Download MATHEMATICS SET 3 Form 4 Topic: Standard Form 1. (7.435 x

MATHEMATICS
SET 3
Form 4
Topic: Standard Form
1. (7.435 x 6.752)/2.314 =
[correct to 3 significant figures]
A
B
C
D
E
19.7
21.6
21.7
21.695
2.59
Answer: C
Solution:
(7.435 x 6.752)/2.314 = 50.20112/2.314
= 21.69452
= 21.7
Topic: Standard Form
2. If x = yz2, find y if x = 17.34 and z = 13.35. Round off the value ok y correct to 2
significant figures.
A
B
C
D
E
0.1
0.09
0.097
0.0972
10.3
Answer: C
Solution:
x = yz2
y = x/z2
y = 17.34/ (13.35)2
y = 17.34/178.2225
y= 0.0972941
y = 0.097
Topic: Quadratic Expressions and Equations
3. Factorise x2 – 3x – 28.
A
B
C
D
E
(x + 3) (x + 8)
(x – 8) (x + 3)
(x – 3) (x + 8)
(x – 4) (x + 7)
(x – 7) (x + 4)
Answer: E
Solution:
x2
||
-3x
-28
||
= (x – 7) (x + 4)
x(4) + x(-7) = -3x
Topic: Quadratic Expressions and Equations
4. Simplify x2 – 16 / (x2 – 2x – 8).
A
B
C
D
E
x+2
x+4
x + 4/x + 2
4x
x + 2/x + 4
Answer: C
Solution:
x2 – 16
_________
x2 – 2x – 8
=
(x – 4) (x + 4)
___________
(x - 4) (x + 2)
x+4
= _____
x+2
Topic: Quadratic Expressions and Equations
5.
0.5 m
0.5 m
DIAGRAM 1
Diagram 1 shows a square mini field that is surrounded by a path 0.5 m width. The area
of the field is 23 m2 bigger than the area of the path. Find the length of a side of the field.
A
B
C
D
E
2m
3m
4m
5m
6m
Answer: E
Solution:
Assume the side of the field = k.
The area of the path = (2)(k + 0.5 + 0.5) + (2)(k)(0.5)
= 2(k + 1)(0.5) + 1k
=k+1+k
= 2k + 1
Given k2 – (2k + 1) = 23
k2 – 2k – 1 – 23 = 0
k2 – 2k – 24 = 0
(k + 4)(k – 6) = 0
k = -4 (ignored)
k=6
Therefore, the length of a side of the field = 6 m
Topic: Sets
6. Given the universal set ε = {x: 10 ≤ x < 20, x are integers} and K = {x: x is a number
where the sum of its digits is more than 3}. Find n (K’).
A
B
C
D
E
3
4
7
8
10
Answer: A
Solution:
K = {13, 14, 15, 16, 17, 18, 19}
K’ = {10, 11, 12}
Therefore, n(K’) = 3
Topic: Sets
7. If the universal set, ε = {all the alphabet from a to z}, G ={vowels}, H = {consonants},
J = {e, x, c, e, l, l, e, n, t}. Find n (H
J)’.
A
B
C
D
E
5
6
9
20
21
Answer: D
Solution:
H
n(H
J = {x, c, l, l, n, t}
J)’ = 26 – 6
= 20
Topic: Sets
8.
ε
•2
•10
X
•1 •6 •3
•13
•9
•8 •7 •12
•11
•5
•16
•4 •14
•p
•15
•17
Y
W
DIAGRAM 1
If ε = {x: x are integers where 1 ≤ x ≤ 18}, find X
A
B
C
D
E
{4, 14, 15, 17}
{4}
{4, 14}
{14}
{4, 11, 14, 16, p}
Answer: D
Solution:
Set W = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Set X = {3, 4, 7, 8, 10, 12, 13, 14}
Set Y = {4, 14, 15, 17}
Set W’ = {10, 11, 12, 13, 14, 15, 17}
X
Y = {4, 14}
X
Y
W’ = {14}
Y
W’.
Topic: Straight Line
9.
DIAGRAM 3
Find the equation of the straight line in Diagram 3.
