Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 4276–4290 of 4910 verified questions
Sort by:
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
Evaluate 2700 000 x 0.03 \div 18 000
A
4.5 x 100
B
4.5 x 101
C
4.5 x 102
D
4.5 x 103
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(2700000 \times 0.03 \div 18000\)
= \(\frac{2700000 \times 0.03}{18000}\)
= \(\frac{81000}{18000} = 4.5 \times 10^{0}\)
Source: Myschool Past Questions Archive (Record #31493)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
The prime factors of 2520 are
A
2, 9, 5
B
2, 9, 7
C
2, 3, 5, 7
D
2, 3, 7, 9
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(2520 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7\) in terms of its prime factors. Hence, the prime factors of 2520 are 2, 3, 5, 7.
Source: Myschool Past Questions Archive (Record #31494)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
If \(12_{e} = X_{7}\), where e = 12, find X.
A
20
B
15
C
14
D
12
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(12_{12} = 1 \times 12^{1} + 2 \times 12^{0} = 12 + 2 = 14_{10}\)
7
14
7
2 rm 0
0 rm 2
\(\therefore 12_{e} = 20_{7}\)
Source: Myschool Past Questions Archive (Record #31495)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\)
A
\(\frac{r}{2}\)
B
2r
C
\(\frac{1}{2r}\)
D
\(\frac{2}{r}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\)
= \(((64r^{-6})^{1/_2})^{1/_3}\)
=\((64r^{-6})^{1/_6}\)
= (\(64)^{1/_6}(r^{-6})^{1/_6}\)
=\(\frac{2}{\text{r}}\)
Source: Myschool Past Questions Archive (Record #31496)
Mathematics
UTME / JAMB 1989
Fractions, Decimals, Percentages & Approximations
Solution
What is the different between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal?
A
10-5
B
7 x 10-4
C
8 x 10-5
D
10-6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
0.007685 = 0.00769 (three significant figures)
0.007685 = 0.0077(4d.p)
the difference = 0.0077 - 0.00769
= 0.00001
= 1.0 x 10-5
Source: Myschool Past Questions Archive (Record #31497)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
If a : b = 5 : 8, x : y = 25 : 16; evaluate \(\frac{a}{x}\) : \(\frac{b}{y}\)
A
125 : 128
B
3 : 5
C
3 : 4
D
2 : 5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
a : b = 5 : 8 = 2.5 : 40
x : y = 25 : 16
\(\frac{a}{x}\) : \(\frac{b}{y}\) = \(\frac{25}{25}\) : \(\frac{40}{16}\)
= 1 : \(\frac{40}{16}\)
= 16 : 40
= 2 : 5
Source: Myschool Past Questions Archive (Record #31498)
Mathematics
UTME / JAMB 1989
Fractions, Decimals, Percentages & Approximations
Solution
Oke deposited N800.00 in the bank at the rate of 12\(\frac{1}{2}\)% simple interest. After some time the total amount was one and half times the principal. For how many years was the money left in the bank?
A
2\(\frac{2}{3}\)
B
4
C
5\(\frac{1}{3}\)
D
8
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
P = N800 (Principal), r = 12\(\frac{1}{2}\)% or 0.125
After sometimes, A = 1.5 x 800 = N1,200
A = P(1 + Tr) = 1200
= 800(1 + 0.125T)
= 1200
= 800(1 + 0.125T)
1200 = 800(1 + 0.125T)
0.125T = 0.5
T = \(\frac{0.5}{0.125}\)
= 4
Source: Myschool Past Questions Archive (Record #31499)
Mathematics
UTME / JAMB 1989
Mensuration (Areas, Perimeters & Volumes)
Solution
If the surface area of a sphere increased by 44%, find the percentage increase in diameter
A
44
B
30
C
22
D
20
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Surface Area of Sphere A = 4\(\pi r^2\)
∴ A = 4\(\pi\)\(\frac{(D)^2}{2}\)
= \(\frac{(D)^2}{2}\)
= \(\pi\)D2
When increased by 44% A = \(\frac{144 \pi D^2}{100}\)
\(\pi\)\(\frac{(12D)^2}{10}\) = \(\pi\)\(\frac{(6D)^2}{5}\)
Increase in diameter = \(\frac{6D}{5}\) - D = \(\frac{1}{5}\)D
Percentage increase = \(\frac{1}{5}\) x \(\frac{1}{100}\)%
= 20%
Source: Myschool Past Questions Archive (Record #31500)
Mathematics
UTME / JAMB 1989
Indices, Logarithms & Surds
Solution
Simplify \(4 - \frac{1}{2 - \sqrt{3}}\)
A
2\(\sqrt{3}\)
B
-2 - \(\sqrt{3}\)
C
-2 + \(\sqrt{3}\)
D
2 - \(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(4 - \frac{1}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3}) - 1}{2 - \sqrt{3}}\)
= \(\frac{8 - 4\sqrt{3} - 1}{2 - \sqrt{3}}\)
= \(\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}\)
Rationalizing,
\((\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}) (\frac{2 + \sqrt{3}}{2 + \sqrt{3}})\)
= \(\frac{14 + 7\sqrt{3} - 8\sqrt{3} - 12}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}\)
= \(\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}\)
Source: Myschool Past Questions Archive (Record #31503)
Mathematics
UTME / JAMB 1989
Indices, Logarithms & Surds
Solution
Find p in terms of q if \(\log_{3} p + 3\log_{3} q = 3\)
A
(\(\frac{3}{q}\))3
B
(\(\frac{q}{3}\))\(\frac{1}{3}\)
C
(\(\frac{q}{3}\))3
D
(\(\frac{3}{q}\))\(\frac{1}{3}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\log_{3} p + 3\log_{3} q = 3\)
\(\log_{3} p + \log_{3} q^{3} = 3\)
\(\implies \log_{3} (pq^{3}) = 3\)
\(pq^{3} = 3^{3} = 27\)
\(\therefore p = \frac{27}{q^{3}}\)
= \((\frac{3}{q})^{3}\)
Source: Myschool Past Questions Archive (Record #31504)
Mathematics
UTME / JAMB 1989
Indices, Logarithms & Surds
Solution
What are the values of y which satisfy the equation \(9^{y} - 4 \times 3^{y} + 3 = 0\) ?
