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Mathematics
WAEC / WASSCE 2000
Fractions, Decimals, Percentages & Approximations
Solution
A car moves at an average speed of 30kmh\(^{-1}\), how long does it take to cover 200 meters?
A
2.4 sec
B
24 sec
C
144 sec
D
240 sec
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Speed = 30 kmh\(^{-1}\)
= \(\frac{30 \times 1000}{3600}\)
= \(\frac{25}{3} ms^{-1}\)
Time = \(\frac{Distance}{Speed}\)
= \(\frac{200}{\frac{25}{3}}\)
= \(\frac{200 \times 3}{25}\)
= 24 seconds
Source: WAEC / JAMB / NECO Classroom Archive (Record #26435)
Mathematics
WAEC / WASSCE 2000
Fractions, Decimals, Percentages & Approximations
Solution
A tap leaks at the rate of 2cm\(^3\) per seconds. How long will it take the tap to fill a container of 45 liters capacity? (1 liters = 1000cm\(^3\))
A
8 hours
B
6hr 15min
C
4hr 25min
D
3hr
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Total capacity of the container = 45 liters = 45 x 1000
= 45000 cm\(^3\)
Time to fill the container = \(\frac{45000}{2}\)
= 22500 seconds
= \(\frac{22500}{3600}\)
= 6 hr 15 mins
Source: WAEC / JAMB / NECO Classroom Archive (Record #26464)
Mathematics
WAEC / WASSCE 2000
Fractions, Decimals, Percentages & Approximations
Solution
A man bought a television set on hire purchase for N25,000, out of which he paid N10,000, if he is allowed to pay the balance in eight equal installments, find the value of each installment.
A
N1250
B
N1578
C
N1875
D
N3125
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Price of television = N25,000
Paid N10, 000
Remainder = N(25,000 - 10,000)
= N15,000
Eight equal installments = \(\frac{15,000}{8}\)
= N1,875 per installment.
Source: WAEC / JAMB / NECO Classroom Archive (Record #26436)
Mathematics
WAEC / WASSCE 2000
Fractions, Decimals, Percentages & Approximations
Solution
If the simple interest on N2000 after 9 months is N60, at what rate per annum is the interest charged?
A
\(2\frac{1}{4}\)%
B
4%
C
5%
D
6%
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(I = \frac{PRT}{100}\)
\(R = \frac{100I}{PT}\)
\(R = \frac{100 \times 60}{2000 \times \frac{3}{4}}\)
\(R = \frac{6000}{1500}\)
= 4%
Source: WAEC / JAMB / NECO Classroom Archive (Record #26471)
Mathematics
WAEC / WASSCE 2000
Variations
Solution
If x varies inversely as y and \(x = \frac{2}{3}\) when y = 9, find the value of y when \(x=\frac{3}{4}\)
A
\(\frac{1}{18}\)
B
\(\frac{8}{81}\)
C
\(\frac{9}{2}\)
D
8
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(x \propto \frac{1}{y}\)
\(x = \frac{k}{y}\)
\(\frac{2}{3} = \frac{k}{9}\)
\(3k = 18 \implies k = 6\)
\(x = \frac{6}{y}\)
When y = \(\frac{3}{4}\),
x = \(\frac{6}{\frac{3}{4}}\)
= \(\frac{6 \times 4}{3}\)
= 8
Source: WAEC / JAMB / NECO Classroom Archive (Record #26452)
Mathematics
WAEC / WASSCE 2000
Variations
Solution
Given that p varies directly as q while q varies inversely as r, which of the following statements is true?
A
r varies directly as p
B
p varies inversely as r
C
p varies directly as r
D
q varies inversely as p
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
P varies directly as q means p = kq some constant k
q varies inversely as r means q = \(\frac{\text{m}}{\text{r}}\) for some constant m
Substituting the expression for q into the equation for p:
p =\(\frac{\text{km}}{\text{r}}\)
This shows that p varies inversely as r.
