PLC Papers
Created For:
Direct and inverse proportion 2
Grade 6
Objective: Write and interpret equations that describe direct and inverse proportion
Question 1.
A is directly proportional to B.
A = 100 when B=25.
Work out the value of A when B = 8.
..............................................
(Total 2 marks)
Question 2.
J is directly proportional to M3.
J = 50 when M = 5.
(a) Find J when M = 8.5
..............................................
(3)
(b) Find M when J = 150
..............................................
(1)
(Total 4 marks)
PiXL PLC 2017 Certification
Question 3.
Y is inversely proportional to √X.
When Y is 9, X is 4.
(a) Find Y when X is 36.
..............................................
(3)
(b) Find X when Y is 36
..............................................
(1)
(Total 4 marks)
Total /10
PiXL PLC 2017 Certification
General iterative processes 2
Grade 7
Objective: Work with general iterative processes
Question 1.
The cubic equation x3 + 2x – 5 = 0 has a solution which lies between 1 and 2.
Use the decimal search method and the table below to find the solution correct to 1dp.
Value of x3 + 2x – 5
x
Positive or Negative?
..............................................
(Total 3 marks)
Question 2.
X n+1 = 1 +
1
�� 2
with X 1 = 1.4
(a) Work out the values of X 2 and X 3
..............................................
(2)
(b) Work out the solution correct to 2 decimal places.
..............................................
(1)
(Total 3 marks)
PiXL PLC 2017 Certification
Question 3.
This iterative process can be used to find approximate solutions to the equation
x3 – 3x – 1 = 0 to 2dp.
Start with a value of x
Work out the value of
3
√1 + 3�
Is your answer to 2 decimal places the
same as your value of x to 2 decimal
places?
Yes
No
This is an approximate solution to
Use your answer as the next value of
x and start again
x3 – 3x – 1 = 0
Use this iterative process to find a solution to 2 decimal places to x3 – 3x – 1 = 0.
Start with x = 2
.............................................
(Total 4 marks)
Total /10
PiXL PLC 2017 Certification
PiXL PLC 2017 Certification
PLC Papers
Created For:
Direct and inverse proportion 2
Grade 6
Solutions
Objective: Write and interpret equations that describe direct and inverse proportion
Question 1.
A is directly proportional to B.
A = 100 when B=25.
Work out the value of A when B = 8.
A ∝ B so A = kB
100 = k x 25
k = 4 (M1)
A = 4 x 8 = 32 (A1)
..............................................
(Total 2 marks)
Question 2.
J is directly proportional to M3.
J = 50 when M = 5.
(a) Find J when M = 8.5
J ∝ M3 so J = kM3 (M1)
50 = k x 53
k = 50 ÷ 125 = 0.4 (M1)
J = 0.4 x 8.53 = 245.65 (A1)
..............................................
(3)
(b) Find M when J = 150
150 = 0.4M3
M3 = 375
M = 7.2 (A1)
..............................................
(1)
(Total 4 marks)
PiXL PLC 2017 Certification
Question 3.
Y is inversely proportional to √X.
When Y is 9, X is 4.
(a) Find Y when X is 36.
Y∝
1
√�
so Y =
�
√�
(M1)
9=
�
√4
k = 18 (M1)
Y=
18
√36
= 3 (A1)
..............................................
(3)
(b) Find X when Y is 36
36 =
18
√�
√� = 0.5
X = 0.25 (A1)
..............................................
(1)
(Total 4 marks)
Total /10
PiXL PLC 2017 Certification
General iterative processes 2
Grade 7
Solutions
Objective: Work with general iterative processes
Question 1.
The cubic equation x3 + 2x – 5 = 0 has a solution which lies between 1 and 2.
Use the decimal search method and the table below to find the solution correct to 1dp.
Value of x3 + 2x – 5
1.1 + 2(1.1) – 5 = -1.469
1.23 + 2(1.2) – 5 = -0.872
1.33 + 2(1.3) – 5 = -0.203
1.43 + 2(1.4) – 5 = 0.544
1.313 + 2(1.31) – 5 = -0.131909
1.323 + 2(1.32) – 5 = -0.060032
1.333 + 2(1.33) – 5 = 0.012637
x
1.1
1.2
1.3
1.4
1.31
1.32
1.33
3
Positive or Negative?
Negative
Negative
Negative
Positive
Negative
Negative
Positive
Correct iterations that identify change of sign (solution) is between 1.3 and 1.4 (M1)
Correct iterations that identify change of sign (solution) is between 1.32 and 1.33 (M1)
So x = 1.3 to 1dp (A1)
..............................................
(Total 3 marks)
Question 2.
X n+1 = 1 +
1
�� 2
with X 1 = 1.4
(a) Work out the values of X 2 and X 3
X 2 = 1.510204082 (B1)
X 3 = 1.438458729 (B1 ft from X 2 )
..............................................
(2)
(b) Work out the solution correct to 2 decimal places.
Using the ANS key to continue to generate terms X 9 = 1.466566356 and X 10 = 1.465003041
Both round to 1.47 (B1)
..............................................
(1)
(Total 3 marks)
PiXL PLC 2017 Certification
Question 3.
This iterative process can be used to find approximate solutions to the equation
x3 – 3x – 1 = 0 to 2dp.
Start with a value of x
Work out the value of
3
√1 + 3�
Is your answer to 2 decimal places the
same as your value of x to 2 decimal
places?
Yes
No
This is an approximate solution to
Use your answer as the next value of
x and start again
x3 – 3x – 1 = 0
Use this iterative process to find a solution to 2 decimal places to x3 – 3x – 1 = 0.
Start with x = 2
Substitutes in x = 2 to get x = 1.912931183 (M1)
Second pass through the flow diagram to give x = 1.888835126 (M1)
Third pass through the flow diagram to give x = 1.88205688 (M1)
Fourth pass through the flow diagram to give x = 1.880141328 so x = 1.88 to 2dp (A1)
.............................................
(Total 4 marks)
Total /10
PiXL PLC 2017 Certification
PiXL PLC 2017 Certification