2005 - JAMB Mathematics Past Questions and Answers - page 1

1
Find the value of m if 13m + 24m = 41m
A
8
B
6
C
5
D
2
correct option: b
If 13m + 24m = 41m
1 * m1 + 3 0 + 2 * m1 + 4 * m0 = 4 * m1 + 1 * m0
1 * m + 3 * 1 + 2 * m + 4 * 1 = 4 * m + 1 * 1
m + 3 + 2m + 4 = 4m + 1
3m + 7 = 4m + 1
4m -3m = 7 - 1
m = 6
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2
If 3214 is divided by 234 and leaves a remainder r, what is the value of r?
A
zero
B
1
C
2
D
3
correct option: c
\(\frac{321_4}{23_4}\=\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\=\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\=\frac{48+8+1}{8+3}\=\frac{57}{11}=5\hspace{1mm}remainder\hspace{1mm}2\∴r=2_{10} \ Now\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\\frac{4}{2} = 2\\frac{4}{0}=0\hspace{1mm}or\hspace{1mm}2\∴r=2\)
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3
Simplify 31/2 - (21/3 * 11/4) + 3/5
A
211/60
B
21/60
C
111/60
D
11/60
correct option: c
31/2 - (21/3 * 11/4) + 3/5
= 7/2 - (7/3 * 5/4) + 3/5
= 7/2 - 35/12 + 3/5
= L.C.M = 60
= (210 - 175 + 36)/60
= 71/60
= 111/60
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4
If the interest on N150.00 for 21/2 years is N4.50, find the interest on N250.00 for 6 months at the same rate
A
N1.50
B
N7.50
C
N15.00
D
N18.00
correct option: a
\(I = N4.50, P = N150,T=2\frac{1}{2}\hspace{1mm}years\I=\frac{P\times T\times R}{100}\4.50=\frac{150 \times 2\frac{1}{2} \times R}{100}\\frac{4.50}{1}=\frac{150 \times 5 \times R}{100\times 2}\4.50\times 4 = 15R\R=\frac{4.50\times5}{15}\R = \frac{6}{5}\Again\hspace{1mm}I\hspace{1mm}=\frac{P\times T \times R}{100}\=\frac{250\times 1 \times 6}{100\times 2\times 5}\=\frac{3}{2}=N1.50\)
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5
Three boys shared some oranges. The first received 1/3 of the oranges and the second received 2/3 of the remaining. If the third boy received the remaining 12 oranges, how many oranges did they share
A
60
B
54
C
48
D
42
correct option: b
Let x = the number of oranges
The 1st received 1/3 of x = 1/3x
∴Remainder = x - 1/3x = 2x/3
The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3
The 3rd received 12 oranges
1/3x + 4x/9 + 12 = x
(3x + 4x + 108)/9 = x
3x + 4x + 108 = 9x
7x + 108 = 9x
9x - 7x = 108
2x = 108
x = 54 oranges
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6
Evaluate \(\frac{(81^{\frac{3}{4}}-27^{\frac{1}{3}})}{3 \times 2^3}\)
A
3
B
1
C
1/3
D
1/8
correct option: b
\(\frac{81^{\frac{3}{4}}-27^{\frac{1}{3}}}{3 \times 2^3} = \frac{(3^{3-\frac{3}{4}}-3^{3-\frac{3}{4}})}{3\times 2^3}\=\frac{3^3 - 3}{3 \times 8}\=\frac{27-3}{24}\=\frac{24}{24}\=1\)
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7
If Log102 = 0.3010 and Log103 = 0.4771, evaluate Log104.5
A
0.9542
B
0.6532
C
0.4771
D
0.3010
correct option: b
Log102 = 0.3010 and Log103 = 0.4771
Log104.5 = Log1041/2
= Log109/2
= Log109 - Log102
= log1032 - Log102
= 2Log103 - Log102
= 2(0.4771) - 0.3010
= 0.9542 - 0.3010
= 0.6532
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8
Simplify \(\frac{(√12-√3)}{(√12+√3)}\)
A
zero
B
1/3
C
3/5
D
1
correct option: b
\(\frac{(\sqrt{12}-\sqrt{3})}{(\sqrt{12}+\sqrt{3})}=\frac{\sqrt{4\times 3}-\sqrt{3}}{\sqrt{4\times 3}+\sqrt{3}}\=\frac{2\sqrt{3}-\sqrt{3}}{2\sqrt{3}+\sqrt{3}}\=\frac{\sqrt{3}}{3\sqrt{3}}\=\frac{1}{3}\)
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9
The venn diagram above shows a class of 40 students with the games they play. How many of the students play two games only?
A
19
B
16
C
15
D
4
correct option: c
Two games played only = 5 + 7 + 3
= 15
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10
If m = 3, p = -3, q = 7 and r = 5/2, evaluate m(p+q+r)
A
19.50
B
19.15
C
18.95
D
18.05
correct option: a
m = 3, p = -3, q = 7 and r = 5/2
m(p+q+r) = 3(-3 + 7 + 5/2)
= 3(4 + 5/2)
= 3(4 + 21/2)
= 3 * 61/2
= 3 * 13/2
= 39/2
= 19.50
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