Year :
1978
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
If y = 2x2 + 9x - 35. Find the range of values for which y < 0. A. 7 < x < \(\frac{5}{2}\) B. -5 < 7 < x C. -7 < x < 5 D. -7 < x < \(\frac{5}{2}\) Detailed Solutiony = 2x2 + 9x - 352x2 + 9x = 35 x2 + \(\frac{9}{2}\) = \(\frac{35}{2}\) x2 + \(\frac{9}{2}\) + \(\frac{81}{16}\) = \(\frac{35}{2}\) = \(\frac{81}{16}\) (x + \(\frac{9}{4}\))2 = \(\frac{361}{16}\) x = \(\frac{-9}{4}\) + \(\frac{\sqrt{361}}{16}\) x = \(\frac{-9}{4}\) + \(\frac{19}{4}\) = 2.5 or -7 -7 < x < \(\frac{5}{2}\) |
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| 12. |
Father reduced the quantity of food bought for the family by 10% when he found that the cost of living had increased 15%. Thus the fractional increase in the family food bill is now A. \(\frac{1}{12}\) B. \(\frac{6}{35}\) C. \(\frac{19}{300}\) D. \(\frac{7}{200}\) E. \(\frac{5}{100}\) Detailed SolutionLet the cost of living = y.The new cost of living = \(y + \frac{15y}{100} = 1.15y\) The food bill now = \((1 - \frac{90}{100})(1.15y)\) = \(1.035y\) The fractional increase in food bill = \((1.035 - 1) \times 100% = 3.5%\) = \(\frac{35}{1000} = \frac{7}{200}\) |
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| 13. |
Given that \(a*b = ab + a + b\) and that \(a ♦ b = a + b = 1\). Find an expression (not involving * or ♦) for (a*b) ♦ (a*c) if a, b, c, are real numbers and the operations on the right are ordinary addition and multiplication of numbers A. ac + ab + bc + b + c + 1 B. ac + ab + a + c + 2 C. ab + ac + a + b + 1 D. ac + bc + ab + b + c + 2 E. ab + ac + 2a + b + c + 1 Detailed SolutionSoln. a*b = ab + a + b,a ♦ b = a + b + 1 a*c = ac + a + c (a*b) ♦ (a*c) = (ab + a + b + ac + a + c + 1) = ab + ac + 2a + b + c + 1 |
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| 14. |
If a circular paper disc is trimmed in such a way that its circumference is reduced in the ratio 2:5, In what ratio is the surface area reduced? A. 8 : 125 B. 2 : 5 C. 8 : 25 D. 4 : 25 E. 4 : 10 Detailed Solutionsurface area of formula = πr\(^2\)If the radius is reduced then let its radius be x. Its area is πx\(^2\) . x : r = 2 : 5, so \(\frac{x}{r}\) = \(\frac{2}{5}\) → 5x = 2r Hence, x = 0.4r. Hence the area of the new circle is π (0.4r)\(^2\) = 0.16π r\(^2\) . The ratio of the two areas is 0.16πr\(^2\) : πr\(^2\) = 0.16 : 1 = 16 : 100 = 4 : 25. |
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| 15. |
If the four interior angles of a quadrilateral are (p + 10)o, (p - 30)o, (2p - 450o, and (p + 15)o, then p is A. 125o B. 82o C. 135o D. 105o E. 60o Detailed SolutionSum of interior angles of a polygon = (2n - 4) x 90where n is the no. of sides sum of interior angles of the quad. = ({2 x 4} -4) x 90 (8 - 4) x 90 = 4 x 90o = 360o (p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o p + 10o + p - 30o + 2p - 45o + p + 15 = 360o 5p = 360o + 50o = 82o |
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| 16. |
Simplify \(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) A. \(\frac{4ab}{a - b}\) B. \(\frac{-4ab}{a^2 - b^2}\) C. \(\frac{-4ab}{a^{-2} - b}\) D. \(\frac{4ab}{a^{-2} - b^{-2}}\) Detailed Solution\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\)applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b) = (a = b + a = b)(a = b - a = b) 2a(-2b) = -4ab = \(\frac{-4ab}{(a + b)(a - b)}\) = \(\frac{-4ab}{a^2 - b^2}\) |
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| 17. |
