2019 - JAMB Mathematics Past Questions and Answers - page 5
Simplify \(\frac{0.0839 \times 6.381}{5.44}\) to 2 significant figures.
Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30
x = 3 or 7, y = 12 or 8
x = 6 or 1, y = 11 or 5
x = 2 or 5, y = 15 or 6
x = 1 or 5, y = 10 or 7
3x + y = 21 ---- (i)
xy = 30 --------- (ii)
From (ii), \(y = \frac{30}{x}\).
Substitute the value of y in (i):
3x + \(\frac{30}{x}\) = 21
\(\implies\) 3x\(^2\) + 30 = 21x
3x\(^2\) - 21x + 30 = 0
3x\(^2\) - 15x - 6x + 30 = 0
3x(x - 5) - 6(x - 5) = 0
(3x - 6)(x - 5) = 0
3x - 6 = 0 \(\implies\) x = 2.
x - 5 = 0 \(\implies\) x = 5.
when x = 2, y = \(\frac{30}{2}\) = 15;
when x = 5, y = \(\frac{30}{5}\) = 6.
Points X and Y are 20km North and 9km East of point O, respectively. What is the bearing of Y from X? Correct to the nearest degree.
24°
56°
127°
156°
\(\tan \theta = \frac{9}{20} = 0.45\)
\(\theta = \tan^{-1} (0.45) \)
= 24.23°
\(\therefore\) The bearing of Y from X = 180° - 24.23°
= 155.77°
= 156° approximately
If \(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\), make T the subject of the formula.
\(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\)
\(P^3 = \frac{Q(R - T)}{15}\)
\(Q(R - T) = 15P^3\)
\(R - T = \frac{15P^3}{Q}\)
\(T = R - \frac{15P^3}{Q}\)
In the diagram above, O is the centre of the circle ABC, < ABO = 26° and < BOC = 130°. Calculate < AOC.
< BAC = \(\frac{130}{2}\) (angle subtended at the centre)
< BAC = 65°
Also, x = 26° (theorem)
y = 65° - 26° = 39°
< AOC = 180° - (39° + 39°)
= 102°
Each of the interior angles of a regular polygon is 140°. Calculate the sum of all the interior angles of the polygon.
1080°
1260°
2160°
1800°
Given that each interior angle = 140°,
each exterior angle = 180° - 140° = 40°
Number of sides of the polygon = \(\frac{360°}{40°}\) = 9
Sum of the angles in the polygon = 140° x 9
= 1260°
A man bought a car newly for ₦1,250,000. He had a crash with the car and later sold it at the rate of ₦1,085,000. What is the percentage gain or loss of the man?
43.7% loss
13.2% gain
13.2% loss
43.7% gain
The cost price of the car = N 1,250.00
The selling price of the car = N 1,085.00
Loss is the difference in the buying and selling price = N (1250 - 1085)
= N 165.00
% loss = \(\frac{165}{1250} \times 100%\)
= 13.2% loss
If the volume of a frustrum is given as \(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\), find \(\frac{\mathrm d V}{\mathrm d R}\).
\(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\)
\(V = \frac{\pi R^2 h}{3} + \frac{\pi Rr h}{3} + \frac{\pi r^2 h}{3}\)
\(\frac{\mathrm d V}{\mathrm d R} = \frac{2 \pi R h}{3} + \frac{\pi r h}{3}\)
= \(\frac{\pi}{3} (2R + r)\)
Express \((0.0439 \div 3.62)\) as a fraction.
\((0.0439 \div 3.62)\)
= 0.01213
\(\approxeq\) 0.012
= \(\frac{12}{1000}\)
If \(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\), find the value of x.
x = -4
x = 2
x = -2
x = 4
\(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\)
\((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\)
\(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\)
\(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\)
By equating the bases, we have
\(2 - 2x + x + 2 + 3x = -4\)
\(4 + 2x = -4 \implies 2x = -4 - 4\)
\(2x = -8\)
\(x = -4\)