Maths Questions IGCSE

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MRSM BALING, KEDAH
Mathematics Form 1 (IGCSE)Homework
Semester 2 2012
Name: Siti Mazenah Dienta Rochani binti Mohamed Tajudin
Matric Number : 12094
Class: 1D ( Jauzi)
Teacher's Name : Mrs. Wan Siti Hajar Wan Hashim
Parent's Name : Mohamed Tajudin bin Alias ( 0166006045)
Date:
Maths is easy and fun Page 1
XTajudin Alias
Manager
TRIGONOMETRY
1.
Calculate the value of cos θ in the following triangle.
Solution:
2.
Calculate the length of the side x, given that sin θ = 0.6.
Solution:
Maths is easy and fun Page 2
3.
Calculate the length of the side x, given that tan θ = 0.4
Solution:
4.
3cm
5cm
Calculate x.
Solution:
Using Pythagoras’ theorem:
x2= 32+52
x = √32+52
= √9+25
= √34
= 5.83 cm
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X
5.
12 m
Calculate y.
Solution:
Using Pythagoras’ theorem:
122 = y2+92
y2 = 122−92
y = √144−81
= √63 = 7.94 m
6.
Calculate a.
Solution:
The missing side is the opposite side and we have the adjacent and angle, hence,
7. Find x and H in the right triangle below.
Maths is easy and fun Page 4
9m
Y
a
34°
1.2 cm
tanΘ = oppadj
tan34Θ = a
1.2a = 1.2 x tan34Θ = 0.809 cm
Solution:
x = 10
tan 51
x = 13 (2 significant digits)
8. Find the lengths of all sides of the right triangle below if its area is 400.
Solution:
Area = (1/2)(2x)(x) = 400
Solution for x:
x = 20x2 = 40 H = x sqrt(5)
H= 20 sqrt(5)
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H = 10
sin 51
H= 8.1 (2 significant digits)
Pythagoras theorem: (2x)2 + (x)2 = H2
9. BH is perpendicular to AC. Find x the length of BC.
Solution:
BH perpendicular to AC, means that triangles ABH and HBC are right triangles. Hence,
tan(39o) = 11 / AH or AH = 11 / tan(39o)
HC = 19  AH = 19  11 / tan(39o)
Solution for x and substitution for HC:
x = sqrt [ 112 + (19  11 / tan(39o) )2 ]
= 12.3 (rounded to 3 significant digits)
10.
The area of a right triangle is 50. One of its angles is 45o. Find the lengths of the sides and hypotenuse of the triangle.
Solution:
The triangle is right and the size one of its angles is 45o; the third angle has a size 45o and therefore the triangle is right and isosceles. Let x be the length of one of the sides and H be the length of the hypotenuse.
Area = (1/2)x2 = 50 , solve for x: x = 10
We now use Pythagoras theorem to find:
H: x2 + x2 = H2
Solution for H:
H = 10 sqrt(2)
Maths is easy and fun Page 6
Pythagoras theorem applied to right triangle HBC: 112 + HC2 = x2
11. ABC is a right triangle with a right angle at A. Find x the length of DC.
Solution:
Since angle A is right, both triangles ABC and ABD are right and therefore we can apply Pythagoras theorem.
142 = 102 + AD2 , 162 = 102 + AC2
Also x = AC  AD
= sqrt( 162  102 )  sqrt( 142  102 )
= 2.69 (rounded to 3 significant digits)
12. Given the following right triangle, find tan A.
Solution:
tan A = 32
tan = 0
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13. Find the missing side of the following triangle.
Solution:
Using Pythagoras theorem,
=
14. Find the area of the following triangle.
Solution:
A = 12
xy
15. The angle of repose for sand is typically about 35°. What is the sine of this angle?
Solution:
1. Type 35 into your calculator
2. Press the sin button.
3. Your calculator should read 0.574.
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√ x2+ y2
ALGEBRAIC EXPRESSIONS
1. Simplify the expression −3 (4 x−2 )−2 (5 x−4 ).
Solution:
