1993 - JAMB Mathematics Past Questions and Answers - page 4
31
If two angles of a triangle are 30o each and the longest side is 10cm. Calculate the length of each of the other sides
A
5cm
B
4cm
C
3√3 cm
D
10\(\frac{3}{3}\)cm
correct option: d
102 = 2x2 - 2x2 cos 120o (Cosine rule)
100 = 242 - 2x2 - 2x2 x -\(\frac{1}{2}\)
100 = 3x2 + x2
= 3x2
x = \(\sqrt{\frac{100}{3}}\)
= \(\frac{10}{\sqrt{3}}\) x \(\frac{3}{\sqrt{3}}\)
x = 10\(\frac{3}{3}\)cm
Users' Answers & Comments100 = 242 - 2x2 - 2x2 x -\(\frac{1}{2}\)
100 = 3x2 + x2
= 3x2
x = \(\sqrt{\frac{100}{3}}\)
= \(\frac{10}{\sqrt{3}}\) x \(\frac{3}{\sqrt{3}}\)
x = 10\(\frac{3}{3}\)cm
32
Quantities in the proportions 1, 4, 6, 7 are to be represented in a pie chart. Calculate the angle of the sector with proportion 7
A
20o
B
80o
C
120o
D
140o
correct option: d
Angle corresponding to 7 in a pie chart will be \(\frac{7 \times 360}{\text{sum of items}}\)
= \(\frac{7 }{18}\) x 360
= 140o
Users' Answers & Comments= \(\frac{7 }{18}\) x 360
= 140o
33
The following marks were obtained by twenty students in an examination: 53, 30, 70, 84, 59, 43, 90, 20, 78, 48, 44, 60, 81, 73, 50, 37, 67, 68, 64, 52. Find the numbers of students who scored at least 50 marks
A
6
B
10
C
13
D
14
correct option: d
Number of students scoring at least 50 marks = Number of students scoring 50 and above
From the table 53, 70, 84, 59, 90, 60, 81, 73, 50, 37, 67, 68, 64, 52. Hence, 14 students scored at least 50 marks
Users' Answers & CommentsFrom the table 53, 70, 84, 59, 90, 60, 81, 73, 50, 37, 67, 68, 64, 52. Hence, 14 students scored at least 50 marks
34
\(\begin{array}{c|c} Weight(s) & 0 -10 & 10 - 20 & 20 - 30 & 40 - 50\ \hline \text{Number of coconuts} & 10 & 27 & 19 & 6 & 2\end{array}\)
Estimate the mode of the frequency distribution above
Estimate the mode of the frequency distribution above
A
13.2g
B
15.0g
C
16.8g
D
17.5g
correct option: c
Mode = a + \(\frac{(b - a)(F_m - F_b)}{2F_m - F_a - F_b}\)
= \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\)
= \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\)
= 10 + \(\frac{170}{25}\)
= 10 + 6.8
= 16.8
Users' Answers & Comments= \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\)
= \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\)
= 10 + \(\frac{170}{25}\)
= 10 + 6.8
= 16.8
35
The mean of the ages of ten secondary school pupils is 16 but when the age of their teacher is added to it the men becomes 19. Find the age of the teacher
A
27
B
35
C
38
D
49
correct option: d
Average age of 110 students = 16
∴ Total age = 16 x 10 = 160 years
Age of teachers = x, total number of people now = 11
mean age = 19
Total age of new group = 19 x 11 = 209
Age of teachers = x = (209 - 160) = 49 yrs
Users' Answers & Comments∴ Total age = 16 x 10 = 160 years
Age of teachers = x, total number of people now = 11
mean age = 19
