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Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
Simplify \((\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}\)
A
\(\frac{2}{3}\)
B
\(\frac{1}{2}\)
C
\(\frac{8}{9}\)
D
\(\frac{1}{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}\)
\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}\)
\((\frac{2^4}{3^4})^{\frac{1}{4}} \div (\frac{4^2}{3^2})^{\frac{1}{2}}\)
\(\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}\)
\(\frac{2}{3} \div \frac{4}{3}\)
\(\frac{2}{3} \times \frac{3}{4}\)
\(\frac{2}{4}\)
\(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #27418)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
If log318 + log33 - log3x = 3, Find x.
A
1
B
2
C
o
D
3
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3
log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3log33
log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = log333
log3(\(\frac{18 \times 3}{X}\)) = log333
\(\frac{18 \times 3}{X}\) = 33
18 x 3 = 27 x X
x = \(\frac{18 \times 3}{27}\)
= 2
Source: Myschool Past Questions Archive (Record #29335)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
Rationalize \(\frac{2 - \sqrt5}{3 - \sqrt5}\)
A
\(\frac{1 - \sqrt5}{2}\)
B
\(\frac{1 - \sqrt5}{4}\)
C
\(\frac{ \sqrt5 - 1}{2}\)
D
\(\frac{1 + \sqrt5}{4}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{2 - \sqrt5}{3 - \sqrt5}\) x \(\frac{3 + \sqrt5}{3 + \sqrt5}\)
\(\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}\) = \(\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}\)
= \(\frac{6 - \sqrt5 - 5}{9 - 5}\)
= \(\frac{1 - \sqrt5}{4}\)
Source: Myschool Past Questions Archive (Record #29353)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
Simplify (\(\sqrt2 + \frac{1}{\sqrt3})(\sqrt2 - \frac{1}{\sqrt3}\))
A
\(\frac{7}{3}\)
B
\(\frac{5}{3}\)
C
\(\frac{5}{2}\)
D
\(\frac{3}{2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
(\(\sqrt2 + \frac{1}{\sqrt3})(\sqrt2 - \frac{1}{\sqrt3}\))
\(\sqrt4 - \frac {\sqrt2}{\sqrt3} + \frac {\sqrt2}{\sqrt3} - \frac {1}{\sqrt9}\)
= 2 - \(\frac {1}{3}\)
= \(\frac {16 - 1}{3}\)
= \(\frac{5}{3}\)
Source: Myschool Past Questions Archive (Record #29358)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
Raial has 7 different posters to be hanged in her bedroom, living room and kitchen. Assuming she has plans to place at least a poster in each of the 3 rooms, how many choices does she have?
A
49
B
170
C
21
D
210
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
The first poster has 7 ways to be arranges, the second poster can be arranged in 6 ways and the third poster in 5 ways.
= 7 x 6 x 5
= 210 ways
or \(\frac{7}{P_3}\) = \(\frac{7!}{(7 - 3)!}\) = \(\frac{7!}{4!}\)
= \(\frac{7 \times 6 \times 5 \times 4!}{4!}\)
= 210 ways
Source: Myschool Past Questions Archive (Record #29369)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
Make R the subject of the formula if T = \(\frac {KR^2 + M}{3}\)
A
\(\sqrt\frac{3T - K}{M}\)
B
\(\sqrt\frac{3T - M}{K}\)
C
\(\sqrt\frac{3T + K}{M}\)
D
\(\sqrt\frac{3T - K}{M}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
T = \(\frac{KR^2 + M}{3}\)
3T = KR2 + M
KR2 = 3T - M
R2 = \(\frac{3T - M}{K}\)
R = \(\sqrt\frac{3T - M}{K}\)
Source: Myschool Past Questions Archive (Record #29375)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
Find the remainder when X3 - 2X2 + 3X - 3 is divided by X2 + 1
A
2X - 1
B
X + 3
C
2X + 1
D
X - 3
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
X2 + 1 \(\frac{X - 2}{\sqrt{X^3 - 2X^2 + 3n - 3}}\)
= \(\frac {- 6X^3 + n}{-2X^2 + 2X - 3}\)
= \(\frac{(-2X^2 - 2)}{2X - 1}\)
