Mathematics (Core), 2004, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2004

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 45 Questions

#	Question	Ans
31.	Factorize completely ac - 2bc - a + 4b² A. (a - 2b)(c - a - 2b) B. (a - 2b)(c + a +2b) C. (a - 2b)(c - a + 2b) D. (a - 2b)(c + a - 2b) Show Content Detailed Solution ac - 2bc - a² + 4b² ac - 2bc - (a² - 4b²) ac - 2bc - (a² -(2b)²) c(a - 2b) - {(a - 2b)(a + 2b)} (a - 2b)(c - (a + 2b)) (a - 2b)(c - a - 2b)	A
32.	Find the sum to infinity of the series 1/2 , 1/6, 1/18, ..... A. 2/3 B. 1/3 C. 3/4 D. 1 Show Content Detailed Solution \(a=\frac{1}{2}\\r=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{6}\times\frac{2}{1}=\frac{1}{3}\\S_\infty = \frac{a}{1-r}\\S_\infty = \frac{\frac{1}{2}}{1-\frac{1}{3}}\\=\frac{\frac{1}{2}}{\frac{3}{2}}\\=\frac{1}{2}\times\frac{3}{2}\\=\frac{3}{4}\)	C
33.	The length L of a simple pendulum varies directly as the square of its period T. If a pendulum with period 4 sec. is 64 cm long, find the length of pendulum whose period is 9 sec A. 96 cm B. 324 cm C. 36 cm D. 144 cm Show Content Detailed Solution \(L \propto T^2\) \(L = kT^2\) \(k = \frac{L}{T^2}\) when T = 4, L = 64. \(k = \frac{64}{4^2} = 4\) \(L = 4T^2\) when T = 9, \(L = 4 \times 9^2 = 324cm\)	B
34.	Three teachers shared a packet of chalk. The first teacher got 2/5 of the chalk and the second teacher received 2/15 of the remainder. What fraction the the third teacher receive? A. 8/15 B. 11/25 C. 12/25 D. 13/25 Show Content Detailed Solution First teacher received \(\frac{2}{5}\) ∴Remainder \(= 1 - \frac{2}{5} = \frac{3}{5}\) 2nd teacher received; \(\frac{2}{15} of \frac{3}{5} = \frac{2}{15} \times \frac{3}{5} = \frac{2}{25}\) 3rd teacher will receive; \(1-\frac{2}{5}-\frac{2}{25}=\frac{25-10-2}{15}\\=\frac{13}{25}\)	D
35.	If 6log_x2 - 3log_x3 = 3log₅0.2, find x. A. 8/3 B. 4/3 C. 3/4 D. 3/8 Show Content Detailed Solution 6log_x2 - 3log_x3 = 3log₅0.2 = logx²⁶ - 3logx³³ = log5^(0.2)3 = logx(64/27) = log5(1/5)³ logx(64/27) = log5(1/125) let logx(64/27) = y ∴x^y = 64/27 and log5(1/125) = y ∴ 5^y = 1/125 5^y = 125^-1 5^y = 5^-3 ∴ y = -3 substitute y = -3 in x^y = 64/27 implies x^-	C
36.	Find P, if 451₆ - P₇ = 305₆ A. 62₇ B. 116₇ C. 611₇ D. 142₇ Show Content Detailed Solution 451₆ - P₇ = 305₆ P₇ = 451₆ - 305₆ P₇ = 142₆ convert 142₆ = 1 * 6² + 4 * 6¹ + 2 * 6⁰ = 36 + 24 + 2 = 62 Convert 62₁₀ to base 7 62/7 = 8 R 6 8/7 = 1 R 1 1/7 = 0 R 1 ∴P₇ = 116₇	B
37.	Given that ³√4^2x = 16, find the value of x A. 4 B. 6 C. 3 D. 2 Show Content Detailed Solution ³√4^2x = 16 this implies that (³√4^2x)³ = (16)³ 4^2x = 4^2*3 4^2x = 4⁶ ∴ 2x = 6 x = 3	C
