2014 - JAMB Mathematics Past Questions and Answers - page 5

41
\(\begin{array}{c|c}
Numbers & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
Frequency & 18 & 22 & 20 & 16 & 10 & 14
\end{array}\)

The table above represents the outcome of throwing a die 100 times. What is the probability of obtaining at least a 4?
A
\(\frac{1}{5}\)
B
\(\frac{1}{2}\)
C
\(\frac{2}{5}\)
D
\(\frac{3}{4}\)
correct option: c

Let E demote the event of obtaining at least a 4

Then n(E) = 16 + 10 + 14 = 40

Hence, prob (E) = (\frac{n(E)}{n(S)})

( = \frac{40}{100})

( = \frac{2}{5})

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42
A number is chosen at random from 10 to 30 both inclusive. What is the probability that the number is divisible by 3?
A
\(\frac{2}{15}\)
B
\(\frac{1}{10}\)
C
\(\frac{1}{3}\)
D
\(\frac{2}{5}\)
correct option: c

Sample space S = {10, 11, 12, ... 30}

Let E denote the event of choosing a number divisible by 3

Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7

Prob (E) = (\frac{n(E)}{n(E)})

Prob (E) = (\frac{7}{21})

Prob (E) = (\frac{1}{3})

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43
From the venn diagram above, the shaded parts represent
A
(P\(\cap\)Q)\(\cup\)(P\(\cap\)R)
B
(P\(\cup\)Q)\(\cap\)(P\(\cap\)R)
C
(P\(\cup\)Q)\(\cup\)(P\(\cup\)R)
D
(P\(\cap\)Q)\(\cup\)(P\(\cup\)R)
correct option: a
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44
In the figure above, KL//NM, LN bisects < KNM. If angles KLN is 54o and angle MKN is 35o, calculate the size of angle KMN.
A
91o
B
89o
C
37o
D
19o
correct option: c

In the diagram above, (\alpha) = 54o(alternate angles; KL||MN) < KNM = 2(\alpha) (LN is bisector of < KNM) = 108o

35o + < KMN + 108o = 180o(sum of angles of (\bigtriangleup))

< KMN + 143o = 180o

< KMN = 180o - 143o

= 37o

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45
From the figure above, what is the value of p?
A
135o
B
90o
C
60o
D
45o
correct option: b

In the figure above, qo = 30o (vertically opposite angles)

(P + 2q)o + 30o = 180o(angles on a straight line)

p + 2 x 30o + 30o = 180o

p + 60o + 30o = 180o

p + 90o = 180o

p = 180o - 90o

= 90o

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46
Find the value of x in the figure above
A
20\(\sqrt{3}\)cm
B
10\(\sqrt{3}\)cm
C
5\(\sqrt{3}\)cm
D
4\(\sqrt{3}\)cm
correct option: b

In the figure above, (\frac{x}{\sin 60^o} = \frac{10}{\sin 30^o}) (Sine rule)

x = (\frac{10 \sin 60^o}{\sin 30^o})

= 10 x (\frac{\sqrt{3}}{2} \times \frac{1}{2})

= 10 x (\frac{\sqrt{3}}{2} \times \frac{2}{1})

= 10(\sqrt{3})cm

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47
in the figure above, what is the equation of the line that passes the y-axis at (0,5) and passes the x-axis at (5,0)?
A
y = x + 5
B
y = -x + 5
C
y = x - 5
D
y = -x - 5
correct option: b

(x1, y1) = (0,5)

(x2, y2) = (5, 0)

Using (\frac{y - y_1}{y_1 - y_1} = \frac{x - x_1}{x_1 - x_1})

(\frac{y - 5}{0 - 5} = \frac{x - 0}{5 - 0})

(\frac{y - 5}{-5} = \frac{x}{5})

5(y - 5) = -5x

y - 5 = -x

x + y = 5

y = -x + 5

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48
The pie chart above shows the monthly distribution of a man's salary on food items. If he spent N8,000 on rice, how much did he spent on yam?
A
N42,000
B
N18,000
C
N16,000
D
N12,000
correct option: c

Angle of sector subtended by yam

= 360o - (70 + 80 + 50)o

= 360o - 200o

= 160o

But (\frac{80^o}{360^o}) x T = 8000

T = (\frac{8000 \times 360^o}{80^o})

= N36,000

Hence the amount spent on yam = (\frac{160^o}{260} \times N36,000)

= N16,000

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