Test GCSE chapitre 3 - Correction Q1. x

−2

0

3

y

7

3

–3

a. the value of y when x = 5 is –7 ; b. the value of y when x = −4 is 11 ; c. the value of x when y = −13 is 8 ; d. the value of x when y = 7 is –2. Q2. Complete the following table for the line y  3 x  1 : x

−4

3x y

−3

−2

−1

0

1

2

3

4

–12

–9

–6

–13

–10

–7

–3

0

3

6

9

12

–4

–1

2

5

8

11

Plot these points on graph paper and hence draw the graph of y  3 x  1 . Use a scale of 1 cm for 2 units on the y-axis and 2 cm for 1 unit on the x-axis.

Q3. Which letters represent the following lines : Line

Letter

a. x = y

B

b. x = 5

C (verticale)

c. y   x

F

d. x = 0

G (axe des y)

e. y  7

E (horizontale)

f. x  y  0

F (y=–x)

g. y  5

A (horizontale)

h. x  y  0

B (y=x)

i. y  0

D (axe des x)

j. x  7  0

H (x=–7)

k. y  x  x  y

G (on simplifie, soit x=0)

Q4. Complete the following table for the line y 

1 x3 2

x

−6

−4

−2

0

2

4

6

1 x 2

–3

–2

–1

0

1

2

3

y

–6

–5

–4

–3

–2

–1

0

Q5. For each of the following lines, give the gradient (la pente) and the coordinates of the point A where the line cuts the y-axis. line

gradient

point A

y  4x  3

4

(0 ; 3)

y  3 x  3

–3

1 y  x4 2 x3  y

line

gradient

point A

y  4  2.5 x

–2.5

(0 ; –4)

(0 ; 3)

3y  4 x  7  0

–4/3

(0 ; 7/3)

–1/2

(0 ; 4)

y  5 x

–5

(0 ;0)

1

(0 ; –3)

1/2

(0 ; 4)

x

1 y4 2

2

y  2x  4  0

(0 ; 4)

y’a pas !

x20

(–2 ; 0)

Q6. What is the gradient and the equation of the lines joining the points : points

gradient

line

points

gradient

line

(3, 5) and (5, 9)

2

y=2x–1

(8, 5) and (6, 4)

–1

y=–x+10

(6, 3) and (10, 5)

0.5

y=0.5x

(−3, −1) and (1, −4)

–0.75

y=–0.75x–3.25

–1

y=–x–2

(0.3, 5.2) and (0.5, 9.1)

19.5

y=19.5x–0.65

–2

y=–2x+18

(1.5, −2) and (−0.5, 0.8)

–1.4

y=–1.4x+0.1

(−6, 4) and (−3, 1) (8, 2) and (4, 10)

Q8. What is the gradient of : Line

Gradient

a. line A

–2/4=–1/2

b. line B

6/2=3

c. line C

–1/4

d. line D

–4/2=–2

e. line E

–2/3

f. line F

–8/3

g. line G

4/1=4

h. line H

3/3=1

i. a line parallel to A

–1/2

j. a line parallel to B

3

k. a line parallel to C

–1/4

Q7. Number of units used

0

100

200

300

Cost using method A

10

10+100*0.25=35

60

85

Cost using method B

40

40+100*0.05=45

50

55

a. Method A : 70 -> 27.5

method B : 70 -> 43.5.

b. Miss Wright used 120 units : to minimize her bill the uses method A. c. They used the abscissa of A, 150 units. Q9. The points (7, 20) and (5, 14) lie on the line y  3 x  1 .

 6 x  5y  430  12x  10y  860  12x  10y  860  12  45  10y  860  10y  320  y  32 Q10.  .     4 x  10y  500  4 x  10y  500  8 x  360  x  45 Q12. Find the equation of the following lines : Line

Equation

a. line A

y  3x  1

b. line B

y  0.5 x  4

c. line C

y  0.2x  7

d. line D

y  0.25x  3

e. line E

y  0.25 x

f. line F

y  2x  6

Q11. The cost of hiring a car is £28 per day plus 25p per mile. a. Find the cost of hiring the car for a day and travelling : 1. 40 miles : 28+40*0.25=38

2. 80 miles : 28+80*0.25=48

b. Write down a formula to give the cost of hiring a car (£ c) for one day and travelling n miles.

c  28  n  0.25 c. Rearrange the formula to make n the subject. c  28  n  0.25  n  0.25  c  28  n 

c  28 c 28    n  4c  112 . 0.25 0.25 0.25

d. How many miles can you travel, during one day, if you have a budget of : 1. £34 : n  4  34  112  4 miles

2. £50 n  4  50  112  88 miles

3. £56.50 : n  4  56.50  112  114 miles