A
B
C
D
E
y = 3/4x + 13/4
y = 3/8x + 13/4
y = 3/8x + 4
y = -3/8x + 13/4
y = 3/4x - 5
Answer: B
Solution:
m = (4 – 1) / [2 – (-6)]
= 3/8
Therefore, y = 3/8x + c -----
1
Substitute (2, 4) into 1
y = 3/8x + c
4 = (3/8) x 2 + c
c=4–¾
= (16 – 3)/4
= 13/4
Hence, y = 3/8x + 13/4
Topic: Straight Line
10. TU is a straight line which is parallel to 2y = 2x – 5. If TU passes through (6, 2), find
the straight line TU.
A
B
C
D
E
y=x-4
y = 2x - 4
y = 3x - 5
y = 1/2x + 4
y=x+4
Answer: A
Solution:
2y = 2x – 5
y = 1x – 5/2
m1 = 1
Since TU is parallel to 2y = 2x – 5
Therefore, mLM = 1
y = mx + c
y = 1x + c ----- 1
Substitute (6, 2) into 1
y = 1x + c
2 = 1(6) + c
c=2–6
= -4
Therefore, y = 1x – 4
y=x–4
Topic: Straight Line
11. In the diagram below, KLMN is a parallelogram. Determine the coordinates of S.
DIAGRAM 4
A
B
C
D
E
(3, 0)
(0, -3)
(0, 3)
(3, -3)
(0, 4)
Answer: C
Solution:
mkl = mmn
mkl = (y2 – y1) / (x2 – x1)
= (6 – 4) / (5-(-2))
= 2/7
S(0, y) (because S is on y-axis)
msm = (5 – y) / (7 – 0) = 2/7
5 – y = 2/7 x 7
y=5–2
=3
Therefore, the coordinates of S = (0, 3)
Topic: Statistics
Question 12 – 14 are based on the information below.
DIAGRAM 5
Diagram 5 shows a frequency polygon representing the length of cockroach in science
experiment.
12. How many cockroaches are used in the experiment?
A
B
C
D
E
10
20
30
50
70
Answer: E
Solution:
Number of cockroaches
= 10 + 20 + 10 + 22
= 70
13. What is the percentage of cockroaches longer than 1.5 cm?
(correct to 3 significant figures).
A
B
C
D
E
14.3%
42.9%
57.1%
85.7%
97.1%
Answer: C
Solution:
Number of cockroaches > 1.5 cm = 40
Percentage = 40/70 x 100%
= 57.14286
= 57.1%
14. Calculate the mean.
A
B
C
D
E
2.0
1.896
1.986
1.8
1.1
Answer: C
Solution:
Mean = [(1 x 10) + (1.5 x 20) + (2 x 10) + (2.5 x 22) + (3.0 x 8)]
_____________________________________________
70
= 10 + 30 + 20 + 55 + 24
__________________
70
= 1.9857
= 1.986
Topic: Probability I
15. The probability to pick a 50 cents coin from a coin box is 4/8. If there are 80 fifty
cents coins and t 1 cents coins, find t.
A
B
C
D
E
20
40
50
80
90
Answer: D
Solution:
80/ (80 + t) = 4/8
8(80) = 4(80 + t)
640 = (320 + 4t)
640 - 320 = 4t
t = 320/4
= 80
Topic: Probability I
16.
In a dart competition, the probability of hitting the shaded region is 2/6 for a particular
participant. Find the area of the unshaded region if the shaded region has an area of 13π
cm2.
A
B
C
D
E
13π
26π
39π
65π
78π
Answer: B
Solution:
P (hit the shaded region) = area of shaded region / total area
2/6 = 13π/total area
2 x total area = 6(13π)
Total area = 78π/2
= 39π
The area of unshaded region = 39π - 13π
=26π
Topic: Probability I
17. The probability that a washing machine produced by a particular manufacturer is of
an acceptable standard is 0.93. How many washing machines are acceptable if the
manufacturer produced 1300 washing machines?
A
B
C
D
E
1209
1211
1309
1311
1409
Answer: A
Solution:
P (washing machine of acceptable standard)
= n (washing machine of acceptable standard) / n(S)
= n (washing machine of acceptable standard) / 1300
= 0.93
n (washing machine of acceptable standard) = 0.93 x 1300
= 1209
Topic: Probability I
18. In a shooting competition, the probability of shooting the target by a shooter is 6/15.
If the particular shooter shoots 120 times, how many times does he achieve shooting
the target?
