Mathematics (Core), 1982, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1982

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 48 Questions

#	Question	Ans
31.	A housewife bought 3 kilograms of garri at N13.00 per kg. She deposited N160. 00 for half a cow and bought 24 oranges at 10k each. She came back home with N20.60. She therefore left home with A. N220.00 B. N222.00 C. N201.40 D. N202.00 E. N180.80 Show Content Detailed Solution 3 kg of garri at N13.00 per kg = N39.00 half a cow = n160.00 24 oranges at 10k each = N2.40 balance = N20.60 Adding all together = N22.00	B
32.	The solution to the simultaneous equations 3x + 5y = 4, 4x + 3y = 5 is A. (\(\frac{-13}{11}, \frac{1}{11}\)) B. (\(\frac{13}{11}, \frac{1}{11}\)) C. (\(\frac{13}{11}, \frac{-1}{11}\)) D. (\(\frac{11}{13}, \frac{1}{11}\)) E. (13, 11) Show Content Detailed Solution 3x + 5y = 4, 4x + 3y = 5 3x + 5y = 4 x 4 4x + 3y = 5 x 3 12x + 20y = 16.....(i) 12x + 9y = 15.......(ii) subtract eqn.(ii) from eqn.(i) 11y = 1 y = \(\frac{1}{11}\) 12x + 20 x \(\frac{1}{11}\) = 16 12x = \(\frac{156}{11}\) x = \(\frac{13}{11}\) = \(\frac{13}{11}, \frac{1}{11}\)	B
33.	The area of the curved surface of the cone generated by the sector of a circle radius 6cm and are length 22cm is (\(\pi\) = \(\frac{22}{7}\)) A. 58 sq.cm B. 34 sq.cm C. 132 sq.cm D. 77 sq.cm E. 66 sq.cm Show Content Detailed Solution Given: length of the arc AOB = 22cm ; L = 6cm Curved surface area of cone = \(\pi\)rl Length of an arc = \(\frac{\theta}{360}\) x 2 \(\pi\)L = 22cm but length of an arc = circumference of the cone = 2\(\pi\) where r is the radius of the cone circle 2\(\pi\)r = 22, r = \(\frac{22}{2\pi}\)r = 11 x \(\frac{7}{22}\) = \(\frac{7}{2}\) curved surface area = \(\pi\)rl = \(\frac{22}{7}\) x \(\frac{7}{2}\) x 6 = 66 sq.cm	E
34.	The square base of a pyramid of side 3cm has height 8cm. If the pyramid is cut into two parts by a plane parallel to the base midway between the base and the vertex, the volumes of the two sections are A. (21cm³, 3cm³) B. (24cm³, 3cm³) C. (24cm³, 21cm³) D. (72cm³, 9cm³) E. (63cm³, 9cm³) Show Content Detailed Solution Vol. of the 1st section is side x height vol. = 3 x 8 = 24cm³ vol. of the second section is 3 x 1 = 3 = 24cm³, 3cm³	B
35.	Simplify f^{\(\frac{1}{2}\)}g²h^{\(\frac{1}{2}\)} \(\div\) f^{\(\frac{5}{2}\)}g^oh^{\(\frac{7}{3}\)} A. (\(\frac{g}{fh}\))² B. f²g²h² C. \(\frac{5}{4}\)g^oh\(\frac{7}{9}\) D. \(\frac{g^2}{f^5h^7}\) E. \(\frac{1}{f^2h^2}\) Show Content Detailed Solution f^{\(\frac{1}{2}\)}g²h^{\(\frac{1}{2}\)} \(\div\) f^{\(\frac{5}{2}\)}g^oh^{\(\frac{7}{3}\)} = f^{\(\frac{1}{2}\)} - \(\frac{5}{2}\) g^{2 - 0} h\(\frac{1}{2}\) - \(\frac{7}{3}\) f^-2 g² h^-2 = \(\frac{g^2}{f^2h^2}\) = (\(\frac{g}{fh}\))²	A
