Solve \(1 + \sqrt[3]{ x - 3} = 4\)
30
6
12
66
Correct answer is A
\(1 + \sqrt [3]{x-3} = 4\)
= \(\sqrt[3]{x - 3} = 4 - 1\)
\(\sqrt[3]{x - 3} = 3 \)
take the cube of both sides
= x - 3 = 27
x = 27 + 3
∴ x = 30
In the diagram, O is the center of the circle QRS and ∠SQR = 28°. Find ∠ORS.
\(56^0\)
\(28^0\)
\(76^0\)
\(62^0\)
Correct answer is D
∠SOR = 2 × 28° = 56° (angle at the centre is twice the angle at the circumference)
From ∆SOR
|OS| = |OR| (radii)
So, ∆SOR is isosceles.
∠ORS = \(\frac{180^0 - 56^0}{2} = \frac{124^0}{2}\) ( base angles of isosceles triangle are equal)
∴ ∠ORS = 62°
6 days
8 days
10 days
12 days
Correct answer is A
12 men working 8 hours a day can complete the work in 4 days
12 men working 1 hour a day will complete the same work in (8 x 4) days [working less hours a day means they have to work for more days]
1 man working 1 hour a day will complete same work in (8 x 4 x 12) days [fewer number of men working means the remaining ones will have to work for more days ]
So, it will take 1 man working 1 hour a day to complete a piece of work in (8 x 4 x 12) days = 384 days
Now,
1 man working 16 hours a day will complete the piece of work in (384/16) days = 24 days [working more hours a day means less days to complete the work]
∴ 4 men working 16 hours a day will complete the piece of work in (24/4) days = 6 days [more men working means fewer days to complete the work]
ALTERNATIVELY
12 x 8 x 4 = 4 x 16 x d
⇒ d = \(\frac{12 \times 8 \times 4}{4 \times 16}\)
⇒ d = \(\frac{384}{64}\)
⇒ d = 6
∴ 4 men working 16 hours a day will complete the piece of work in 6 days
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
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