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.
VIP = 80, Regular = 100
VIP = 60, Regular = 120
VIP = 60, Regular = 100
VIP = 80, Regular = 120
Correct answer is D
Let \(x\) = number of VIP tickets sold and
\(y\) = number of regular tickets sold
Total number of tickets sold = 200
⇒ \(x\) + \(y\) = 200 ---- (i)
If it costs ₦1,200 for a VIP ticket, then it costs ₦1200x for \(x\) number of VIP tickets sold and
If it costs ₦700 for a regular ticket, then it costs ₦700\(y\) for \(y\) number of VIP tickets sold
The total amount realised from the sale of tickets = ₦180,000
⇒ 1200\(x\) + 700\(y\) = 180000 ----- (ii)
From equation (i)
\(x\) = 200 - \(y\) ----- (iii)
Substitute (200 - \(y\)) for \(x\) in equation (ii)
⇒ 1200(200 - \(y\)) + 700\(y\) = 180000
⇒ 240000 - 1200\(y\) + 700\(y\) = 180000
⇒ 240000 - 500\(y\) = 180000
Collect like terms
⇒ 240000 - 180000 = 500\(y\)
⇒ 60000 = 500\(y\)
⇒ \(y = \frac{60000}{500} = 120\)
Substitute 120 for \(y\) in equation (iii)
⇒ \(x = 200 - 120\)
⇒ \(x = 80\)
∴ The total number of VIP tickets sold is 80 and regular is 120
16 cm
8 cm
5 cm
10 cm
Correct answer is D
|AP| = |PB| = \(x\) (The perpendicular to a chord bisects the chord if drawn from the center of the circle.)
From ∆OPB
Using Pythagoras theorem
⇒ \(13^2 = 12^2 + x^2\)
⇒ \(169 = 144 + x^2\)
⇒ \(169 - 144 = x^2\)
⇒ \(x^2 = 25\)
⇒ \(x = \sqrt25 = 5 cm\)
∴ Length of the chord |AB| = \(x + x = 5 + 5 = 10 cm\)
θ = 223\(^o\), 305\(^o\)
θ = 210\(^o\), 330\(^o\)
θ = 185\(^o\), 345\(^o\)
θ = 218\(^o\), 323\(^o\)
Correct answer is D
On the \(y\)-axis, each box is \(\frac{1 - 0}{5} = \frac{1}{5}\) = 0.2unit
On the \(x\)-axis, each box is \(\frac{90 - 0}{6} = \frac{90}{6} = 15^o\)
⇒ \(θ_1 = 180^o + (2.5\times15^o) = 180^o + 37.5^o = 217.5^o ≃ 218^o \)(2 and half boxes were counted to the right of 180\(^o\))
⇒ \(θ_2 = 270^o + (3.5\times15^o) = 270^o + 52.5^o = 322.5^o ≃ 323^o \)(3 and half boxes were counted to the right of 270\(^o\))
∴ \(θ = 218^o, 323^o\)
62 km
97 km
389 km
931 km
Correct answer is A
AB = \(\frac{θ}{360}\times 2\pi Rcos\propto\) (distance on small circle)
= 64 - 56 = 8\(^o\)
\(\propto = 86^o\)
⇒ AB = \(\frac{8}{360}\) x 2 x 3.142 x 6370 x cos 86
⇒ AB = \(\frac{22,338.29974}{360}\)
∴ AB = 62km (to the nearest km)
The perimeter of an isosceles right-angled triangle is 2 meters. Find the length of its longer side.
