Paper 1  Objectives  47 Questions
JAMB Exam
Year: 2005
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
Find the value of m if 13_{m} + 24_{m} = 41_{m} A. 8 B. 6 C. 5 D. 2
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Detailed SolutionIf 13_{m} + 24_{m} = 41_{m}1 * m^{1} + 3 ^{0} + 2 * m^{1} + 4 * m^{0} = 4 * m^{1} + 1 * m^{0} 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 

2. 
If 321_{4} is divided by 23_{4} and leaves a remainder r, what is the value of r? A. zero B. 1 C. 2 D. 3
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Detailed Solution\(\frac{321_4}{23_4}\\=\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\\=\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\\=\frac{48+8+1}{8+3}\\=\frac{57}{11}=5\hspace{1mm}remainder\hspace{1mm}2\\∴r=2_{10} \\ Now\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\\\frac{4}{2} = 2\\\frac{4}{0}=0\hspace{1mm}or\hspace{1mm}2\\∴r=2\) 

3. 
Simplify 3^{1}/_{2}  (2^{1}/_{3} * 1^{1}/_{4}) + ^{3}/_{5} A. 2^{11}/_{60} B. 2^{1}/_{60} C. 1^{11}/_{60} D. 1^{1}/_{60}
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Detailed Solution3^{1}/_{2}  (2^{1}/_{3} * 1^{1}/_{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} = 1^{11< } 

4. 
If the interest on N150.00 for 2^{1}/_{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
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Detailed Solution\(I = N4.50, P = N150,T=2\frac{1}{2}\hspace{1mm}years\\I=\frac{P\times T\times R}{100}\\4.50=\frac{150 \times 2\frac{1}{2} \times R}{100}\\\frac{4.50}{1}=\frac{150 \times 5 \times R}{100\times 2}\\4.50\times 4 = 15R\\R=\frac{4.50\times5}{15}\\R = \frac{6}{5}\\Again\hspace{1mm}I\hspace{1mm}=\frac{P\times T \times R}{100}\\=\frac{250\times 1 \times 6}{100\times 2\times 5}\\=\frac{3}{2}=N1.50\) 

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
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Detailed SolutionLet x = the number of orangesThe 1^{st} received ^{1}/_{3} of x = ^{1}/_{3}x ∴Remainder = x  ^{1}/_{3}x = ^{2x}/_{3} The 2^{nd} received ^{2}/_{3} of ^{2x}/_{3} = ^{2}/_{3} * ^{2x}/_{3} = ^{4x}/_{3} The 3^{rd} received 12 oranges ∴^{1}/_{3 } 

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
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Detailed Solution\(\frac{81^{\frac{3}{4}}27^{\frac{1}{3}}}{3 \times 2^3} = \frac{(3^{3\frac{3}{4}}3^{3\frac{3}{4}})}{3\times 2^3}\\=\frac{3^3  3}{3 \times 8}\\=\frac{273}{24}\\=\frac{24}{24}\\=1\) 

7. 
If Log_{10}2 = 0.3010 and Log_{10}3 = 0.4771, evaluate Log_{10}4.5 A. 0.9542 B. 0.6532 C. 0.4771 D. 0.3010
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Detailed SolutionLog_{10}2 = 0.3010 and Log_{10}3 = 0.4771Log_{10}4.5 = Log_{10}4^{1}/_{2} = Log_{10}^{9}/_{2} = Log_{10}9  Log_{10}2 = log_{10}3^{2}  Log_{10}2 = 2Log_{10}3  Log_{10}2 = 2(0.4771)  0.3010 = 0.9542  0.3010 = 0.6532 

8. 
Simplify \(\frac{(√12√3)}{(√12+√3)}\) A. zero B. 1/3 C. 3/5 D. 1
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Detailed Solution\(\frac{(\sqrt{12}\sqrt{3})}{(\sqrt{12}+\sqrt{3})}=\frac{\sqrt{4\times 3}\sqrt{3}}{\sqrt{4\times 3}+\sqrt{3}}\\=\frac{2\sqrt{3}\sqrt{3}}{2\sqrt{3}+\sqrt{3}}\\=\frac{\sqrt{3}}{3\sqrt{3}}\\=\frac{1}{3}\) 

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
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Detailed Solution= 15 

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
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Detailed Solutionm = 3, p = 3, q = 7 and r = 5/2m(p+q+r) = 3(3 + 7 + ^{5}/_{2}) = 3(4 + ^{5}/_{2}) = 3(4 + 2^{1}/_{2}) = 3 * 6^{1}/_{2} = 3 * ^{13}/_{2} = ^{39}/_{2} = 19.50 
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