Source: WAEC / JAMB / NECO Classroom Archive (Record #26460)
Mathematics
WAEC / WASSCE 2000
Algebraic Expressions, Polynomials & Factorisation
Solution
Given that x = -\(\frac{1}{2}\) and y = 4 evaluate 3x\(^2y+xy^2\)
A
-5
B
-1
C
4
D
11
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(x = -\frac{1}{2}\), y = 4
3x\(^2\)y + xy\(^2\)
3\(\left[-\frac{1}{2}\right]^2 \times 4 + \left(\frac{-1}{2}\right)(4)^2\)
3\(\times \frac{1}{4} \times 4 -\frac{1}{2} \times 16\)
3 - 8 = -5
Source: WAEC / JAMB / NECO Classroom Archive (Record #26450)
Mathematics
WAEC / WASSCE 2000
Algebraic Expressions, Polynomials & Factorisation
Solution
Make t the subject of formula \(k = m\sqrt{\frac{t-p}{r}}\)
A
\(\frac{rk^2 + p}{m^2}\)
B
\(\frac{rk^2+pm^2}{m^2}\)
C
\(\frac{rk^2-p}{m^2}\)
D
\(\frac{rk^2-p^2}{m^2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(k = m\sqrt{\frac{t - p}{r}}\)
\(\frac{k}{m} = \sqrt{\frac{t - p}{r}}\)
\((\frac{k}{m})^2 = \frac{t - p}{r}\)
\(rk^2 = m^2 (t - p)\)
\(\therefore m^2 t = rk^2 + m^2 p\)
\(t = \frac{rk^2 + m^2 p}{m^2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #26457)
Mathematics
WAEC / WASSCE 2000
Linear, Simultaneous & Quadratic Equations
Solution
Find the equation whose roots are -8 and 5
A
\(x^2 + 13x + 40=0\)
B
\(x^2 - 13x - 40=0\)
C
\(x^2 - 3x +40=0\)
D
\(x^2 + 3x - 40=0\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Equation with roots -8 and 5: (x + 8)(x - 5) = 0
\(x^2 - 5x + 8x - 40 = 0\)
\(x^2 + 3x - 40 = 0\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #26456)
Mathematics
WAEC / WASSCE 2000
Linear, Simultaneous & Quadratic Equations
Solution
Solve the equation 3 + 5x - 2x\(^2\) = 0
A
\(-\frac{1}{2},-3\)
B
2, 3
C
-2, 3
D
\(-\frac{1}{2},3\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
3 + 5x - 2x\(^2\) = 0
3 + 6x - x - 2x\(^2\) = 0
3(1 + 2x) - x(1 + 2x) = 0
(3 - x)(1+2x) = 0
3 - x = 0 or 1 + 2x = 0
x = 3 or x = \(\frac{-1}{2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #26470)
Mathematics
WAEC / WASSCE 2000
Linear, Simultaneous & Quadratic Equations
Solution
(a) Given the expression \(y = ax^{2} - bx - 12\) , find the values of x when a = 1, b = 2 and y = 3.
(b) If \(\sqrt{x^{2} + 1} = \frac{5}{4}\), find the positive value of x.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(y = ax^{2} - bx - 12\)
When a = 1, b = 2 and y = 3.
\(3 = x^{2} - 2x - 12\)
\(x^{2} - 2x - 12 - 3 = 0 \implies x^{2} - 2x - 15 = 0\)
\(x^{2} - 5x + 3x - 15 = 0\)
\((x - 5)(x + 3) = 0\)
\(\text{x = 5 or -3}\)
(b) \(\sqrt{x^{2} + 1} = \frac{5}{4}\)
Squaring both sides,
\(x^{2} + 1 = \frac{25}{16}\)
\(x^{2} = \frac{25}{16} - 1 = \frac{9}{16}\)
\(x = \sqrt{\frac{9}{16}} = \pm \frac{3}{4}\)
The positive value of x = \(\frac{3}{4}\).
