Paper 1  Objectives  50 Questions
JAMB Exam
Year: 2001
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
Evaluate 21.05347  1.6324 x 0.43 to 3 decimal places A. 20.980 B. 20.351 C. 20.981 D. 20.352
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Detailed SolutionHint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first. 

2. 
Simplify \((\sqrt[3]{64a^{3}})^{1}\) A. 4a B. 1/8a C. 8a D. 1/4a
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Detailed Solution\((\sqrt[3]{64a^{3}})^{1} = (\sqrt[3]{(4a)^{3}})^{1}\)= \((4a)^{1} \) = \(\frac{1}{4a}\) 

3. 
Given that \(p = 1 + \sqrt{2}\) and \(q = 1  \sqrt{2}\), evaluate \(\frac{p^{2}  q^{2}}{2pq}\). A. 2(2+√2) B. 2(2+√2) C. 2√2 D. 2√2
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Detailed Solution\(\frac{p^{2}  q^{2}}{2pq} = \frac{(p + q)(p  q)}{2pq}\)= \(\frac{(1 + \sqrt{2}  (1  \sqrt{2}))(1 + \sqrt{2} + 1  \sqrt{2})}{2(1 + \sqrt{2})(1  \sqrt{2})}\) = \(\frac{(2\sqrt{2})(2)}{2}\) = \(2\sqrt{2}\) 

4. 
A car dealer bought a secondhand car for N250,000 and spent N70,000 refurbishing it. He then sold the car for N400,000. What is the percentage gain? A. 60% B. 32% C. 25% D. 20%
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Detailed SolutionTotal cost = N(250,000 + 70,000) = N320,000Selling price = N400,000 (given) Gain = SP  CP = N(400,000  320,000) = N80,000 Gain % = gain/CP x 100 = (80,000/320,000) x 100 Gain % = 25% 

5. 
If \(x = \frac{y}{2}\),evaluate\(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2}  \frac{x^{2}}{y^{2}}\right)\) A. 5/8 B. 5/2 C. 5/32 D. 5/16
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Detailed Solution\(x = \frac{y}{2} \)\(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2}  \frac{x^{2}}{y^{2}}\right)\) \(\frac{x^3}{y^3} + \frac{1}{2} = (\frac{y}{2})^{3} \div y^{3} + \frac{1}{2}\) = \(\frac{y^{3}}{8} \times \frac{1}{y^3} + \frac{1}{2}\) = \(\frac{1}{8} + \frac{1}{2}\) = \(\frac{5}{8}\) \(\frac{1}{2}  \frac{x^2}{y^2} = \frac{1}{2}  (\frac{y}{2})^{2} \div y^2)\) = \(\frac{1}{2}  \frac{y^2}{4} \times \frac{1}{y^2}\) = \(\frac{1}{2}  \frac{1}{4}\) = \(\frac{1}{4}\) \(\therefore \left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2}  \frac{x^{2}}{y^{2}}\right) = \frac{5}{8} \div \frac{1}{4}\) = \(\frac{5}{2}\) 

6. 
Find the principal which amounts to N5,500 at a simple interest in 5 years at 2% per annum. A. N4,900 B. N5,000 C. N4,700 D. N4,800
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Detailed SolutionPrincipal, P = Amount, A  Interest, I.A = P + I I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10 But A = P + I, => 5500 = P + (P/10) => 55000 = 10P + P => 55000 = 11P Thus P = 55000/11 = N5,000 

7. 
Evaluate \(\frac{(0.14^2 \times 0.275)}{7(0.02)}\) to 3 decimal places. A. 0.039 B. 0.385 C. 0.033 D. 0.038
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Detailed Solution\(\frac{(0.14)^{2} \times 0.275}{7(0.02)} = \frac{(0.14)^{2} \times 0.275}{0.14}\)= \(0.14 \times 0.275\) = \(0.0385 \approxeq 0.039\) 

8. 
Divide: \(a^{3x}  26a^{2x} + 156a^{x}  216\) by \(a^{2x}  24a^{x} + 108\). A. a^{x}  2 B. a^{x} + 2 C. a^{x}  8 D. a^{x}  6
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Detailed Solution\(\frac{a^{3x}  26a^{2x} + 156a^{x}  216}{a^{2x}  24a^{x} + 108}\)Let \(a^{x} = z\) \(\therefore = \frac{z^{3}  26z^{2} + 156z  216}{z^{2}  24y + 108} ... (i)\) Dividing (i) above, we get \(z  2\) = \(a^{x}  2\) 

9. 
If two graphs y = px\(^2\) + q and y = 2x\(^2\) 1 intersect at x = 2, find the value of p in terms q. A. \(\frac{q8}{7}\) B. \(\frac{7q}{4}\) C. \(\frac{8q}{2}\) D. \(\frac{7+q}{8}\)
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Detailed Solution\(y = px^{2} + q ... (i)\)\(y = 2x^{2}  1 ... (ii)\) At x = 2, (i): \(y = p(2^{2}) + q = 4p + q\) (ii): \(y = 2(2^{2})  1 = 7\) \(\therefore \text{The coordinates of the point of intersection = (2, 7)}\) (i): \(7 = 4p + q \implies p = \frac{7  q}{4}\) 

10. 
Find the integral values of x and y satisfying the inequality 3y + 5x \(\leq\) 15, given that y > 0, y < 3 and x > 0. A. (1,1), (1,2), (1,3) B. (1,1), (2,1), (1,3) C. (1,1), (3,1), (2,2) D. (1,1), (1,2), (2,1)
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Detailed SolutionHint: Sketch the inequality graph for the 3 conditions given and read out your points from the coordinates. 
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