= −12 x+6 10 x+8= −12 x−10 x+6+8= −22 x+14
2. Simplify the following algebraic expression.
2x + 5 + 10x  9
Solution:
= (10x  2x) + (5  9) = 8x  4
3. Evaluate 8x + 7 given that x  3 = 10.
Solution:
= 8 (13 )+7=104+7=111
4. Simplify the following algebraic expression.
3(x + 7) + 2(x + 4) + 5x
Solution:
= 3x + 21  2x + 8 + 5x = (3x  2x + 5x) + (21 + 8) = 6x + 29
5. Evaluate: 18 + 4(6 ÷ 2)2
Solution:
= 18 + 4(6 ÷ 2)2 = 18 + 4(3)2 = 18 + 4*9 = 18 + 36 = 18
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6. Simplify: 12x3  3(2x3 + 4x 1)  5x + 7 .
Solution:
=12x3  3(2x3 + 4x 1)  5x + 7 = 12x3  6 x3  12 x + 3  5x + 7 = 6 x3  17 x + 10
7. Solve the equation: 5x + 20 = 25
Solution:
= −5 x=25−20= −5 x=5= x=−1
8. Solve 4 y+8 x−16 y
Solution:
= 4 y−16 y+8 x= −12 y+8 x
9. Simplify 3x + 1 + 8x + 9.
Solution:
= 11x+1010. Simplify 2x + 5y  7x + 8y.
Solution:
= 2 x−7 x+5 y+8 y= −5 x+13 y
11. Calculate 3(4y+9)4
Solution:
= 12 y+27−4= 12 y+23
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¿3 x+8 x+1+9
12. Simplify 7(4+5)+8s
Solution:
= 28+35+8 s= 63+8 s
13. Evaluate 3x+5(74)
Solution:
= 3 x+35−20= 3 x+15
14. 5y+6+4y6x
Solution:
= 5 y+4 y−6 x+6= 9 y−6x+6
15. 4(7y6) x 8
Solution:
= = 28y  192
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28 y−24 x8
INTEGERS
1. 3 + (2)
Solution:
= −3−2= 5
2. 12 + (7) + (5)
Solution:
= 12 + (7) = 19 = 19 + (5) = 24
3. 8(3)
Solution:
= 8 x 3= 24
4. (4) 5
Solution:
= 1024
5. Add 17 + (3) + (2) + 5.
Solution:
= 17−3−2+5= 12 + 5= 17
6. Multiply (–9) (6).
Solution:
= 9 x 6= 54
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7. Divide (–22) ÷ (–11).
Solution:
= 22 ÷ 11= 2
8. Simplify 8 – 4×2 + 6.
Solution:
= 8  8 + 6= 6
9. Simplify 5 × (13 – 6) 3 + 96 ÷ 6.
Solution:
= 5 x21+16= 1731
10. Simplify 8 – 3 + 2 – 7 – 5.
Solution:
= 5 + (55)= 5 + (10)= 5
11. Multiply (–7) (+9).
Solution:
= 7 x 9= 63
12. Find (–9) 4.
Solution:
= 9 x 4= 6561
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13. Add 9 + 6.
Solution:
= 3
14. Divide (–48) ÷ (+6).
Solution:
= 48 ÷ 6= 8
15. Simplify 5×2 2 + 4 – 3.
Solution:
= 5 x 4 + 1= 21
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LINEAR EQUATION
1. Solve for x.
x  4 = 10
Solution:
X = 10+4X = 14
2. Solve x.
Solution:
=
= =
= = = = x = 3
3. Solve x.
2(3x  7) + 4 (3 x + 2) = 6 (5 x + 9) + 3
Solution:
= = = =
=
=
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4. 2x  4 = 10
Solution:
x = 2 x=10+4X = 2 x=14X = 7
5. 5x  6 = 3x  8
Solution:
X = 5 x−3 x=−8+6X = 2 x=−2X = 1
6. Solve x.
Solution:
X = 12( 34
x+ 56 )=12(5x−125
3)
X = 12( 34
x+ 56 )=12(5 x−125
3)
X= 12( 34
x)+12( 56 )=12 (5 x )−12
1253
X = 3(3x) + 2(5) = 60x  500X = 9x + 10 = 60x  500X = 10 = 51x  500X= 510 = 51x
X = 51051
X= 10
7. Find x if: 2x + 4 = 10
Solution:
X = 2x + 4  4 = 10  4X = 2x = 6X = 2x / 2 X = 6 / 2
X = 3
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8. Find x if: 3x  4 = 10.