38.	Evaluate \(\frac{\frac{1}{10}\times\frac{2}{3}+\frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}}\) A. \(\frac{7}{12}\) B. \(\frac{19}{35}\) C. \(\frac{2}{25}\) D. \(\frac{19}{60}\) Show Content Detailed Solution \(\frac{\frac{1}{10}\times\frac{2}{3}+\frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}}\\Numerator \hspace{1mm}\frac{1}{10}\times\frac{2}{3}+\frac{1}{4} = \frac{1}{5}+\frac{1}{4}\\=\frac{4+15}{60}=\frac{19}{60}\\denominator\hspace{1mm}= \frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}=\frac{1}{2}\times\frac{5}{3}-\frac{1}{4}\\=\frac{5}{6}-\frac{1}{4}\\=\frac{10-3}{12}\\=\frac{7}{12}\\\frac{Numerator}{denominator}=\frac{\frac{19}{60}}{\frac{7}{12}}\\=\frac{19}{60}\times\frac{12}{7}=\frac{19}{35}\)	B
39.	The shaded region in the Venn diagram above is A. P^c ∪ (Q ∩ R) B. P^c ∩ (Q ∪ R) C. P ∩ Q D. P^c ∩ (Q ∩ R) Show Content Detailed Solution The shaded portion shows the elements in Q and R, but not in P, Q n R n P'	D
40.	Simplify \(\frac{1}{\sqrt{3}+2}\) in the form \(a+b\sqrt{3}\) A. 2 -√3 B. -2 - √3 C. 2 + √3 D. -2 + √3 Show Content Detailed Solution \(\frac{1}{\sqrt{3}+2}=\frac{1}{\sqrt{3}+2}\times \frac{\sqrt{3}-2}{\sqrt{3}-2}\\=\frac{\sqrt{3}-2}{(\sqrt{3})^{2} -2^{2}}\\=\frac{\sqrt{3}-2}{3-4}=\frac{\sqrt{3}-2}{1}\\=-\sqrt{3}+2\\=2-\sqrt{3}\)	A

31.	Factorize completely ac - 2bc - a + 4b² A. (a - 2b)(c - a - 2b) B. (a - 2b)(c + a +2b) C. (a - 2b)(c - a + 2b) D. (a - 2b)(c + a - 2b) Show Content Detailed Solution ac - 2bc - a² + 4b² ac - 2bc - (a² - 4b²) ac - 2bc - (a² -(2b)²) c(a - 2b) - {(a - 2b)(a + 2b)} (a - 2b)(c - (a + 2b)) (a - 2b)(c - a - 2b)	A
32.	Find the sum to infinity of the series 1/2 , 1/6, 1/18, ..... A. 2/3 B. 1/3 C. 3/4 D. 1 Show Content Detailed Solution \(a=\frac{1}{2}\\r=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{6}\times\frac{2}{1}=\frac{1}{3}\\S_\infty = \frac{a}{1-r}\\S_\infty = \frac{\frac{1}{2}}{1-\frac{1}{3}}\\=\frac{\frac{1}{2}}{\frac{3}{2}}\\=\frac{1}{2}\times\frac{3}{2}\\=\frac{3}{4}\)	C
33.	The length L of a simple pendulum varies directly as the square of its period T. If a pendulum with period 4 sec. is 64 cm long, find the length of pendulum whose period is 9 sec A. 96 cm B. 324 cm C. 36 cm D. 144 cm Show Content Detailed Solution \(L \propto T^2\) \(L = kT^2\) \(k = \frac{L}{T^2}\) when T = 4, L = 64. \(k = \frac{64}{4^2} = 4\) \(L = 4T^2\) when T = 9, \(L = 4 \times 9^2 = 324cm\)	B
