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Mathematics
UTME / JAMB 1988
Algebra
Solution
The solution of the quadratic equation px2 + qx + b = 0 is
A
\(\sqrt{\frac{-b \pm b^2 - 4ac}{2a}}\)
B
\(\frac{-b \pm \sqrt{ p^2 - 4pb}}{2a}\)
C
\(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\)
D
\(\frac{-q \pm \sqrt{ p^2 - 4bp}}{2p}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
px2 + qx + b = 0
Using almighty formula
\(\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}\).........(i)
Where a = p, b = q and c = b
substitute for this value in equation (i)
= \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\)
Source: Myschool Past Questions Archive (Record #31730)
Mathematics
WAEC / WASSCE 1988
Algebra
Solution
Illustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time :
\(- x + 5y \leq 10 ; 3x - 4y \leq 8\) and \(x > -1\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Draw the table of value such that \(-x + 5y \leq 10 \implies -x + 5y = 10\)
\(\implies 5y = 10 + x \)
\(y = 2 + \frac{x}{5}\)
and \(3x - 4y \leq 8 \implies 3x - 4y = 8\)
\(y = -2 + \frac{3x}{4}\)
Table of value for \(-x + 5y = 10\)
x
-4
0
4
y
1.2
2
2.8
Table of value for \(3x - 4y = 8\)
x
-4
0
4
y
-5
-2
1
Source: Myschool Past Questions Archive (Record #54714)
Mathematics
WAEC / WASSCE 1988
Algebra
Solution
The table below is for the relation \(y = 2 + x - x^{2}\)
x
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
y
-4
-1.75
0
1.25
2
2.25
2
1.25
0
-1.75
-4
(a) Using a scale of 2cm to 1 unit on each axis, draw the graph of the relation in the interval \(-2 \leq x \leq 3\).
(b) From your graph, find the greatest value of y and the value of x for which this occurs.
(c) Using the same scale and axes, draw the graph of \(y = 1 - x\)
(d) Use your graphs to solve the equation \(1 + 2x - x^{2} = 0\)
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
(b) The greatest value of y = 2.25 and this occurs at x = 0.5.
(c) Table of values of \(y = 1 - x\)
x
-2
2
y
3
-1
(d) \(1 + 2x - x^{2} = 0\)
\(1 + 1 + 2x - x - x^{2} = 1 - x\)
\(2 + x - x^{2} = 1 - x\)
The solution of \(1 + 2x - x^{2} = 0\) occurs at the point of intersection of \(y = 2 + x - x^{2}\) and \(y = 1 - x\). This is at points A (x = -0.8) and B (x = 2.8).
Source: Myschool Past Questions Archive (Record #54715)
Mathematics
UTME / JAMB 1987
Algebra
Solution
If P varies inversely as V and V varies directly as R2, find the relationship between P and R given that R = 7 when P = 2
A
P = 98R2
B
PR2 = 98
C
P = \(\frac{1}{98R2
D
P = \(\frac{PR2}{98}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
P = \(\frac{1}{v}\) and vR2 = P = \(\frac{k}{v}\)......(i)
and v KR2 .......(ii)
(where k is constant)
Subst. for v in equation (i) = p = \(\frac{1^2}{KR}\).....(ii)
when r = 7, p = 2
2 = \(\frac{k}{7^2}\)
k = 2 x 49
= 98
Subt. foe k in ....(iii)
P = \(\frac{98}{R^2}\)
PR2 = 98
Source: Myschool Past Questions Archive (Record #31890)
Mathematics
UTME / JAMB 1987
Algebra
Solution
Find the values of m which make the following quadratic function a perfect square. x2 + 2(m + 1)x + m + 3
A
-1, 1
B
-1, 2
C
1, -2
D
2, -2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x2 + 2(m + 1)x + m + 3 from (A + B)2 = a2 + 2AB + b2
2B = \(\frac{2(m + 1)}{2}\) = m + 1 .......(i)
Also B2 = m + 3..........(ii)
Subst. for B in equation ..........(iii)
= (m + 1)2 = m + 3 = m2 + 2m + 1 = m + 3
= m2 + 2m - m + 1 - 3 = 0
m2 + m - 2 = 0
(m + 2)(m - 1) = 0 or m - 1 = 0
m = -2 or 1
Source: Myschool Past Questions Archive (Record #31899)
Mathematics
UTME / JAMB 1987
Algebra
Solution
Find the values of y which satisfy the simultaneous equations x + y = 5, x2 - 2y2 = 1
A
-12, -2
B
-12, 12
C
-12, +2
D
2, -2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x + y = 5.......(i)
x2 - 2y2 = 1.......(ii)
x = 5 - y.........(iii)
Subst. for x in eqn.(ii) = (5 - y)2 - 2y2 = 1
25- 10y + y2 - 2y2 = 1
25 - 1 = y2 + 10y
y2 + 10y2 - 24 = 0
(y + 12)(y - 2) = 0
Then Either y + 12 = 0 or y - 2 = 0
= (-12, +2)
Source: Myschool Past Questions Archive (Record #31901)
Mathematics
UTME / JAMB 1987
Algebra
Solution
Solve the inequality x - 1 > 4(x + 2)
A
x > -3
B
x < -3
C
2 < x < 3
D
-3 < x < -2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x - 1 > 4(x + 2) = x - 1 > 4x + 8
4x + 8 < x - 1 = 4x - x < -1 -8
= 3x < -9
∴ x < -3
Source: Myschool Past Questions Archive (Record #31909)
Mathematics
UTME / JAMB 1987
Algebra
Solution
The minimum value of y in the equation y = x\(^2\) - 6x + 8 is
A
8
B
3
C
7
D
-1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = x\(^2\) - 6x + 8
\(\frac{dy}{dx}\) = 2x - 6
\(\frac{dy}{dx}\) = 0
2x - 6 = 0
x = 3
\(\therefore\) y = 3\(^2\) - 6(3) + 8
= 9 - 18 + 8
= -1
Source: Myschool Past Questions Archive (Record #31914)
Mathematics
UTME / JAMB 1987
Algebra
Solution
Find the eleventh term of the progression 4, 8, 16.....
