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Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
John gives one-third of his money to Janet who has N105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money john has at first
A
N45.00
B
N48.00
C
N52.00
D
N60.00
E
N58.00
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Let x be John's money, Janet already had N105, \(\frac{1}{3}\) of x was given to Janet
Janet now has \(\frac{1}{3}\)x + 105 = \(\frac{x + 315}{3}\)
John's money left = \(\frac{2}{3}\)x = \(\frac{1}{4} \times\frac{ x + 315}{3}\)
cross multiplying
24x = 3x + 945
∴ x = 45
Source: Myschool Past Questions Archive (Record #32209)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
Write h in terms of a, b, c, d if a = \(\frac{b(1 - ch)}{1 - dh}\)
A
h = \(\frac{a - b}{ad}\)
B
h = \(\frac{1 - b}{ad - bc}\)
C
h = \(\frac{(a - b)^2}{ad - bc}\)
D
h = \(\frac{a - b}{ad - bc}\)
E
h = \(\frac{b - a}{ab - dc}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
a = \(\frac{b(1 - ch)}{1 - dh}\)
a = \(\frac{b - bch}{1 - dh}\)
a - adh = b - bch
a - b = -bch + adh
a - b = adh - bch
a - b = h(ad - bc)
h = \(\frac{a - b}{ad - bc}\)
Source: Myschool Past Questions Archive (Record #32216)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
22\(\frac{1}{2}\)% of the Nigerian Naira is equal to 17\(\frac{1}{10}\)% of a foreign currency M. What is the conversion rate of the M to the Naira?
A
1M = 1\(\frac{15}{57}\)N
B
1M = 38\(\frac{1}{4}\)N
C
1M = 1\(\frac{18}{57}\)N
D
1M = 384\(\frac{3}{4}\)N
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
N = 22\(\frac{1}{2}\)%, M = 17\(\frac{1}{10}\)%
M = \(\frac{171}{10}\)%, N = \(\frac{45}{2}\)
\(\frac{45}{2}\) x \(\frac{10}{171}\)
= \(\frac{225}{171}\)
= 1 \(\frac{54}{171}\)
= 1 \(\frac{18}{57}\)
Source: Myschool Past Questions Archive (Record #32218)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
If \(e^{x} = 1 + x + \frac{x^{2}}{1.2} + \frac{x^{3}}{1.2.3} + ... \), find \(\frac{1}{e^{\frac{1}{2}}}\)
A
1 - \(\frac{x}{2}\) + \(\frac{x^2}{1.2^3}\) + \(\frac{x^3}{2^4.3}\) + .........
B
1 - \(\frac{x}{2}\) + \(\frac{x^2}{1.2^3}\) + \(\frac{x^4}{2.4.3}\) + ..
C
1 + \(\frac{x}{2}\) + \(\frac{x^2}{1.2}\) + \(\frac{x^3}{1.2.3}\) + \(\frac{x^4}{1.23.4}\) + .........
D
1 - x + \(\frac{x^2}{1.2^3}\) + \(\frac{x^3}{2^4.3}\) + .........
E
1 + \(\frac{x}{2}\) + \(\frac{x^2}{1.2^3}\) + \(\frac{x^4}{1.2.6}\) + .........