Source: WAEC / JAMB / NECO Classroom Archive (Record #53462)
Mathematics
WAEC / WASSCE 2000
Linear, Simultaneous & Quadratic Equations
Solution
(a) Copy and complete the following table of values for the relation \(y = x^{2} - 2x - 5\)
x
-3
-2
-1
0
1
2
3
4
5
y
-2
-6
-2
3
10
(b) Draw the graph of the relation \(y = x^{2} - 2x - 5\); using a scale of 2 cm to 1 unit on the x- axis, and 2 cm to 2 units on the y- axis.
(c) Using the same axes, draw the graph of \(y = 2x + 3\).
(d) Obtain in the form \(ax^{2} + bx + c = 0\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.
(e) From your graphs, determine the roots of the equation obtained in (d) above.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
x
-3
-2
-1
0
1
2
3
4
5
\(x^{2}\)
9
4
1
0
1
4
9
16
25
\(-2x\)
6
4
2
0
-2
-4
-6
-8
-10
-5
-5
-5
-5
-5
-5
-5
-5
-5
-5
y
10
3
-2
-5
-6
-5
-2
3
10
(b) see graph above
(c)
x
-3
-2
-1
0
1
2
3
4
5
\(2x\)
-6
-4
-2
0
2
4
6
8
10
-3
-3
-3
-3
-3
-3
-3
-3
-3
-3
y
-9
-7
-5
-3
-1
1
3
5
7
(d) \(x^{2} - 2x - 5 = 2x - 3\)
\(x^{2} - 2x - 2x - 5 + 3 = 0\)
\(x^{2} - 4x - 2 = 0\)
(e) x = -0.8 or 4.3
Source: WAEC / JAMB / NECO Classroom Archive (Record #53477)
Mathematics
WAEC / WASSCE 2000
Inequalities & Linear Programming
Solution
Form an inequality for a distance d meters which is more than 18m, but not more than 23m
A
18 ≤ d ≤ 23
B
18 < d ≤ 23
C
18 ≤ d < 23
D
d 23
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
more than means greater than WHILE not more than means less than or equal to
therefore the answer is 18 < d \le 23 meaning d is more than 8 but can be equal to or less than 23
Source: WAEC / JAMB / NECO Classroom Archive (Record #26455)
Mathematics
UTME / JAMB 2000
Inequalities & Linear Programming
Diagram
Solution
The solution of the simultaneous inequalities \(2x - 2 \leq\) y and \( 2y - 2 \leq\) x is represented by
Use the diagrams above as the OPTIONS
Attached Figure / Diagram
A
A
B
B
C
C
D
D
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(2x - 2 \leq\) y = \( y \geq\) 2x - 2
when x = 0 , y = -2 and when y = 0, x = 1
\(2y - 2 \leq\) x
when x = 0, y = 1 and when y = 0, x = 1
Use the diagram above for the OPTIONS
Source: WAEC / JAMB / NECO Classroom Archive (Record #69880)
Mathematics
WAEC / WASSCE 2000
Algebraic Expressions, Polynomials & Factorisation
Solution
Simplify \(\frac{1}{x-3}-\frac{3(x-1)}{x^2 - 9}\)
A
\(\frac{x-1}{x-3}\)
B
\(\frac{-2}{x+3}\)
C
\(\frac{x-1}{x+3}\)
D
\(\frac{4x}{x^2-9}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{1}{x-3}-\frac{3(x-1)}{x^2 - 9}\)
\(\frac{1}{x-3}-\frac{3(x-1)}{(x-3)(x+3)}\)
\(\frac{x+3-3x+3}{(x-3)(x+3)};\frac{-2x+6}{(x-3)(x+3)}\)
\(\frac{-2(x-3)}{(x-3)(x+3)}=\frac{-2}{x+3}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #26454)