Mathematics (Core), 1997, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1997

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

21 - 30 of 45 Questions

#	Question	Ans
21.	Determine x + y if \(\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}\) \(\begin{pmatrix} x \\ y \end{pmatrix}\) = \(\begin{pmatrix}-1 \\ 8 \end{pmatrix}\) A. 3 B. 4 C. 7 D. 12 Show Content Detailed Solution \(\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix}\) \(\begin{pmatrix} 2x - 3y \\ -x + 4y \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix}\) \(2x - 3y = -1 ... (i)\) \(-x + 4y = 8 ... (ii)\) From (ii), x = 4y - 8. \(2(4y - 8) - 3y = -1 \implies 8y - 16 - 3y = -1\) \(5y = -1 + 16 = 15 \implies y = 3\) \(x = 4(3) - 8 = 12 - 8 = 4\) \(\therefore x + y = 3 + 4 = 7\)	C
22.	Find the non-zero positive value of x which satisfies the equation \(\begin{vmatrix} x & 1 & 0 \\ 1 & x & x \\ 0 & 1 & x\end{vmatrix}\) = 0 A. 2 B. 4 C. √3 D. √2 E. 1 Show Content Detailed Solution x(x² - 1) - x = 0 = x³ - 2x = 0 x(x² - 2) = 0 x = 2	A
23.	Each of the base angles of a isosceles triangle is 58° and the verticles of the triangle lie on a circle. Determine the angle which the base of the triangle subtends at the centre of the circle. A. 128^o B. 16^o C. 64^o D. 58^o Show Content Detailed Solution Base angle of the triangle = 58°. The third angle = 180° - (58° + 58°) = 64° Angle subtended at the centre = \(2 \times 64° = 128°\) (angle at the centre is twice the angle at the circumference).	A
24.	A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc A. 22cm B. 44cm C. 110cm D. 220cm Show Content Detailed Solution Diameter = 42cm Length of the arc = \(\frac{\theta}{360°} \times \pi d\) = \(\frac{60}{360} \times \frac{22}{7} \times 42cm\) = \(22cm\)	A
25.	An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.(\(\pi\) = \(\frac{22}{7}\)) A. 22cm² B. 44cm² C. 66cm² D. 88cm² Show Content Detailed Solution Area of a sector = \(\frac{\theta}{360°} \times \pi r^{2}\) r = 6cm; \(\theta\) = 70°. Area of the sector = \(\frac{70}{360} \times \frac{22}{7} \times 6^{2}\) = \(22 cm^{2}\)	A
26.	A cone with the sector angle of 45° is cut out of a circle of radius r of the cone. A. \(\frac{r}{16}\) cm B. \(\frac{r}{6}\) cm C. \(\frac{r}{8}\) cm D. \(\frac{r}{2}\) cm Show Content Detailed Solution The formula for the base radius of a cone formed from the sector of a circle = \(\frac{r \theta}{360°}\) = \(\frac{r \times 45°}{360°}\) = \(\frac{r}{8} cm\)	C
27.	The angle between the positive horizontal axis and a given line is 135^o. Find the equation of the line if it passes through the point (2,3) A. x - y = 1 B. x + y = 1 C. x + y = 5 D. x - y = 5 Show Content Detailed Solution m = tan 135^o = -tan 45^o = -1 \(\frac{y - y_1}{x - x_1}\) = m \(\frac{y - 3}{x - 2}\) = -1 = y - 3 = -(x - 2) = -x + 2 x + y = 5	C
28.	A point P moves so that its equidistant from point L and M. If LM is16cm, find the distance of P from LM when P is 10cm from L A. 12cm B. 10cm C. 8cm D. 6cm Show Content Detailed Solution p from LM = \(\sqrt{10^2 - 8^2}\) = \(\sqrt{36}\) = 6cm	D
29.	Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2 A. 3√10 B. 3√5 C. √26 D. √13 Show Content Detailed Solution 2x - y .....(i) x + y.....(ii) from (i) y = 2x - 4 from (ii) y = -x + 2 2x - 4 = -x + 2 x = 2 y = -x + 2 = -2 + 2 = 0 x₁ = 2₁ y₄ = 0₁ x₂ = 4₁ y₂ = 3 Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\) = \(\sqrt{(3 - 0)^2}\) + (4 - 2)² = \(\	D
30.	The angle of elevation of a building from a measuring instrument placed on the ground is 30^o. If the building is 40m high, how far is the instrument from the foot of the building? A. \(\frac{20}{√3}\)m B. \(\frac{40}{√3}\)m C. 20√3m D. 40√3m Show Content Detailed Solution \(\frac{40}{x}\) = tan 30^o x = \(\frac{40}{tan 36}\) = \(\frac{40}{1\sqrt{3}}\) = 40√3m	D