36.	Determine the mean monthly salary of 50 employees of a company from the following frequency distribution. \(\begin{array}{c\|c} \text{Monthly salary} & \text{Frequency}\\\hline N200.00 & 10\\ N325.00 & 5\\N100.00 & 20\\N120.00 & 2\\ N60.00 & 10\\ N80.00 & 3\end{array}\) A. N215.30 B. N134.10 C. N143.10 D. N80.00 E. N50.30 Show Content Detailed Solution \(\bar{x}\) = \(\frac{fx}{N}\) = \(\frac{6705}{50}\) = 1341 = N134.10	B
37.	If the function y = 5x is graphed, what would be its intercept on the y-axis? A. 5 B. \(\frac{1}{5}\) C. 1 D. 2 E. zero Show Content Detailed Solution y = 5x, when x = 0 y = 0, when x = 1 y = 5 intercept on y axis is 0	E
38.	A side of a rhombus is 2cm in length. An angle of the rhombus is 60^o. What is the length of the diagonal facing this angle? A. 2cm B. 8cm C. 4cm D. 2\(\sqrt{3}\)cm E. 16cm Show Content Detailed Solution \(\frac{x}{2}\) = sin60^o = cos30^o x = 2 sin^o = 2 x \(\frac{\sqrt{3}}{2}\) = \(\sqrt{3}\) length of the diagonal = 2 x \(\sqrt{2}\) = 2\(\sqrt{3}\)	D
39.	A variable y is inversely proportional to x², when y = 10, x = 2. What is y when x = 10? A. 2 B. 4 C. 100 D. 0.4 E. 0.1 Show Content Detailed Solution y \(\alpha\) \(\frac{1}{x^2}\) y = \(\frac{k}{x^2}\) k = x²y = (2)² x 10 = 40 y = \(\frac{40}{x^2}\) = \(\frac{40}{(10)^2}\) = \(\frac{40}{100}\) = 0.4	D
40.	Two distinct sectors i the same circle substend 100^o and 30^o respectively at the centre of the circle. Their corresponding arcs are in ratio A. 10:3 B. 1:100 C. 3:1 D. 5:2 Show Content Detailed Solution The distinct sectors i the same circle substend 100^o and 30^o respectively at the centre of the circle. Their corresponding arcs are in ratio 100:30 = 10:3	A

31.	A housewife bought 3 kilograms of garri at N13.00 per kg. She deposited N160. 00 for half a cow and bought 24 oranges at 10k each. She came back home with N20.60. She therefore left home with A. N220.00 B. N222.00 C. N201.40 D. N202.00 E. N180.80 Show Content Detailed Solution 3 kg of garri at N13.00 per kg = N39.00 half a cow = n160.00 24 oranges at 10k each = N2.40 balance = N20.60 Adding all together = N22.00	B
32.	The solution to the simultaneous equations 3x + 5y = 4, 4x + 3y = 5 is A. (\(\frac{-13}{11}, \frac{1}{11}\)) B. (\(\frac{13}{11}, \frac{1}{11}\)) C. (\(\frac{13}{11}, \frac{-1}{11}\)) D. (\(\frac{11}{13}, \frac{1}{11}\)) E. (13, 11) Show Content Detailed Solution 3x + 5y = 4, 4x + 3y = 5 3x + 5y = 4 x 4 4x + 3y = 5 x 3 12x + 20y = 16.....(i) 12x + 9y = 15.......(ii) subtract eqn.(ii) from eqn.(i) 11y = 1 y = \(\frac{1}{11}\) 12x + 20 x \(\frac{1}{11}\) = 16 12x = \(\frac{156}{11}\) x = \(\frac{13}{11}\) = \(\frac{13}{11}, \frac{1}{11}\)	B
33.	The area of the curved surface of the cone generated by the sector of a circle radius 6cm and are length 22cm is (\(\pi\) = \(\frac{22}{7}\)) A. 58 sq.cm B. 34 sq.cm C. 132 sq.cm D. 77 sq.cm E. 66 sq.cm Show Content Detailed Solution Given: length of the arc AOB = 22cm ; L = 6cm Curved surface area of cone = \(\pi\)rl Length of an arc = \(\frac{\theta}{360}\) x 2 \(\pi\)L = 22cm but length of an arc = circumference of the cone = 2\(\pi\) where r is the radius of the cone circle 2\(\pi\)r = 22, r = \(\frac{22}{2\pi}\)r = 11 x \(\frac{7}{22}\) = \(\frac{7}{2}\) curved surface area = \(\pi\)rl = \(\frac{22}{7}\) x \(\frac{7}{2}\) x 6 = 66 sq.cm	E
