F2 Uniform Test 2017
Solution and Marking Scheme(Revised)
1.
Solution:
(a)
In the figure, AB // CD. E is a point lying
on BC such that BD = BE. Find x, y and z.
C
1M
n = 15
1A
∴
D
x 46°
E
(n – 2) × 180° = 2340° (∠ sum of polygon)
The number of sides of the polygon is 15.
(b) The size of each interior angle of a regular
y
15-sided polygon is
2340° ÷ 15 = 156°
z
A
39°
.1M
Suppose that AB and BC are two sides of a
B
regular 15-sided polygon.
Then, we have AB = BC
(5 marks)
and
Solution:
x = ∠ABC (alt. ∠s, AB // CD)
∴
x = 39°
∠BAC = ∠BCA (base ∠s, isos. ∆)
1A
∠BAC + ∠BCA + 156° = 180° (∠ sum of ∆)
y = 46° + x (ext. ∠ of ∆)
1M
∴
∠BAC = 12°
∴
1A
∴
It is possible that it is a regular polygon. 1A
y = 85°
∠BDE = y (base ∠s, isos. ∆)
1M
y + y + z = 180° (∠ sum of ∆)
∴
2.
1M
z = 10°
3.
1A
The sum of the interior angles of an
n-sided polygon is 2340°.
(a) Find the number of sides of the
polygon.
(b) Suppose that AB and BC are two
sides of the polygon and ∠BAC = 12°.
Is it possible that it is a regular
polygon? Explain your answer.
(5 marks)
In the figure, ABCDE, BGH and CFG are
straight lines.
(a) Find x.
(b) Find y.
(5 marks)
(a)
In ∆BCG ,
90° + 155° + x = 360° (sum of ext. ∠s of polygon) 1M
(b)
x = 115°
1A
∠BCG + x = 180° (adj. ∠s on st. line)
1M
∠BCG + 115° = 180°
∠BCG = 65°
1A
In ∆CDF ,
133° + ∠BCG + y = 360°
(sum of ext. ∠s
of polygon)
133° + 65° + y = 360°
y = 162°
1A
4.
Solve
Substitute x = 1 into (1):
the
simultaneous
equations
2n − 6m
5m + n + 2 =
= 6 by the method
4
1M
−5(1) + 3 y = 1
y=2
of substitution.
∴
(5 marks)
The solution is x = 1, y = 2.
2A
Solution:
6.
Rewriting the given equations, we have
5m + n + 2 = 6.....................(1)

 2n − 6m
= 6.......................(2)
 4

The difference between two numbers is 25
and the larger number is 10 more than
twice the smaller number. Find the two
numbers.
(5 marks)
From (2), we have
Solution:
Let x and y be the larger number and the smaller
2n − 6m = 24
number respectively.
2(n − 3m) = 24
n = 3m + 12............(3)
 x − y = 25............................(1)

 x − 2 y = 10..........................(2)
1M
2M
(1) − (2):
Substitute (3) into (1):
( x − y ) − ( x − 2 y ) = 25 − 10
5m + 3m + 12 + 2 = 6
8m = −8
m = −1
1M
y = 15
Substitute y = 15 into (1):
Substitute m = –1 into (3):
1M
x − 15 = 25
x = 40
n = 3(−1) + 12 = 9
∴
5.
The solution is m = −1, n = 9.
Solve
the
− 5 x + 3 y = 1

8 x − 5 y = −2
simultaneous
by
the
2A
equations
method
of
1M
∴
7.
The two numbers are 40 and 15.
2A
In the figure, P is the point of intersection
of the graphs of the equations 3x – 4y + 24
= 0 and 4x + 2y – 1 = 0. Find the
coordinates of P.
elimination.
(5 marks)
y
4x + 2y – 1 = 0
3x – 4y + 24 = 0
Solution:
−5 x + 3 y = 1............................(1)

8 x − 5 y = −2...........................(2)
(1) × 5: −25 x + 15 y = 5............(3)
P
1M
O
x
(2) × 3: 24 x − 15 y = −6............(4)
(3) + (4):
(5 marks)
(−25 x + 15 y ) + (24 x − 15 y ) = 5 + (−6)
−x = −1
x =1
1M
Solution:
3 x − 4 y + 24 = 0.................(1)

4 x + 2 y − 1 = 0..................(2)
(a) Complete the following table for the
equation y = x.
1M
From (2), we have
x
2y = 1 – 4x
y=
0
1
y
1 − 4x
……….(3)
2
(b) Draw the graph of y = x in the same
coordinate plane above.
(c) Solve the simultaneous equations
1M
Substitute (3) into (1):
 1 − 4x 
3x – 4 
 + 24 = 0
 2 
y = x
graphically.

3 x + 2 y = 6
3x – 2 + 8x + 24 = 0
11x = −22
x = −2
(5 marks)
1A
Solution:
Substitute x = −2 into (3):
y=
=
∴
8.
2
3(−2) + 24
4
(a)
1M
0
1
2
y
0
1
2
1A
9
2
9

The coordinates of P are  − 2,  .
2

x
(b)
1A
The figure shows the graph of the
equation 3x + 2y = 6.
2A
(c)
9.
The approximate solution is x = 1.2, y = 1.2. 2A
The histogram shows the weights of
students in S.2A.
(a) Write down the class boundaries of
the 3rd class interval.
(b) Find the difference in the numbers of
students between the highest and the
lowest frequency.
(5 marks)
Solution:
(a)
(b)
Lower class boundary = 54.5 kg
2A
Upper class boundary = 59.5 kg
2A
The difference = 15 − 2 = 13
1A
10. The cumulative frequency polygon shows
the speeds of cars passing a police radar
station within 10 minutes.
(a)
There are 200 cars.
2A
(b)
Number of cars = 200 − 175 = 25
2A
(c)
Number of cars = 200 − 160 = 40
1A
End of Paper
(a) How many cars are there?
(b) How many cars are faster than 90
km/h?
(c) If the speed limit is 80 km/h, how
many cars exceed the speed limit?
(5 marks)