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
7 significant figures
3 significant figures
4 significant figures
5 significant figures
Correct answer is D
The two trailing zeros in the number are not significant, but the other five are, making it a five-figure number.
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)
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