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Mathematics
NECO SSCE 2001
Number and Numeration
Solution
From the top of a storey-buliding, the angle of depression of a mango on a tree 20m away from the base of the storey building is 68^\circ. if the mango is 4.5 m above the ground, what is the height of the storey building to the nearest metre?
A
55m
B
54m
C
50m
D
49m
E
45m
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the figure above,
Tan\(\theta\) = \(\frac{\text{O}}{\text{A}}\)
Tan 68º = \(\frac{\text{x}}{20}\)
x = Tan 68º x 20 = 49.501m to 3 d.p
Height of Storey building = 49.501 + 4.5 = 54. 001 ≈ 54m to the nearest metre.
Source: Myschool Past Questions Archive (Record #41792)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
If n{A} = 6, n{B} = 5 and n{A ∩ B} = 2, find n{A ∪ B}
A
7
B
8
C
9
D
13
E
15
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
n(A U B) = n(A) + n(B) - n(A n B)
n(A U B) = 6 + 5 - 2 = 9
Source: Myschool Past Questions Archive (Record #41811)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
If 3 boys shared x oranges among themselves in the ratio 3:5:8, and smallest share was 45 oranges, find the value of x
A
75
B
120
C
165
D
195
E
240
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Total ratio = 3 + 5 + 8 = 16
Total orange shared = x
Smallest share received = 45
The orange shared = \(\frac{3}{16} \times\) x = 45
3x = 16 x 45
x = 16 x 15 = 240 oranges
Source: Myschool Past Questions Archive (Record #41815)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
If the volume of a cube is (8 x \(10^3\))cm\(^3\), what is the surface area of the cube?
A
2,400 cm\(^2\)
B
1,600 cm\(^2\)
C
120 cm\(^2\)
D
24 cm\(^2\)
E
16 cm\(^2\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
VOLUME of CUBE = \(L^3\)
8000 = \(L^3\)
L = \(\sqrt[3]{8000}\) = 20
S. A. CUBE = 6\(L^2\)
S. A. CUBE = 6 x 20 x 20 = 2400\(cm^2\)
Source: Myschool Past Questions Archive (Record #41836)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
Simplify \(\sqrt{6a^{-2}} \times \sqrt{24a^6}\)
A
± 16a\(^2\)
B
± 16a
C
± 14a\(^2\)
D
± 12a\(^2\)
E
± 12a
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\sqrt{6a^{-2}} \times \sqrt{24a^6}\)
= \(\sqrt{6a^{-2} \times 24a^6}\)
= \(\sqrt{144a^4}\) = \pm12a\(^2\).
Source: Myschool Past Questions Archive (Record #41869)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
The surface area of a sphere is 616m2. what is the volume of-the sphere (correct to two significant figures)? (take \pi = 3.142)
A
14 cm\(^3\)
B
210 cm\(^3\)
C
1,400 cm\(^3\)
D
2,100 cm\(^3\)
E
4,300 cm\(^3\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(Surface area (A) = 4\pi r^{2} = 616m^{2}\)
solving, we have
\(r = 7m\)
\(Volume = \frac{4}{3}\pi r^{3}\)
\(Volume = \frac{4\times 3.142 \times 7^{3}}{3}\)
\(Volume = \frac{4310.824}{3}\)
= \(1436.941m^{3} \approxeq 1400m^{3}(2 sig. fig)\)
Source: Myschool Past Questions Archive (Record #41880)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
The radius of a sphere is 21cm. If s and v represent the curved surface area and volume respectively, what is s : v? (take \pi = 3.142)
A
1:03
B
1:07
C
3:01
D
3:07
E
7:01
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
C.S.A = 5,544
Volume = 38,808
S : V = 1 : 07
Source: Myschool Past Questions Archive (Record #41887)
Mathematics
NECO SSCE 2001
Number and Numeration
The first term of a G.P is 7. Find its 4th term, if its common ratio is 3.
A
567
B
280
C
189
D
63
E
21
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #41895)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
Convert (\(101.011)_2\) to base 10
A
63
B
43
C
10
D
\(\frac{77}{8}\)
E
\(\frac{43}{8}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Convert (\(101.011)_2\) to base 10
\(1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3}\)
1 x 4 + 0 x 2 + 1 x 1 + 0 x \(\frac{1}{2}\) + 1 x \(\frac{1}{4}\) + 1 x \(\frac{1}{8}\)
4 + 1 + \(\frac{1}{4}\) + \(\frac{1}{8}\)
= \(\frac{43}{8}\)
Source: Myschool Past Questions Archive (Record #41910)
Mathematics
NECO SSCE 2001
Number and Numeration
If t varies inversely as the cube of s, and t = 24, when s = 3, find t when s = 2
A
8
B
24
C
27
D
81
E
648
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #41968)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Solution
(a) Simplify : \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\)
(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years' time, the ratio of their ages will be 7 : 2. Find their present ages.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\)
\(3^{-5n} \times (3^{2})^{n - 1} \times (3^{3})^{n + 1}\)
\(3^{-5n} \times 3^{2n - 2} \times 3^{3n + 3}\)
\(3^{-5n + 2n - 2 + 3n + 3}\)
= \(3^{1} = 3\)
(b) Let the daughter's age be c and the woman's age be d.
