Mathematics (Core), 1988, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1988

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

11 - 20 of 49 Questions

#	Question	Ans
11.	The solutions of x2 - 2x - 1 = 0 are the points of intersection of two graphs. if one of the graphs is y = 2 + x - x2, find the second graph A. y = 1 - x B. y = 1 + x C. y = x - 1 D. y = 3x + 3 Show Content Detailed Solution The second graph is \((x^{2} - 2x - 1) + (2 + x - x^{2})\) = \(1 - x\)	A
12.	If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference A. 4 B. 8 C. 6\(\frac{2}{3}\) D. 9\(\frac{1}{3}\) Show Content Detailed Solution Let the first term and common difference = a & d respectively. \(T_{n} = \text{nth term} = a + (n - 1) d\) (A.P) Given: \(T_{4} = -6 \implies a + 3d = -6 ... (i)\) \(T_{8} + T_{9} = 72\) \(\implies a + 7d + a + 8d = 72 \implies 2a + 15d = 72 ... (ii)\) From (i), \(a = -6 - 3d\) \(\therefore\) (ii) becomes \(2(-6 - 3d) + 15d = 72\) \(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\) \(d = \frac{84}{9} = 9\frac{1}{3}\)	D
13.	If 7 and 189 are the first and fourth terms of geometric progression respectively, find the sum of the first three terms of the progression A. 182 B. 91 C. 63 D. 28 Show Content Detailed Solution \(T_{n} = ar^{n - 1}\) (nth term of a G.P) \(T_{4} = ar^{3} = 189\) \(7 \times r^{3} = 189 \implies r^{3} = 27\) \(r = \sqrt[3]{27} = 3\) \(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) \(S_{3} = \frac{7(3^{3} - 1)}{3 - 1} \) = \(\frac{7 \times 26}{2} = 91\)	B
14.	Find correct to one decimal place, 0.24633 \(\div\) 0.0306 A. 0.8 B. 1.8 C. 8.0 D. 8.1 Show Content Detailed Solution \(\frac{0.24633}{0.03060}\) multiplying throughout by 100,000 = \(\frac{24633}{3060}\) = 8.05 = 8.1	D
15.	In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12^o. How many students are offering Physics? A. 18 B. 15 C. 10 D. 5 Show Content Detailed Solution No of students offering Physics are \(\frac{12}{360}\) x 150 = 5	D
16.	\(\begin{array}{c\|c} Scores(x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency(f) & 7 & 11 & 6 & 7 & 7 & 5 & 3\end{array}\) In the distribution above, the mode and median respectively are A. 1,3 B. 1, 2 C. 3, 3 D. 0, 2 Show Content Detailed Solution From the distribution, Mode = 1 and Median = \(\frac{2 + 2}{2}\) = 2 = 1, 2	B
17.	If x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y A. 27 B. 30 C. 33 D. 90 Show Content Detailed Solution Prime numbers between 1 and 6 are 2, 3 and 5 x = 2 + 3 = 5 = 10 H.C.F. of 6, 9, 15 = 3 ∴ y = 3 X x y = 10 x 3 = 30	B
18.	Four interior angles of a pentagon are 90^o - x^o, 90^o + x^o, 110^o - 2x^o, 110^o + 2x^o. Find the fifth interior angle A. 110^o B. 120^o C. 130^o D. 140^o Show Content Detailed Solution Let the fifth interior angle be y: sum of interior angle of a pentagon = (2 x 5 - 4) x 90^o = 6 x 90^o = 540^o (90 - x) + (90 + x) + (110 - 2x) + (110 + 2x) + y = 540^o 400^o + y = 540^o y = 540 - 400^o y = 140^o	D
19.	For which of the following exterior angles is a regular polygon possible? i. 35° ii. 18° iii. 15° A. i and ii B. ii only C. ii and iii D. iii only Show Content Detailed Solution for a regular polygon to be possible, it must have all sides angles equal. \(\frac{360}{18}\) = 20 sides and \(\frac{360}{15}\) = 24 sides (ii) and (iii) are right	C
20.	Simplify \(\frac{1\frac{1}{2}}{2 \div \frac{1}{4} \text{of } 32}\) A. \(\frac{3}{256}\) B. \(\frac{3}{32}\) C. 6 D. 85 Show Content Detailed Solution \(\frac{1\frac{1}{2}}{2 \div \frac{1}{4} \text{ of }32}\) = \(\frac{\frac{3}{2}}{2 \div \frac{1}{4} \times 32}\) \(\frac{3}{2} \times 4 = 6\)	C

