Mathematics questions and answers

Mathematics Questions and Answers

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.

51.

200 tickets were sold for a show. VIP tickets costs ₦1,200 and ₦700 for regular. Total amount realised from the sale of the tickets was ₦180,000. Find the number of VIP tickets sold and the the number of regular ticket sold.

A.

VIP = 80, Regular = 100

B.

VIP = 60, Regular = 120

C.

VIP = 60, Regular = 100

D.

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

52.

Give the number of significant figures of the population of a town which has approximately 5,020,700 people

A.

7 significant figures

B.

3 significant figures

C.

4 significant figures

D.

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.

53.

A circle has a radius of 13 cm with a chord 12 cm away from the centre of the circle. Calculate the length of the chord.

A.

16 cm

B.

8 cm

C.

5 cm

D.

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| = \(x  + x  = 5 + 5 = 10 cm\)

54.

Use the graph of sin (θ) above to estimate the value of θ when sin (θ) = -0.6 for \(0^o ≤ θ ≤ 360^o\)

A.

θ = 223\(^o\), 305\(^o\)

B.

θ = 210\(^o\), 330\(^o\)

C.

θ = 185\(^o\), 345\(^o\)

D.

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\(^o\))

∴ \(θ = 218^o, 323^o\)

55.

A ship sets sail from port A (86\(^o\)N, 56\(^o\)W) for port B (86\(^o\)N, 64\(^o\)W), which is close by. Find the distance the ship covered from port A to port B, correct to the nearest km.

[Take \(\pi\) = 3.142 and R = 6370 km]

A.

62 km

B.

97 km

C.

389 km

D.

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)