\(c + d = 46 .... (1)\)
In 4 years time, the daughter's age = c + 4
The woman's age = d + 4
\(\frac{d + 4}{c + 4} = \frac{7}{2}\)
\(2(d + 4) = 7(c + 4) \implies 2d + 8 = 7c + 28\)
\(2d - 7c = 28 - 8 \implies 2d - 7c = 20 ... (2)\)
\(c + d = 46 \implies d = 46 - c\)
\(\therefore 2(46 - c) - 7c = 20\)
\(92 - 2c - 7c = 20 \implies 92 - 20 = 9c\)
\(72 = 9c \implies c = 8\)
\(d = 46 - c \implies d = 46 - 8 = 38\)
Therefore, the daughter is 8 years old and the woman is 38 years.
Source: Myschool Past Questions Archive (Record #53384)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Diagram
Solution
The diagram above shows the bar charts representing the number of vehicles manufactured by a company in January, February and March, 1992.
(a) How many vehicles were produced in February?
(b) What fraction of the vehicles manufactured in February were cars?
(c) How many buses were produced altogether from January to March, 1992?
(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
VEHICLES
Jan
Feb
Mar
CAR
10
40
5
LORRY
15
25
35
BUS
20
30
50
TOTAL
45
95
90
(a) Vehicles produced in February = 40 + 25 + 30 = 95
(b) Fraction of the vehicles in February that were cars = \(\frac{40}{95} = \frac{8}{19}\)
(c) Total number of buses from January to March = 20 + 30 + 50 = 100
(d) Ratio of lorries in February to that in March = \(25 : 35 \equiv 5 : 7\).
Source: Myschool Past Questions Archive (Record #53386)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Solution
(a) In a market survey, 100 traders sell fruits, 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three types of fruits. Each of the 100 traders sells at least one of the three fruits.
(i) Represent the information in a Venn diagram ; (ii) Find the number that sell oranges and mangoes only.
(b) Find the value of x for which \(312_{four} + 52_{x} = 96_{ten}\)
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i)
(ii) \(21 + 4 + 5 + 10 + y + 35 - y + 32 - y = 100\)
\(107 - y = 100\)
\(y = 107 - 100 = 7\)
Therefore, 7 traders sell oranges and mangoes alone.
(b) \(312_{four} + 52_{x} = 96_{ten}\)
\(312_{4} = 3 \times 4^{2} + 1 \times 4^{1} + 2 \times 4^{0}\)
= \(48 + 4 + 2 = 54_{10}\)
\(52_{x} = 5 \times x^{1} + 2 \times x^{0}\)
= \((5x + 2)_{10}\)
\(\therefore 54 + (5x + 2) = 96\)
\(56 + 5x = 96 \implies 5x = 96 - 56 = 40\)
\(x = 8\)
\(\therefore 312_{4} + 52_{8} = 96_{10}\)
Source: Myschool Past Questions Archive (Record #53395)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Solution
(a) The probabilities that three boys pass an examination are \(\frac{2}{3}, \frac{5}{8}\) and \(\frac{3}{4}\) respectively. Find the probability that :
(i) all three boys pass ; (ii) none of the boys pass ; (iii) only two of the boys pass.
(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) P(A) = \(\frac{2}{3}\) ; P(B) = \(\frac{5}{8}\) ; P(C) = \(\frac{3}{4}\).
\(P(A \cap B \cap C) = P(A) \times P(B) \times P(C)\)
= \(\frac{2}{3} \times \frac{5}{8} \times \frac{3}{4} = \frac{5}{16}\)
(ii) P(A') = \(1 - \frac{2}{3} = \frac{1}{3}\)
P(B') = \(1 - \frac{5}{8} = \frac{3}{8}\)
P(C') = \(1 - \frac{3}{4} = \frac{1}{4}\)
\(P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')\)
= \(\frac{1}{3} \times \frac{3}{8} \times \frac{1}{4} \)
= \(\frac{1}{32}\)
(iii) P(A and B pass, C fails) = \(\frac{2}{3} \times \frac{5}{8} \times \frac{1}{4} = \frac{5}{48}\)
P( A and C pass, B fails) = \(\frac{2}{3} \times \frac{3}{4} \times \frac{3}{8} = \frac{3}{16}\)
P(B and C pass, A fails) = \(\frac{5}{8} \times \frac{3}{4} \times \frac{1}{3} = \frac{5}{32}\)
P(only two of the boys pass) = \(\frac{5}{48} + \frac{3}{16} + \frac{5}{32} = \frac{43}{96}\)
(b) Let the cost price be x
% gain = \(\frac{\text{actual gain}}{\text{cost price}} \times 100%\)
\(20 = \frac{36,000 - x}{x} \times 100%\)
\(\frac{36,000 - x}{x} = \frac{1}{5}\)
\(5(36,000 - x) = x \implies 180,000 - 5x = x\)
\(180,000 = x + 5x = 6x\)
\(x = N30,000\)
Discount (5%) : \(\frac{5}{100} \times N36,000 = N1,800\)
Selling price : \(N(36,000 - 1,800) = N34,200\)
Actual gain : \(\frac{34,200 - 30,000}{30,000} \times 100% \)
= 14%
Source: Myschool Past Questions Archive (Record #53399)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Diagram
Solution
The number line represents
Attached Figure / Diagram
A
0 > x \(\geq\) -7
B
-7 \(\leq\) x < -1
C
-7 < x \(\leq\) -1
D
-7 < x < -1
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
From the diagram provided, x lies between -7 and or equal to 2, i.e. -7 < x \(\leq\) -1
Source: Myschool Past Questions Archive (Record #54933)