A
-1 and 0
B
-1 and 1
C
1 and 3
D
0 and 1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(9^{y} - 4 \times 3^{y} + 3 = 0\)
\(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\)
\((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\)
Let \(3^{y}\) be r. Then,
\(r^{2} - 4r + 3 = 0\)
Solving the equation,
\(r^{2} - 3r - r + 3 = 0\)
\(r(r - 3) - 1(r - 3) = 0\)
\((r - 3)(r - 1) = 0\)
\(\therefore \text{r = 3 or 1}\)
Recall, \(3^{y} = r\)
\(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\)
\(\implies \text{y = 1 or 0}\)
Source: Myschool Past Questions Archive (Record #31507)
Mathematics
UTME / JAMB 1989
Algebraic Expressions, Polynomials & Factorisation
Solution
Make R the subject of the formula S = \(\sqrt{\frac{2R + T}{2RT}}\)
A
R = \(\frac{T}{(TS^2 + 1)}\)
B
R = \(\frac{T}{2(TS^2 - 1)}\)
C
R = \(\frac{T}{2(TS^2 + 1)}\)
D
R = \(\frac{R}{2(TS^2 + 1)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
S = \(\sqrt{\frac{2R + T}{2RT}}\)
Squaring both sides,
\(S^{2} = \frac{2R + T}{2RT}\)
\(S^{2} (2RT) = 2R + T\)
\(2S^{2} RT - 2R = T\)
\(R = \frac{T}{2TS^{2} - 2}\)
= \(\frac{T}{2(TS^2 - 1)}\)
Source: Myschool Past Questions Archive (Record #31509)
Mathematics
UTME / JAMB 1989
Fractions, Decimals, Percentages & Approximations
Solution
The cost of dinner for a group of students is partly constant and partly varies directly as the number of students. If the cost is N74.00 when the number of students is 20 and N96.00 when the number of students is 30, find the cost when there are 15 students
A
N68.50
B
N63.00
C
N60.00
D
N52.00
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
where C = cost, S = number of students, a and K are constants.
C = a + kS. If C = 74, S = 20
C = 96, S = 30, C= ? S = 15
74 = a + 20k......(1)
96 = a + 30k......(2)
subtract equation (1) from (2)
96 = a + 30k
-
74 = a + 20k
--------------
22 = 10k
k = 2.2
find a
74 = a + 44
a = 30
C = 30 + 2.2S
when S = 15, C = 30 + 2.2 x 15
= 30 + 33
= N63
Source: Myschool Past Questions Archive (Record #31516)
Mathematics
UTME / JAMB 1989
Linear, Simultaneous & Quadratic Equations
Solution
If \(f(x) = 2x^2 - 5x + 3\), find f(x + 1).
A
2x2 - x
B
2x2 - x + 10
C
4x2 + 3x + 2
D
4x2 + 3x + 12
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(f(x) = 2x^2 - 5x + 3\)
\(f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3\)
= \(2(x^2 + 2x + 1) - 5x - 5 + 3\)
= \(2x^2 + 4x + 2 - 5x - 2\)
= \(2x^2 - x\)
Source: Myschool Past Questions Archive (Record #31519)
Mathematics
UTME / JAMB 1989
Linear, Simultaneous & Quadratic Equations
Solution
Solve for a positive number x such that \(2^{(x^3 - x^2 - 2x)} = 1\)
A
4
B
3
C
2
D
1
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(2^{(x^3 - x^2 - 2x)} = 1\)
\(x^3 - x^2 - 2x = 0\)
\(x(x^2 - x - 2) = 0\)
\(x^2 - x - 2 = 0\)
\((x + 1)(x - 2) = 0\)
x = 2 is the positive answer.
Source: Myschool Past Questions Archive (Record #31522)