Mathematics (Core), 1989, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1989

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

1 - 10 of 45 Questions

#	Question	Ans
1.	Which of the following is in descending order? A. \(\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{20}\) B. \(\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}\) C. \(\frac{9}{10} \frac{17}{20} \frac{4}{5} \frac{3}{4}\) D. \(\frac{4}{5} \frac{9}{10} \frac{17}{20} \frac{3}{4}\) Show Content Detailed Solution \(\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{20}\) = \(\frac{18, 16, 15, 17}{20}\) \(\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}\) = \(\frac{16, 18, 15, 17}{20}\) \(\frac{9}{10} \frac{17}{20} \frac{4}{5} \frac{3}{4}\) = \(\frac{18, 17, 16, 15}{20}\) ∴ \(\frac{9}{10}; \frac{17}{20}; \frac{4}{5}; \frac{3}{4}\) is in descending order.	C
2.	Evaluate 2700 000 x 0.03 ÷ 18 000 A. 4.5 x 10⁰ B. 4.5 x 10¹ C. 4.5 x 10² D. 4.5 x 10³ Show Content Detailed Solution \(2700000 \times 0.03 \div 18000\) = \(\frac{2700000 \times 0.03}{18000}\) = \(\frac{81000}{18000} = 4.5 \times 10^{0}\)	A
3.	The prime factors of 2520 are A. 2, 9, 5 B. 2, 9, 7 C. 2, 3, 5, 7 D. 2, 3, 7, 9 Show Content Detailed Solution \(2520 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7\) in terms of its prime factors. Hence, the prime factors of 2520 are 2, 3, 5, 7.	C
4.	If \(12_{e} = X_{7}\), where e = 12, find X. A. 20 B. 15 C. 14 D. 12 Show Content Detailed Solution \(12_{12} = 1 \times 12^{1} + 2 \times 12^{0} = 12 + 2 = 14_{10}\) \(\therefore 12_{e} = 20_{7}\)	A
5.	Simplify \(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\) A. \(\frac{r}{2}\) B. 2r C. \(\frac{1}{2r}\) D. \(\frac{2}{r}\) Show Content Detailed Solution \(\sqrt[3]{(64r^6)^{\frac{1}{2}}}\) = \(\sqrt[3]{((8r^{3})^{2})^{\frac{1}{2}}}\) =\(2r\)	B
6.	What is the different between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal? A. 10^-5 B. 7 x 10^-4 C. 8 x 10^-5 D. 10^-6 Show Content Detailed Solution 0.007685 = 0.00769 (three significant figures) 0.007685 = 0.0077(4d.p) the difference = 0.0077 - 0.00769 = 0.00001 = 1.0 x 10^-5	A
7.	If a : b = 5 : 8, x : y = 25 : 16; evaluate \(\frac{a}{x}\) : \(\frac{b}{y}\) A. 125 : 128 B. 3 : 5 C. 3 : 4 D. 2 : 5 Show Content Detailed Solution a : b = 5 : 8 = 2.5 : 40 x : y = 25 : 16 \(\frac{a}{x}\) : \(\frac{b}{y}\) = \(\frac{25}{25}\) : \(\frac{40}{16}\) = 1 : \(\frac{40}{16}\) = 16 : 40 = 2 : 5	D
8.	Oke deposited N800.00 in the bank at the rate of 12\(\frac{1}{2}\)% simple interest. After some time the total amount was one and half times the principal. For how many years was the money left in the bank? A. 2\(\frac{2}{3}\) B. 4 C. 5\(\frac{1}{3}\) D. 8 Show Content Detailed Solution P = N800 (Principal), r = 12\(\frac{1}{2}\)% or 0.125 After sometimes, A = 1.5 x 800 = N1,200 A = P(1 + Tr) = 1200 = 800(1 + 0.125T) = 1200 = 800(1 + 0.125T) 1200 = 800(1 + 0.125T) 0.125T = 0.5 T = \(\frac{0.5}{0.125}\) = 4	B
9.	If the surface area of a sphere increased by 44%, find the percentage increase in diameter A. 44 B. 30 C. 22 D. 20 Show Content Detailed Solution Surface Area of Sphere A = 4\(\pi r^2\) ∴ A = 4\(\pi\)\(\frac{(D)^2}{2}\) = \(\frac{(D)^2}{2}\) = \(\pi\)D² When increased by 44% A = \(\frac{144 \pi D^2}{100}\) \(\pi\)\(\frac{(12D)^2}{10}\) = \(\pi\)\(\frac{(6D)^2}{5}\) Increase in diameter = \(\frac{6D}{5}\) - D = \(\frac{1}{5}\)D Percentage increase = \(\frac{1}{5}\) x \(\frac{1}{100}\)% = 20%	D
10.	Simplify \(4 - \frac{1}{2 - \sqrt{3}}\) A. 2\(\sqrt{3}\) B. -2 - \(\sqrt{3}\) C. -2 + \(\sqrt{3}\) D. 2 - \(\sqrt{3}\) Show Content Detailed Solution \(4 - \frac{1}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3}) - 1}{2 - \sqrt{3}}\) = \(\frac{8 - 4\sqrt{3} - 1}{2 - \sqrt{3}}\) = \(\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}\) Rationalizing, \((\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}) (\frac{2 + \sqrt{3}}{2 + \sqrt{3}})\) = \(\frac{14 + 7\sqrt{3} - 8\sqrt{3} - 12}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}\) = \(\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}\)	D

