2011 JAMB MATHEMATICS PAST QUESTIONS AND ANSWER

To solve this, take the 35th root of both sides after dividing by 2. $778/2 = 389$ Then $q = \sqrt[35]{389}$ This equals 2.

Multiply numerators together and denominators together, then simplify. This gives $\frac{3 \times 5 \times 2 \times 11 \times 3 \times 2}{2 \times 6 \times 3 \times 15 \times 4 \times 27} = \frac{5}{23}$

Using Simple Interest formula: $I = \frac{PRT}{100}$ where P=5000, R=4, T=9/12. $I = \frac{5000 \times 4 \times \frac{9}{12}}{100} = 150$

Let M=5k, N=4k, Q=3k. Then $\frac{2N-Q}{M} = \frac{8k-3k}{5k} = \frac{5k}{5k} = 1$

First, simplify the indices: $\left(\frac{16}{81}\right)^{\frac{1}{4}} = \frac{2}{3}$ and $\left(\frac{9}{16}\right)^{-\frac{1}{2}} = \frac{4}{3}$. Then $\frac{2}{3} \div \frac{4}{3} = \frac{2}{4} = \frac{1}{2}$

Using log laws: $\log_3 18 + \log_3 3 = \log_3 54$. Then $\log_3 54 – \log_3 x = 3$ Therefore $\log_3 \frac{54}{x} = 3$ so $\frac{54}{x} = 27$ giving x = 2

Multiply numerator and denominator by $3+\sqrt{5}$ to rationalize: $\frac{(2-\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})} = \frac{1-\sqrt{5}}{4}$

Using difference of squares formula and $\sqrt[3]{1}=1$: $(2-\sqrt{2}+1)(2-\sqrt{2}-1) = (3-\sqrt{2})(1-\sqrt{2}) = \frac{7}{3}$

Using the multiplication principle and ensuring at least one poster per room: Total ways = $3^7 – 3 \times 2^7 + 3$ = 210

Multiply both sides by 3: $3T = \frac{3K}{R^2}+M$. Then $3T-M = \frac{3K}{R^2}$ Therefore $R = \sqrt{\frac{3K}{3T-M}}$

Using polynomial division, the remainder will be of degree less than the divisor. Therefore, it will be of the form ax + b. The remainder is 2X – 1

This is a difference of squares: $9y^2 – 16x^2 = (3y+4x)(3y-4x)$

Using substitution method: From second equation, $x = -3y$. Substitute in first equation: $6y – 5y = 3$ giving $y = 3$. Therefore $x = -9$

Let $x = k\sqrt{y}$. When $x = 81$ and $y = 9$: $81 = k\sqrt{9}$ so $k = 27$. Therefore when $y = \frac{1}{79}$, $x = 27\sqrt{\frac{1}{79}} = \frac{27}{9} = 3$

Let $T = \frac{k}{R^3}$. When $R = 3$ and $T = \frac{2}{81}$: $\frac{2}{81} = \frac{k}{27}$ so $k = \frac{2}{3}$. Therefore when $R = 2$, $T = \frac{2}{3 \times 8} = \frac{1}{12}$

Expand: $-6x – 18 \leq 4x – 8$. Collect terms: $-10x \leq 10$ Therefore $x \geq -1$

Rearrange: $x^2 + 2x – 15 > 0$ Factor: $(x+5)(x-3) > 0$ Therefore $x < -5$ or $x > 3$

Using arithmetic sequence formula: $S_n = \frac{n}{2}(a_1 + a_n)$ where $n = 18$, $a_1 = 3$, $a_{18} = 54$. Therefore $S_{18} = \frac{18}{2}(3 + 54) = 513$

If $a_2 = 4$ and $a_4 = 16$, then $r^2 = 4$ so $r = 2$. Therefore $a_1 = 2$. Sum = $2 + 4 + 8 + 16 + 32 = 62$

Substitute into formula: $3 \oplus -\frac{2}{3} = 3(-\frac{2}{3}) + 3 + (-\frac{2}{3}) = -2 + 3 – \frac{2}{3} = \frac{1}{3}$

Expand both determinants: $(2)(3x) – (5)(3) = (4)(2x) – (3)(1)$ $6x – 15 = 8x – 3$ Therefore $x = 6$

Using expansion by first row: $4\begin{vmatrix} 3 & 1 \ -1 & 3 \end{vmatrix} – 2\begin{vmatrix} 2 & 1 \ -1 & 3 \end{vmatrix} + (-1)\begin{vmatrix} 2 & 3 \ -1 & -1 \end{vmatrix} = 25$

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

For n-sided polygon, each interior angle = $\frac{(n-2) \times 180°}{n}$. For n=12: $\frac{10 \times 180°}{12} = 150°$

Using Pythagorean theorem: If d is distance from center and r is radius, chord length = $2\sqrt{r^2-d^2}$ = $2\sqrt{49-25}$ = $2\sqrt{24}$ = $2\sqrt{4 \times 6}$ = $4\sqrt{6}$ cm

Tank volume = $3 \times 4 \times 5 = 60$ cm³. Cube volume = $3^3 = 27$ cm³. Remaining volume = $60 – 27 = 33$ cm³

The perpendicular bisector is the set of all points equidistant from X and Y. This is a fundamental property of perpendicular bisectors.

Using midpoint formula: $\frac{x+8}{2} = 5$ and $\frac{y+6}{2} = 8$. Therefore $x = 2$ and $y = 10$

Slope of given line = $\frac{5}{2}$. Perpendicular slope = $-\frac{2}{5}$. Using point-slope form: $y – 2 = -\frac{2}{5}(x – 4)$ gives $5y + 2x – 18 = 0$

Using Pythagorean theorem: $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$. Therefore $\cos \theta – \sin \theta = \frac{4}{5} – \frac{3}{5} = \frac{1}{5}$

From Z to X is South-East. Using tangent: $\tan \theta = 1$, so $\theta = 45°$. Therefore bearing = $180° + 45° = 225°$

Using product rule: $2(3x + 1) + 3(2x + 1) = 6x + 2 + 6x + 3 = 12x + 5$

The class with highest frequency is 6-8 with frequency 5. Modal class midpoint = 7

Differentiate: $\frac{dy}{dx} = 3x^2 + 2x – 1$. Set equal to zero and solve quadratic: $x = -1$ or $x = \frac{1}{3}$. Second derivative test confirms $x = -1$ is minimum

$\int_0^1 (3 – 2x)dx = [3x – x^2]_0^1 = (3 – 1) – (0 – 0) = 2$

Using double angle formula and integration rules: $\int \cos^4 x dx = \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C = -\frac{1}{4}\sin 4x + k$

Let n be least number. Then $n + (n+1) + (n+2) + (n+3) = 34$. Solve: $4n + 6 = 34$ gives $n = 7$

Arrange in order and count to middle (total frequency = 23). Median = 3. Range = 5 – 0 = 5

Using formula $\sigma = \sqrt{\frac{\sum f(x-\bar{x})^2}{N}}$ where $\bar{x} = 7$, we get $\sigma = \sqrt{6}$

Total number of letters = 7. Use permutation formula: $7!$