A
B
C
D
E
9
40
44
48
52
Answer: D
Solution:
6/15 x 120
=6x8
= 48
Topic: Circle III
19.
DIAGRAM 6
From the diagram 6, EF and EG are tangents to the circle. Calculate the value of s.
A
B
C
D
E
30°
50°
130°
140°
170°
Answer: B
Solution:
ΔFOG = 360° - 230°
= 130°
ΔEFO = ΔEGO = 90°
Therefore, ΔFEG = 360° – ΔFOG – ΔEFO - ΔEGO
= 360° – 130° – 90°- 90°
= 50°
Topic: Circle III
20.
DIAGRAM 7
TUV is a tangent to the circle. Find the value of s.
A
B
C
D
E
15°
38°
40°
53°
59°
Answer: E
Solution:
ΔPRU = 78°
ΔQRU = 180° – 43°
= 137° (cyclical rectangle)
s = 137° – 78°
= 59°
Topic: Circle III
21.
DIAGRAM 8
In Diagram 8, FGH and HIJ are two common tangents. What is the major angle of FDJ?
A
B
C
D
E
104°
114°
128°
214°
232°
Answer: E
Solution:
ΔDFH = ΔDJH = 90°
Hence the minor angle of FDJ
= 360° – 90° – 90° – 52°
= 128°
Therefore, the major angle of FDJ
= 360° – 128°
= 232°
Topic: Circle III
22.
DIAGRAM 9
KLM is a common tangent to the two circles centered O and S. Calculate the length of
OS.
A
B
C
D
E
15.56 cm
14.56 cm
15.65 cm
16.34 cm
15.75 cm
Answer: A
Solution:
OS = OL + LS
sin 40° = 4/OL
OL = 4/sin 40°
= 6.2229
sin 40° = 6/LS
LS = 6/sin 40°
= 9.3343
Therefore,OP = 6.229 + 9.3343
= 15.5633 cm
= 15.56 cm
Topic: Trigonometry II
23.
DIAGRAM 10
From Diagram 10, tan θ = 3/4. Find the length of MN.
A
B
C
D
E
3 cm
4 cm
5 cm
7 cm
9 cm
Answer: C
Solution:
tan θ= KN/JN
3/4 = 6/KJ
Therefore, KJ = (4 x 6) / 3
= 8 cm
Hence JN = 10 cm (Phytagoras Theorem)
JM = 5 cm (Phytagoras Theorem)
MN = 10 – 5
= 5 cm
Topic: Trigonometry II
24. Evaluate 3 sin 30° – 4 cos 45° + tan 45°.
__
A
-5/2 - 2√2
__
B
5/2 + 2√2
__
C
5/2 - 2√2
__
D
3/2 - 2√2
__
E
5/2 - √2
Answer: C
Solution:
3 sin 30° – 4 cos 45° + tan 45°
_
= 3(0.5) + 4(√2 / 2) + 1
_
= 3/2 + 1 - 2√2
_
= 5/2 - 2√2
Topic: Trigonometry II
25.
DIAGRAM 11
In Diagram 11, given that cos ΔEFD = 4/7. Calculate the perimeter of the triangle DEF.
A
B
C
D
E
42.66 cm
52.66 cm
52.60 cm
55.66 cm
59.66 cm
Answer: B
Solution:
Cos ΔEFD = EF/DF = 9/DF = 4/7 (given)
DF = 9(10) / 4
= 22.5 cm
__________
DE = √ (22.5)2 - 102
= 20.1556
Therefore, perimeter = 10 + 22.5 + 20.1556
= 52.66 cm
Topic: Trigonometry II
26. Given tan θ = 2.156 and θ is an reflex angle. Find θ.
A
B
C
D
E
245°5’
245°7’
244°7’
244°5’
245°8’
Answer: B
Solution:
θ = tan-1 2.156
= 65°7’
since 180° ≤ θ ≤ 360°
θ = 180° + 65°7’
= 245°7’
Topic: Angles of Elevation and Depression
27.
C
θ
A
B
AC is horizontal line. If A is the observer and C is the object, find the angle of elevation.
A
B
C
D
E
ΔACB
ΔABC
ΔBCA
ΔCAB
ΔCBA
Answer: D
Solution:
The angle of elevation is ΔCAB.
Topic: Angles of Elevation and Depression
28.
