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 ( 016-6006045)
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:
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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.
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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)
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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(7-4)
Solution:
= 3 x+35−20= 3 x+15
14. 5y+6+4y-6x
Solution:
= 5 y+4 y−6 x+6= 9 y−6x+6
15. 4(7y-6) 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 + (-5-5)= 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= 12-9X= 3
11. Solve 4 x = 36..Solution:
X = 44
x=364
X = 9
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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 x-coordinate goes up by 1, the y-coordinate goes down by 3.
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-10-8-6-4-202468
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 y-intercept 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).
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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 y-intercept of this equation.
y = 2(x-4)
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 y-intercept of the following equation: 3 y=2(3−2 x)
Answer:
Y = 1 12
8. What is the y-intercept 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 y-intercept 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 y-intercept?
Answer:
= Slope = 2, y-intercept = 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