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Mathematics
UTME / JAMB 2006
Plane Geometry, Polygons & Circle Theorems
Solution
If the locus of the points which are equidistant from point P and Q meets line PQ at point N, then PN equals
A
NQ
B
1/4NQ
C
2NQ
D
1/2NQ
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Locus of point equidistant from points P and Q is a perpendicular bisector of the straight joining P and Q
∴PN = NQ
Source: Myschool Past Questions Archive (Record #8907)
Mathematics
UTME / JAMB 2006
Trigonometry, Elevation & Bearings
Diagram
Solution
In the diagram above, PQ = 10 cm, PS = 8 cm and ∠PSR is 60o while ∠SRQ is a right angle. Find SR
Attached Figure / Diagram
A
14 cm
B
14√3 cm
C
10 cm
D
10√3 cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In ΔPST; cos 60 = ST/8
ST = 8cos 60
ST = 8 x 1/2 = 4
TR = 10 cm (opp. sides of a rectangle PQRT)
SR = ST + TR
SR = 4 + 10
SR = 14 cm
Source: Myschool Past Questions Archive (Record #8983)
Mathematics
UTME / JAMB 2006
Inequalities & Linear Programming
Diagram
Solution
The solution set of the shaded area above is
Attached Figure / Diagram
A
y ≥ 0, y ≥ x and y + x ≤ 4
B
y ≤ x, y + x ≤ 4
C
y + x ≥ 4, y ≤ x
D
y ≤ x, y + x ≤ 4 and y ≥ 0
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = x implies y \le x
y + x = 4 implies y = 4 – x
∴y = 4 – x
∴y \le x, y + x \le 4 and y \ge 0
Source: Myschool Past Questions Archive (Record #9008)
Mathematics
UTME / JAMB 2006
Binary Operations, Matrices & Determinants
Solution
A binary operation \(\oplus\) defined on the set of real number is such that x\(\oplus\)y = xy/6 for all x, y ∈ R. Find the inverse of 20 under this operation when the identity element is 6
A
1/12
B
10/3
C
1/20
D
9/5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x ⊕ y = xy/6 = xe/6 = x
xe = 6x
e = 6
x ⊕ y = xx1/6 where x1 = 20
x x 20/6 = 6
x = 36/20
x = 9/5
Source: Myschool Past Questions Archive (Record #9014)
Mathematics
UTME / JAMB 2006
Variations
Solution
If p varies inversely as the cube of q and q varies directly as the square of r, what is the relationship between p and r?
A
p varies directly as r3
B
p varies inversely as r6
C
p varies inversely as 6√P
D
p varies directly as r6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
P ∝ 1/q3
P = K/q3
q3 = K/P
q = K/p1/3
But q ∝ r2
q = Kr2
K/p1/3 = Kr2
r2 = K/p1/3 x 1/K
r2 = 1/p1/3
r = (1/p1/3)2
r = 1/p1/6
r ∝ 1/p1/6 = p = \(\frac{1}{r^6}\)
∴ p varies inversely as \(r^6\)
Source: Myschool Past Questions Archive (Record #9035)
Mathematics
UTME / JAMB 2006
Binary Operations, Matrices & Determinants
Solution
A binary operation * on the set of rational numbers is defined as \(x \ast y = \frac{x^2 - y^2}{2xy}\). Find \(-5 \ast 3\)
A
\(\frac{-8}{15}\)
B
\(\frac{8}{15}\)
C
\(\frac{17}{15}\)
D
\(\frac{-17}{15}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(x \ast y = \frac{x^2 - y^2}{2xy}\)
=\(\frac{(x+y)(x-y)}{2xy}\)
-5 \(\ast 3 =\frac{(-5+3)(-5-3)}{2(-5\times3)}\)
=\(\frac{-2 \times -8}{2(-5\times3)}\)
=\(\frac{-8}{15}\)
Source: Myschool Past Questions Archive (Record #9036)
Mathematics
UTME / JAMB 2006
Algebraic Expressions, Polynomials & Factorisation
Solution
Find the value of k if the expression kx3 + x2 - 5x - 2 leaves a remainder 2 when it is divided by 2x + 1
A
10
B
8
C
-10
D
-8
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
F(x) = Q x D + R
Kx3 + x2 - 5x – 2 = Q(2x+1)+R
If 2x+1 = 0 implies x = -1/2
∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2
= Q(2(-1/2) + 1) +2
K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2
-k/8 + 1/4 + 5/2 - 2 = 0+2
(-k+2+20-16) / 8 = 2
(-k+6) / 8 = 2
-k+6 = 2*8
-k+6 = 16
-k = 16-6
-k = 10
∴k = -10
Source: Myschool Past Questions Archive (Record #9037)
Mathematics
UTME / JAMB 2006
Inequalities & Linear Programming
Solution
Solve the inequalities for which \(\frac{x+4}{3}-\frac{x-3}{2} < 4\)
A
x < 7
B
x > -7
C
x < -7
D
x > 7
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{x+4}{3}-\frac{x-3}{2}\) < 4
\(\frac{(2(x+4))-(3(x-3))}{6}\) < 4
2(x+4) – 3(x-3) < 4 x 6
2x + 8 – 3x+9 < 24
-x + 17 < 24
-x < 24 – 17
-x < 7
x > -7
Source: Myschool Past Questions Archive (Record #9038)
Mathematics
UTME / JAMB 2006
Plane Geometry, Polygons & Circle Theorems
Solution
The cost of renovating a 6 m square room is N540. What is the cost of renovating a 9 m square room?
