Mathematics (Core), 1991, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1991

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

41 - 48 of 48 Questions

#	Question	Ans
41.	A crate of soft drinks contains 10 bottles of Coca-cola, 8 of Fanta and 6 of sprite. If one bottle is selected at random, what is the probability that it is NOT a Cocacola bottle? A. \(\frac{5}{12}\) B. \(\frac{1}{3}\) C. \(\frac{3}{4}\) D. \(\frac{7}{12}\) Show Content Detailed Solution Coca-cola = 10 bottles, Fanta = 8 bottles Sprite = 6 bottles, Total = 24 P(cola-cola) = \(\frac{10}{24}\) P(not coca-cola) = 1 - \(\frac{10}{24}\) \(\frac{24 - 10}{24}\) = \(\frac{14}{24}\) = \(\frac{7}{12}\)	D
42.	In the figure above, Find the value of x A. 130^o B. 110^o C. 100^o D. 90^o	A
43.	PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PM = 5cm, PN = 12cm and PQ = 4.8cm, calculate the respective lengths of PR and PT in centimeters A. 7.3, 5.9 B. 7.7, 12.5 C. 12.5, 7.7 D. 5.9, 7.3 Show Content Detailed Solution \(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR}\) \(\frac{4.8}{12} = \frac{5}{PR}\) PR = \(\frac{5 \times 12}{4.8} = \frac{50}{4}\) = 12.5 \(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT}\) \(\frac{PT}{12} = \frac{5}{PR}\) PT² = 60 PT = \(\sqrt{60}\) = 7.746 = 7.7	C
44.	PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PNR = 110^o and PMQ = 55^o, find MPQ A. 40^o B. 30^o C. 25^o D. 15^o Show Content Detailed Solution PQM = 110^o(Ext. < of a cyclic quad) MPQ = 180^o - (110 + 55) = 180^o - 165^o = 15^o	D
45.	In the figure above, find the value of y A. 28^o B. 122^o C. 150^o D. 152^o Show Content Detailed Solution ADB = 180^o - 152^o = 28^o 28^o + y + 30^o = 180^o y = 180^o - 58^o = 122^o	B
46.	In this figure, PQ = PR = PS and SRT = 68^o. Find QPS A. 136^o B. 124^o C. 12^o D. 68^o Show Content Detailed Solution Since PQRS is quadrilateral 2y + 2x = QPS = 360^o i.e. 2(y + x) + QPS = 360^o QPS = 360^o - 2(y + x) But x + y + 68^o = 180^o x + y = 180^o - 68^o = 180^o x + y = 180^o - 68^o = 112^o	A
47.	In the figure, PQRS is a square of sides 8cm. What is the area of \(\bigtriangleup\)UVW? A. cm B. 40 sq.cm C. 50 sq.cm D. 10 sq.cm	D
48.	Find the volume of the figure A. \(\frac{a \pi^2}{3} (x + 3y) \) B. a\(\pi ^2\)y C. \(\frac{a \pi ^2}{3}\)(y + x) D. (\(\frac{1}{3} a \pi ^2 x + y\))	A

41.	A crate of soft drinks contains 10 bottles of Coca-cola, 8 of Fanta and 6 of sprite. If one bottle is selected at random, what is the probability that it is NOT a Cocacola bottle? A. \(\frac{5}{12}\) B. \(\frac{1}{3}\) C. \(\frac{3}{4}\) D. \(\frac{7}{12}\) Show Content Detailed Solution Coca-cola = 10 bottles, Fanta = 8 bottles Sprite = 6 bottles, Total = 24 P(cola-cola) = \(\frac{10}{24}\) P(not coca-cola) = 1 - \(\frac{10}{24}\) \(\frac{24 - 10}{24}\) = \(\frac{14}{24}\) = \(\frac{7}{12}\)	D
42.	In the figure above, Find the value of x A. 130^o B. 110^o C. 100^o D. 90^o	A
43.	PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PM = 5cm, PN = 12cm and PQ = 4.8cm, calculate the respective lengths of PR and PT in centimeters A. 7.3, 5.9 B. 7.7, 12.5 C. 12.5, 7.7 D. 5.9, 7.3 Show Content Detailed Solution \(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR}\) \(\frac{4.8}{12} = \frac{5}{PR}\) PR = \(\frac{5 \times 12}{4.8} = \frac{50}{4}\) = 12.5 \(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT}\) \(\frac{PT}{12} = \frac{5}{PR}\) PT² = 60 PT = \(\sqrt{60}\) = 7.746 = 7.7	C
44.	PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PNR = 110^o and PMQ = 55^o, find MPQ A. 40^o B. 30^o C. 25^o D. 15^o Show Content Detailed Solution PQM = 110^o(Ext. < of a cyclic quad) MPQ = 180^o - (110 + 55) = 180^o - 165^o = 15^o	D

45.	In the figure above, find the value of y A. 28^o B. 122^o C. 150^o D. 152^o Show Content Detailed Solution ADB = 180^o - 152^o = 28^o 28^o + y + 30^o = 180^o y = 180^o - 58^o = 122^o	B
46.	In this figure, PQ = PR = PS and SRT = 68^o. Find QPS A. 136^o B. 124^o C. 12^o D. 68^o Show Content Detailed Solution Since PQRS is quadrilateral 2y + 2x = QPS = 360^o i.e. 2(y + x) + QPS = 360^o QPS = 360^o - 2(y + x) But x + y + 68^o = 180^o x + y = 180^o - 68^o = 180^o x + y = 180^o - 68^o = 112^o	A
47.	In the figure, PQRS is a square of sides 8cm. What is the area of \(\bigtriangleup\)UVW? A. cm B. 40 sq.cm C. 50 sq.cm D. 10 sq.cm	D
48.	Find the volume of the figure A. \(\frac{a \pi^2}{3} (x + 3y) \) B. a\(\pi ^2\)y C. \(\frac{a \pi ^2}{3}\)(y + x) D. (\(\frac{1}{3} a \pi ^2 x + y\))	A