Paper 1  Objectives  48 Questions
JAMB Exam
Year: 1991
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
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#  Question  Ans 

1. 
Simplify 3\(\frac{1}{3}\)  1\(\frac{1}{4}\) x \(\frac{2}{3}\) + 1\(\frac{2}{5}\) A. 2 B. 3 C. 4 D. 5
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Detailed Solution3  \(\frac{1}{3}\)  (\(\frac{5}{4}\) x \(\frac{2}{3}\)) + 1\(\frac{2}{5}\)= \(\frac{10}{3}\)  \(\frac{5}{6}\) + \(\frac{7}{5}\) = \(\frac{100  25 + 42}{30}\) = \(\frac{117}{30}\) = 3.9 \(\approx\) 4 

2. 
If 2257 is the result of subtracting 4577 from 7056 in base n, find n A. 8 B. 9 C. 10 D. 11
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Detailed Solution\(\begin{array}{cc} 7056 \\ \text{ 4577} \\\hline 2257 \end{array}\)By trial and error method Let the base to 8 i.e. Let n = 8 and it is easily verified that the subtraction holds. The subtraction does not hold when other values of n are tried n = 8 

3. 
Find correct to 3 decimal places (\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\))  (0.05 x 2.05) A. 99.998 B. 98.999 C. 89.899 D. 9.998
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Detailed Solution(\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\))  (0.05 x 2.05)(\(\frac{1}{0.05}\) x 5.005)  (0.05 x 2.05) \(\frac{5.005}{0.05}\)  0.1025 100.1  0.1025 = 99.9975 99.998 to 3 decimal place. 

4. 
express 62 \(\div\) 3 as a decimal correct to 3 significant figures A. 20.6 B. 20.667 C. 20.67 D. 20.7
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Detailed Solution\(\frac{62}{3}\) = 20.6666.....= 20.7 to 3 sig. fig. 

5. 
Factory P produces 20,000 bags of cement per day while factory Q produces 15,000 bags per day. If P reduces production by 5% and Q increases production by 5%, determine the effective loss in the number of bags produced per day by the two factories A. 250 B. 750 C. 1000 D. 1250
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Detailed SolutionP produces 20,000 bagsP reduces production by 5% i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\) = 1000 Q produces 15000 bags Q increases production by 5% i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\) = 750 Effective loss by the two = 1000  750 = 250 bags 

6. 
Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe? A. N10.80 B. N10.67 C. N2.80 D. N2.67
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Detailed SolutionI = \(\frac{PRT}{100}\)= \(\frac{10 \times 2 \times 4}{100}\) = \(\frac{4}{5}\) = 0.8 Total amount = N10.80 He pays N8.00 Remainder = 10.80  8.00 = N2.80 

7. 
If 3 gallons of spirit containing 20% water are added to 5 gallons of another spirit containing 15% water, what percentage of the mixture is water? A. 2\(\frac{4}{5}\)% B. 16\(\frac{7}{8}\)% C. 18\(\frac{1}{8}\)% D. 18\(\frac{7}{8}\)%
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Detailed Solution% of water in the mixture= \(\frac{\text{Total Amount of water}}{\text{Total quantity of spirit}}\) x \(\frac{100}{1}\) \(\frac{3(\frac{20}{100}) + 5 (\frac{15}{100})}{3 + 5}\) x \(\frac{100}{1}\) = \(\frac{\frac{6}{10} + \frac{75}{100}}{8}\) x \(\frac{100}{1}\) = \(\frac{0.6 + 0.75}{8}\) x \(\frac{100}{1}\) = \(\frac{1.35}{8}\) x \(\frac{100}{1}\) = \(\frac{33.75}{2}\) = 16.875 = 16\(\frac{7}{8}\) 

8. 
What is the product of \(\frac{27}{5^1}\)(3)3 and \(\frac{(1)^{1}}{5}\)? A. 5 B. 3 C. 1 D. \(\frac{1}{25}\)
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Detailed Solution\(\frac{27}{5^1}\)(3)3 x \(\frac{(1)^{1}}{5}\) = \(\frac{27}{5}\) x \(\frac{1}{3^3}\) x \(\frac{1}{\frac{1}{5}}\)= \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{1}{5}\) = \(\frac{1}{25}\) 

9. 
Simplify 2log \(\frac{2}{5}\)  log\(\frac{72}{125}\) + log 9 A. 1  4 log3 B. 1 + 2 log 3 C. 1 + 5 log2 D. 1  2log 2
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Detailed Solution2log \(\frac{2}{5}\)  log\(\frac{72}{125}\) + log 9[\(\frac{2}{5}\))^{2} x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\) = log \(\frac{72}{125}\) = log \(\frac{5}{2}\) = log \(\frac{10}{4}\) = log 10  log 4 = log_{10} 10  log_{10} 22 = 1  2 log2 

10. 
Simplify \(\frac{1}{1 + \sqrt{5}}\)  \(\frac{1}{1  \sqrt{5}}\) A.  \(\frac{1}{2}\sqrt{5}\) B. \(\frac{1}{2}\sqrt{5}\) C.  \(\frac{1}{4}\sqrt{5}\) D. 5
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Detailed Solution\(\frac{1}{1 + \sqrt{5}}\)  \(\frac{1}{1  \sqrt{5}}\)= \(\frac{1  \sqrt{5}  1  \sqrt{5}}{(1 + \sqrt{5}) (1  \sqrt{5}}\) = \(\frac{2\sqrt{5}}{1  5}\) = \(\frac{2\sqrt{5}}{ 4}\) = \(\frac{1}{2}\sqrt{5}\) 
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