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Mathematics
UTME / JAMB 1997
Algebraic Expressions, Polynomials & Factorisation
Solution
Make F the subject of the formula t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)
A
\(\frac{gv-t^2}{gt^2}\)
B
\(\frac{gt^2}{gv-t^2}\)
C
\(\frac{v}{\frac{1}{t^2} - \frac{1}{g}}\)
D
\(\frac{gv}{t^2 - g}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)
t2 = \(\frac{v}{\frac{1}{f} + \frac{1}{g}}\)
= \(\frac{vfg}{ftg}\)
\(\frac{1}{f} + \frac{1}{g}\) = \(\frac{v}{t^2}\)
= (g + f)t2 = vfg
gt2 = vfg - ft2
gt2 = f(vg - t2)
f = \(\frac{gt^2}{gv-t^2}\)
Source: Myschool Past Questions Archive (Record #30807)
Mathematics
UTME / JAMB 1997
Linear, Simultaneous & Quadratic Equations
Solution
What value of g will make the expression 4x2 - 18xy + g a perfect square?
A
9
B
\(\frac{9y^2}{4}\)
C
81y^2
D
\(\frac{18y^2}{4}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
4x2 - 18xy + g = g \(\to\) (\(\frac{18y}{4}\))2
= \(\frac{18y^2}{4}\)
Source: Myschool Past Questions Archive (Record #30808)
Mathematics
UTME / JAMB 1997
Algebraic Expressions, Polynomials & Factorisation
Solution
Find the value of k if \(\frac{5 + 2r}{(r + 1)(r - 2)}\) expressed in partial fraction is \(\frac{k}{r - 2}\) + \(\frac{L}{r + 1}\) where K and L are constants
A
3
B
2
C
1
D
-1
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
5 + 2r = k(r + 1) + L(r - 2)
but r - 2 = 0 and r = 2
9 = 3k
k = 3
Source: Myschool Past Questions Archive (Record #30809)
Mathematics
UTME / JAMB 1997
Inequalities & Linear Programming
Solution
Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1
A
x < - \(\frac{3}{4}\)
B
x > - \(\frac{4}{3}\)
C
x > - \(\frac{3}{4}\)
D
x > - 12
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{f(x)}{g(x)}\) 2x + 4
= 6x - 2x > 4 - 7
= 4x > -3
∴ x > -\(\frac{3}{4}\)
Source: Myschool Past Questions Archive (Record #30814)
Mathematics
UTME / JAMB 1997
Inequalities & Linear Programming
Solution
Find the range of values of x which satisfies the inequality 12x2 < x + 1
A
-\(\frac{1}{4}\) < x < \(\frac{1}{3}\)
B
\(\frac{1}{4}\) < x < -\(\frac{1}{3}\)
C
-\(\frac{1}{3}\) < x < \(\frac{1}{4}\)
D
-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
12x2 2 - 4x + 3x - 1 0, 3x - 1 -\(\frac{1}{4}\)x 0
-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #30815)
Mathematics
UTME / JAMB 1997
Sequences & Series (AP & GP)
Solution
Sn is the sum of the first n terms of a series given by Sn = n\(^2\) - 1. Find the nth term
A
4n + 1
B
4n - 1
C
2n + 1
D
2n - 1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(S_{n} = n^{2} - 1\)
\(T_{n} = S_{n} - S_{n - 1}\)
\(S_{n - 1} = (n - 1)^{2} - 1\)
= \(n^{2} - 2n + 1 - 1\)
= \(n^{2} - 2n\)
\(S_{n} - S_{n - 1} = (n^{2} - 1) - (n^{2} - 2n)\)
= \(2n - 1\)
Source: Myschool Past Questions Archive (Record #30816)
Mathematics
UTME / JAMB 1997
Sequences & Series (AP & GP)
Solution
The nth term of a sequence is given \(3^{1 - n}\), find the sum of the first three terms of the sequence.
A
\(\frac{13}{9}\)
B
1
C
\(\frac{1}{3}\)
D
\(\frac{1}{9}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Tn = \(3^{1 - n}\)
where n = 1, 2, 3.
\(S_3 = 3^{1 - 1} + 3^{1 - 2} + 3^{1 - 3}\)
= 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\)
= \(\frac{13}{9}\)
Source: Myschool Past Questions Archive (Record #30817)
Mathematics
UTME / JAMB 1997
Binary Operations, Matrices & Determinants
Solution
Two binary operations \(\ast\) and \(\oplus\) are defines as m \(\ast\) n = mn - n - 1 and m \(\oplus\) n = mn + n - 2 for all real numbers m, n.
