Mathematics (Core), 2001, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2001

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

1 - 10 of 50 Questions

#	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 Show Content Detailed Solution Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.	D
2.	Simplify \((\sqrt[3]{64a^{3}})^{-1}\) A. 4a B. 1/8a C. 8a D. 1/4a Show Content Detailed Solution \((\sqrt[3]{64a^{3}})^{-1} = (\sqrt[3]{(4a)^{3}})^{-1}\) = \((4a)^{-1} \) = \(\frac{1}{4a}\)	D
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 Show Content Detailed Solution \(\frac{p^{2} - q^{2}}{2pq} = \frac{(p + q)(p - q)}{2pq}\) = \(\frac{(1 + \sqrt{2} - (1 - \sqrt{2}))(1 + \sqrt{2} + 1 - \sqrt{2})}{2(1 + \sqrt{2})(1 - \sqrt{2})}\) = \(\frac{(2\sqrt{2})(2)}{-2}\) = \(-2\sqrt{2}\)	D
4.	A car dealer bought a second-hand 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% Show Content Detailed Solution Total cost = N(250,000 + 70,000) = N320,000 Selling 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%	C
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 Show Content Detailed Solution \(x = \frac{y}{2} \) \(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\) \(\frac{x^3}{y^3} + \frac{1}{2} = (\frac{y}{2})^{3} \div y^{3} + \frac{1}{2}\) = \(\frac{y^{3}}{8} \times \frac{1}{y^3} + \frac{1}{2}\) = \(\frac{1}{8} + \frac{1}{2}\) = \(\frac{5}{8}\) \(\frac{1}{2} - \frac{x^2}{y^2} = \frac{1}{2} - (\frac{y}{2})^{2} \div y^2)\) = \(\frac{1}{2} - \frac{y^2}{4} \times \frac{1}{y^2}\) = \(\frac{1}{2} - \frac{1}{4}\) = \(\frac{1}{4}\) \(\therefore \left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right) = \frac{5}{8} \div \frac{1}{4}\) = \(\frac{5}{2}\)	B
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 Show Content Detailed Solution Principal, 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	B
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 Show Content Detailed Solution \(\frac{(0.14)^{2} \times 0.275}{7(0.02)} = \frac{(0.14)^{2} \times 0.275}{0.14}\) = \(0.14 \times 0.275\) = \(0.0385 \approxeq 0.039\)	A
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 Show Content 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 \(z - 2\) = \(a^{x} - 2\)	A
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. \(\frac{q-8}{7}\) B. \(\frac{7-q}{4}\) C. \(\frac{8-q}{2}\) D. \(\frac{7+q}{8}\) Show Content 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): \(7 = 4p + q \implies p = \frac{7 - q}{4}\)	B
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) Show Content Detailed Solution Hint: Sketch the inequality graph for the 3 conditions given and read out your points from the co-ordinates.	D

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 Show Content Detailed Solution Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.	D
2.	Simplify \((\sqrt[3]{64a^{3}})^{-1}\) A. 4a B. 1/8a C. 8a D. 1/4a Show Content Detailed Solution \((\sqrt[3]{64a^{3}})^{-1} = (\sqrt[3]{(4a)^{3}})^{-1}\) = \((4a)^{-1} \) = \(\frac{1}{4a}\)	D
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 Show Content Detailed Solution \(\frac{p^{2} - q^{2}}{2pq} = \frac{(p + q)(p - q)}{2pq}\) = \(\frac{(1 + \sqrt{2} - (1 - \sqrt{2}))(1 + \sqrt{2} + 1 - \sqrt{2})}{2(1 + \sqrt{2})(1 - \sqrt{2})}\) = \(\frac{(2\sqrt{2})(2)}{-2}\) = \(-2\sqrt{2}\)	D
4.	A car dealer bought a second-hand 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% Show Content Detailed Solution Total cost = N(250,000 + 70,000) = N320,000 Selling 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%	C
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 Show Content Detailed Solution \(x = \frac{y}{2} \) \(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\) \(\frac{x^3}{y^3} + \frac{1}{2} = (\frac{y}{2})^{3} \div y^{3} + \frac{1}{2}\) = \(\frac{y^{3}}{8} \times \frac{1}{y^3} + \frac{1}{2}\) = \(\frac{1}{8} + \frac{1}{2}\) = \(\frac{5}{8}\) \(\frac{1}{2} - \frac{x^2}{y^2} = \frac{1}{2} - (\frac{y}{2})^{2} \div y^2)\) = \(\frac{1}{2} - \frac{y^2}{4} \times \frac{1}{y^2}\) = \(\frac{1}{2} - \frac{1}{4}\) = \(\frac{1}{4}\) \(\therefore \left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right) = \frac{5}{8} \div \frac{1}{4}\) = \(\frac{5}{2}\)	B

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 Show Content Detailed Solution Principal, 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	B
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 Show Content Detailed Solution \(\frac{(0.14)^{2} \times 0.275}{7(0.02)} = \frac{(0.14)^{2} \times 0.275}{0.14}\) = \(0.14 \times 0.275\) = \(0.0385 \approxeq 0.039\)	A
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 Show Content 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 \(z - 2\) = \(a^{x} - 2\)	A
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. \(\frac{q-8}{7}\) B. \(\frac{7-q}{4}\) C. \(\frac{8-q}{2}\) D. \(\frac{7+q}{8}\) Show Content 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): \(7 = 4p + q \implies p = \frac{7 - q}{4}\)	B
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) Show Content Detailed Solution Hint: Sketch the inequality graph for the 3 conditions given and read out your points from the co-ordinates.	D