How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
Find the volume of a cone which has a base radius of 5 cm and slant height of 13 cm.
\(300\pi\) cm\(^3\)
\(325\pi\) cm\(^3\)
\(\frac{325}{3}\pi\) cm\(^3\)
\(100\pi\) cm\(^3\)
Correct answer is D
Volume of a cone = \(\frac{1}{3}\pi r^2 h\)
r = 5 cm
l = 13 cm
Using Pythagoras theorem
⇒ \(13^2 = 5^2 + h^2\)
⇒ \(169 = 25 + h^2\)
⇒ \(169 - 25 = h^2\)
⇒ \(h^2 = 144\)
⇒ \(h = \sqrt144 = 12 cm\)
∴ Volume of the cone = \(\frac{1}{3} \times\pi\times 5^2 x 12 = 100\pi\) cm\(^3\)
Which inequality describes the graph above?
\(4y + 5x ≥ 20\)
\(5y + 4x ≤ 20\)
\(4y + 5x ≤ 20\)
\(5y + 4x ≥ 20\)
Correct answer is B
First, we find the equation of the boundary line using the two intercepts.The slope is
m = \(\frac{4 - 0}{0 - 5} = {4}{5}\)
The y-intercept is 4
The slope-intercept form of the equation is therefore
y = -\(\frac{4}{5} x + 4\)
\(\implies y + \frac{4}{5} x = 4\)
Multiply both sides by \(\frac{5}{4}\)
\(\implies\frac{5}{4}(y +\frac{4}{5} x) = 4\times\frac{5}{4}\)
\(\implies\frac{5}{4} y + x = 5\)
Multiply both sides by 4
⇒ \(5y + 4x = 20\)
The inequality is therefore either \(5y + 4x ≤ 20\) or \(5y + 4x ≥ 20.\)
Using the test point (0, 0) -The origin
⇒ 5(0) + 4(0) ≤ 20
⇒ 0 ≤ 20 (True)
∴ The inequality is \(5y + 4x ≤ 20\)
Find the value of x in the diagram above
10 units
15 units
5 units
20 units
Correct answer is A
Intersecting Chords Theorem states that If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal.
⇒ AE * EB = CE * ED
⇒ 6 * \(x\) = 4 * (\(x\) + 5)
⇒ 6\(x\) = 4\(x\) + 20
⇒ 6\(x\) - 4\(x\) = 20
⇒ 2\(x\) = 20
∴ \(x = \frac{20}{2}\) = 10 units
Calculate the area of the composite figure above.
6048 m\(^2\)
3969 m\(^2\)
4628 m\(^2\)
5834 m\(^2\)
Correct answer is B
Area of the composite figure = Area of semi circle + Area of rectangle + Area of triangle
Area of semi circle = \(\frac{1}{2}\pi r^2 = \frac{1}{2}\times\pi\times\frac{d^2}{4} = \frac{1}{2}\times\frac{22}{7}\times\frac{42^2}{4} = 693 m^2\)
Area of rectangle = l x b = 42 x 60 =2520 m\(^2\)
Area of triangle = \(\frac{1}{2}\times b \times h = \frac{1}{2}\times 36 \times 42 = 756 m^2\)
∴ Area of the composite figure = 693 + 2520 + 756 = 3969 m\(^2\)
Solve the logarithmic equation: \(log_2 (6 - x) = 3 - log_2 x\)
\(x\) = 4 or 2
\(x\) = -4 or -2
\(x\) = -4 or 2
\(x\) = 4 or -2
Correct answer is A
\(log_2 (6 - x) = 3 - log_2 x\)
⇒ \(log_2 (6 - x) = 3 log_2 2 - log_2 x\) (since \(log_2\) 2 = 1)
⇒ \(log_2 (6 - x) = log_2 2^3 - log_2 x\) \((a log\) c = \(log\) c\(^a)\)
⇒ \(log_2 (6 - x) = log_2 8 - log_2 x\)
⇒\(log_2 (6 - x) = log_2 \frac{8}{x}\) (\(log\) a - \(log\) b = \(log \frac{a}{b})\)
⇒ \(6 - x = \frac{8}{x}\)
⇒ \(x (6 - x) = 8\)
⇒ \(6x - x^2 = 8\)
⇒ \(x^2 - 6x + 8 = 0\)
⇒ \(x^2 - 4x - 2x + 8 = 0\)
⇒ \(x (x - 4) - 2(x - 4) = 0\)
⇒ \((x - 4)(x - 2) = 0\)
⇒ \(x - 4 = 0 or x - 2 = 0\)
∴ x = 4 or 2
20
300
50
60
Correct answer is D
Let number of children's ticket at ₦250.00 each = \(x\)
∴ Number of adult tickets at ₦520.00 each = 5\(x\)
Then,
Total amount of money received from children's tickets = 250\(x\)
Total amount of money received from adult tickets = 520(5\(x\))
⇒ 250\(x\) + 520(5\(x\)) = 171,000
⇒ 250\(x\) + 2600\(x\) = 171,000
⇒ 2850\(x\) = 171,000
⇒ \(x = \frac{171,000}{2850} = 60\)
∴ 60 tickets were sold at ₦250.00 and 300 tickets were sold at ₦520.00
The line \(3y + 6x\) = 48 passes through the points A(-2, k) and B(4, 8). Find the value of k.
