2021 - JAMB Mathematics Past Questions and Answers - page 2

Interest, I = \(\frac{PRT}{100}\)

Hence, R = \(\frac{100 \times 1}{100 \times 5}\)

= \(\frac{100 \times 7.50}{500 \times 5}\)

= \(\frac{750}{500}\)

= \(\frac{3}{2}\)

= 1.5%

When the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

Next, when the second child takes \(\frac{3}{4}\) of the remainder

=> \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\)

= \(\frac{3}{4}\) x \(\frac{3}{4}\)

= \(\frac{9}{16}\)

Fraction remaining = \(\frac{3}{4}\) - \(\frac{9}{16}\)

= \(\frac{12 - 9}{16}\)

= \(\frac{3}{16}\)

\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\)

 = \(\frac{275 \times 64}{2500 \times 800}\)

= \(\frac{88}{10^4}\)

88 x 10-4 = 88 x 10-1 x 10-4

= 8.8 x 10\(^{-3}\)

\(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\)

= \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\)

= 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)

Let T = total profit.

The first brother receives \(\frac{1}{3}\) T

balance, 1 - \(\frac{1}{3}\)

= \(\frac{2}{3}\)T

The seconds brother receives the remnant: \(\frac{2}{3}\)

\(\frac{2}{3}\) x \(\frac{2}{3}\) x \(\frac{4}{9}\)

The third brother receives the left over: \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T

= \(\frac{2}{9}\)T

The third brother receives \(\frac{2}{9}\)T => N12000

If \(\frac{2}{9}\)T = N12, 000

T = \(\frac{12 000}{\frac{2}{9}}\) = N54, 000

PQ\(^2\) = (x2 - x1)\(^2\)  + (y2 - y1)\(^2\) 

= 12\(^2\)  + 5\(^2\) 
= 144 + 25
= 169

PQ = √169 = 13

PQ = Diameter = 2 * Radius, r

r = PQ/2 = 6.5

Length of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r

= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)

= \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm

Perimeter of sector = 8.8 + 2r

= 8.8 + 2(10.5)

= 8.8 + 21

= 29.8cm

For a rhombus,

• The diagonals bisect each other at right angles.
• The diagonals divide into four congruent right-angled triangles.

Hence,  6cm splits into 3cm each and 8cm to 4cm each

Hypotenuse(hyp) = Adjacent(adj) + Opposite(opp)

hyp\(^2\) = adj\(^2\) + opp\(^2\)

hyp\(^2\) = 3\(^2\) + 4\(^2\)

hyp\(^2\) = 25

hyp = 5

Length (L) = 5cm since a rhombus is a quadrilateral with 4 equal lengths

Interior angles of a regular polygon = \(\frac{(n - 2)180}{n}\)

140 = \(\frac{(n - 2)180}{n}\)

140n = 180n - 360

40n = 360

n = 9 sides

Volume of a cylinder = πr\(^2\)h

Convert 3m to cm by multiplying by 100

Volume of exterior = \(π \times 13^2 \times 300\)

Volume of interior = \(π \times 10^2 \times 300\)

Difference: volume of the external cylinder - volume of the internal cylinder

Total volume (v) = \(π (169 - 100) \times 300\)

V = \(π \times 69 \times 300\)

V = 20700πcm\(^3\)