Maths Questions IGCSE
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Transcript of Maths Questions IGCSE
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)
XTajudin Alias Manager
Date:
Maths is easy and fun
Page 1
TRIGONOMETRY
1.
Calculate the value of cos in the following triangle. Solution:
2.
Calculate the length of the side x, given that sin = 0.6. Solution:
Maths is easy and fun
Page 2
3.
Calculate the length of the side x, given that tan = 0.4 Solution:
4.X
3cm5cm
Calculate x.Solution:
Using Pythagoras theorem: = x= = = = 5.83 cm
Maths is easy and fun
Page 3
5.9m 12 m
Y Calculate y.Solution:
Using Pythagoras theorem: = = = = = 7.94 m
6.
a
34 1.2 cm
Calculate a.Solution:
The missing side is the opposite side and we have the adjacent and angle, hence, tan = tan34 =
a = 1.2 x tan34 = 0.809 cm Maths is easy and fun Page 4
7. Find x and H in the right triangle below.
Solution:
x= x = 13 (2 significant digits)
H= H= 8.1 (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) Pythagoras theorem: (2x)2 + (x)2 = H2
Maths is easy and fun
Page 5
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(39 ) 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:o
Pythagoras theorem applied to right triangle HBC: 112 + HC2 = x2
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
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 = tan = 0
Maths is easy and fun
Page 7
13. Find the missing side of the following triangle.
Solution:
Using Pythagoras theorem, =
14. Find the area of the following triangle.
Solution: A=
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.
Maths is easy and fun
Page 8
ALGEBRAIC EXPRESSIONS 1. Simplify the expressionSolution:
(
)
(
).
= = =

2. Simplify the following algebraic expression. 2x + 5 + 10x  9Solution:
= (10x  2x) + (5  9) = 8x  4 3. Evaluate 8x + 7 given that x  3 = 10.Solution:
= ( ) =104+7 =111 4. Simplify the following algebraic expression. 3(x + 7) + 2(x + 4) + 5xSolution:
= 3x + 21  2x + 8 + 5x = (3x  2x + 5x) + (21 + 8) = 6x + 29 5. Evaluate: 18 + 4(6 2)2Solution:
= = = = =
18 + 4(6 2)2 18 + 4(3)2 18 + 4*9 18 + 36 18
Maths is easy and fun
Page 9
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 = 25Solution:
= = =
8. SolveSolution:
= =
9. Simplify 3x + 1 + 8x + 9.Solution:
= 11x+10
10. Simplify 2x + 5y  7x + 8y.Solution:
= =
11. Calculate 3(4y+9)4Solution:=
=
Maths is easy and fun
Page 10
12. Simplify 7(4+5)+8sSolution:
= =
13. Evaluate 3x+5(74)Solution:
= = 14. 5y+6+4y6xSolution:
==
15. 4(7y6) x 8Solution:
= = 28y  192
Maths is easy and fun
Page 11
INTEGERS
1.
3 + (2)
Solution:
= = 5 2. 12 + (7) + (5)Solution:
= 12 + (7) = 19 = 19 + (5) = 24 3. 8(3)Solution:
= 8 x 3 = 24 4. (4) 5Solution:
= 1024 5. Add 17 + (3) + (2) + 5.Solution:
= = 12 + 5 = 17 6. Multiply (9) (6).Solution:
= 9 x 6 = 54
Maths is easy and fun
Page 12
7. Divide (22) (11).Solution:
= 22 11 = 2 8. Simplify 8 42 + 6.Solution:=8
8+6
=6 9. Simplify 5 (13 6) 3 + 96 6.Solution:
= = 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
Maths is easy and fun
Page 13
13. Add 9 + 6.Solution:
= 3 14. Divide (48) (+6).Solution:
= 48 6 = 8 15. Simplify 52 2 + 4 3.Solution:
=5x4+1 = 21
Maths is easy and fun
Page 14
LINEAR EQUATION1. Solve for x.
x  4 = 10Solution:
X = 10+4 X = 14 2. Solve x.
Solution:
= = = = = ==3. Solve x. 2(3x  7) + 4 (3 x + 2) = 6 (5 x + 9) + 3Solution: = =
x = 3
= = = =
Maths is easy and fun
Page 15
4.
2x  4 = 10
Solution:
x=X = X =7 5. 5x  6 = 3x  8Solution:
X= X= X = 1 6. Solve x.
Solution:
X= X=
( (
) )
( ( )
) )
( X= ( ) ( ) X = 3(3x) + 2(5) = 60x  500 X = 9x + 10 = 60x  500 X = 10 = 51x  500 X= 510 = 51x X= X= 10 7. Find x if: 2x + 4 = 10Solution:
X = 2x + 4  4 = 10  4 X = 2x = 6 X = 2x / 2 X=6/2 X =3
Maths is easy and fun
Page 16
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 + 4y X= 4x = 8 + 4y X= 4x / 4 = (8 + 4y) / 4 X= 2 + y 10. Find x if: x + 32 = 12Solution:
X= X= x + 9 = 12 X= x + 9  9 X= 129 X= 3 11. Solve 4 x = 36. .Solution:
X= 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:
Maths is easy and fun
Page 18
14. Find x forSolution:
.
15. Solve x 4(3 x 2) ( x + 6)= 5 x + 8.Solution:
Maths is easy and fun
Page 19
STRAIGHT LINE GRAPH 1. What are the coordinate of Point A marked on the following grid?
Answer :
= 1, 12. Find the equation of the straight line shown.
2
1
10 8 6 4 2 0 2 0 4 6 8 10
1
2
3
4
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. So the gradient is
gradient
3 3 1
As the yintercept is 4, the equation of the line must be y = 3x + 4.
Maths is easy and fun
Page 20
3. Draw the lines y = 3x 2 and y = 6 2x.Solution:
Table of values for y = 3x 2x y 0 30  2 = 2 1 31  2 =1 2 32 2 =4 3 33  2 =7
Work out the y value corresponding to each x value using the formula y = 3x  2.
Then, plot the points (0, 2), (1, 1), (2, 4) and (3, 7) and join them up. Table of values for y = 6 2xx y 0 6  20 =6 1 6  21 =4 2 6  22 =2 3 6  23 =0
We plot the points (0, 6), (1, 4), (2, 2) and (3, 0).8 6 4 2 0 2 4 6 8
8 6
4
2

2
4
6
8
Maths is easy and fun
Page 21
4. Write down the straight line equation from the information given. M=3 C=4Answer:
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:
y= 7. Write down the yintercept of the following equation:Answer:
(
)
Y=1 8. What is the yintercept of this graph?
Answer:
=3
Maths is easy and fun
Page 22
9. Complete the table of values for y = x + 3, and then draw the graph.Solution:x Y= x + 3
3 0
2 1
1 2
0 3
1 4
2 5
3 6
10. Write down coordinate P.
Answer:
= 2, 1
Maths is easy and fun
Page 23
11. Find the yintercept of the straight line that passes through (1, 5) and (5, 21).Solution:
y = mx + c 5=41+c 5=4+c 1=c m = (y2 y1) (x2  x1) m = (21  5) (5 1) m = 16 4 m=4 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  6m = (y2 y1) (x2  x1) m = (4  12) (1  3) m = 8 4 m=2
13. For the straight line y = 2x + 3, what are: a) the slope b) the yintercept?Answer:= Slope
= 2, yintercept =