A force of 5 units acts on a particle in the direction to the east and another force of 4 units acts on the particle in the direction north-east. The resultants of the two forces is A. \(\sqrt{3}\) units B. 3 units C. \(\sqrt{41 + 20 \sqrt{2}}\) units D. \(\sqrt{41 - 20 \sqrt{2}}\) units Detailed SolutionForce to the east = 5 unitsforce to the North - east = 4 units. Resultant of the two forces is the square root 52 + 42 = 41 and plus the sum of its resistance 5 x 4\(\sqrt{2}\) = 20\(\sqrt{2}\) = \(\sqrt{41 + 20 \sqrt{2}}\) units |
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| 18. |
The minimum point on the curve y = x2 - 6x + 5 is at A. (1, 5) B. (2, 3) C. (-3, -4) D. (3, -4) Detailed SolutionGiven the curve \(y = x^{2} - 6x + 5\)At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\) \(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\) \(2x - 6 = 0 \implies x = 3\) Since 3 > 0, it is a minimum point. When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\) Hence, the turning point has coordinates (3, -4). |
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| 19. |
If \(3x - \frac{1}{4})^{\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is A. x > \(\frac{1}{3}\) B. x < \(\frac{1}{3}\) C. x < \(\frac{1}{4}\) D. x < \(\frac{9}{16}\) E. x > \(\frac{9}{16}\) Detailed Solution\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\) = \(3x - 2 > \frac{1}{4} - x\) = \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\) \(x > \frac{9}{16}\) |
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| 20. |
A canal has rectangular cross section of width10cm and breadth 1m. If water of uniform density 1 gm cm-3 flows through it at a constant speed of1000mm per minute, the adjacent sea is A. 100000 B. 1000000 C. 120000 D. 30000 E. 350000 Detailed SolutionThe canal's width = 10cm = 100mm (given)The speed of water = 1000mm 10mm = 1cm 1000mm = 100cm The adjacent sea must give speed x width = 1000 x 100 = 100,000 |
| 11. |
If y = 2x2 + 9x - 35. Find the range of values for which y < 0. A. 7 < x < \(\frac{5}{2}\) B. -5 < 7 < x C. -7 < x < 5 D. -7 < x < \(\frac{5}{2}\) Detailed Solutiony = 2x2 + 9x - 352x2 + 9x = 35 x2 + \(\frac{9}{2}\) = \(\frac{35}{2}\) x2 + \(\frac{9}{2}\) + \(\frac{81}{16}\) = \(\frac{35}{2}\) = \(\frac{81}{16}\) (x + \(\frac{9}{4}\))2 = \(\frac{361}{16}\) x = \(\frac{-9}{4}\) + \(\frac{\sqrt{361}}{16}\) x = \(\frac{-9}{4}\) + \(\frac{19}{4}\) = 2.5 or -7 -7 < x < \(\frac{5}{2}\) |
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| 12. |
Father reduced the quantity of food bought for the family by 10% when he found that the cost of living had increased 15%. Thus the fractional increase in the family food bill is now A. \(\frac{1}{12}\) B. \(\frac{6}{35}\) C. \(\frac{19}{300}\) D. \(\frac{7}{200}\) E. \(\frac{5}{100}\) Detailed SolutionLet the cost of living = y.The new cost of living = \(y + \frac{15y}{100} = 1.15y\) The food bill now = \((1 - \frac{90}{100})(1.15y)\) = \(1.035y\) The fractional increase in food bill = \((1.035 - 1) \times 100% = 3.5%\) = \(\frac{35}{1000} = \frac{7}{200}\) |
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| 13. |
Given that \(a*b = ab + a + b\) and that \(a ♦ b = a + b = 1\). Find an expression (not involving * or ♦) for (a*b) ♦ (a*c) if a, b, c, are real numbers and the operations on the right are ordinary addition and multiplication of numbers A. ac + ab + bc + b + c + 1 B. ac + ab + a + c + 2 C. ab + ac + a + b + 1 D. ac + bc + ab + b + c + 2 E. ab + ac + 2a + b + c + 1 Detailed SolutionSoln. a*b = ab + a + b,a ♦ b = a + b + 1 a*c = ac + a + c (a*b) ♦ (a*c) = (ab + a + b + ac + a + c + 1) = ab + ac + 2a + b + c + 1 |