2 - \(\sqrt2\)
-4 + 3\(\sqrt2\)
It cannot be determined
-2 + 2\(\sqrt2\) m
Correct answer is D
Perimeter of a triangle = sum of all sides
⇒ \(P = y + x + x = 2\)
⇒ \(y + 2x = 2\)
⇒ \(y= 2 - 2x\)-----(i)
Using Pythagoras theorem
\(y^2 = x^2 + x^2\)
⇒ \(y^2 = 2x^2\)
⇒ \(y = \sqrt2x^2\)
⇒ \(y = x\sqrt2\)-----(ii)
Equate \(y\)
⇒ \(2 - 2x = x\sqrt2\)
Square both sides
⇒ \((2 -2x) ^2 = (x\sqrt2)^2\)
⇒ \(4 - 8x + 4x^2 = 2x^2\)
⇒ \(4 - 8x + 4x^2 - 2x^2 = 0\)
⇒ \(2x^2 - 8x + 4 = 0\)
⇒ \(x = \frac{-(-8)\pm\sqrt(-8)^2 - 4\times2\times4}{2\times2}\)
⇒ \(x = \frac{8\pm\sqrt32}{4}\)
⇒ \(x = \frac{8\pm4\sqrt2}{4}\)
⇒ \(x = 2\pm\sqrt2\)
⇒ \(x = 2 + \sqrt2\) or \(2 - \sqrt2\)
∴ \(x = 2 - \sqrt2\) (for \(x\) has to be less than its perimeter)
∴ \(y = 2 - 2x = 2 - 2(2 - \sqrt2) = -2 + 2 \sqrt2\)
∴ The length of the longer side = -2 + 2\(\sqrt2\)m
270 km
200 km
360 km
450 km
Correct answer is B
Speed = \(\frac{Distance}{Time}\)
⇒ Time = \(\frac{Distance}{Time}\)
Let D = distance between the two airports
∴ Time taken to get to the airport = \(\frac{D}{120}\) and Time taken to return =\( \frac{D}{150}\)
Since total time of flight= 3hours,
⇒ \(\frac{D}{120} + \frac{D}{150}\) = 3
⇒ \(\frac{15D + 12D}{1800}\) = 3
⇒ \(\frac{27D}{1800}\) = 3
⇒ \(\frac{3D}{200} = \frac{3}{1}\)
⇒ 3D = 200 x 3
∴ D =\(\frac{ 200\times3}{3}\)= 200km
PQRS is a cyclic quadrilateral. Find \(x\) + \(y\)
50
60
15
0
Correct answer is D
∠PQR + ∠PSR = 180o (opp. angles of cyclic quad. are supplementary)
⇒ 5\(x\) - \(y\) + 10 + (-2\(x\) + 3\(y\) + 145) = 180
⇒ 5\(x\) - \(y\) + 10 - 2\(x\) + 3\(y\) + 145 = 180
⇒ 3\(x\) + 2\(y\) + 155 = 180
⇒ 3\(x\) + 2\(y\) = 180 - 155
⇒ 3\(x\) + 2\(y\) = 25 ----- (i)
∠QPS + ∠QRS = 180o (opp. angles of cyclic quad. are supplementary)
⇒ -4\(x\) - 7\(y\) + 150 + (2\(x\) + 8\(y\) + 105) = 180
⇒ -4\(x\) - 7\(y\) + 75 + 2\(x\) + 8\(y\) + 180 = 180
⇒ -2\(x\) + \(y\) + 255 = 180
⇒ -2\(x\) + y = 180 - 255
⇒ -2\(x\) + \(y\) = -75 ------- (ii)
⇒ \(y\) = -75 + 2\(x\) -------- (iii)
Substitute (-75 + 2\(x\)) for \(y\) in equation (i)
⇒ 3\(x\) + 2(-75 + 2\(x\)) = 25
⇒ 3\(x\) - 150 + 4\(x\) = 25
⇒ 7\(x\) = 25 + 150
⇒ 7\(x\) = 175
⇒ \(x = \frac{175}{7} = 25\)
From equation (iii)
⇒ \(y\) = -75 + 2(25) = -75 + 50
⇒ \(y\) = -25
∴ \(x\) + \(y\) = 25 + (-25) = 0
10 cm and 15 cm
8 cm and 12 cm
6 cm and 9 cm
12 cm and 18 cm
Correct answer is A
Area of trapezium = \(\frac{1}{2}(a + b) h\)
⇒ \(\frac{1}{2} (a + b)\times 16 = 200\)
⇒ 8(a + b) = 200
⇒ a + b = \(\frac{200}{8}\) = 25 -----(i)
⇒ a : b = 2 : 3
⇒ \(\frac{a}{b} = \frac{2}{3}\)
⇒ 3a = 2b
⇒ a = \(\frac{2b}{3}\) -------(ii)
Substitute \(\frac{2b}{3}\) for a in equation (i)
⇒ \(\frac{2b}{3}\) + b = 25
\(\frac{5b}{3}\) = 25
⇒ b = 25 ÷ \(\frac{5}{3} = 25\times\frac{3}{5} = 15cm\)
From equation (ii)
⇒ a = \(\frac{2 \times 15}{3} = 2\times5 = 10cm\)
∴ Lengths of each parallel sides are 10cm and 15cm
Study the given histogram above and answer the question that follows.
What is the total number of students that scored at most 50 marks?
380
340
360
240
Correct answer is C
Total number of students that scored at most 50 marks = 100 + 80 + 60 + 40 + 80 = 360
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