Paper 1  Objectives  49 Questions
JAMB Exam
Year: 1988
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Did you apply the MTN Ghana Bright Scholarships 2022 for Ghana Undergraduate Students?
Eat well during in test days for good grades. How can you plan your exam diet with good grades in mind?
There's no secret to passing WAEC or BECE apart from the merit of hard work and adequate preparation
#  Question  Ans 

1. 
Tope bought X oranges at N5.00 each and some mangoes at N4.00 each. if she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of values of X. A. 4≤X≤5 B. 5≤X≤8 C. 5≤X≤10 D. 8≤X≤10
Show Content
Detailed SolutionNumber of oranges = X; Costing N5X.Number of mangoes = 2X; Costing N8X. \(\therefore 65 \leq 5X + 8X \leq 130\) \(5 \leq X \leq 10\) 

2. 
If (1P03)4 = 11510 find P A. o B. 1 C. 2 D. 3
Show Content
Detailed Solution1 x 43 + P x 42 + 0 x 4 + 3 = 1151016p + 67 = 115 p = \(\frac{48}{16}\) = 3 

3. 
If \(\frac{1}{p}\) = \(\frac{a^2 + 2ab + b^2}{a  b}\) and \(\frac{1}{q}\) = \(\frac{a + b}{a^2  2ab + b^2}\) Find \(\frac{p}{q}\) A. \(\frac{a + b}{a  b}\) B. \(\frac{1}{a^2  b^2}\) C. \(\frac{a  b}{a + b}\) D. a^{2}  b^{2}
Show Content
Detailed Solution\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a  b}\)\(\frac{1}{q} = \frac{a + b}{a^{2}  2ab + b^{2}}\) \(\frac{1}{p} = \frac{(a + b)^{2}}{a  b}\) \(\frac{1}{q} = \frac{a + b}{(a  b)^{2}}\) \(\therefore p = \frac{a  b}{(a + b)^{2}}\) \(\frac{p}{q} = p \times \frac{1}{q} = \frac{a  b}{(a + b)^{2}} \times \frac{a + b}{(a  b)^{2}}\) = \(\frac{1}{(a + b)(a  b)}\) = \(\frac{1}{a^{2}  b^{2}}\) 

4. 
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3 A. \(\frac{1}{3}\) B. \(\frac{2}{3}\) C. \(\frac{8}{27}\) D. \(\frac{4}{9}\)
Show Content
Detailed Solution\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\)When y = 8, x = 1 \(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\) \(k = 2\) \(\therefore x = \frac{2}{\sqrt[3]{y}}\) When x = 3, \(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\) \(y = (\frac{2}{3})^{3} = \frac{8}{27}\) 

5. 
If a = 3, b = 2, c = 4, evaluate \(\frac{a^3  b^3  c^{\frac{1}{2}}}{b  a  c}\) A. 37 B. \(\frac{37}{5}\) C. \(\frac{37}{5}\) D. 37
Show Content
Detailed Solution\(\frac{a^{3}  b^{3}  c^{\frac{1}{2}}}{b  a  c} = \frac{(3)^{3}  (2)^{3}  4^{\frac{1}{2}}}{2  (3)  4}\)= \(\frac{37}{1} = 37\) 

6. 
If (g(y)) = \(\frac{y  3}{11}\) + \(\frac{11}{y^2  9}\). what is g(y + 3)? A. \(\frac{y}{11} + \frac{11}{y(y + 6)}\) B. \(\frac{y}{11} + \frac{11}{y(y + 3)}\) C. \(\frac{y + 30}{11} + \frac{11}{y(y + 3)}\) D. \(\frac{y + 3}{11} + \frac{11}{y(y  6)}\)
Show Content
Detailed Solution\(g(y) = \frac{y  3}{11} + \frac{11}{y^{2}  9}\)\(\therefore g(y + 3) = \frac{(y + 3)  3}{11} + \frac{11}{(y + 3)^{2}  9}\) \(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y + 9  9}\) \(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y}\) = \(\frac{y}{11} + \frac{11}{y(y + 6)}\) 

7. 
Factorize completely \((x^2 + x)^2  (2x + 2)^2\) A. (x + 1)(x + 2)(x  2) B. (x + 1) 2(x + 2) (x  2) C. (x + 1)^{2} (x + 2)^{2} D. (x + 1)^{2} (x + 2)(x  2)
Show Content
Detailed Solution\((x^{2} + x)^{2}  (2x + 2)^{2}\)= \((x^{2} + x + 2x + 2)(x^{2} + x  (2x + 2))\) = \((x^{2} + 3x + 2)(x^{2}  x  2)\) = \(((x + 1)(x + 2))((x + 1)(x  2))\) = \((x + 1)^{2} (x + 2)(x  2)\) 

8. 
Simplify \(\frac{x  y}{x^{\frac{1}{3}}  x^{\frac{1}{3}}}\) A. x^{2} + xy + y^{2} B. x^{\(\frac{2}{3}\)} + x ^{\(\frac{1}{3}\)} + y^{\(\frac{2}{3}\)} C. x^{\(\frac{2}{3}\)}  x^{\(\frac{1}{3}\)}y^{\(\frac{2}{3}\)} D. y^{\(\frac{2}{3}\)} 
B 
9. 
What is the solution of the equation x2  x  1 + 0? A. x = 1.6 and x = 0.6 B. x = 1.6 and x = 0.6 C. x = 1.6 and x = 0.6 D. x = 1.6 and x = 0.6
Show Content
Detailed Solution\(x^{2}  x  1 = 0\)Using the quadratic formula, \(x = \frac{b \pm \sqrt{b^{2}  4ac}}{2a}\) a = 1, b = 1, c = 1. \(x = \frac{(1) \pm \sqrt{(1)^{2}  4(1)(1)}}{2(1)}\) \(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\) \(x = \frac{1 + 2.24}{2} ; x = \frac{1  2.24}{2}\) \(x = \frac{3.24}{2}; x = \frac{1.24}{2}\) \(x = 1.62 ; x = 0.61 \) \(x \approxeq 1.6; 0.6\) 

10. 
For what values of x is the curve y = \(\frac{x^2 + 3}{x + 4}\) decreasing? A. 3 < x \(\leq\) 0 B. 3 \(\geq\) x < 0 C. 0 < x < 3 D. 0 \(\leq\) x \(\leq\) 3 
D 
Preview displays only 10 out of the 49 Questions