In the diagram above, UV is a tree on a horizontal plane. T is a point on the horizontal
plane. If ΔTUV = 90, UV = 4 m and TU = 7 m, calculate the angle of elevation of the
point V from the point T.
27°15’
55°9’
29°45’
60°15’
34°60’
A
B
C
D
E
Answer: C
Solution:
tan ΔUTV = 4/7
Therefore UTV = tan-1 4/7
= 29°45’
Topic: Angles of Elevation and Depression
29.
In the diagram above, GH is a vertical pole and H is a point on the same horizontal plane.
If the angle of elevation of F from H is 10 and the distance GH = 28 m, find the length of
the pole.
A
B
C
D
E
2.354 m
2.623 m
4.892 m
4.927 m
4.937 m
Answer: E
Solution:
F
10°
G
tan 10° = GH/28
GH = 28 tan 10°
= 4.937 m
H
Topic: Angles of Elevation and Depression
30.
DIAGRAM 12
Diagram 12 shows a pyramid on a square base of side 8 cm. If MD = MG = MF = ME =
14 cm, find the height MN of the pyramid.
A
B
C
D
E
6.432 cm
12.8063 cm
13.1432 cm
14.2442 cm
15.3543 cm
Answer: B
Solution:
_______
DF = √82 + 82
= 11.3137
DN = 11.3137/2
= 5.6569
By using the Pythagoras Theorem,
____________
2
MN = √142 – 5.65692
= 164
____
Hence, MN = √164
= 12.8063 cm
Form 5
Topic: Number Bases
31. The value of digit 5, in base ten, in the number 952610 is
A
B
C
D
E
5
50
10
25
500
Answer: E
Solution:
5 x 102 = 500
Topic: Number Bases
32. Express 2210 as a number in base two.
A
B
C
D
E
1012
10102
101102
110002
100002
Answer: C
Solution:
Therefore, 2210 = 101102
Topic: Number Bases
33. It is given that s8 = 10110102, find the value of s.
A
B
C
D
E
113
131
121
112
132
Answer: E
Solution:
001
1
011
3
010
2
Therefore, s8 = 1328
Topic: Graphs of Functions II
Question 34 – 35 are based on the following table which shows the corresponding values
of x and y of the function y = 2x2 + x + 5.
x
y
-3
k
-2
11
34. The value of k is
A
B
C
D
E
6
8
14
20
21
Answer: D
Solution:
Replace x = -3 in the equation,
y = 2x2 + x + 5
y = 2(-3)2 + (-3) + 5
y = 2(9) – 3 + 5
y = 20
35. Find the value of m.
A
B
C
D
E
5
10
15
17
19
Answer: C
-1
6
0
5
1
8
2
m
Solution:
Replace x = 2 in the equation,
y = 2x2 + x + 5
y = 2(2) + 2 + 5
y = 2(4) + 2 + 5
y = 15
Topic: Graphs of Functions II
36.
DIAGRAM 13
Diagram 13 shows the graph of y = c + mx – x2. Find the value of m.
A
B
C
D
E
5/2
3/2
1
11/2
-5/2
Answer: A
Solution:
y = c + mx – x2
y = 6 + mx – x2 (c = 6 because the curve cuts the y-axis at (0, 6))
Using point (3, 0),
0 = 6 + m(4) – 42
4m = 10
m = 10/4
m = 5/2
Topic: Transformation III
37.
DIAGRAM 14
Diagram 14 shows six trapezoids on a Cartesian plane. Given that G is the translation 3
-2
which of the trapezoids A, B, C, D and E is the image of trapezoid H under the combined
transformation G2?
Answer: E
Solution:
E is the image of trapezoid H under the combined transformation G2.
Topic: Transformation III
38.
DIAGRAM 15
Diagram 15 shows a point F on a Cartesian plane. Transformation T is a reflection about
the line y = 2 and transformation U is a clockwise rotation of 90° about the centre (5, -2).
Find the coordinates of the image of point F under transformation TU.
A
B
C
D
E
(11, 4)
(7, 4)
(11, 2)
(11, -4)
(11, 2)
Answer: A
Solution:
The coordinates of the image of point F under transformation TU is (11, 4).
Topic: Transformation III
39.
DIAGRAM 16
Diagram 16 shows two regular pentagons JKLMN and PQRST. PQRST is the image of
JKLMN under transformation V. Given that the area of JKLMN is 25 cm2, calculate the
area of PQRST.
