Mathematics (Core)
Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1978
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1978
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Five revision strategies that are strange but effective for memorization for high exams performance
These are the three things to do in test preparation for higher grades and excellent performance.
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
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| 1. |
A rectangular picture 6cm by 8cm is enclosed by a frame \(\frac{1}{2}\)cm wide. Calculate the area of the frame A. 15sq.cm B. 20sq.cm C. 13sq.cm D. 16sq.cm E. 17sq.cm
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Detailed SolutionArea of the rectangular picture = L x B = 8 x 6= 48 sq.cm. Area of the whole surface (which is gotten by adding \(\frac{1}{2}\) on every side to the original picture size) is 9 x 7 = 63 sq. cm area of the frame is 63 - 48 = 15 sq. cm |
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| 2. |
The sum of \(3\frac{7}{8}\) and \(1\frac{1}{3}\) is less than the difference between \(\frac{1}{8}\) and \(1\frac{2}{3}\) by: A. 3\(\frac{2}{3}\) B. 5\(\frac{1}{4}\) C. 6\(\frac{1}{2}\) D. 8 E. 8\(\frac{1}{8}\)
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Detailed Solution\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\)= \(4\frac{29}{24}\) \(\equiv 5\frac{5}{24}\) \(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\) = \(\frac{40 - 3}{24}\) = \(\frac{37}{24}\) \(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\) = \(\frac{88}{24}\) = 3\(\frac{2}{3}\) |
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| 3. |
Multiply (x + 3y + 5) by (2x2 + 5y + 2) A. 2x2 + 3yx2 + 10xy + 15y2 + 13y + 10x2 + 2x + 10 B. 2x3 + 6yx2 + 5xy + 15y2 + 31y + 5x2 + 2x + 10 C. 2x3 + 6xy2 + 5xy + 15y2 + 12y + 10x2 + 2x = 10 D. 2x2 + 6xy2 + 5xy + 15y2 + 13y + 10x2 + 2x + 10 E. 2x3 + 2yx2 + 10xy + 10y2 + 31y + 5x2 + 10
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Detailed Solution\((x + 3y + 5)(2x^{2} + 5y + 2)\)= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\) = \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\) |
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| 4. |
Arrange \(\frac{3}{5}\),\(\frac{9}{16}\), \(\frac{34}{59}\) and \(\frac{71}{97}\) in ascending order of magnitude. A. \(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{71}{97}\) B. \(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{3}{5}\) , \(\frac{71}{97}\) C. \(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{71}{97}\), \(\frac{34}{59}\) D. \(\frac{9}{16}\), \(\frac{3}{5}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
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Detailed Solution\(\frac{3}{5}\) = 0.60, \(\frac{9}{16}\) = 0.56, \(\frac{34}{59}\) = 0.58, \(\frac{71}{97}\) = 0.73Hence, \(\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}\) |
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| 5. |
The sum of the progression is 1 + x + x2 + x3 + ...... A. \(\frac{1}{1 - x}\) B. \(\frac{1}{1 + x}\) C. \(\frac{1}{x - 1}\) D. \(\frac{1}{x}\)
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Detailed SolutionSum of n terms of Geometric progression is \(S_{n} = \frac{a(1 - r^n)}{1 - r}\)In the given series, a (the first term) = 1 and r (the common ratio) = x. \(S_{n} = \frac{1(1 - x^{n})}{1 - x}\) a = 1, and as n tends to infinity \(S_{n} = \frac{1}{1 - x}\) |
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| 6. |
The number of telephone calls N between two cities A and B varies directly as the population P\(_{A}\), P\(_B\) respectively and inversely as the square of the distance D between A and B. Which of the following equations represents this relation? A. N = \(\frac{kp_A}{D^2} + {cp_B}{D^2}\) B. N = \(\frac{k P_{A} P_{B} }{D^2}\) C. N = \(\frac{kD_AP_D}{B^2}\) D. N = \(\frac{kD^2_AP_D}{B}\)
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Detailed Solution\(N \propto P_{A}\); \(N \propto P_{B}\); \(N \propto \frac{1}{D^{2}}\)\(\therefore N \propto \frac{P_{A} P_{B}}{D^{2}}\) \(N = \frac{k P_{A} P_{B}}{D^{2}}\) |
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| 7. |
Find the square root of 170 - 20\(\sqrt{30}\) A. 2 \(\sqrt{10}\) - 5\(\sqrt{3}\) B. 2 \(\sqrt{5}\) - 5\(\sqrt{6}\) C. 5 \(\sqrt{10}\) - 2\(\sqrt{3}\) D. 3 \(\sqrt{5}\) - 8\(\sqrt{6}\)
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Detailed Solution\(\sqrt{170 - 20 \sqrt{30}} = \sqrt{a} - \sqrt{b}\)Squaring both sides, \(170 - 20\sqrt{30} = a + b - 2\sqrt{ab}\) Equating the rational and irrational parts, we have \(a + b = 170 ... (1)\) \(2 \sqrt{ab} = 20 \sqrt{30}\) \(2 \sqrt{ab} = 2 \sqrt{30 \times 100} = 2 \sqrt{3000} \) \(ab = 3000 ... (2)\) From (2), \(b = \frac{3000}{a}\) \(a + \frac{3000}{a} = 170 \implies a^{2} + 3000 = 170a\) \(a^{2} - 170a + 3000 = 0\) \(a^{2} - 20a - 150a + 3000 = 0\) \(a(a - 20) - 150(a - 20) = 0\) \(\text{a = 20 or a = 150}\) \(\therefore b = \frac{3000}{20} = 150 ; b = \frac{3000}{150} = 20\) \(\sqrt{170 - 20\sqrt{30}} = |
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| 8. |
If x\(^2\) + 4 = 0, then x ? A. 4 B. -4 C. 2 D. -2 E. None of the above
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Detailed Solutionx\(^2\) + 4 = 0x\(^2\) = -4 You can't apply a square root to a negative number. So the answer is None of the above. |
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| 9. |
What is the number whose logarithm to base 10 is 2.3482? A. 223 B. 2228 C. 2.235 D. 22.37 E. 0.2229
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Detailed Solution\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\)N : b Look for the antilog of .3482 which gives 222.9 |
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| 10. |
Five years ago, a father was 3 times as old as his son, now their combined ages amount to 110years. thus, the present age of the father is A. 75 years B. 60 years C. 98 years D. 80 years E. 105 years
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Detailed SolutionLet the present ages of father be x and son = yfive years ago, father = x - 5 son = y - 5 x - 5 = 3(y - 5) x - 5 = 3y - 15 x - 3y = -15 + 5 = -10 ......(i) x + y = 110......(ii) eqn(ii) - eqn(i) 4y = 120 y = 30 sub for y = 30 in eqn(i) x -10 = 3(30) x -10 = 90 x = 80 |
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