Mathematics (Core), 2006, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2006

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

21 - 30 of 50 Questions

#	Question	Ans
21.	Evaluate \(\frac{2}{6-5\sqrt{3}}\) A. \(-\left(\frac{12}{39}-\frac{10\sqrt{3}}{39}\right)\) B. \(\frac{12}{39}-\frac{10\sqrt{3}}{39}\) C. \(-\left(\frac{12}{39}+\frac{10\sqrt{3}}{39}\right)\) D. \(\frac{12}{39}+\frac{10\sqrt{3}}{39}\) Show Content Detailed Solution \(\frac{2}{6-5\sqrt{3}} = \frac{2}{6-5\sqrt{3}} \times \frac{6+5\sqrt{3}}{6+5\sqrt{3}}\\ =\frac{2(6+5\sqrt{3})}{(6-5\sqrt{3})(6+5\sqrt{3})}\\ =\frac{12+10-\sqrt{3}}{36-25(3)}\\ =\frac{12+10-\sqrt{3}}{36-75}\\ =\frac{12+10-\sqrt{3}}{-39}\\ =-\left(\frac{12}{39}-\frac{10\sqrt{3}}{39}\right)\)	A
22.	Compute 110011₂ + 11111₂ A. 1001010₂ B. 1010010₂ C. 1000110₂ D. 1000100₂ Show Content Detailed Solution 110011₂ + 11111₂ = 1010010₂	B
23.	Simplify \((25)^{\frac{-1}{2}} \times (27)^{\frac{1}{3}} + (121)^{\frac{-1}{2}} \times (625)^{\frac{-1}{4}}\) A. ³⁴/₅₅ B. ⁹/₁₁ C. ¹⁴/₅ D. ³/₂₇₅ Show Content Detailed Solution (25)\(^{\frac{-1}{2}}\) x (27)\(^{\frac{1}{3}}\) + (121)\(^{\frac{-1}{2}}\) x (625)\(^{\frac{-1}{4}}\) 5\(^{2 \times \frac{-1}{2}}\) x 3\(^{3 \times \frac{1}{3}}\) + 11\(^{2 \times \frac{-1}{2}}\) x 5\(^{4 \times \frac{-1}{4}}\) 5\(^{-1}\) x 3\(^1\) x 11\(^{-1}\) x 5\(^{-1}\) \(\frac{1}{5} \times \frac{3}{1} + \frac{1}{11} \times \frac{1}{5}\) \(\frac{3}{5} + \frac{1}{55} = \frac{33+1}{55}\) = \(\frac{34}{55}\)	A
24.	Convert 2232\(_4\) to base six A. 450₆ B. 254₆ C. 553₆ D. 540₆ Show Content Detailed Solution 1st convert to base 10 2232\(_4\) = 2 x 4\(^3\) + 2 x 4\(^2\) + 3 x 4\(^1\) + 2 x 4\(^0\) = 2 x 64 + 2 x 16 + 3 x 4 + 2 x 1 = 128 + 32 + 12 + 2 = 174 convert to base 6 6/174 6/29 R 0 6/4 R 5 6/0 R 4 = 450\(_6\)	A
25.	In the diagram above, QR is in the diameter of the semicircle QR. Find the area of the figure to the nearest whole number. [\(\pi = \frac{22}{7}\)] A. 89 cm² B. 70 cm² C. 90 cm² D. 80 cm² Show Content Detailed Solution Area of rectangle PQRS = \(10 \times 7 = 70cm^2\) Area of semi-circle: \(\frac{\pi r^{2}}{2}\) r = \(\frac{7}{2} cm\) Area of semi-circle = \(\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) = \(\frac{77}{4} = 18.75 cm^{2}\) Area of figure = \((70 + 18.75) cm^{2}\) = 88.75 cm\(^{2} \approxeq\) 89 cm\(^{2}\)	A
26.	If tan θ = ⁵/₄, find sin²θ - cos²θ. A. ⁵/₄ B. ⁴¹/₉ C. ⁹/₄₁ D. 1 Show Content Detailed Solution Tan θ = ⁵/₄ x² = 5² + 4² = 25 + 16 = 41 x = √41 sin²θ - cos²θ = (⁵/_√41)² - (⁴/_√41)² = ²⁵/₄₁ - ¹⁶/₄₁ =	C
27.	PQ and RS are two parallel lines. If the coordinates of P, Q, R, S are (1,q), (3,2), (3,4), (5,2q) respectively, find the value of q A. 3 B. 4 C. 1 D. 2 Show Content Detailed Solution Gradient PQ, P(1,q) and Q(3,2) \(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\) Gradient of RS : R(3,4) and S(5,2q) \(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\) = q-2 Since PQ and RS are parallel, their gradients are equal \(∴ \frac{(2-q)}{2} = q-2\) 2-q = 2(q-2) 2-q = 2q-4 2+4 = 2q+q 6 = 3q q = 2	D
28.	In triangle XYZ, ∠XYZ = 15^o, ∠XZY = 45^o and lXYl = 7 cm. Find lYZl. A. 14√2 cm B. \(7\left(\frac{\sqrt{6}}{2}\right)\) C. 7√2 cm D. 7 cm Show Content Detailed Solution ∠yxz = 180 - (45 + 15) = 180 - 60 = 120^o Using sine rule \(\frac{x}{sinx}=\frac{7}{sinz}\\ \frac{x}{sin 120}=\frac{7}{sin 45}\\ x=\frac{7 sin 120}{sin45}\\ x=\frac{7sin(180-120)}{sin 45}\\ x=\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x =\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x = 7\left(\frac{\sqrt{6}}{2}\right)\)	B
29.	In the diagram above, find the value of x A. 55^o B. 50^o C. 45^o D. 40^o Show Content Detailed Solution y = 55 (alternate angles) x = x (same reason) 260 + y + x = 360 (∠s at a point) 260 + 55 + x = 360 315 + x = 360 x = 360 - 315 x = 45^o	C
30.	In the diagram above, POQ is a diameter of the circle PQRS. If ∠PSR = 145°, find x° A. 55^o B. 45^o C. 35^o D. 25^o Show Content Detailed Solution ∠PSQ = 90° (∠ in semi circle) ∠PSQ + ∠QSR = 145° 90 + ∠QSR = 145 ∠QSR = 145 - 90 ∠QSR = 55° BUt x° = ∠QSR (∠s in the same segment) ∴x° = 55°	A

