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
Good jobs available to people without a college degree, how to get good job without a college degree?
Adequate preparation and effective revision strategies tips to get a high score 320 in JAMB in 2022
8 Tips To Be More Productive in Test Revision for good grades and higher performance in exams
#  Question  Ans 

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
Show Content
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 

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\(\frac{1}{8}\)
Show Content
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\(\frac{2}{3}\) 

3. 
Multiply (x + 3y + 5) by (2x2 + 5y + 2) A. 2x^{2} + 3yx^{2} + 10xy + 15y^{2} + 13y + 10x^{2} + 2x + 10 B. 2x^{3} + 6yx^{2} + 5xy + 15y^{2} + 31y + 5x^{2} + 2x + 10 C. 2x^{3} + 6xy^{2} + 5xy + 15y^{2} + 12y + 10x^{2} + 2x = 10 D. 2x^{2} + 6xy^{2} + 5xy + 15y^{2} + 13y + 10x^{2} + 2x + 10 E. 2x^{3} + 2yx^{2} + 10xy + 10y^{2} + 31y + 5x^{2} + 10
Show Content
Detailed Solution\((x + 3y + 5)(2x^{2} + 5y + 2)\)= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\) = \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\) 

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. \(\frac{9}{16}\), \(\frac{3}{5}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
Show Content
Detailed Solution\(\frac{3}{5}\) = 0.60, \(\frac{9}{16}\) = 0.56, \(\frac{34}{59}\) = 0.58, \(\frac{71}{97}\) = 0.73Hence, \(\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}\) 

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. \(\frac{1}{x}\)
Show Content
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 \(S_{n} = \frac{1}{1  x}\) 

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 = \(\frac{kD^2_AP_D}{B}\)
Show Content
Detailed Solution\(N \propto P_{A}\); \(N \propto P_{B}\); \(N \propto \frac{1}{D^{2}}\)\(\therefore N \propto \frac{P_{A} P_{B}}{D^{2}}\) \(N = \frac{k P_{A} P_{B}}{D^{2}}\) 

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\(\sqrt{6}\)
Show Content
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}} = 

8. 
If x\(^2\) + 4 = 0, then x ? A. 4 B. 4 C. 2 D. 2 E. None of the above
Show Content
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. 

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
Show Content
Detailed Solution\(\begin{array}{cc} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\)N : b Look for the antilog of .3482 which gives 222.9 

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
Show Content
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 
Preview displays only 10 out of the 48 Questions