2013 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

To convert to base 3, divide by 3 repeatedly until quotient becomes 0. From bottom up, read remainders: $2710 \div 3 = 903 r1$, $903 \div 3 = 301 r0$, $301 \div 3 = 100 r1$, $100 \div 3 = 33 r1$, $33 \div 3 = 11 r0$, $11 \div 3 = 3 r2$, $3 \div 3 = 1 r0$, $1 \div 3 = 0 r1$. Reading from bottom: $10110013_3$

Let’s solve this step by step: If 5 parts = 40 apples, then 1 part = 8 apples. Therefore, 2 parts (smallest share) = 16 apples. The ratio 5:3:2 means the shares are 40:24:16.

First multiply numerator: $1.25 \times 0.025 = 0.03125$. Then divide by 0.05: $0.03125 \div 0.05 = 0.625$. Rounding to 1 decimal place gives 0.6.

Using Simple Interest formula: $I = PRT/100$. Here $600 = 3000 \times 8 \times T/100$. Solving for T: $T = (600 \times 100)/(3000 \times 8) = 2.5$ years

Converting to same base: $\frac{3^{-5n}}{(3^2)^{1-n}} \times (3^3)^{n+1}$ = $3^{-5n} \times 3^{-2(1-n)} \times 3^{3(n+1)}$ = $3^{-5n-2+2n+3n+3}$ = $3^{-2n+1}$ = $3$

Using log rules: $\log_{10}\frac{4}{3} = \log_{10}4 – \log_{10}3$ = $0.6021 – 0.4014$ = $0.2007$

Simplify under the outer root: $\sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}$. $\sqrt{12}\sqrt{15} = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$. Therefore $\sqrt{7\sqrt{3} – 6\sqrt{5}} = 5$

Set P contains odd numbers {1,3,5,7,9,11,13,15,17,19}. Set Q contains prime numbers {2,3,5,7,11,13,17,19,23}. The intersection contains numbers that are both odd and prime: {3,5,7,11,13,17,19}

Let $S = \sqrt{t^2-4t+4}$. Square both sides: $S^2 = t^2-4t+4 = (t-2)^2$. Therefore $t-2 = \pm S$. Solving for t: $t = S + 2$

If x – 4 is a factor, then x = 4 should give zero when substituted into $x^2 – x – k$. $16 – 4 – k = 0$. Therefore $k = 12$

Using polynomial remainder theorem, substitute p = 3: $6(3^3) – 3^2 – 47(3) + 30$ = $162 – 9 – 141 + 30$ = $42$

Using $P = k\frac{mu}{q}$. Find k using initial values: $4 = k\frac{3 \times 2}{1}$, so $k = \frac{2}{3}$. Then substitute new values: $P = \frac{2}{3} \times \frac{6 \times 4}{\frac{8}{5}} = \frac{2}{3} \times \frac{6 \times 4 \times 5}{8} = 10$

If $r \propto \frac{1}{\sqrt{st}}$, then $r = \frac{k}{\sqrt{st}}$. Square both sides: $r^2 = \frac{k^2}{st}$. Therefore $s = \frac{k^2}{r^2t}$, meaning s varies directly as $r^2$ and inversely as t

Expand: $3x + 6 > 6x + 18$. Subtract 3x from both sides: $6 > 3x + 18$. Subtract 18 from both sides: $-12 > 3x$. Divide by 3: $-4 > x$ or $x < -4$

The solution is the set of all numbers whose distance from 2 is less than 3. This means $-3 < x - 2 < 3$. Adding 2: $-1 < x < 5$

The numerator forms an AP with d = 3 starting at 4: $a_n = 4 + 3(n-1)$. The denominator is n+1. Therefore nth term = $\frac{3n+1}{n+1}$

Work from inside out. First calculate 3 * 4 = 3 + 2(4) = 11. Then calculate 2 * 11 = 2 + 2(11) = 24

First multiply P by 2: $2P = \begin{pmatrix} 10 & 4 \ 6 & 2 \end{pmatrix}$. Then add Q: $2P + Q = \begin{pmatrix} 14 & 7 \ 8 & 7 \end{pmatrix}$

For 2×2 matrix, inverse = $\frac{1}{ad-bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix}$. Here determinant = 20-18 = 2. Therefore inverse = $\frac{1}{2}\begin{pmatrix} 4 & -6 \ -3 & 5 \end{pmatrix}$

Use product rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$. Here u = x and v = sin x. Therefore $\frac{dy}{dx} = x\cos x + \sin x$

Use chain rule: $\frac{dy}{dx} = 3(2x + 2)^2 \times \frac{d}{dx}(2x + 2) = 3(2x + 2)^2 \times 2 = 6(2x + 2)^2$

Area = $\pi r^2$. Rate of change = $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$ = $2\pi(7)(0.02)$ = $0.88\pi$ $\approx$ 0.88cm²s$^{-1}$

Sum terms and divide by 4: $\frac{(t + 2) + (2t – 4) + (3t + 2) + 2t}{4}$ = $\frac{8t + 0}{4}$ = $2t$

Let x be the seventh number. $\frac{2+4+8+14+16+18+x}{7} = 10$. Therefore x = 8. Numbers are now 2,4,8,8,14,16,18. Mode is 8

Total frequency = 15. Median position = 8th position. Cumulative frequencies: 3,8,9,10,12,15. 8th position falls in the 25 age group

Standard deviation is square root of variance. Given variance = 4, standard deviation = $\sqrt{4}$ = 2

Range = highest value – lowest value = 10 – 3 = 7

This is a combination problem: $C(5,2) = \frac{5!}{2!(5-2)!}$ = $\frac{5 \times 4}{2 \times 1}$ = 10

This is a permutation: $P(5,3) = \frac{5!}{(5-3)!}$ = $\frac{5!}{2!}$ = 60

Numbers divisible by both 2 and 3 (i.e., by 6) in range 1-25: 6,12,18,24. Total count = 4. Probability = $\frac{4}{25}$

Total fruits = 24. Probability of banana = $\frac{8}{24}$ = $\frac{1}{3}$

Based on the shape and position of the parabola, with x-intercepts and y-intercept visible, the quadratic equation is y = x² – x – 2

Based on the angle properties of circles, angle in a semi-circle is 90°. Using angle properties, x = 30°

Using angle properties of parallel lines and transversal, corresponding angles are equal. x = 110°

Adding the heights of all bars: 160 + 120 + 200 + 160 + 160 + 160 = 960 minutes

The sector for Mathematics is 90° out of 360°. Therefore fraction = $\frac{90}{360}$ = $\frac{1}{4}$. $\frac{1}{4}$ of 80 = 20 students

Sum of angles in quadrilateral = 360°. (3y + 10) + (2y + 30) + (y + 20) + 4y = 360. 10y + 60 = 360. y = 30

Convert to cm: 720 × 420. Area of floor = 302,400 cm². Area of tile = 900 cm². Number of tiles = 336

For 90° at center, chord = r$\sqrt{2}$ = 8$\sqrt{2}$ cm

Area of sector = $\frac{\theta}{360°} \times \pi r^2$. Here θ = 120°, r = $\frac{\sqrt{3}}{2}$. Area = $\frac{1}{3} \times \pi \times \frac{3}{4}$ = $\frac{\pi}{4}$ cm²