2013 - JAMB Mathematics Past Questions and Answers - page 4
31
What is the probability that an integer x \((1 \leq x \leq 25)\) chosen at random is divisible by both 2 and 3?
A
\(\frac{1}{25}\)
B
\(\frac{1}{5}\)
C
\(\frac{4}{25}\)
D
\(\frac{3}{4}\)
correct option: c
\((1 \leq x \leq 25)\) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}
Number N of x divisible by both 2 and 3 is 4.
n(\(\varepsilon\)) = 25
= \(\frac{N}{n(\varepsilon)}\)
= \(\frac{4}{25}\)
Users' Answers & CommentsNumber N of x divisible by both 2 and 3 is 4.
n(\(\varepsilon\)) = 25
= \(\frac{N}{n(\varepsilon)}\)
= \(\frac{4}{25}\)
32
A basket contains 9 apples, 8 bananas and 7 oranges. A fruit is picked from the basket, find the probability that it is neither an apple nor an orange.
A
\(\frac{3}{8}\)
B
\(\frac{1}{3}\)
C
\(\frac{7}{24}\)
D
\(\frac{2}{3}\)
correct option: b
n(apples) = 9
n(bananas) = 9
n(oranges) = 9
n(\(\varepsilon\)) = 24
Hence Prob(not apple, nor orange) = Prob(banana) = \(\frac{8}{24}\) = \(\frac{1}{3}\)
Users' Answers & Commentsn(bananas) = 9
n(oranges) = 9
n(\(\varepsilon\)) = 24
Hence Prob(not apple, nor orange) = Prob(banana) = \(\frac{8}{24}\) = \(\frac{1}{3}\)
33
The graph is shown is correctly represented by
A
y = x2 - x - 2
B
y = x2 - 3x + 2
C
y = x2 - x - 1
D
y = x2 + x - 2
correct option: a
The graph crosses the x-axis at x = -1 and x = 2
Thus, x + 1 = 0 and x - 2 = 0
x2 - 2x + x - 2 = 0
x2 - x - 2 = 0
giving y = x2 - x - 2
Users' Answers & CommentsThus, x + 1 = 0 and x - 2 = 0
x2 - 2x + x - 2 = 0
x2 - x - 2 = 0
giving y = x2 - x - 2
34
In the diagram given, find the value of x.
A
30o
B
40o
C
45o
D
15o
correct option: b
In the diagram above, < CDE = < CED = 15o
(base < s of isos. \(\bigtriangleup\))
< ECD = 180o - (15 + 15)o
= 180o - 30o = 150o
But x + 110o = 150o
(Sum of opp. interior < s of a\(\bigtriangleup\) = opp. exterior < )
x = 150o - 110o = 40o
Users' Answers & Comments(base < s of isos. \(\bigtriangleup\))
< ECD = 180o - (15 + 15)o
= 180o - 30o = 150o
But x + 110o = 150o
(Sum of opp. interior < s of a\(\bigtriangleup\) = opp. exterior < )
x = 150o - 110o = 40o
36
The bar chart above shows the allotment of time(in minutes) per week for selected subjects in a certain school. What is the total time allocated to the six subjects per week?
A
460mins
B
720mins
C
960mins
D
200mins
37
The pie chart above shows the statistical distribution of 80 students in five subjects in an examination. Calculate how many student offer Mathematics.
A
30
B
11
C
50
D
20
correct option: b
5xo + (16x - 24)o + 5xo + (4x + 12)o + (6x + 12)o = 360o
360xo - 24 + 12 + 12 = 360o
36xo = 360o
xo = \(\frac{360^0}{36}\)
= 10o
Thus, the angle of sector representing Mathematics is 5 x 10o = 50o. Hence the number of students who offer mathematics is
\(\frac{55^o}{36} \times 80 \approx 11\)
Users' Answers & Comments360xo - 24 + 12 + 12 = 360o
36xo = 360o
xo = \(\frac{360^0}{36}\)
= 10o
Thus, the angle of sector representing Mathematics is 5 x 10o = 50o. Hence the number of students who offer mathematics is
\(\frac{55^o}{36} \times 80 \approx 11\)