Mathematics (Core), 1990, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1990

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

1 - 10 of 44 Questions

#	Question	Ans
1.	Simplify \(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) A. -7\(\frac{7}{8}\) B. \(\frac{-2}{7}\) C. \(\frac{-10}{21}\) D. \(\frac{10}{21}\) Show Content Detailed Solution \(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) \(\frac{19}{4}\) - \(\frac{25}{4}\)............(A) \(\frac{21}{5}\) x \(\frac{5}{4}\).............(B) Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\) A = \(\frac{19}{4}\) - \(\frac{25}{4}\) = \(\frac{-6}{4}\) B = \(\frac{21}{5}\) x \(\frac{5}{4}\) = \(\frac{105}{20}\) = \(\frac{21}{4}\) But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\) = \(\frac{-6}{4}\) x \(\f	B
2.	The H.C.F. of a²bx + ab²x and a²b - b² is A. b B. a + b C. b(a \(\div\) b) D. abx(a² - b²) Show Content Detailed Solution a²bx + ab²x; a²b - b² abx(a + b); b(a² - b²) b(a + b)(a + b) ∴ H.C.F. = (a + b)	B
3.	Correct 241.34(3 x 10-\(^3\))\(^2\) to 4 significant figures A. 0.0014 B. 0.001448 C. 0.0022 D. 0.002172 Show Content Detailed Solution first work out the expression and then correct the answer to 4 s.f = 241.34..............(A) (3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f)	D
4.	At what rate would a sum of N100.00 deposited for 5 years raise an interest of N7.50? A. \(\frac{1}{2}\)% B. 2\(\frac{1}{2}\)% C. 1.5% D. 25% Show Content Detailed Solution Interest I = \(\frac{PRT}{100}\) ∴ R = \(\frac{100 \times 1}{100 \times 5}\) = \(\frac{100 \times 7.50}{500 \times 5}\) = \(\frac{750}{500}\) = \(\frac{3}{2}\) = 1.5%	C
5.	Three children shared a basket of mangoes in such a way that the first child took \(\frac{1}{4}\) of the mangoes and the second \(\frac{3}{4}\) of the remainder. What fraction of the mangoes did the third child take? A. \(\frac{3}{16}\) B. \(\frac{7}{16}\) C. \(\frac{9}{16}\) D. \(\frac{13}{16}\) Show Content Detailed Solution You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket. If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Next, the second child takes \(\frac{3}{4}\) of the remainder which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\) = \(\frac{3}{4}\) x \(\frac{3}{4}\) = \(\frac{9}{16}\) the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\) = \(\frac{12 - 9}{16}\) = \(\frac{3}{16}\)	A
6.	Simplify and express in standard form \(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) A. 8.8 x 10^-1 B. 8.8 x 10^-2 C. 8.8 x 10^-3 D. 8.8 x 10³ Show Content Detailed Solution \(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\) = \(\frac{88}{10^4}\) 88 x 10-4 = 88 x 10-1 x 10-4 = 8.8 x 10-3	C
7.	Three brothers in a business deal share the profit at the end of a contact. The first received \(\frac{1}{3}\) of the profit and the second \(\frac{2}{3}\) of the remainder. If the third received the remaining N12000.00 how much profit did they share? A. N60 000.00 B. N54 000.00 C. N48 000.00 D. N42 000.00 Show Content Detailed Solution use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T remaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) x \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\)	B
8.	Simplify \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) A. 9r² B. 12\(\sqrt{3r}\) C. 13r D. \(\sqrt{13r}\) Show Content Detailed Solution \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) Simplifying from the innermost radical and progressing outwards we have the given expression \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) = \(\sqrt{160r^2 + \sqrt{71r^4 + 10r^4}}\) = \(\sqrt{160r^2 + \sqrt{81r^4}}\) \(\sqrt{160r^2 + 9r^2}\) = \(\sqrt{169r^2}\) = 13r	C
9.	Simplify \(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) A. 4\(\sqrt{3}\) B. \(\frac{4}{\sqrt{3}}\) C. 3\(\sqrt{3}\) D. \(\frac{\sqrt{3}}{4}\) Show Content Detailed Solution \(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) = \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)	A
10.	Simplify 3 log₆9 + log₆12 + log₆64 - log₆72 A. 5 B. 7776 C. log₆31 D. (7776)₆ Show Content Detailed Solution 3 log₆9 + log₆12 + log₆64 - log₆72 = log₆9³ + log₆12 + log₆64 - log₆72 log₆729 + log₆12 + log₆64 - log₆72 log₆(729 x 12 x 64) = log₆776 = log₆6₅ = 5 log₆6 = 5 N.B: log₆6 = 1	A