Total age of new group = 19 x 11 = 209
Age of teachers = x = (209 - 160) = 49 yrs
36
\(\begin{array}{c|c} Class & Frequency\ \hline 1 - 5 & 2\6 - 10 & 4\11 - 15 & 5\16 - 20 & 2 \ 21 - 25 & 3\26 - 30 & 2\31 - 35 & 1\36 - 40 & 1 \end{array}\)
Find the median of the observation in the table given
Find the median of the observation in the table given
A
11.5
B
12.5
C
14.0
D
14.5
correct option: d
Median = L1 + (\(\frac{Ef}{fm}\)) - fo
\(\frac{\sum f}{2}\)
= \(\frac{20}{2}\)
= 10, L1 = 10.5, fo = 6, fm = 5
Median = 10.5 + \(\frac{(10 - 6)}{5}\)5
= 10.5 + 4
= 14.5
Users' Answers & Comments\(\frac{\sum f}{2}\)
= \(\frac{20}{2}\)
= 10, L1 = 10.5, fo = 6, fm = 5
Median = 10.5 + \(\frac{(10 - 6)}{5}\)5
= 10.5 + 4
= 14.5
37
A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime
A
\(\frac{2}{11}\)
B
\(\frac{5}{11}\)
C
\(\frac{6}{11}\)
D
\(\frac{8}{11}\)
correct option: a
Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors
itself and 1
The prime numbers among the group are 23, 29. Probability of choosing a prime number
= \(\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}\)
= \(\frac{2}{11}\)
Users' Answers & Commentsitself and 1
The prime numbers among the group are 23, 29. Probability of choosing a prime number
= \(\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}\)
= \(\frac{2}{11}\)
38
Calculate the standard deviation of the following data: 7, 8, 9, 10, 11, 12, 13.
A
2
B
4
C
10
D
11
correct option: a
\(\begin{array}{c|c} x & x - x & (x - x)^2\ \hline 7 & -3 & 9\8 & -2 & 4 \9 & -1 & 1\10 & 0 & 1\11 & 1 & 1\ 12 & 2 & 4\13 & 3 & 9\ \hline & & 28\end{array}\)
S.D = \(\sqrt{\frac{\sum(x - x)^2}{N}}\)
= \(\sqrt{\frac{\sum d^2}{N}}\)
= \(\sqrt{\frac{28}{7}}\)
= \(\sqrt{4}\)
= 2
Users' Answers & CommentsS.D = \(\sqrt{\frac{\sum(x - x)^2}{N}}\)
= \(\sqrt{\frac{\sum d^2}{N}}\)
= \(\sqrt{\frac{28}{7}}\)
= \(\sqrt{4}\)
= 2
39
The chances of three independent events X, Y, Z occurring are \(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{1}{4}\) respectively. What are the chances of Y and Z only occurring?
A
\(\frac{1}{8}\)
B
\(\frac{1}{24}\)
C
\(\frac{1}{12}\)
D
\(\frac{1}{4}\)
correct option: c
Chance of x = \(\frac{1}{2}\)
Change of Y = \(\frac{2}{3}\)
Chance of Z = \(\frac{1}{4}\)
Chance of Y and Z only occurring
= Pr (Y ∩ Z ∩ Xc)
where Xc = 1 - Pr(X)
1 = \(\frac{1}{2}\) = 1\(\frac{1}{2}\)
= Pr(Y) x Pr(Z) x Pr(Xc)
= \(\frac{2}{3}\) x \(\frac{1}{4}\) x \(\frac{1}{2}\)
= \(\frac{1}{12}\)
Users' Answers & CommentsChange of Y = \(\frac{2}{3}\)
Chance of Z = \(\frac{1}{4}\)
Chance of Y and Z only occurring
= Pr (Y ∩ Z ∩ Xc)
where Xc = 1 - Pr(X)
1 = \(\frac{1}{2}\) = 1\(\frac{1}{2}\)
= Pr(Y) x Pr(Z) x Pr(Xc)
= \(\frac{2}{3}\) x \(\frac{1}{4}\) x \(\frac{1}{2}\)
= \(\frac{1}{12}\)
40
A
x \(\cap\) z
B
xo \(\cap\) y \(\cap\) z
C
x \(\cap\) yo \(\cap\) z
D
x \(\cap\) y \(\cap\) zo