2019 - JAMB Mathematics Past Questions and Answers - page 4

\(p \propto \frac{1}{\sqrt{q}}\)

\(\implies p = \frac{k}{\sqrt{q}}\)

when p = 3, q = 16.

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

\(k = 3 \times 4 = 12\)

\(\therefore p = \frac{12}{\sqrt{q}}\)

at p = 4,

\(4 = \frac{12}{\sqrt{q}} \implies \sqrt{q} = \frac{12}{4}\)

\(\sqrt{q} = 3 \implies q = 3^2 \)

\(q = 9\)

For 200 mangoes @4 for N2.50

it \(\implies\) that the total cost price = \(\frac{200}{4} \times N 2.50\)

= N 125.00

Since there are 30 mangoes that got spoilt, \(\implies\) Left over = 200 - 30

= 170 mangoes 

170 mangoes at 2 for N 2.40

\(\implies\) Total selling point = \(\frac{170}{2} \times N 2.40\)

= N 204.00

Profit : N (204.00 - 125.00) = N 79.00

% profit = \(\frac{79}{125} \times 100%\)

profit = 63.2%.

 

the S.I = \(\frac{PRT}{100}\)

\(\implies\) N 4890 = \(\frac{8550 \times 3 \times x}{100}\)

\(x = \frac{4890 \times 100}{8550 \times 3}\)

\(x = 19.06%\)

\(x \approxeq 19%\)

81\(^{\frac{-3}{4}}\) x 25\(^{\frac{1}{2}}\) x 243\(^{\frac{2}{5}}\)

= \((\sqrt[4]{81})^{-3} \times \sqrt{25} \times (\sqrt[5]{243})^2\)

= \(\frac{5 \times 3^2}{3^{-3}}\)

= \(\frac{5}{3}\)

(\frac{(0.5436)^3}{0.017 \times 0.219}\)

= \(\frac{0.16063}{0.017 \times 0.219}\)

= 43.1 (to 3 s.f)

\(s = (4t + 3)(t - 2)\)

\(\frac{\mathrm d s}{\mathrm d t} = (4t + 3)(1) + (t - 2)(4)\)

= \(4t + 3 + 4t - 8\)

= 8t - 5

\(\frac{\mathrm d s}{\mathrm d t} (t = 5 secs) = 8(5) - 5\)

= 40 - 5 

= 35 units per second

given that there are 4 angles, it means that the polygon is a quadrilateral.

The sum of the angles in a quadrilateral = 360°

\(\therefore\) 2x + 5x + x + 4x = 360°

12x = 360°

x = 30°

Actual weight = 0.18g

Error = 0.21g - 0.18g = 0.03g

% error = \(\frac{0.03}{0.18} \times 100%\) = 16.7%

% error = 16.7%

\((\frac{6}{0.32} \div \frac{2}{0.084})^{-1}\)

= \((\frac{600}{32} \div \frac{2000}{84})^{-1}\)

= \((\frac{600}{32} \times \frac{84}{2000})^{-1}\)

= \((\frac{63}{80})^{-1}\)

= \(\frac{80}{63}\)

= 1.3 (to 1 decimal place)

2\(^{x + y}\) = 16 ; 4\(^{x - y}\) = \(\frac{1}{32}\).

\(\implies 2^{x + y} = 2^4\)

\(x + y = 4 ... (1)\)

\(2^{2(x - y)} = 2^{-5} \)

\(2^{2x - 2y} = 2^{-5}\)

\(\implies 2x - 2y = -5 ... (2)\)

Solving the equations (1) and (2) simultaneously gives:

x = \(\frac{3}{4}\) and y = \(\frac{13}{4}\)