A
B
C
D
E
50 cm2
100 cm2
225 cm2
75 cm2
115 cm2
Answer: B
Solution:
Transformation V is an enlargement about the centre (6, 8) with scale factor 2.
Therefore, the area of PQRST = 25 x 22
= 100 cm2
Topic: Matrices
40. Given that s 6 = 3
3 10
3
A
B
C
D
E
6
2s + t
, find the value of t.
1
2
4
-4
16
Answer: C
Solution:
s=3
2s + t = 10
2(3) + t = 10
t = 10 – 6
=4
Topic: Matrices
41. It is given that p + 2 p – 4 = 16 . Find the value of p and q.
q
q–2
11
A
B
C
D
E
p = 8 and q = 5
p = -8 and q = -5
p = 5 and q = 8
p = -5 and q = 8
p = 5 and q = -8
Answer: A
Solution:
p + 2 p – 4 = 16
q
q–2
11
p + 2p – 8 = 16
q
2q – 4
11
Therefore, p + 2p – 8 = 16
3p = 24
p=8
and
q + 2q – 4 = 11
3q = 15
q=5
Topic: Matrices
42. It is given that matrix K = u -1
6 v
KL = 1 0 .
0 1
A
B
C
D
E
and matrix L = 1/9 v 1 such that
-6 3
u = -1 and v = -3
u = 1 and v = 3
u = 3 and v = -1
u = -3 and v = 1
u = 3 and v = 1
Answer: E
Solution:
u -1 -1 = 1/9 v 1
6 v
-6 3
1/(uv + 6)
v 1
-6 u
= 1/9
v 1
-6 3
Therefore, u = 3 and
uv + 6 = 9
3v = 3
v=1
Topic: Variations
43. If J varies directly as L to the power of 2 and J = 36 when L = 3, find the value of L
when J = 144.
A
B
C
D
E
2
4
18
6
36
Answer: D
Solution:
J α L2
J = kL2
k = J/L2
k = 36/32
k=4
When J = 144,
144 = 4(L2)
L2 = 36
L=6
Topic: Variations
__
44. If D α 1/√E and D = 3 when E = 144, find the value of D when E = 36.
A
B
C
D
E
6
8
10
12
14
Answer: A
Solution:
__
D α 1/√E
__
D = k/√E
___
k = 3 x √144
k = 3 x 12
= 36
__
Therefore, D = 36/√E
When E = 9,
__
D = 36/√36
D = 36/6
D=6
Topic: Variations
45. It is known that X varies directly as Y and inversely as the square of Z. If X = 3 when
Y = 15 and Z = 5, then the relation of X, Y and Z is
A
B
C
D
E
X = 5Z2/Y
X = 5Y/Z2
X = 4Y/Z2
X = 4Z2/Y
X = Y/Z2
Answer: B
Solution:
X α Y/Z2
X = kY/Z2
2 = k(15)/52
k = (3 x 52)/15
k=5
Therefore, X = 5Y/Z2.
Topic: Gradient and Area under a Graph
Question 46 – 48 are based on the following distance-time graph that shows the journey
of a bus departed from a place 20 km from town M, and then returned to town M over a
period of 120 minutes.
DIAGRAM 17
46. Calculate the distance, in km, traveled by the bus in the first 50 minutes.
A
B
C
D
E
10 km
15 km
20 km
25 km
30 km
Answer: D
Solution:
Distance traveled = 45 – 20
= 25 km
47. Calculate the distance, in km, traveled by the bus over the period of 120 minutes.
A
B
C
D
E
50 km
60 km
80 km
130 km
140 km
Answer: C
Solution:
Distance traveled = 50 + 30
= 80 km
48. Find the average speed, in km-1, of the bus in the first 40 minutes of its journey.
A
B
C
D
E
30 km h-1
20 km h-1
60 km h-1
50 km h-1
70 km h-1
Answer: A
Solution:
Average speed = 20/(40/60)
= 20 x (60/40)
= 30 km h-1
Topic: Probability II
49. A box contains 18 yellow marbles and 24 silver marbles. A marble is drawn at
random from the box. What is the probability of drawing a yellow marbles?
A
B
C
D
E
1
4/7
3/7
5/7
2/7
Answer: C
Solution:
P (yellow marbles) = n (yellow marbles) / n(S)
= 18 / (18 + 24)
= 18 / 42
=3/7
Topic: Probability II
50.
