Mathematics Past Questions

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Past Questions

2,359 questions
Question 1 of 20 2,359 total questions
Question 1051
jamb 1983

Find x if (x\(_4\))\(^2\) = 100100\(_2\)

A6
B12
C100
D210
E110
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Question 1052
jamb 1983
The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.
A47
B50
C48\(\frac{1}{2}\)
D48
E49
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Question 1053
jamb 1983

The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x

A2
B6
C18
D12
E5
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Question 1054
jamb 1983
Simplify \(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)
A\(\frac{x}{(x - 3)(x - 7)}\)
B\(\frac{x}{(x + 3)(x - 7)}\)
C\(\frac{x}{(x + 3)(x + 7)}\)
D\(\frac{x}{(x - 3)(x + 7)}\)
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Question 1055
jamb 1983
y varies partly as the square of x and partly as the inverse of the square root of x.Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4
Ay = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
By = x2 + \(\frac{1}{\sqrt{x}}\)
Cy = x2 + \(\frac{1}{x}\)
Dy = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\)
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Question 1056
jamb 1983

The value of (0.03)\(^3\) - (0.02)\(^3\) is

A0.019
B0.0019
C0.00019
D0.000019
E0.000035
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Question 1057
jamb 1983

Make T the subject of the equation \(\frac{av}{1 - v}\) = \(\sqrt[3]{\frac{2v + T}{a + 2T}}\)

AT = \(\frac{3av}{1 - v}\)
BT = \(\frac{1 + v}{2a^2v^3}\)
CT = \(\frac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 + (1 - v)^2}\)
D\(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
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Question 1058
jamb 1983

If (x + 2) and (x - 1) are factors of the expression \(Lx^{3} + 2kx^{2} + 24\), find the values of L and k.

A1 = -6, k = -9
B1 = -2, k = 1
Ck = -1, 1 = -2
D1 = 0, k = 1
Ek = 0,1 = 6
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Question 1059
jamb 1983

If 0.0000152 x 0.00042 = A x 10\(^{B}\), where 1 \(\leq\) A < 10, find A and B

AA = 9, B = 6.38
BA = 6.38, B = -9
CA = -6.38, B = -9
DA = -9, B = -6.38
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Question 1060
jamb 1983

Given that cos z = L , whrere z is an acute angle, find an expression for \(\frac{\cot z - \csc z}{\sec z + \tan z}\)

A\(\frac{1 - L}{1 + L}\)
B\(\frac{L^2 \sqrt{3}}{1 + L}\)
C\(\frac{1 + L^3}{L^2}\)
D\(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)
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Question 1061
jamb 1983

In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS

A2\(\sqrt{3}\)cm
B4\(\sqrt{6}\)cm
C2\(\sqrt{6}\)cm
D8\(\sqrt{6}\)cm
E8cm
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Question 1062
jamb 1983
Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12
Ax = 5, y = 2
Bx = 2, y = 5
Cx = 5, y = -2
Dx = -2, y = -5
Ex = -5, y = -2
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Question 1063
jamb 1983

If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?

A4:1 increase
B2:1 increase
C1:1 no change
D1:4 decrease
E3:4 decrease
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Question 1064
jamb 1983
Given a regular hexagon, calculate each interior angle of the hexagon
A60o
B30o
C120o
D45o
E135o
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Question 1065
jamb 1983
A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received N5,000 more than X. What is the total profit of the company for the year?
AN20, 000
BN25,000
CN30,000
DN15,000
EN45, 000
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Question 1066
jamb 1983

If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8, then (M, D) is

A(6, 5)
B(5, 8)
C(5, 7)
D(5, 5)
E(7, 5)
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Question 1067
jamb 1982

TP and TQ are tangents to a circle centre 0, and r is a point on the circumference of the circle as shown in the figure. If angle PTQ = 45º, what is the magnitude of the angle PRO?

A45\(^0\)
B135\(^0\)
C40\(^0\)
D67\(\frac{1}{2}\)\(^0\)
E52\(\frac{1}{2}\)\(^0\)
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Question 1068
jamb 1982

In the figure, QRST is a rectangle; PT// QM, angle P = 60o. Find angle MUR

A150o
B30o
C60o
D120o
E90o
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Question 1069
jamb 1982
PQRS is a parallelogram with area 50 square cm and the side PQ is 10cm long. T is a point on RS and TF is the altitude of the triangle TPQ. Find TF
A10cm
B12.5cm
C8cm
D6cm
E5cm
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Question 1070
jamb 1982
If K is a constant, which of the following equations best describes the parabola?
Ay = kx2
Bx = y2 - k
Cy = k - x2
Dx2 = y2 - k
Ey = (k - x)2
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