21.	Evaluate \(\frac{2}{6-5\sqrt{3}}\) A. \(-\left(\frac{12}{39}-\frac{10\sqrt{3}}{39}\right)\) B. \(\frac{12}{39}-\frac{10\sqrt{3}}{39}\) C. \(-\left(\frac{12}{39}+\frac{10\sqrt{3}}{39}\right)\) D. \(\frac{12}{39}+\frac{10\sqrt{3}}{39}\) Show Content Detailed Solution \(\frac{2}{6-5\sqrt{3}} = \frac{2}{6-5\sqrt{3}} \times \frac{6+5\sqrt{3}}{6+5\sqrt{3}}\\ =\frac{2(6+5\sqrt{3})}{(6-5\sqrt{3})(6+5\sqrt{3})}\\ =\frac{12+10-\sqrt{3}}{36-25(3)}\\ =\frac{12+10-\sqrt{3}}{36-75}\\ =\frac{12+10-\sqrt{3}}{-39}\\ =-\left(\frac{12}{39}-\frac{10\sqrt{3}}{39}\right)\)	A
22.	Compute 110011₂ + 11111₂ A. 1001010₂ B. 1010010₂ C. 1000110₂ D. 1000100₂ Show Content Detailed Solution 110011₂ + 11111₂ = 1010010₂	B
23.	Simplify \((25)^{\frac{-1}{2}} \times (27)^{\frac{1}{3}} + (121)^{\frac{-1}{2}} \times (625)^{\frac{-1}{4}}\) A. ³⁴/₅₅ B. ⁹/₁₁ C. ¹⁴/₅ D. ³/₂₇₅ Show Content Detailed Solution (25)\(^{\frac{-1}{2}}\) x (27)\(^{\frac{1}{3}}\) + (121)\(^{\frac{-1}{2}}\) x (625)\(^{\frac{-1}{4}}\) 5\(^{2 \times \frac{-1}{2}}\) x 3\(^{3 \times \frac{1}{3}}\) + 11\(^{2 \times \frac{-1}{2}}\) x 5\(^{4 \times \frac{-1}{4}}\) 5\(^{-1}\) x 3\(^1\) x 11\(^{-1}\) x 5\(^{-1}\) \(\frac{1}{5} \times \frac{3}{1} + \frac{1}{11} \times \frac{1}{5}\) \(\frac{3}{5} + \frac{1}{55} = \frac{33+1}{55}\) = \(\frac{34}{55}\)	A
24.	Convert 2232\(_4\) to base six A. 450₆ B. 254₆ C. 553₆ D. 540₆ Show Content Detailed Solution 1st convert to base 10 2232\(_4\) = 2 x 4\(^3\) + 2 x 4\(^2\) + 3 x 4\(^1\) + 2 x 4\(^0\) = 2 x 64 + 2 x 16 + 3 x 4 + 2 x 1 = 128 + 32 + 12 + 2 = 174 convert to base 6 6/174 6/29 R 0 6/4 R 5 6/0 R 4 = 450\(_6\)	A
25.	In the diagram above, QR is in the diameter of the semicircle QR. Find the area of the figure to the nearest whole number. [\(\pi = \frac{22}{7}\)] A. 89 cm² B. 70 cm² C. 90 cm² D. 80 cm² Show Content Detailed Solution Area of rectangle PQRS = \(10 \times 7 = 70cm^2\) Area of semi-circle: \(\frac{\pi r^{2}}{2}\) r = \(\frac{7}{2} cm\) Area of semi-circle = \(\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) = \(\frac{77}{4} = 18.75 cm^{2}\) Area of figure = \((70 + 18.75) cm^{2}\) = 88.75 cm\(^{2} \approxeq\) 89 cm\(^{2}\)	A