21.	Determine x + y if \(\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}\) \(\begin{pmatrix} x \\ y \end{pmatrix}\) = \(\begin{pmatrix}-1 \\ 8 \end{pmatrix}\) A. 3 B. 4 C. 7 D. 12 Show Content Detailed Solution \(\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix}\) \(\begin{pmatrix} 2x - 3y \\ -x + 4y \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix}\) \(2x - 3y = -1 ... (i)\) \(-x + 4y = 8 ... (ii)\) From (ii), x = 4y - 8. \(2(4y - 8) - 3y = -1 \implies 8y - 16 - 3y = -1\) \(5y = -1 + 16 = 15 \implies y = 3\) \(x = 4(3) - 8 = 12 - 8 = 4\) \(\therefore x + y = 3 + 4 = 7\)	C
22.	Find the non-zero positive value of x which satisfies the equation \(\begin{vmatrix} x & 1 & 0 \\ 1 & x & x \\ 0 & 1 & x\end{vmatrix}\) = 0 A. 2 B. 4 C. √3 D. √2 E. 1 Show Content Detailed Solution x(x² - 1) - x = 0 = x³ - 2x = 0 x(x² - 2) = 0 x = 2	A
23.	Each of the base angles of a isosceles triangle is 58° and the verticles of the triangle lie on a circle. Determine the angle which the base of the triangle subtends at the centre of the circle. A. 128^o B. 16^o C. 64^o D. 58^o Show Content Detailed Solution Base angle of the triangle = 58°. The third angle = 180° - (58° + 58°) = 64° Angle subtended at the centre = \(2 \times 64° = 128°\) (angle at the centre is twice the angle at the circumference).	A
24.	A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc A. 22cm B. 44cm C. 110cm D. 220cm Show Content Detailed Solution Diameter = 42cm Length of the arc = \(\frac{\theta}{360°} \times \pi d\) = \(\frac{60}{360} \times \frac{22}{7} \times 42cm\) = \(22cm\)	A
25.	An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.(\(\pi\) = \(\frac{22}{7}\)) A. 22cm² B. 44cm² C. 66cm² D. 88cm² Show Content Detailed Solution Area of a sector = \(\frac{\theta}{360°} \times \pi r^{2}\) r = 6cm; \(\theta\) = 70°. Area of the sector = \(\frac{70}{360} \times \frac{22}{7} \times 6^{2}\) = \(22 cm^{2}\)	A

26.	A cone with the sector angle of 45° is cut out of a circle of radius r of the cone. A. \(\frac{r}{16}\) cm B. \(\frac{r}{6}\) cm C. \(\frac{r}{8}\) cm D. \(\frac{r}{2}\) cm Show Content Detailed Solution The formula for the base radius of a cone formed from the sector of a circle = \(\frac{r \theta}{360°}\) = \(\frac{r \times 45°}{360°}\) = \(\frac{r}{8} cm\)	C
27.	The angle between the positive horizontal axis and a given line is 135^o. Find the equation of the line if it passes through the point (2,3) A. x - y = 1 B. x + y = 1 C. x + y = 5 D. x - y = 5 Show Content Detailed Solution m = tan 135^o = -tan 45^o = -1 \(\frac{y - y_1}{x - x_1}\) = m \(\frac{y - 3}{x - 2}\) = -1 = y - 3 = -(x - 2) = -x + 2 x + y = 5	C
28.	A point P moves so that its equidistant from point L and M. If LM is16cm, find the distance of P from LM when P is 10cm from L A. 12cm B. 10cm C. 8cm D. 6cm Show Content Detailed Solution p from LM = \(\sqrt{10^2 - 8^2}\) = \(\sqrt{36}\) = 6cm	D
29.	Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2 A. 3√10 B. 3√5 C. √26 D. √13 Show Content Detailed Solution 2x - y .....(i) x + y.....(ii) from (i) y = 2x - 4 from (ii) y = -x + 2 2x - 4 = -x + 2 x = 2 y = -x + 2 = -2 + 2 = 0 x₁ = 2₁ y₄ = 0₁ x₂ = 4₁ y₂ = 3 Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\) = \(\sqrt{(3 - 0)^2}\) + (4 - 2)² = \(\	D
30.	The angle of elevation of a building from a measuring instrument placed on the ground is 30^o. If the building is 40m high, how far is the instrument from the foot of the building? A. \(\frac{20}{√3}\)m B. \(\frac{40}{√3}\)m C. 20√3m D. 40√3m Show Content Detailed Solution \(\frac{40}{x}\) = tan 30^o x = \(\frac{40}{tan 36}\) = \(\frac{40}{1\sqrt{3}}\) = 40√3m	D