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| 14. |
If a circular paper disc is trimmed in such a way that its circumference is reduced in the ratio 2:5, In what ratio is the surface area reduced? A. 8 : 125 B. 2 : 5 C. 8 : 25 D. 4 : 25 E. 4 : 10 Detailed Solutionsurface area of formula = πr\(^2\)If the radius is reduced then let its radius be x. Its area is πx\(^2\) . x : r = 2 : 5, so \(\frac{x}{r}\) = \(\frac{2}{5}\) → 5x = 2r Hence, x = 0.4r. Hence the area of the new circle is π (0.4r)\(^2\) = 0.16π r\(^2\) . The ratio of the two areas is 0.16πr\(^2\) : πr\(^2\) = 0.16 : 1 = 16 : 100 = 4 : 25. |
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| 15. |
If the four interior angles of a quadrilateral are (p + 10)o, (p - 30)o, (2p - 450o, and (p + 15)o, then p is A. 125o B. 82o C. 135o D. 105o E. 60o Detailed SolutionSum of interior angles of a polygon = (2n - 4) x 90where n is the no. of sides sum of interior angles of the quad. = ({2 x 4} -4) x 90 (8 - 4) x 90 = 4 x 90o = 360o (p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o p + 10o + p - 30o + 2p - 45o + p + 15 = 360o 5p = 360o + 50o = 82o |
| 16. |
Simplify \(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) A. \(\frac{4ab}{a - b}\) B. \(\frac{-4ab}{a^2 - b^2}\) C. \(\frac{-4ab}{a^{-2} - b}\) D. \(\frac{4ab}{a^{-2} - b^{-2}}\) Detailed Solution\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\)applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b) = (a = b + a = b)(a = b - a = b) 2a(-2b) = -4ab = \(\frac{-4ab}{(a + b)(a - b)}\) = \(\frac{-4ab}{a^2 - b^2}\) |
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| 17. |
A force of 5 units acts on a particle in the direction to the east and another force of 4 units acts on the particle in the direction north-east. The resultants of the two forces is A. \(\sqrt{3}\) units B. 3 units C. \(\sqrt{41 + 20 \sqrt{2}}\) units D. \(\sqrt{41 - 20 \sqrt{2}}\) units Detailed SolutionForce to the east = 5 unitsforce to the North - east = 4 units. Resultant of the two forces is the square root 52 + 42 = 41 and plus the sum of its resistance 5 x 4\(\sqrt{2}\) = 20\(\sqrt{2}\) = \(\sqrt{41 + 20 \sqrt{2}}\) units |
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| 18. |
The minimum point on the curve y = x2 - 6x + 5 is at A. (1, 5) B. (2, 3) C. (-3, -4) D. (3, -4) Detailed SolutionGiven the curve \(y = x^{2} - 6x + 5\)At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\) \(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\) \(2x - 6 = 0 \implies x = 3\) Since 3 > 0, it is a minimum point. When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\) Hence, the turning point has coordinates (3, -4). |
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| 19. |
If \(3x - \frac{1}{4})^{\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is A. x > \(\frac{1}{3}\) B. x < \(\frac{1}{3}\) C. x < \(\frac{1}{4}\) D. x < \(\frac{9}{16}\) E. x > \(\frac{9}{16}\) Detailed Solution\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\) = \(3x - 2 > \frac{1}{4} - x\) = \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\) \(x > \frac{9}{16}\) |
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| 20. |
A canal has rectangular cross section of width10cm and breadth 1m. If water of uniform density 1 gm cm-3 flows through it at a constant speed of1000mm per minute, the adjacent sea is A. 100000 B. 1000000 C. 120000 D. 30000 E. 350000 Detailed SolutionThe canal's width = 10cm = 100mm (given)The speed of water = 1000mm 10mm = 1cm 1000mm = 100cm The adjacent sea must give speed x width = 1000 x 100 = 100,000 |