DIAGRAM 17
Diagram 17 shows a group of letter cards in a box. Sherri took out all the cards marked
with the letter S and replaced them with several cards marked with the letter T. A card is
then drawn at random from the box and the probability of drawing a card with the letter I
is 1/8. Calculate the number of cards with the letter T in the box.
A
B
C
D
E
3
5
9
11
13
Answer: E
Solution:
Let the number of cards with the letter T added = p
n (S) = 14 – 2 + p
= 12 + p
n (letter I) = 3
Hence,
3/(12 + p) = 1/8
12 + p = 24
p = 12
Therefore, number of cards with letter T = 12 + 1
= 13
Topic: Probability II
51. Naji has 60 bags of graded tomatoes. 25 bags are grade S tomatoes and the rest are
grade B and grade C tomatoes. If a bag is picked at random, the probability of
picking a bag of grade B tomatoes is 1/4. Calculate the probability of picking a bag of
grade C tomatoes.
A
B
C
D
E
2/3
1/3
1/5
2/5
1/4
Answer: B
Solution:
Probability of picking a grade S bag = 25/60
= 5/12
Probability of picking a grade C bag = 1 – 5/12 – 1/4
= 4/12
= 1/3
Topic: Bearing
52.
North
K
L
31°
M
DIAGRAM 18
In Diagram 18, K, L and M are three points on a horizontal plane. The bearing of the
point M from the point K is
A
B
C
D
E
31°
45°
59°
111°
121°
Answer: E
Solution:
ΔLKM = 180° – 90° – 31°
= 59°
Therefore, the bearing of the point M from the point K is = 180° – ΔLKM
= 180° - 59°
= 121°
Topic: Bearing
53. Given that the bearing of point X from point Y is 232°, then the bearing of point Y
from point X is
A
B
C
D
E
52°
62°
128°
138°
148°
Answer: A
Solution:
Therefore, the bearing of point Y from point X is 52°.
Topic: Bearing
54.
DIAGRAM 19
In Diagram 19, S, T and U are three points on a horizontal plane. The point S is due west
of T. Find the bearing of S from U.
A
B
C
D
E
45°
115°
135°
225°
240°
Answer: D
Solution:
Therefore, the bearing of S from U = 180° + 32° + 13°
= 225°
Topic: Earth as a Sphere
Question 55 – 56 are based on Diagram 20 where O is a centre of the earth and G, H and
K are the three places on the equator such that GH is a diameter of the equator.
DIAGRAM 20
55. It is given that ΔKOG = 45°, the longitude of G is
A
B
C
D
E
50°W
45°W
45°E
135°E
135°W
Answer: C
Solution:
The longitude of G is 45°E.
56. The longitude of H is
A
B
C
D
E
45°W
115°W
125°W
135°W
180°W
Answer: D
Solution:
The longitude of H is = (180 – 45) °W
= 135°W
Topic: Earth as a Sphere
57. What is the distance, in nautical miles, between the points Q (32°N, 115°W) and
R (69°S, 115°E) along a meridian?
A
B
C
D
E
6060 n.m
2220 n.m
3030 n.m
1920 n.m
4140 n.m
Answer: A
Solution:
Q and R lie on the same meridian. Difference between latitudes 32°N and 69°S
= 32° + 69°
= 101°
Therefore, the distance between points Q and R = 101° x 60’
= 6060’
Topic: Earth as a Sphere
Question 58 – 60 are based on Diagram 21 where J, K, M, U, T and L are places on the
surface of the earth. O is the centre of the earth.
DIAGRAM 21
58. Which of the following points lie on the same parallel of latitude as point U?
A
B
C
D
E
J
U
K
M
L
Answer: B
Solution:
U and T lie on the same parallel of latitude.
59. The location of the place J is
A
B
C
D
E
(53°N, 70°W)
(0°, 18°W)
(53°S, 18°W)
(53°N, 18°W)
(53°S, 70°E)
Answer: A
Solution:
J has latitude 53°N and longitude 70°W.
60. The location of the place U is
A
B
C
D
E
(0°, 18°W)
(36°N, 70°W)
(53°N, 18°W)
(53°N, 70°W)
(36°S, 18°W)
Answer: E
Solution:
U has latitude 36°S and longitude 18°W.
60 Questions