1.	Which of the following is in descending order? A. \(\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{20}\) B. \(\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}\) C. \(\frac{9}{10} \frac{17}{20} \frac{4}{5} \frac{3}{4}\) D. \(\frac{4}{5} \frac{9}{10} \frac{17}{20} \frac{3}{4}\) Show Content Detailed Solution \(\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{20}\) = \(\frac{18, 16, 15, 17}{20}\) \(\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}\) = \(\frac{16, 18, 15, 17}{20}\) \(\frac{9}{10} \frac{17}{20} \frac{4}{5} \frac{3}{4}\) = \(\frac{18, 17, 16, 15}{20}\) ∴ \(\frac{9}{10}; \frac{17}{20}; \frac{4}{5}; \frac{3}{4}\) is in descending order.	C
2.	Evaluate 2700 000 x 0.03 ÷ 18 000 A. 4.5 x 10⁰ B. 4.5 x 10¹ C. 4.5 x 10² D. 4.5 x 10³ Show Content Detailed Solution \(2700000 \times 0.03 \div 18000\) = \(\frac{2700000 \times 0.03}{18000}\) = \(\frac{81000}{18000} = 4.5 \times 10^{0}\)	A
3.	The prime factors of 2520 are A. 2, 9, 5 B. 2, 9, 7 C. 2, 3, 5, 7 D. 2, 3, 7, 9 Show Content Detailed Solution \(2520 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7\) in terms of its prime factors. Hence, the prime factors of 2520 are 2, 3, 5, 7.	C
4.	If \(12_{e} = X_{7}\), where e = 12, find X. A. 20 B. 15 C. 14 D. 12 Show Content Detailed Solution \(12_{12} = 1 \times 12^{1} + 2 \times 12^{0} = 12 + 2 = 14_{10}\) \(\therefore 12_{e} = 20_{7}\)	A
5.	Simplify \(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\) A. \(\frac{r}{2}\) B. 2r C. \(\frac{1}{2r}\) D. \(\frac{2}{r}\) Show Content Detailed Solution \(\sqrt[3]{(64r^6)^{\frac{1}{2}}}\) = \(\sqrt[3]{((8r^{3})^{2})^{\frac{1}{2}}}\) =\(2r\)	B

6.	What is the different between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal? A. 10^-5 B. 7 x 10^-4 C. 8 x 10^-5 D. 10^-6 Show Content Detailed Solution 0.007685 = 0.00769 (three significant figures) 0.007685 = 0.0077(4d.p) the difference = 0.0077 - 0.00769 = 0.00001 = 1.0 x 10^-5	A
7.	If a : b = 5 : 8, x : y = 25 : 16; evaluate \(\frac{a}{x}\) : \(\frac{b}{y}\) A. 125 : 128 B. 3 : 5 C. 3 : 4 D. 2 : 5 Show Content Detailed Solution a : b = 5 : 8 = 2.5 : 40 x : y = 25 : 16 \(\frac{a}{x}\) : \(\frac{b}{y}\) = \(\frac{25}{25}\) : \(\frac{40}{16}\) = 1 : \(\frac{40}{16}\) = 16 : 40 = 2 : 5	D
8.	Oke deposited N800.00 in the bank at the rate of 12\(\frac{1}{2}\)% simple interest. After some time the total amount was one and half times the principal. For how many years was the money left in the bank? A. 2\(\frac{2}{3}\) B. 4 C. 5\(\frac{1}{3}\) D. 8 Show Content Detailed Solution P = N800 (Principal), r = 12\(\frac{1}{2}\)% or 0.125 After sometimes, A = 1.5 x 800 = N1,200 A = P(1 + Tr) = 1200 = 800(1 + 0.125T) = 1200 = 800(1 + 0.125T) 1200 = 800(1 + 0.125T) 0.125T = 0.5 T = \(\frac{0.5}{0.125}\) = 4	B
9.	If the surface area of a sphere increased by 44%, find the percentage increase in diameter A. 44 B. 30 C. 22 D. 20 Show Content Detailed Solution Surface Area of Sphere A = 4\(\pi r^2\) ∴ A = 4\(\pi\)\(\frac{(D)^2}{2}\) = \(\frac{(D)^2}{2}\) = \(\pi\)D² When increased by 44% A = \(\frac{144 \pi D^2}{100}\) \(\pi\)\(\frac{(12D)^2}{10}\) = \(\pi\)\(\frac{(6D)^2}{5}\) Increase in diameter = \(\frac{6D}{5}\) - D = \(\frac{1}{5}\)D Percentage increase = \(\frac{1}{5}\) x \(\frac{1}{100}\)% = 20%	D
10.	Simplify \(4 - \frac{1}{2 - \sqrt{3}}\) A. 2\(\sqrt{3}\) B. -2 - \(\sqrt{3}\) C. -2 + \(\sqrt{3}\) D. 2 - \(\sqrt{3}\) Show Content Detailed Solution \(4 - \frac{1}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3}) - 1}{2 - \sqrt{3}}\) = \(\frac{8 - 4\sqrt{3} - 1}{2 - \sqrt{3}}\) = \(\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}\) Rationalizing, \((\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}) (\frac{2 + \sqrt{3}}{2 + \sqrt{3}})\) = \(\frac{14 + 7\sqrt{3} - 8\sqrt{3} - 12}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}\) = \(\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}\)	D