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Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If y varies directly as the square root of x and y = 3 when x = 16. Calculate y when x = 64
A
12
B
6
C
3
D
5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
y ∝ \sqrtx
y = K\sqrtx
K = y/\sqrtx = 3/\sqrt16
= 3/4
y = 3/4\sqrtx
= 3/4\sqrt64 when x = 64
= 3/4 x 8/1
= 6
Source: Myschool Past Questions Archive (Record #9796)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If x * y = x + y\(^2\), find then value of (2*3)*5
A
36
B
25
C
11
D
55
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x * y = x + y\(^2\)
2 * 3 = 2 + 3\(^2\)
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 5\(^2\)
= 11 + 25
= 36
Source: Myschool Past Questions Archive (Record #9802)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If p and q are two non zero numbers and 18(p+q) = (18+p)q, which of the following must be true?
A
q = 18
B
p <1
C
p = 18
D
q < 1
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
If 18(p + q) = (18 + p)q
then 18p + 18q = 18q + pq
18p = pq \(\implies\) q = 18.
Source: Myschool Past Questions Archive (Record #9803)
Mathematics
UTME / JAMB 2010
Algebra
Solution
If \(\begin{vmatrix} x & 3 \\ 2 & 7 \end{vmatrix} = 15\), find the value of x
A
3
B
4
C
5
D
2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\begin{vmatrix} x & 3 \\ 2 & 7 \end{vmatrix} = 15\)
x x 7 - 2 x 3 = 15
7x - 6 = 15
7x = 15+6
7x = 21
x = 21/7
x = 3
Source: Myschool Past Questions Archive (Record #9804)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If two smaller sides of a right angled triangle are 4cm and 5cm, find its area
A
10 cm2
B
6 cm2
C
24 cm2
D
8 cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of a triangle = 1/2 base x height
= 1/2 x 4 x 5
= 10 cm2
Source: Myschool Past Questions Archive (Record #9951)
Mathematics
UTME / JAMB 2010
Geometry and Trigonometry
Solution
An arc subtends an angle of 50º at the center of circle of radius 6cm. Calculate the area of the sector formed
A
100/7
B
80/7
C
110/7
D
90/7
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of a sector = \(\frac{\theta}{360}\) x \(\pi\)r\(^2\)
= \(\frac{50}{360} \times \frac{22}{7}\) x 36
= \(\frac{110}{7}\)
Source: Myschool Past Questions Archive (Record #9952)
Mathematics
UTME / JAMB 2010
Algebra
Solution
What is the locus of point that is equidistant from points P(1,3) and Q(3,5)?
A
y = -χ + 6
B
y = -χ - 6
C
y = χ + 6
D
y = χ - 6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Locus of a point equidistant from two fixed points is the perpendicular bisector of the straight line joining the points Gradients of PQ = (5-3 / 3-1) = 2/2 = 1
Gradient bisector = -1
Equation of perpendicular bisector
y - y1 = m(χ - χ1
y - 4 = -χ + 2
y = -χ + 6
Source: Myschool Past Questions Archive (Record #9954)
Mathematics
UTME / JAMB 2010
Geometry and Trigonometry
Diagram
Solution
If the area of \DeltaPQR above is 12\(\sqrt{3}\) cm\(^2\), find the value of q?
Attached Figure / Diagram
A
6 cm
B
7 cm
C
8 cm
D
5 cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of a triangle = \(\frac{1}{2}\) ab Sin\theta
12\(\sqrt{3}\) = \(\frac{1}{2}\) x 8 x q sin 60
12\(\sqrt{3}\) = 4 x q x \(\frac{\sqrt{3}}{2}\)
12\(\sqrt{3}\) = 2q\(\sqrt{3}\)
q = \(\frac{12\sqrt{3}}{2\sqrt{3}}\)
q = 6cm.
Source: Myschool Past Questions Archive (Record #9958)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If y = (2x + 1)\(^3\) find \(\frac{\text{d}}{\text{dx}}\)
A
3(2x+1)2
B
6(2x+1)
C
3(2x+1)
D
6(2x+1)2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = (2x + 1)\(^3\)
\(\frac{dy}{dx}\) = 3(2x + 1)\(^{3-1}\) x 2
= 3(2x + 1)\(^2\)x 2
= 6(2x + 1)\(^2\)
Source: Myschool Past Questions Archive (Record #9959)
Mathematics
UTME / JAMB 2010
Geometry and Trigonometry
Solution
Find ∫(sin x + 2)dx
A
cos x + x2 + K
B
cos x 2x + K
C
-cos x + x2 + K
D
-cos x + 2x + K
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
∫(sin χ + 2)dχ = (∫sin χ + ∫2)dx
= -cos χ + 2χ + K
Source: Myschool Past Questions Archive (Record #10307)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
Marks
2
3
4
5
6
7
8
No. of students
3
1
5
2
4
2
3
from the table above, if the pass mark is 5, how many students failed the test?
A
6
B
2
C
7
D
9
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Marks
Number of students
2
3
3
1
4
5
Total
9
Source: Myschool Past Questions Archive (Record #10316)
Mathematics
UTME / JAMB 2010
Statistics and Probability
Solution
Marks
1
2
3
4
5
Frequency
2
2
8
4
4
The table above show the marks obtained in a given test.
How many student took the test?
A
16
B
13
C
20
D
15
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
∑f = 2 + 2 + 8 + 4 + 4
= 20
Source: Myschool Past Questions Archive (Record #10323)
Mathematics
UTME / JAMB 2010
Statistics and Probability
Solution
Marks
1
2
3
4
5
Frequency
2
2
8
4
4
The table above show the marks obtained in a given test.
Find the mean mark
A
3.1
B
3.0
C
3.2
D
3.3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Mark(χ)
f
fχ
1
2
2
2
2
4
3
8
24
4
4
16
5
4
20
∑f = 20
∑fχ = 66
\(Mean = \frac{\sum fx}{\sum f}\)
= \(\frac{66}{20}\)
= 3.3
Source: Myschool Past Questions Archive (Record #10347)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
Find r, if 6r7\(_8\) = 511\(_9\)
A
3
B
2
C
6
D
5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
6r7\(_8\) = 511\(_9\)
6 x 8\(^2\) + r x 8\(^1\) + 7 x 8\(^0\) = 5 x 9\(^2\) + 1 x 9\(^1\) + 1 x 9\(^0\)
6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1
384 + 8r + 7 = 405 + 9 + 1
8r = 415 - 391
8r = 24
r = \(\frac{24}{8}\)
= 3
Source: WAEC / JAMB / NECO Classroom Archive (Record #29544)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
Simplify (\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
A
\(\frac{1}{5}\)
B
\(\frac{1}{4}\)
C
\(\frac{1}{36}\)
D
\(\frac{1}{25}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
Applying the rule of BODMAS, we have:
(\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\)
(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)
\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)
= \(\frac{1}{36}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #29546)