A
N1215
B
N720
C
N1620
D
N810
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Cost of removing a 6m\(^2\) room = N540
Cost of renovating a 1m\(^2\) room = 540 / 6
Cost of removing a 9m\(^2\) room
= (540 / 6) * 9
= N810
Source: Myschool Past Questions Archive (Record #9039)
Mathematics
UTME / JAMB 2006
Plane Geometry, Polygons & Circle Theorems
Solution
The sum of the first n positive integers is
A
1/2 n(n-1)
B
n(n+1)
C
n(n-1)
D
1/2 n(n+1)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)
Source: Myschool Past Questions Archive (Record #9040)
Mathematics
UTME / JAMB 2006
Algebraic Expressions, Polynomials & Factorisation
Solution
If \(T = 2\pi \sqrt{\frac{l}{g}}\), make g the subject of the formula
A
(4π l2) / T
B
(4π2l) / T2
C
(4π2l2) / T2
D
(2π√l) / T
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(T = 2\pi \sqrt{\frac{l}{g}}\)
\(\frac{T}{2\pi} = \sqrt{\frac{l}{g}}\)
\((\frac{T}{2\pi})^{2} = \frac{l}{g}\)
\(\therefore g = \frac{4 \pi^{2} l}{T^{2}}\)
Source: Myschool Past Questions Archive (Record #9041)
Mathematics
UTME / JAMB 2006
Inequalities & Linear Programming
Solution
If y = x\(^2\) - x - 12, find the range of values of x for which y \( \geq \) 0
A
x 4
B
x \( \leq \) -3 or x \( \geq \) 4
C
-3 < x \( \geq \) 4
D
-3 \( \leq \) x \( \leq \) 4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
y = x\(^2\) - x - 12
= (x - 4)(x + 3)
∴ x = 4 or x = -3
Checking the cases for y \( \geq \) 0
We check values on the range x - 4 \(\geq\) 0; x + 3 \(\leq\) 0; x - 4 \(\leq\) 0 and x + 3 \(\geq\) 0 for the range which satisfies the inequality x\(^2\) - x - 12 \(\geq\) 0.
We find that the inequality is satisfied on the range x \(\leq\) -3 and x \(\geq\) 4.
Source: Myschool Past Questions Archive (Record #9042)
Mathematics
UTME / JAMB 2006
Indices, Logarithms & Surds
Solution
How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1 - 2\(^8\)?
A
12
B
10
C
9
D
8
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
3, - 6, +12, -24
a = 3, r = -2
\(Sn = \frac{a(1-r^n)}{1-r}\\
∴1-2^8 = \frac{3(1-(-2^{n}))}{1-(-2)}\\
1-2^8 = \frac{3(1-(-2^{n}))}{3}\)
1 - 2\(^8\) = 1 - (-2)\(^n\)
Since 1 - 2\(^8\) = 1 - (-2)\(^n\)
Source: Myschool Past Questions Archive (Record #9043)
Mathematics
UTME / JAMB 2006
Binary Operations, Matrices & Determinants
Solution
Find p, q for which \(\begin{pmatrix} 2p & 8 \\ 3 & -5q \end{pmatrix}\)\(\begin{pmatrix} 1 \\ 2\end{pmatrix}\) = \(\begin{pmatrix}24 \\ -17\end{pmatrix}\)
A
-4
B
45
C
4,2
D
2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
2p x 1 + 8 x 2 = 24
\(\to\) 4p = 24 - 8 = 16,
p = 4
3 x 1 + -5q x 2 = -17
\(\to\) -10q = -17 - 3
-10q = -20
q = 2
Source: WAEC / JAMB / NECO Classroom Archive (Record #30481)
Mathematics
UTME / JAMB 2006
Binary Operations, Matrices & Determinants
Solution
If x = \(\begin{pmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 0 & 1\end{pmatrix}\) and y = \(\begin{pmatrix} -1 & 1 & 2 \\ 0 & -1 & -1 \\ 2 & -1 & 1\end{pmatrix}\)
find 2x - y
A
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & -1 \\ -4 & 1 & 1\end{pmatrix}\)
B
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
C
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
D
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & 1 & 1 \\ -4 & -1 & 1\end{pmatrix}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
2(1) - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4
2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1
2(1) - 2 = 0 2(0) - (-1) = 1 2(1) - (1) = 1
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #30522)