1.	Simplify \(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) A. -7\(\frac{7}{8}\) B. \(\frac{-2}{7}\) C. \(\frac{-10}{21}\) D. \(\frac{10}{21}\) Show Content Detailed Solution \(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) \(\frac{19}{4}\) - \(\frac{25}{4}\)............(A) \(\frac{21}{5}\) x \(\frac{5}{4}\).............(B) Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\) A = \(\frac{19}{4}\) - \(\frac{25}{4}\) = \(\frac{-6}{4}\) B = \(\frac{21}{5}\) x \(\frac{5}{4}\) = \(\frac{105}{20}\) = \(\frac{21}{4}\) But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\) = \(\frac{-6}{4}\) x \(\f	B
2.	The H.C.F. of a²bx + ab²x and a²b - b² is A. b B. a + b C. b(a \(\div\) b) D. abx(a² - b²) Show Content Detailed Solution a²bx + ab²x; a²b - b² abx(a + b); b(a² - b²) b(a + b)(a + b) ∴ H.C.F. = (a + b)	B
3.	Correct 241.34(3 x 10-\(^3\))\(^2\) to 4 significant figures A. 0.0014 B. 0.001448 C. 0.0022 D. 0.002172 Show Content Detailed Solution first work out the expression and then correct the answer to 4 s.f = 241.34..............(A) (3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f)	D
4.	At what rate would a sum of N100.00 deposited for 5 years raise an interest of N7.50? A. \(\frac{1}{2}\)% B. 2\(\frac{1}{2}\)% C. 1.5% D. 25% Show Content Detailed Solution Interest I = \(\frac{PRT}{100}\) ∴ R = \(\frac{100 \times 1}{100 \times 5}\) = \(\frac{100 \times 7.50}{500 \times 5}\) = \(\frac{750}{500}\) = \(\frac{3}{2}\) = 1.5%	C
5.	Three children shared a basket of mangoes in such a way that the first child took \(\frac{1}{4}\) of the mangoes and the second \(\frac{3}{4}\) of the remainder. What fraction of the mangoes did the third child take? A. \(\frac{3}{16}\) B. \(\frac{7}{16}\) C. \(\frac{9}{16}\) D. \(\frac{13}{16}\) Show Content Detailed Solution You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket. If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Next, the second child takes \(\frac{3}{4}\) of the remainder which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\) = \(\frac{3}{4}\) x \(\frac{3}{4}\) = \(\frac{9}{16}\) the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\) = \(\frac{12 - 9}{16}\) = \(\frac{3}{16}\)	A

6.	Simplify and express in standard form \(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) A. 8.8 x 10^-1 B. 8.8 x 10^-2 C. 8.8 x 10^-3 D. 8.8 x 10³ Show Content Detailed Solution \(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\) = \(\frac{88}{10^4}\) 88 x 10-4 = 88 x 10-1 x 10-4 = 8.8 x 10-3	C
7.	Three brothers in a business deal share the profit at the end of a contact. The first received \(\frac{1}{3}\) of the profit and the second \(\frac{2}{3}\) of the remainder. If the third received the remaining N12000.00 how much profit did they share? A. N60 000.00 B. N54 000.00 C. N48 000.00 D. N42 000.00 Show Content Detailed Solution use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T remaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) x \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\)	B
8.	Simplify \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) A. 9r² B. 12\(\sqrt{3r}\) C. 13r D. \(\sqrt{13r}\) Show Content Detailed Solution \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) Simplifying from the innermost radical and progressing outwards we have the given expression \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) = \(\sqrt{160r^2 + \sqrt{71r^4 + 10r^4}}\) = \(\sqrt{160r^2 + \sqrt{81r^4}}\) \(\sqrt{160r^2 + 9r^2}\) = \(\sqrt{169r^2}\) = 13r	C
9.	Simplify \(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) A. 4\(\sqrt{3}\) B. \(\frac{4}{\sqrt{3}}\) C. 3\(\sqrt{3}\) D. \(\frac{\sqrt{3}}{4}\) Show Content Detailed Solution \(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) = \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)	A
10.	Simplify 3 log₆9 + log₆12 + log₆64 - log₆72 A. 5 B. 7776 C. log₆31 D. (7776)₆ Show Content Detailed Solution 3 log₆9 + log₆12 + log₆64 - log₆72 = log₆9³ + log₆12 + log₆64 - log₆72 log₆729 + log₆12 + log₆64 - log₆72 log₆(729 x 12 x 64) = log₆776 = log₆6₅ = 5 log₆6 = 5 N.B: log₆6 = 1	A