16
20
8
-2
Correct answer is B
The line: \(3y + 6x\) = 48
Divide through by 3
⇒ y + 2\(x\) = 16
⇒ y = -2\(x\) + 16
∴ The gradient of the line = -2
The points: A(-2, k) and B (4, 8)
m =\(\frac{y2 - y1}{x2 - x1} = \frac{8 - k}{4 - (-2)}\)
⇒ m =\(\frac[8 - k}{4 + 2} = {8 - k}{6}\)
Since the line passes through the points
∴ -2 = \(\frac{8 - k}{6}\)
⇒ \(\frac{-2}[1} = \frac{8 - k]{6}\)
⇒ 8 - k = -12
⇒ k = 8 + 12
∴ k = 20
Find the value of the angle marked x in the diagram above
60\(^0\)
45\(^0\)
90\(^0\)
30\(^0\)
Correct answer is A
\(PR^2 = PQ^2 + RQ^2 - 2(PQ)(RQ)cos Q\)
\(\implies cos Q = \frac{PQ^2 + RQ^2 - PR^2}{2(PQ)(RQ)}\)
\(\implies cos Q = \frac{8^2 + 5^2 - 7^2}{2\times8\times5}\)
\(\implies cos Q = \frac{64 + 25 - 49}{80}\)
\(\implies cos Q = \frac{40}{80} = 0.5\)
\(\implies Q = cos^{-1} (0.5) = 60^0\)
\(\therefore x = 60^0\)
\(^{-1}/_3\)
2
\(^{4}/_3\)
\(^{2}/_9\)
Correct answer is A
\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\)
\(T_n = ar^n - 1\)
∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i)
\(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii)
= 2a = 3(1 - r)
= 2a = 3 - 3r
∴ a = \(\frac{3 - 3r}{2}\)
Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i)
= \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\)
= \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\)
= 3(3r - 3r\(^2\)) = -4
= 9r - 9r\(^2\) = -4
= 9r\(^2\) - 9r - 4 = 0
= 9r\(^2\) - 12r + 3r - 4 = 0
= 3r(3r - 4) + 1(3r - 4) = 0
= (3r - 4)(3r + 1) = 0
∴ r = \(\frac{4}{3} or - \frac{1}{3}\)
For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.
∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)
15 cm
19 cm
13 cm
21 cm
Correct answer is C
Let the length of the longer side = \(x\) cm
∴ The length of the shorter side = (\(x\) - 6) cm
If we increase each side's length by 2 cm, it becomes
(\(x\) + 2) cm and (\(x\) - 4) cm respectively
Area of a rectangle = L x B
\(A_1 = x(x - 6) = x^2 - 6x\)
\(A_2 = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8\)
\(A_1 + 68 = A_2\) (Given)
⇒ \(x^2 - 6x + 68 = x^2 - 2x - 8\)
⇒ \(x^2 - x^2 - 6x + 2x\) = -8 - 68
⇒ -4\(x\) = -76
⇒ \(x\) = \(\frac{-76}{-4}\) = 19cm
∴ The length of the shorter side = \(x\) - 6 = 19 - 6 = 13 cm
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