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(e^{x} = 1 + x + \frac{x^{2}}{1.2} + \frac{x^{3}}{1.2.3} + ...\)
\(\frac{1}{e^{\frac{1}{2}}} = e^{-\frac{1}{2}}\)
\(e^{-\frac{1}{2}} = 1 - \frac{x}{2} + \frac{x^{2}}{1.2^{3}} - \frac{x^{3}}{1.2^{4}.3} + ... \)
Source: Myschool Past Questions Archive (Record #32220)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2})(\sqrt{3} + \sqrt{2})\) is equal to
A
1
B
(\(\sqrt{2} + 4\sqrt{2}\))
C
(6\(\sqrt{2}\)
D
8
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2})(\sqrt{3} + \sqrt{2})\)
\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2}) = \sqrt{3} - \sqrt[4]{6} + \sqrt[4]{6} - \sqrt{2}\)
= \(\sqrt{3} - \sqrt{2}\)
\((\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2}) = 3 + \sqrt{6} - \sqrt{6} - 2\)
= \(3 - 2 = 1\)
Source: Myschool Past Questions Archive (Record #32221)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
The factors of 9 - (x2 - 3x - 1)2 are
A
-(x - 4)(x + 1) (x - 1)(x - 2)
B
(x - 4)(x - 2) (x - 1)(x + 1)
C
-(x - 2)(x + 1) (x - 2) (x - 1)
D
(x - 2)(x + 2) (x - 1)(x + 1)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
9 - (x2 - 3x - 1)2 = [3 - (x2 - 3x - 1)] [3 + (x2 - 3x - 1)]
= (3 - x2 + 3x + 1)(3 + x2 - 3x - 1)
= (4 + 3x - x2)(x2 - 3x + 2)
= (4 - x)(1 + x)(x - 1)(x - 2)
= -(x - 4)(x + 1) (x - 1)(x - 2)
Source: Myschool Past Questions Archive (Record #32223)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
If 32y + 6(3y) = 27. Find y
A
3
B
-1
C
2
D
-3
E
1
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
32y + 6(3y) = 27
This can be rewritten as (3y)2 + 6(3y) = 27
Let 3y = x
x2 + 6x - 27 = 0
(x + 9)(x - 3) = 0
when x - 3 = 0, x = 3
sub. for x in 3y = x
3y = 3
log33 = y
y = 1
Source: Myschool Past Questions Archive (Record #32224)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
Factorize abx2 + 8y - 4bx - 2axy
A
(ax - 4)(bx - 2y)
B
(ax + b)(x - 8y)
C
(ax - 2y)(bx - 4)
D
(bx - 4)(ax - 2y)
E
(abx - 4)(x - 2y)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
abx2 + 8y - 4bx - 2axy = (abx2 - 4bx) + (8y - 2axy)
= bx(ax - 4) 2y(ax - 4) 2y(ax - 4)
= (bx - 2y)(ax - 4)
Source: Myschool Past Questions Archive (Record #32225)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
If f(x - 2) = 4x2 + x + 7, find f(1)
A
12
B
27
C
7
D
46
E
17
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
f(x - 2) = 4x2 + x + 7
x - 2 = 1, x = 3
f(x - 2) = f(1)
= 4(3)2 + 3 + 7
= 36 + 10
= 46
Source: Myschool Past Questions Archive (Record #32231)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
What is the probability that a number chosen at random from the intergers between 1 and 10 inclusive is either a prime or a multiple of 3?
A
\(\frac{7}{10}\)
B
\(\frac{3}{5}\)
C
\(\frac{4}{5}\)
D
\(\frac{3}{10}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10)
Prime: (2, 3, 5, 7)
multiples of 3: (3, 6, 9)
Using: ∣A∪B ∣= ∣A∣ + ∣B∣ − ∣A∩B∣ = 4 + 3 − 1 =6.
Or
Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6)
Probability = \(\frac{6}{10}\)
= \(\frac{3}{5}\)
Source: Myschool Past Questions Archive (Record #32265)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?
A
5kg
B
16kg
C
19kg
D
6kg
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
A solid sphere of radius \( 4\, \text{cm} \) has a mass of \( 64\, \text{kg} \). What will be the mass of a shell of the same metal whose internal and external radii are \( 2\, \text{cm} \) and \( 3\, \text{cm} \) respectively?
Calculate the density of the solid sphere:
The volume \( V \) of a solid sphere is given by the formula:
\(V = \frac{4}{3} \pi r^3\)
For the solid sphere with radius \( r = 4\, \text{cm} \):
\(V = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \, \text{cm}^3\)
To find the density \( \rho \):
\(\rho = \frac{\text{mass}}{\text{volume}} = \frac{64\, \text{kg}}{\frac{256}{3} \pi \, \text{cm}^3}\)
Converting mass to grams (since \( 1\, \text{kg} = 1000\, \text{g} \)):
\(\rho = \frac{64000\, \text{g}}{\frac{256}{3} \pi} = \frac{64000 \times 3}{256 \pi} = \frac{192000}{256 \pi} = \frac{750}{\pi} \, \text{g/cm}^3\)
Calculate the volume of the shell:
The volume \( V_{\text{shell}} \) of the shell is the difference between the volume of the outer sphere and the inner sphere.