26.	If tan θ = ⁵/₄, find sin²θ - cos²θ. A. ⁵/₄ B. ⁴¹/₉ C. ⁹/₄₁ D. 1 Show Content Detailed Solution Tan θ = ⁵/₄ x² = 5² + 4² = 25 + 16 = 41 x = √41 sin²θ - cos²θ = (⁵/_√41)² - (⁴/_√41)² = ²⁵/₄₁ - ¹⁶/₄₁ =	C
27.	PQ and RS are two parallel lines. If the coordinates of P, Q, R, S are (1,q), (3,2), (3,4), (5,2q) respectively, find the value of q A. 3 B. 4 C. 1 D. 2 Show Content Detailed Solution Gradient PQ, P(1,q) and Q(3,2) \(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\) Gradient of RS : R(3,4) and S(5,2q) \(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\) = q-2 Since PQ and RS are parallel, their gradients are equal \(∴ \frac{(2-q)}{2} = q-2\) 2-q = 2(q-2) 2-q = 2q-4 2+4 = 2q+q 6 = 3q q = 2	D
28.	In triangle XYZ, ∠XYZ = 15^o, ∠XZY = 45^o and lXYl = 7 cm. Find lYZl. A. 14√2 cm B. \(7\left(\frac{\sqrt{6}}{2}\right)\) C. 7√2 cm D. 7 cm Show Content Detailed Solution ∠yxz = 180 - (45 + 15) = 180 - 60 = 120^o Using sine rule \(\frac{x}{sinx}=\frac{7}{sinz}\\ \frac{x}{sin 120}=\frac{7}{sin 45}\\ x=\frac{7 sin 120}{sin45}\\ x=\frac{7sin(180-120)}{sin 45}\\ x=\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x =\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x = 7\left(\frac{\sqrt{6}}{2}\right)\)	B
29.	In the diagram above, find the value of x A. 55^o B. 50^o C. 45^o D. 40^o Show Content Detailed Solution y = 55 (alternate angles) x = x (same reason) 260 + y + x = 360 (∠s at a point) 260 + 55 + x = 360 315 + x = 360 x = 360 - 315 x = 45^o	C
30.	In the diagram above, POQ is a diameter of the circle PQRS. If ∠PSR = 145°, find x° A. 55^o B. 45^o C. 35^o D. 25^o Show Content Detailed Solution ∠PSQ = 90° (∠ in semi circle) ∠PSQ + ∠QSR = 145° 90 + ∠QSR = 145 ∠QSR = 145 - 90 ∠QSR = 55° BUt x° = ∠QSR (∠s in the same segment) ∴x° = 55°	A