11.	The solutions of x2 - 2x - 1 = 0 are the points of intersection of two graphs. if one of the graphs is y = 2 + x - x2, find the second graph A. y = 1 - x B. y = 1 + x C. y = x - 1 D. y = 3x + 3 Show Content Detailed Solution The second graph is \((x^{2} - 2x - 1) + (2 + x - x^{2})\) = \(1 - x\)	A
12.	If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference A. 4 B. 8 C. 6\(\frac{2}{3}\) D. 9\(\frac{1}{3}\) Show Content Detailed Solution Let the first term and common difference = a & d respectively. \(T_{n} = \text{nth term} = a + (n - 1) d\) (A.P) Given: \(T_{4} = -6 \implies a + 3d = -6 ... (i)\) \(T_{8} + T_{9} = 72\) \(\implies a + 7d + a + 8d = 72 \implies 2a + 15d = 72 ... (ii)\) From (i), \(a = -6 - 3d\) \(\therefore\) (ii) becomes \(2(-6 - 3d) + 15d = 72\) \(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\) \(d = \frac{84}{9} = 9\frac{1}{3}\)	D
13.	If 7 and 189 are the first and fourth terms of geometric progression respectively, find the sum of the first three terms of the progression A. 182 B. 91 C. 63 D. 28 Show Content Detailed Solution \(T_{n} = ar^{n - 1}\) (nth term of a G.P) \(T_{4} = ar^{3} = 189\) \(7 \times r^{3} = 189 \implies r^{3} = 27\) \(r = \sqrt[3]{27} = 3\) \(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) \(S_{3} = \frac{7(3^{3} - 1)}{3 - 1} \) = \(\frac{7 \times 26}{2} = 91\)	B
14.	Find correct to one decimal place, 0.24633 \(\div\) 0.0306 A. 0.8 B. 1.8 C. 8.0 D. 8.1 Show Content Detailed Solution \(\frac{0.24633}{0.03060}\) multiplying throughout by 100,000 = \(\frac{24633}{3060}\) = 8.05 = 8.1	D
15.	In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12^o. How many students are offering Physics? A. 18 B. 15 C. 10 D. 5 Show Content Detailed Solution No of students offering Physics are \(\frac{12}{360}\) x 150 = 5	D

16.	\(\begin{array}{c\|c} Scores(x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency(f) & 7 & 11 & 6 & 7 & 7 & 5 & 3\end{array}\) In the distribution above, the mode and median respectively are A. 1,3 B. 1, 2 C. 3, 3 D. 0, 2 Show Content Detailed Solution From the distribution, Mode = 1 and Median = \(\frac{2 + 2}{2}\) = 2 = 1, 2	B
17.	If x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y A. 27 B. 30 C. 33 D. 90 Show Content Detailed Solution Prime numbers between 1 and 6 are 2, 3 and 5 x = 2 + 3 = 5 = 10 H.C.F. of 6, 9, 15 = 3 ∴ y = 3 X x y = 10 x 3 = 30	B
18.	Four interior angles of a pentagon are 90^o - x^o, 90^o + x^o, 110^o - 2x^o, 110^o + 2x^o. Find the fifth interior angle A. 110^o B. 120^o C. 130^o D. 140^o Show Content Detailed Solution Let the fifth interior angle be y: sum of interior angle of a pentagon = (2 x 5 - 4) x 90^o = 6 x 90^o = 540^o (90 - x) + (90 + x) + (110 - 2x) + (110 + 2x) + y = 540^o 400^o + y = 540^o y = 540 - 400^o y = 140^o	D
19.	For which of the following exterior angles is a regular polygon possible? i. 35° ii. 18° iii. 15° A. i and ii B. ii only C. ii and iii D. iii only Show Content Detailed Solution for a regular polygon to be possible, it must have all sides angles equal. \(\frac{360}{18}\) = 20 sides and \(\frac{360}{15}\) = 24 sides (ii) and (iii) are right	C
20.	Simplify \(\frac{1\frac{1}{2}}{2 \div \frac{1}{4} \text{of } 32}\) A. \(\frac{3}{256}\) B. \(\frac{3}{32}\) C. 6 D. 85 Show Content Detailed Solution \(\frac{1\frac{1}{2}}{2 \div \frac{1}{4} \text{ of }32}\) = \(\frac{\frac{3}{2}}{2 \div \frac{1}{4} \times 32}\) \(\frac{3}{2} \times 4 = 6\)	C