34.	Three teachers shared a packet of chalk. The first teacher got 2/5 of the chalk and the second teacher received 2/15 of the remainder. What fraction the the third teacher receive? A. 8/15 B. 11/25 C. 12/25 D. 13/25 Show Content Detailed Solution First teacher received \(\frac{2}{5}\) ∴Remainder \(= 1 - \frac{2}{5} = \frac{3}{5}\) 2nd teacher received; \(\frac{2}{15} of \frac{3}{5} = \frac{2}{15} \times \frac{3}{5} = \frac{2}{25}\) 3rd teacher will receive; \(1-\frac{2}{5}-\frac{2}{25}=\frac{25-10-2}{15}\\=\frac{13}{25}\)	D
35.	If 6log_x2 - 3log_x3 = 3log₅0.2, find x. A. 8/3 B. 4/3 C. 3/4 D. 3/8 Show Content Detailed Solution 6log_x2 - 3log_x3 = 3log₅0.2 = logx²⁶ - 3logx³³ = log5^(0.2)3 = logx(64/27) = log5(1/5)³ logx(64/27) = log5(1/125) let logx(64/27) = y ∴x^y = 64/27 and log5(1/125) = y ∴ 5^y = 1/125 5^y = 125^-1 5^y = 5^-3 ∴ y = -3 substitute y = -3 in x^y = 64/27 implies x^-	C

36.	Find P, if 451₆ - P₇ = 305₆ A. 62₇ B. 116₇ C. 611₇ D. 142₇ Show Content Detailed Solution 451₆ - P₇ = 305₆ P₇ = 451₆ - 305₆ P₇ = 142₆ convert 142₆ = 1 * 6² + 4 * 6¹ + 2 * 6⁰ = 36 + 24 + 2 = 62 Convert 62₁₀ to base 7 62/7 = 8 R 6 8/7 = 1 R 1 1/7 = 0 R 1 ∴P₇ = 116₇	B
37.	Given that ³√4^2x = 16, find the value of x A. 4 B. 6 C. 3 D. 2 Show Content Detailed Solution ³√4^2x = 16 this implies that (³√4^2x)³ = (16)³ 4^2x = 4^2*3 4^2x = 4⁶ ∴ 2x = 6 x = 3	C
38.	Evaluate \(\frac{\frac{1}{10}\times\frac{2}{3}+\frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}}\) A. \(\frac{7}{12}\) B. \(\frac{19}{35}\) C. \(\frac{2}{25}\) D. \(\frac{19}{60}\) Show Content Detailed Solution \(\frac{\frac{1}{10}\times\frac{2}{3}+\frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}}\\Numerator \hspace{1mm}\frac{1}{10}\times\frac{2}{3}+\frac{1}{4} = \frac{1}{5}+\frac{1}{4}\\=\frac{4+15}{60}=\frac{19}{60}\\denominator\hspace{1mm}= \frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}=\frac{1}{2}\times\frac{5}{3}-\frac{1}{4}\\=\frac{5}{6}-\frac{1}{4}\\=\frac{10-3}{12}\\=\frac{7}{12}\\\frac{Numerator}{denominator}=\frac{\frac{19}{60}}{\frac{7}{12}}\\=\frac{19}{60}\times\frac{12}{7}=\frac{19}{35}\)	B
39.	The shaded region in the Venn diagram above is A. P^c ∪ (Q ∩ R) B. P^c ∩ (Q ∪ R) C. P ∩ Q D. P^c ∩ (Q ∩ R) Show Content Detailed Solution The shaded portion shows the elements in Q and R, but not in P, Q n R n P'	D
40.	Simplify \(\frac{1}{\sqrt{3}+2}\) in the form \(a+b\sqrt{3}\) A. 2 -√3 B. -2 - √3 C. 2 + √3 D. -2 + √3 Show Content Detailed Solution \(\frac{1}{\sqrt{3}+2}=\frac{1}{\sqrt{3}+2}\times \frac{\sqrt{3}-2}{\sqrt{3}-2}\\=\frac{\sqrt{3}-2}{(\sqrt{3})^{2} -2^{2}}\\=\frac{\sqrt{3}-2}{3-4}=\frac{\sqrt{3}-2}{1}\\=-\sqrt{3}+2\\=2-\sqrt{3}\)	A