Mathematics (Core), 1995, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1995

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 46 Questions

#	Question	Ans
31.	Two variables x and y are such that \(\frac{dy}{dx}\) = 4x - 3 and y = 5 when x = 2. Find y in terms of x A. 2x² - 3x + 5 B. 2x² - 3x + 3 C. 2x² - 3x D. 4 Show Content Detailed Solution ∫dy = ∫(4x - 3)dx, y = 2x² - 3x + C when y = 5, x = 2, C = 3 y = 2x² - 3x + 3	B
32.	Find the area bounded by the curve y = 3x\(^2\) - 2x + 1, the ordinates x = 1 and x = 3 and the x-axis. A. 24 B. 22 C. 21 D. 20 Show Content Detailed Solution \(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x + 1\) \(y = \int_{1} ^{3} (3x^{2} - 2x + 1) \mathrm d x\) \(y = [x^{3} - x^{2} + x]_{1} ^{3}\) = \([3^{3} - 3^{2} + 3] - [1^{3} - 1^{2} + 1]\) = \(21 - 1 = 20\)	D
33.	\(\begin{array}{c\|c} \text{Age in years} & 13 & 14 & 15 & 16 & 17 \\ \hline \text{No. of students} & 3 & 10 & 30 & 42 & 15\end{array}\) The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is A. 27^o B. 30^o C. 54^o D. 108^o Show Content Detailed Solution Number of 15- year old pupils = 30. Total number of students: 3 + 10 + 30 + 42 + 15 = 100. Angle = \(\frac{30}{100} \times 360°\) = \(108°\)	D
34.	The pie chart shows the distribution of students in a secondary school class. If 30 students offered French, how many offered C.R.K? A. 25 B. 15 C. 10 D. 8 Show Content Detailed Solution Total number of students (y) using the subject French ( 90 / 360 )º x Y = 30 Y = 30 x 4 Y = 120 History- 90º + French- 90º + Economics- 150º + CRK = 360º CRK = 360º - 330º = 30º Number that offered CRK: ( 30 / 360 )º x 120 = 10	C
35.	\(\begin{array}{c\|c} class& 1 - 3 & 4 - 6 & 7 - 9\\ \hline Frequency & 5 & 8 & 5\end{array}\) Find the standard deviation of the data using the table above A. 5 B. \(\sqrt{6}\0 C. \(\frac{5}{3}\) D. \(\sqrt{5}\) Show Content Detailed Solution \(\begin{array}{c\|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 10 & 10 & -3 & 9 & 90\\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \\ \hline & 18 & & 90 & & & 450 \end{array}\) x = \(\frac{\sum fx}{\sum f}\) = \(\frac{90}{18}\) = 5 S.D = \(\frac{\sum f(x - x)^2}{\sum f}\) = \(\frac{450}{18}\) = \(\sqrt{25}\) = 5	A
36.	The variance of the scores 1, 2, 3, 4, 5 is A. 1, 2 B. 1, 4 C. 2.0 D. 3.0 Show Content Detailed Solution Mean = \(\frac{15}{5} = 3\) \(Variance = \frac{\sum (x - \bar{x})^{2}}{n}\) = \(\frac{10}{5}\) = \(2.0\)	C
37.	\(\begin{array}{c\|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\) Find the mode of t A. 3.2 B. 3.3 C. 3.7 D. 4.2 Show Content Detailed Solution Mode = a + (b - a)(f_m - F_b) 2F_m - F_a - F_b = 3.0 + \(\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10}\) = 3 + \(\frac{(6.4)(11)}{30 - 14}\) = 3 + \(\frac{4.4}{16}\) = 3 + 0.275 = 3.275 = 3.3cm	B
38.	\(\begin{array}{c\|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\) The median of the A. 4.0 B. 3.4 C. 3.2 D. 3.0 Show Content Detailed Solution Median = \(\frac{a + b}{fm}\) (\(\frac{1}{2} \sum f - CF_b\)) = 2.95 + \(\frac{0.5}{15}\)(2.-7) = 2.95 + \(\frac{0.5}{15}\) x 13 = 2.95 + 0. 43 = 3.38 = 3.4	B
39.	Let p be a probability function on set S, where S = (a₁, a₂, a₃, a₄). Find P(a₁) if P(a₂) = \(\frac{1}{3}\), p(a₃) = \(\frac{1}{6}\) and p(a₄) = \(\frac{1}{5}\) A. \(\frac{7}{3}\) B. \(\frac{2}{3}\) C. \(\frac{1}{3}\) D. \(\frac{3}{10}\)	D
40.	A die has four of it's faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive time, the top face will be white in both cases? A. \(\frac{2}{3}\) B. \(\frac{1}{9}\) C. \(\frac{4}{9}\) D. \(\frac{1}{36}\) Show Content Detailed Solution \(\begin{array}{c\|c} & W & W & W & W & B & B \\ \hline W & WW & WW & WW & WW & WB & WB \\ W & WW & WW & WW & WW & WB & WB\\W & WW & WW & WW & WW & WB & WB\\ B & BW & BW & BW & BW & BB & BB \\ B & BW & BW & BW & BW & BB & BB\end{array}\) P(WW) = \(\frac{16}{36}\) = \(\frac{4}{9}\)	C