A
213
B
212
C
211
D
210
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
a = 4, r = \(\frac{4}{2}\)
\(\frac{8}{4}\) = 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212
Source: Myschool Past Questions Archive (Record #31921)
Mathematics
UTME / JAMB 1986
Algebra
Solution
The ages of Tosan and Isa differ by 6 and the product of their ages is 187. Write their ages in the form (x, y), where x > y.
A
(12, 6)
B
(23, 17)
C
(17, 11)
D
(18, 12)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x - y = 6.......(i)
xy = 187.......(ii)
From equation (i), x(6 + y)
sub. for x in equation (ii) = y(6 + y)
= 187
y2 + 6y = 187
y2 + 6y - 187 = 0
(y + 17)(y - 11) = 0
y = -17 or y = 11
y cannot be negative, y = 11
Sub. for y in equation(i) = x - 11
= 16
x = 6 + 11
= 17
∴(x, y) = (17, 11)
Source: Myschool Past Questions Archive (Record #32030)
Mathematics
UTME / JAMB 1986
Algebra
Solution
If x varies directly as y3 and x = 2 when y = 1, find x when y = 5
A
2
B
10
C
125
D
250
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x \(\alpha\) y3
x = ky3
k = \(\frac {x}{y^3}\)
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250
Source: Myschool Past Questions Archive (Record #32063)
Mathematics
UTME / JAMB 1986
Algebra
Solution
Find all real numbers x which satisfy the inequality \(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)
A
x < 11
B
x < -1
C
x > 6
D
x > 11
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)
= \(\frac{x + 1}{3}\) - 1 > \(\frac{x + 4}{5}\)
\(\frac{x + 1}{3}\) - \(\frac{x + 4}{5}\) - 1 > 0
= \(\frac{5x + 5 - 3x - 12}{15}\)
= 2x - 7 > 15
= 2x > 22
= x > 11
Source: Myschool Past Questions Archive (Record #32080)
Mathematics
UTME / JAMB 1986
Algebra
Solution
Solve the equation 3x\(^2\) + 6x - 2 = 0
A
x = -1 \(\pm\) \(\frac{\sqrt{3}}{3}\)
B
x = -1 \(\pm\) \(\frac{\sqrt{15}}{3}\)
C
x = -2 \(\pm\) 2
D
x = 3 \(\pm\) \(\frac{\sqrt{3}}{15}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
3x2 + 6x - 2 = 0
Using almighty formula i.e. x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2a}\)
a = 3, b = 6, c = -2
x = \(\frac{-6 \pm \sqrt{6^2 - 4(3)(-2)}}{2(3)}\)
x = \(\frac{-6 \pm \sqrt{36 + 24}}{6}\)
x = \(\frac{-6 \pm \sqrt{60}}{6}\)
x = \(\frac{-6 \pm \sqrt{4 \times 15}}{6}\)
x = \(-1 \pm \frac{\sqrt{15}}{3}\)
Source: Myschool Past Questions Archive (Record #32086)
Mathematics
UTME / JAMB 1986
Algebra
Solution
If \(5^{(x + 2y)} = 5\) and \(4^{(x + 3y)} = 16\), find \(3^{(x + y)}\).
A
7
B
1
C
3
D
27
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(5^{(x + 2y)} = 5\)
∴ x + 2y = 1.....(i)
\(4^{(x + 3y)} = 16 = 4^2\)
x + 3y = 2 .....(ii)
x + 2y = 1.....(i)
x + 3y = 2......(ii)
y = 1
Substitute y = 1 into equation (i)
\(x + 2y = 1 \implies x + 2(1) = 1\)
\(x + 2 = 1 \implies x = -1\)
\(\therefore 3^{(x + y)} = 3^{(-1 + 1)}\)
\(3^{0} = 1\)
Source: Myschool Past Questions Archive (Record #32111)
Mathematics
UTME / JAMB 1986
Algebra
Solution
Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0
A
1 and 4
B
-2 and 2
C
0 and 1
D
-1 and 0
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
16x - 5 x 4x + 4 = 0
(4x)2 - 5(4x) + 4 = 0
let 4x = y
y2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
y = 4 or 1
4x = 4
x = 1
4x = 1
i.e. 4x = 4o, x = 0
∴ x = 1 or 0
Source: Myschool Past Questions Archive (Record #32117)