34.	The square base of a pyramid of side 3cm has height 8cm. If the pyramid is cut into two parts by a plane parallel to the base midway between the base and the vertex, the volumes of the two sections are A. (21cm³, 3cm³) B. (24cm³, 3cm³) C. (24cm³, 21cm³) D. (72cm³, 9cm³) E. (63cm³, 9cm³) Show Content Detailed Solution Vol. of the 1st section is side x height vol. = 3 x 8 = 24cm³ vol. of the second section is 3 x 1 = 3 = 24cm³, 3cm³	B
35.	Simplify f^{\(\frac{1}{2}\)}g²h^{\(\frac{1}{2}\)} \(\div\) f^{\(\frac{5}{2}\)}g^oh^{\(\frac{7}{3}\)} A. (\(\frac{g}{fh}\))² B. f²g²h² C. \(\frac{5}{4}\)g^oh\(\frac{7}{9}\) D. \(\frac{g^2}{f^5h^7}\) E. \(\frac{1}{f^2h^2}\) Show Content Detailed Solution f^{\(\frac{1}{2}\)}g²h^{\(\frac{1}{2}\)} \(\div\) f^{\(\frac{5}{2}\)}g^oh^{\(\frac{7}{3}\)} = f^{\(\frac{1}{2}\)} - \(\frac{5}{2}\) g^{2 - 0} h\(\frac{1}{2}\) - \(\frac{7}{3}\) f^-2 g² h^-2 = \(\frac{g^2}{f^2h^2}\) = (\(\frac{g}{fh}\))²	A

36.	Determine the mean monthly salary of 50 employees of a company from the following frequency distribution. \(\begin{array}{c\|c} \text{Monthly salary} & \text{Frequency}\\\hline N200.00 & 10\\ N325.00 & 5\\N100.00 & 20\\N120.00 & 2\\ N60.00 & 10\\ N80.00 & 3\end{array}\) A. N215.30 B. N134.10 C. N143.10 D. N80.00 E. N50.30 Show Content Detailed Solution \(\bar{x}\) = \(\frac{fx}{N}\) = \(\frac{6705}{50}\) = 1341 = N134.10	B
37.	If the function y = 5x is graphed, what would be its intercept on the y-axis? A. 5 B. \(\frac{1}{5}\) C. 1 D. 2 E. zero Show Content Detailed Solution y = 5x, when x = 0 y = 0, when x = 1 y = 5 intercept on y axis is 0	E
38.	A side of a rhombus is 2cm in length. An angle of the rhombus is 60^o. What is the length of the diagonal facing this angle? A. 2cm B. 8cm C. 4cm D. 2\(\sqrt{3}\)cm E. 16cm Show Content Detailed Solution \(\frac{x}{2}\) = sin60^o = cos30^o x = 2 sin^o = 2 x \(\frac{\sqrt{3}}{2}\) = \(\sqrt{3}\) length of the diagonal = 2 x \(\sqrt{2}\) = 2\(\sqrt{3}\)	D
39.	A variable y is inversely proportional to x², when y = 10, x = 2. What is y when x = 10? A. 2 B. 4 C. 100 D. 0.4 E. 0.1 Show Content Detailed Solution y \(\alpha\) \(\frac{1}{x^2}\) y = \(\frac{k}{x^2}\) k = x²y = (2)² x 10 = 40 y = \(\frac{40}{x^2}\) = \(\frac{40}{(10)^2}\) = \(\frac{40}{100}\) = 0.4	D
40.	Two distinct sectors i the same circle substend 100^o and 30^o respectively at the centre of the circle. Their corresponding arcs are in ratio A. 10:3 B. 1:100 C. 3:1 D. 5:2 Show Content Detailed Solution The distinct sectors i the same circle substend 100^o and 30^o respectively at the centre of the circle. Their corresponding arcs are in ratio 100:30 = 10:3	A