+2−1x−2=1(1)+2−1(1)−2
Evaluate
=13−1−1=13+1=113
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Substitute x by −5 in the given expression
|−x+1−6|+x2−1=|−(−5)+1−6|+(−5)2−1
Evaluate
=|5+1−6|+25−1=|−1|+25−1 =1+25−1=25
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Substitute a and b by 2 and −2 respectively in the given expression
2a−√b2=2(2)−√(−2)2
Evaluate
=4−√4=4−2=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)
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a)
Use the distributive property on the expression 3(x+2)
3(x+2)+x−12=3x+3×2+x−12 =3x+6+x−12
Group like terms
=(3x+x)+(6−12)
Simplify
=4x−6
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)+3x−4=0.2×5x+0.2×10+3x−4
Simplify
=x+2+3x−4 =(x+3x)+(2−4) =4x−2
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
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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
24−20x=4×6−4×5x
Use the distributive property in reverse to factor 4 out.
=4(6−5x)
c)
The greatest common factor of the coefficients 8, 4 and 32 is equal to 4, hence.
8b−4a+32=4×2b−4×a+4×8
Use the distributive property in reverse to factor the 4 out
=4(2b−a+8)
13 - Equation with One Variable and Related Problems
- Solve the equations
a)
Given the equation 3(x−2)=3
Expand the expression 3(x−2) using the distributive property
3x−6=3
Add 6 to both sides
3x−6+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(9−x)=−(x+5)
Expand the parentheses in both sides of the equation using the distributive property
18−2x=−x−5
Add 2x to both sides of the equation and simplify
18−2x+2x=−x−5+2x
18=x−5
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+1−1=−15−1
4x=−16
x=−4
e)
Given the equation x−x2=3
Multiply all terms by the denominator 2.
x×2−x2×2=3×2
Simplify and solve for x.
2x−x=6
x=6
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Given
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 =396−96=300m2.
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Let x be the original number.
"10 is subtracted from twice a number" is written as : 2x−10
"the result is multiplied by half" is written as : 12(2x−10)
"the answer is 5" is written as : 12(2x−10)=5
Multiply both sides of the equation and simplify
12(2x−10)×2=5×2
2x−10=10
The original number is equal to 10
14 - Inequality with One Variable
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a)
Given the inequality x+2<4
Subtract 2 from both sides of the inequality and simplify
x+2−2<4−2
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+2×3≥2
2x+6≥2
Subtract 6 from both sides of the inequality and simplify
2x+6−6≥2−6
2x≥−4
Divide both sides of the inequality by 2 and simplify
2x2≥−42
x≥−2
c) Given the inequality −3x+2≤11
Subtract 2 from both sides of the inequality and simplify
−3x+2−2≤11−2
−3x≤9
Divide both sides of the inequality by −3 and change the symbol of the inequality because −3 is negative.
−3x−3≥9−3
Simplify
x≥−3
d) Given the inequality 4x+12≥x+3
Multiply both sides of the inequality by the denominator 2
4x+12×2≥(x+3)×2
Simplify
4x+1≥2x+6
Subtract 1 from both sides of the inequality and simplify
4x+1−1≥2x+6−1
4x≥2x+5
Subtract 2x from both sides of the inequality and simplify
4x−2x≥2x+5−2x
2x≥5
Divide both sides of the inequality by 2 and simplify
x≥5/2
15 - Functions
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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.
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Graph (3) is a line and is therefore the graph of a linear function.
- 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.
a) The function corresponding to graph (1) has a higher rate of change because it increases faster as x increases.
b)
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=7−13−0=2
Rate of change of Graph (2) : r2=Change in yChange in x=8−38−0=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
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
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Note that ∠AOC=∠AOB+∠BOC
Substitute the known angles by their sizes.
79∘=31∘+∠BOC
Hence
∠BOC=79∘−31∘=48∘
Note that angles ∠BOC and ∠EOF are vertical and therefore have equal sizes. Hence
∠EOF=48∘
A square has 4 lines of symmetry as shown below.
Angles m∠1 and m∠2 are supplementary and therefore their sum is equal to 180∘. Hence
40∘+m∠2=180∘
Solve for m∠2
m∠2=180∘−40∘=140∘
m∠1 and m∠3 are vertical and therefore have equal measure, hence
m∠3=m∠1=40∘
m∠2 and m∠4 are vertical and therefore have equal measure, hence
m∠4=m∠2=140∘
m∠1 and m∠5 are corresponding angles and therefore have equal measure, hence
m∠5=m∠1=40∘
m∠2 and m∠6 are corresponding angles and therefore have equal measure, hence
m∠6=m∠2=140∘
m∠4 and m∠8 are corresponding angles and therefore have equal measure, hence
m∠8=m∠4=140∘
m∠3 and m∠7 are corresponding angles and therefore have equal measure, hence
m∠7=m∠3=40∘
17 - Perimeter and Area of Planar Figures
Radius: r=Diameter÷2=20÷2=10 cm
Area=π×r2=3.14×102=3.14×100=314cm2
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=202−162=144
b=√144=12cm
Area of the right triangle is equal to: 12×16×12=96cm2
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
A=12(¯FG+¯ED)ׯHE
¯FG=12+16−4=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
We decompose the given shape into basic shapes whose areas are easily calculated using formulas.
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
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
Because of the symmetry of the rectangular prism, the volume of the triangular prism is equal to half the volume of the rectangular 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
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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
-
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 =97−31=66
c)
Start with the class 30−39 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 30−39 is given by
(30+10)−(39+10) = 40−49
and continue until all data values are covered as shown in the frequency table below
d)
A histogram is made using the number of student in the vertical axis and the classes on the horizontal axis as shown below.
e)
The scores in the three classes 3−39 , 40−49 and 50−59 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 3−39
2 students in the class 40−49
3 students in the class 50−59
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
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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
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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+x5≥90
Multiply both sides of the above inequality by 5
83+94+97+93+x5×5≥90×5
Simplify
83+94+97+93+x≥450
Solve for x
x≥450−(83+94+97+93)
x≥83
Mark needs to score at least 83 in the fifth quiz in order to have an average which is at least 90
21 - Probabilities
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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
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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
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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
If 5 said blue was their favorite color and 6 said brown was their favorite number, then
20−5−6=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
- Grade 8 Math Topics
- Fractions Questions and Problems with Solutions