Find the value of 3 \(\oplus\) (4 \(\ast\) 5)
A
60
B
57
C
54
D
42
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
m \(\ast\) n = mn - n - 1, m \(\oplus\) n = mn + n - 2
3 \(\oplus\) (4 \(\ast\) 5) = 3 \(\oplus\) (4 x 5 - 5 - 1) = 3 \(\oplus\) 14
3 \(\oplus\) 14 = 3 x 14 + 14 - 2
= 54
Source: Myschool Past Questions Archive (Record #30818)
Mathematics
UTME / JAMB 1997
Binary Operations, Matrices & Determinants
Solution
If X \(\ast\) Y = X + Y - XY, find x when (x \(\ast\) 2) + (x \(\ast\) 3) = 68
A
24
B
22
C
-12
D
-21
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x \(\ast\) y = x + y - xy
(x \(\ast\) 2) + (x \(\ast\) 3) = 68
= x + 2 - 2x + x + 3 - 3x = 68
-3x = 63
x = -21
Source: Myschool Past Questions Archive (Record #30819)
Mathematics
UTME / JAMB 1997
Plane Geometry, Polygons & Circle Theorems
Solution
Determine x + y if \(\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}\) \(\begin{pmatrix} x \\ y \end{pmatrix}\) = \(\begin{pmatrix}-1 \\ 8 \end{pmatrix}\)
A
3
B
4
C
7
D
12
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix}\)
\(\begin{pmatrix} 2x - 3y \\ -x + 4y \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix}\)
\(2x - 3y = -1 ... (i)\)
\(-x + 4y = 8 ... (ii)\)
From (ii), x = 4y - 8.
\(2(4y - 8) - 3y = -1 \implies 8y - 16 - 3y = -1\)
\(5y = -1 + 16 = 15 \implies y = 3\)
\(x = 4(3) - 8 = 12 - 8 = 4\)
\(\therefore x + y = 3 + 4 = 7\)
Source: Myschool Past Questions Archive (Record #30820)
Mathematics
UTME / JAMB 1997
Linear, Simultaneous & Quadratic Equations
Solution
Find the non-zero positive value of x which satisfies the equation \(\begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x\end{vmatrix}\) = 0
A
2
B
4
C
√3
D
√2
E
1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x(x\(^2\) - 1) - x = 0
= x\(^3\) - 2x = 0
x(x\(^2\) - 2) = 0
x\(^2\) - 2 = 0
x\(^2\) = 2
x = \(\sqrt{2}\)
Source: Myschool Past Questions Archive (Record #30821)
Mathematics
UTME / JAMB 1997
Plane Geometry, Polygons & Circle Theorems
Solution
Each of the base angles of a isosceles triangle is 58^\circ and the verticles of the triangle lie on a circle. Determine the angle which the base of the triangle subtends at the centre of the circle.
A
128o
B
16o
C
64o
D
58o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Base angle of the triangle = 58^\circ.
The third angle = 180^\circ - (58^\circ + 58^\circ)
= 64^\circ
Angle subtended at the centre = \(2 \times 64^\circ = 128^\circ\) (angle at the centre is twice the angle at the circumference).
Source: Myschool Past Questions Archive (Record #30822)
Mathematics
UTME / JAMB 1997
Plane Geometry, Polygons & Circle Theorems
Solution
A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc
A
22cm
B
44cm
C
110cm
D
220cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Diameter = 42cm
Length of the arc = \(\frac{\theta}{360^\circ} \times \pi d\)
= \(\frac{60}{360} \times \frac{22}{7} \times 42cm\)
= \(22cm\)
Source: Myschool Past Questions Archive (Record #30823)
Mathematics
UTME / JAMB 1997
Mensuration (Areas, Perimeters & Volumes)
Solution
An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.(\(\pi\) = \(\frac{22}{7}\))
A
22cm2
B
44cm2
C
66cm2
D
88cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of a sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\)
r = 6cm; \(\theta\) = 70^\circ.
Area of the sector = \(\frac{70}{360} \times \frac{22}{7} \times 6^{2}\)
= \(22 cm^{2}\)
Source: Myschool Past Questions Archive (Record #30824)
Mathematics
UTME / JAMB 1997
Mensuration (Areas, Perimeters & Volumes)
Solution
A cone with a sector angle of 45^\circ is cut out of a circle of radius r cm. Find the base radius of the cone.
A
\(\frac{r}{16}\) cm
B
\(\frac{r}{6}\) cm
C
\(\frac{r}{8}\) cm
D
\(\frac{r}{2}\) cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The formula for the base radius of a cone formed from the sector of a circle = \(\frac{r \theta}{360^\circ}\)
= \(\frac{r \times 45^\circ}{360^\circ}\)
= \(\frac{r}{8} cm\)
Source: Myschool Past Questions Archive (Record #30825)