31.	Two variables x and y are such that \(\frac{dy}{dx}\) = 4x - 3 and y = 5 when x = 2. Find y in terms of x A. 2x² - 3x + 5 B. 2x² - 3x + 3 C. 2x² - 3x D. 4 Show Content Detailed Solution ∫dy = ∫(4x - 3)dx, y = 2x² - 3x + C when y = 5, x = 2, C = 3 y = 2x² - 3x + 3	B
32.	Find the area bounded by the curve y = 3x\(^2\) - 2x + 1, the ordinates x = 1 and x = 3 and the x-axis. A. 24 B. 22 C. 21 D. 20 Show Content Detailed Solution \(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x + 1\) \(y = \int_{1} ^{3} (3x^{2} - 2x + 1) \mathrm d x\) \(y = [x^{3} - x^{2} + x]_{1} ^{3}\) = \([3^{3} - 3^{2} + 3] - [1^{3} - 1^{2} + 1]\) = \(21 - 1 = 20\)	D
33.	\(\begin{array}{c\|c} \text{Age in years} & 13 & 14 & 15 & 16 & 17 \\ \hline \text{No. of students} & 3 & 10 & 30 & 42 & 15\end{array}\) The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is A. 27^o B. 30^o C. 54^o D. 108^o Show Content Detailed Solution Number of 15- year old pupils = 30. Total number of students: 3 + 10 + 30 + 42 + 15 = 100. Angle = \(\frac{30}{100} \times 360°\) = \(108°\)	D
34.	The pie chart shows the distribution of students in a secondary school class. If 30 students offered French, how many offered C.R.K? A. 25 B. 15 C. 10 D. 8 Show Content Detailed Solution Total number of students (y) using the subject French ( 90 / 360 )º x Y = 30 Y = 30 x 4 Y = 120 History- 90º + French- 90º + Economics- 150º + CRK = 360º CRK = 360º - 330º = 30º Number that offered CRK: ( 30 / 360 )º x 120 = 10	C
35.	\(\begin{array}{c\|c} class& 1 - 3 & 4 - 6 & 7 - 9\\ \hline Frequency & 5 & 8 & 5\end{array}\) Find the standard deviation of the data using the table above A. 5 B. \(\sqrt{6}\0 C. \(\frac{5}{3}\) D. \(\sqrt{5}\) Show Content Detailed Solution \(\begin{array}{c\|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 10 & 10 & -3 & 9 & 90\\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \\ \hline & 18 & & 90 & & & 450 \end{array}\) x = \(\frac{\sum fx}{\sum f}\) = \(\frac{90}{18}\) = 5 S.D = \(\frac{\sum f(x - x)^2}{\sum f}\) = \(\frac{450}{18}\) = \(\sqrt{25}\) = 5	A

36.	The variance of the scores 1, 2, 3, 4, 5 is A. 1, 2 B. 1, 4 C. 2.0 D. 3.0 Show Content Detailed Solution Mean = \(\frac{15}{5} = 3\) \(Variance = \frac{\sum (x - \bar{x})^{2}}{n}\) = \(\frac{10}{5}\) = \(2.0\)	C
37.	\(\begin{array}{c\|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\) Find the mode of t A. 3.2 B. 3.3 C. 3.7 D. 4.2 Show Content Detailed Solution Mode = a + (b - a)(f_m - F_b) 2F_m - F_a - F_b = 3.0 + \(\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10}\) = 3 + \(\frac{(6.4)(11)}{30 - 14}\) = 3 + \(\frac{4.4}{16}\) = 3 + 0.275 = 3.275 = 3.3cm	B
38.	\(\begin{array}{c\|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\) The median of the A. 4.0 B. 3.4 C. 3.2 D. 3.0 Show Content Detailed Solution Median = \(\frac{a + b}{fm}\) (\(\frac{1}{2} \sum f - CF_b\)) = 2.95 + \(\frac{0.5}{15}\)(2.-7) = 2.95 + \(\frac{0.5}{15}\) x 13 = 2.95 + 0. 43 = 3.38 = 3.4	B
39.	Let p be a probability function on set S, where S = (a₁, a₂, a₃, a₄). Find P(a₁) if P(a₂) = \(\frac{1}{3}\), p(a₃) = \(\frac{1}{6}\) and p(a₄) = \(\frac{1}{5}\) A. \(\frac{7}{3}\) B. \(\frac{2}{3}\) C. \(\frac{1}{3}\) D. \(\frac{3}{10}\)	D
40.	A die has four of it's faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive time, the top face will be white in both cases? A. \(\frac{2}{3}\) B. \(\frac{1}{9}\) C. \(\frac{4}{9}\) D. \(\frac{1}{36}\) Show Content Detailed Solution \(\begin{array}{c\|c} & W & W & W & W & B & B \\ \hline W & WW & WW & WW & WW & WB & WB \\ W & WW & WW & WW & WW & WB & WB\\W & WW & WW & WW & WW & WB & WB\\ B & BW & BW & BW & BW & BB & BB \\ B & BW & BW & BW & BW & BB & BB\end{array}\) P(WW) = \(\frac{16}{36}\) = \(\frac{4}{9}\)	C