Solution:
X = 3x  4 + 4 = 10 + 4 X =3x X = 6 X = 3x / 3 X = 6 / 3 X = 2
9. Find x if: 4x  4y = 8.
Solution:
X= 4x  4y + 4y = 8 + 4yX= 4x = 8 + 4yX= 4x / 4 = (8 + 4y) / 4X= 2 + y
10. Find x if: x + 32 = 12
Solution:
X= x+32=12X= x + 9 = 12X= x + 9  9X= 129X= 3
11. Solve 4 x = 36..Solution:
X = 44
x=364
X = 9
Maths is easy and fun Page 17
12. Work out 6(8 – 2 x )+ 25 = 5(2 – 3 x ).
Solution:
13. Explain 7(2 x – 3) – 4( x + 5)= 8( x – 1)+ 3.
Solution:
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14. Find x for .
Solution:
15. Solve x – 4(3 x – 2) – (– x + 6)= – 5 x + 8.
Solution:
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STRAIGHT LINE GRAPH
1. What are the coordinate of Point A marked on the following grid?
Answer :
= 1, 1
2. Find the equation of the straight line shown.
Solution:
The line passes through the points (0, 4) and (1, 1).So when the xcoordinate goes up by 1, the ycoordinate goes down by 3.
Maths is easy and fun Page 20
10864202468
10
2 1 0 1 2 3 4
8
6
4
2
0
2
4
6
8
So the gradient is
gradient =−3
1=−3
As the yintercept is 4, the equation of the line must be y = 3x + 4.
3. Draw the lines y = 3x – 2 and y = 6 – 2x.
Solution:
Table of values for y = 3x – 2
x 0 1 2 3y 3×0  2
= 23×1  2
= 13×2 2
= 43×3  2
= 7
Then, plot the points (0, 2), (1, 1), (2, 4) and (3, 7) and join them up.
Table of values for y = 6 – 2x
x 0 1 2 3y 6  2×0
= 66  2×1
= 46  2×2
= 26  2×3
= 0
We plot the points (0, 6), (1, 4), (2, 2) and (3, 0).
Maths is easy and fun Page 21
Work out the y value corresponding to each x value using the formula
y = 3x  2.
2 4 6 8
4. Write down the straight line equation from the information given.
M=3 C=4
Answer:
y = 3x+4
5. Write down the gradient and yintercept of this equation.
y = 2(x4)
Answer:
= 2, 8
6. Find the equation of line that passes trough point (6, 2), (4, 7)
Answer:
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y = −52
x−17
7. Write down the yintercept of the following equation: 3 y=2(3−2 x)
Answer:
Y = 1 12
8. What is the yintercept of this graph?
Answer:
= 3
9. Complete the table of values for y = x + 3, and then draw the graph.
Solution:
x 3 2 1 0 1 2 3Y= x + 3 0 1 2 3 4 5 6
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10. Write down coordinate P.
Answer:
= 2, 1
11. Find the yintercept of the straight line that passes through (1, 5) and (5, 21).
Solution:
y = mx + c
5 = 4 × 1 + c
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m = (y2 – y1) ÷ (x2  x1)
m = (21  5) ÷ (5 – 1)
m = 16 ÷ 4
5 = 4 + c
1 = c
12. Find the equation of the straight line that passes through (3, 12) and (1,4).
Solution:
y = mx + c
12 = 2 × 3 + c
12 = 6 + c
6 = c
= y = 2x  6
13. For the straight line y = 2x + 3, what are:
a) the slopeb) the yintercept?
Answer:
= Slope = 2, yintercept = 3
14. Find the equation of this linear graph.
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m = (y2 – y1) ÷ (x2  x1)
m = (4  12) ÷ (1  3) m = 8 ÷ 4
m = 2
Solution:
y = 3x – 5
15. Find the equation of this linear graph.
Solution:
= y =  13
x + 3
~THANK YOU! ~Maths is easy and fun Page 26
m = 6 ÷ 2 = 3
m = 3 ÷ 9 = −13
Maths is easy and fun Page 27