Outer sphere = (radius \( 3\, \text{cm} \)):
\(V_{\text{outer}} = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36 \pi \, \text{cm}^3\)
Inner sphere = (radius \( 2\, \text{cm} \)):
\(V_{\text{inner}} = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \, \text{cm}^3\)
Calculate the volume of the shell:
\(V_{\text{shell}} = V_{\text{outer}} - V_{\text{inner}} = 36 \pi - \frac{32}{3} \pi\)
Converting \( 36 \) to a fraction with a denominator of 3:
\(V_{\text{outer}} = \frac{108}{3} \pi\)
Now subtract:
\(V_{\text{shell}} = \left(\frac{108}{3} \pi - \frac{32}{3} \pi\right) = \frac{76}{3} \pi \, \text{cm}^3\)
Calculate the mass of the shell:
Using the density \( \rho = \frac{750}{\pi} \, \text{g/cm}^3 \):
\(\text{Mass of shell} = \text{Density} \times \text{Volume} = \frac{750}{\pi} \times \frac{76}{3} \pi\)
The \( \pi \) cancels out: \(\text{Mass of shell} = \frac{750 \times 76}{3} = \frac{57000}{3} \approx 19000 \, \text{g} \approx 19 \, \text{kg}\)
Source: Myschool Past Questions Archive (Record #32269)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
PRSQ is a trapezium of area 14cm2 in which PQ||RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm2
A
7.0
B
6.0
C
5.2
D
5.0
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area of trapezium = 14cm2
Area of trapezium = \(\frac{1}{2}\)(a + b)h
14 = \(\frac{1}{2}\)(4 + 3)h
14 = \(\frac{7}{2}\)h
h = \(\frac{14 \times 2}{7}\)
= 4
Area of SQR = \(\frac{1}{2}\)(3 x 4)
= \(\frac{12}{2}\)
= 6.0
Source: Myschool Past Questions Archive (Record #32274)
Mathematics
UTME / JAMB 1985
Number and Numeration
Solution
If a{\(\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}\)} = 6x. Find a in its simplest form
A
x2 - 1
B
x2 + 1
C
x2 + 4
D
x2 - 4
E
1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
a{\(\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}\)} = a{\(\frac{(x + 1)(x + 2)- (x - 1)(x - 2)}{(x - 2)(x + 2)}\)}
= 6
\(\frac{6x}{x^2 - 4}\) = 6x
a = x2 - 4
Source: Myschool Past Questions Archive (Record #32289)
Mathematics
UTME / JAMB 1985
Number and Numeration
Diagram
Solution
In the figure, MNQP is a cyclic quadrilateral. MN and Pq are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = 86o and NQP = 122o find (xo, yo)
Attached Figure / Diagram
A
28o, 36o
B
36o, 28o
C
43o, 61o
D
61o, 43o
E
36o, 43o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
yo = 180o - (86o + 58o)
180 - 144 = 36o
xo = 180 - (94 + 58)
180 -152 = 28
(xo, yo) = (28o, 36o)
Source: Myschool Past Questions Archive (Record #32801)
Mathematics
UTME / JAMB 1985
Number and Numeration
Diagram
Solution
In the figure, GHIJKLMN is a cube of side a. Find the length of HN.
Attached Figure / Diagram
A
\(\sqrt{3a}\)
B
3a
C
3a2
D
a\(\sqrt{2}\)
E
a\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
HJ2 = a2 + a2 = 2a2
HJ = \(\sqrt{2a^2} = a \sqrt{2}\)
HN2 = a2 + (a\(\sqrt{2}\))2 = a2 + 2a2 = 3a2
HN = \(\sqrt{3a^2}\)
= a\(\sqrt{3}\)cm
Source: Myschool Past Questions Archive (Record #32814)