Remainder is 2X - 1
Source: Myschool Past Questions Archive (Record #29376)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
Factorize completely 9y2 - 16X2
A
(3y - 2x)(3y + 4x)
B
(3y + 4x)(3y + 4x)
C
(3y + 2x)(3y - 4x)
D
(3y - 4x)(3y + 4x)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
9y2 - 16x2
= 32y2 - 42x2
= (3y - 4x)(3y +4x)
Source: Myschool Past Questions Archive (Record #29378)
Mathematics
UTME / JAMB 2011
Algebra
Solution
Solve for x and y respectively in the simultaneous equations -2x - 5y = 3. x + 3y = 0
A
-3, -9
B
9, -3
C
-9,3
D
3, -9
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
-2x -5y = 3
x + 3y = 0
x = -3y
-2 (-3y) - 5y = -3
6y - 5y = 3
y = 3
but, x = -3y
x = -3(3)
x = -9
therefore, x = -9, y = 3
Source: Myschool Past Questions Archive (Record #29379)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
If x varies directly as square root of y and x = 81 when y = 9, Find x when y = 1\(\frac{7}{9}\)
A
20\(\frac{1}{4}\)
B
27
C
2\(\frac{1}{4}\)
D
36
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x \(\alpha\sqrt y\)
x = k\(\sqrt y\)
81 = k\(\sqrt9\)
k = \(\frac{81}{3}\)
= 27
therefore, x = 27\(\sqrt y\)
y = 1\(\frac{7}{9}\) = \(\frac{16}{9}\)
x = 27 x \(\sqrt{\frac{16}{9}}\)
= 27 x \(\frac{4}{3}\)
dividing 27 by 3
= 9 x 4
= 36
Source: Myschool Past Questions Archive (Record #29381)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
T varies inversely as the cube of R. When R = 3, T = \(\frac{2}{81}\), find T when R = 2
A
\(\frac{1}{18}\)
B
\(\frac{1}{12}\)
C
\(\frac{1}{24}\)
D
\(\frac{1}{6}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
T \(\alpha \frac{1}{R^3}\)
T = \(\frac{k}{R^3}\)
k = TR3
= \(\frac{2}{81}\) x 33
= \(\frac{2}{81}\) x 27
dividing 81 by 27
k = \(\frac{2}{2}\)
therefore, T = \(\frac{2}{3}\) x \(\frac{1}{R^3}\)
When R = 2
T = \(\frac{2}{3}\) x \(\frac{1}{2^3}\) = \(\frac{2}{3}\) x \(\frac{1}{8}\)
= \(\frac{1}{12}\)
Source: Myschool Past Questions Archive (Record #29385)
Mathematics
UTME / JAMB 2011
Algebra
Solution
Solve the inequality -6(x + 3) \(\leq\) 4(x - 2)
A
x \(\leq\) 2
B
x \(\geq\) -1
C
x \(\geq\) -2
D
x \(\leq\) -1
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
-6(x + 3) \(\leq\) 4(x - 2)
-6(x +3) \(\leq\) 4(x - 2)
-6x -18 \(\leq\) 4x - 8
-18 + 8 \(\leq\) 4x +6x
-10 \(\leq\) 10x
10x \(\geq\) -10
x \(\geq\) -1
Source: Myschool Past Questions Archive (Record #29386)
Mathematics
UTME / JAMB 2011
Algebra
Solution
Find the sum of the first 18 terms of the series 3, 6, 9,..., 36.
A
505
B
513
C
433
D
635
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
3, 6, 9,..., 36.
a = 3, d = 3, i = 36, n = 18
Sn = \(\frac{n}{2}\) [2a + (n - 1)d
S18 = \(\frac{18}{2}\) [2 x 3 + (18 - 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513
Source: Myschool Past Questions Archive (Record #29400)
Mathematics
UTME / JAMB 2011
Algebra
Solution
The seconds term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms
A
60
B
62
C
54
D
64
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
T2 = 4, T4 = 16
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, \(\frac{T_4}{T_r}\) = \(\frac{ar^3}{ar}\) = \(\frac{16}{4}\)
r2 = 4 and r = 2
but ar = 4
a = \(\frac{4}{r}\) = \(\frac{4}{2}\)
a = 2
Sn = \(\frac{a(r^n - 1)}{r - 1}\)
S5 = \(\frac{2(2^5 - 1)}{2 - 1}\)
= \(\frac{2(32 - 1)}{2 - 1}\)
= 2(31)
= 62
Source: Myschool Past Questions Archive (Record #29413)
Mathematics
UTME / JAMB 2011
Number and Numeration
Solution
A binary operation \(\oplus\) om real numbers is defined by x \(\oplus\) y = xy + x + y for two real numbers x and y. Find the value of 3 \(\oplus\) - \(\frac{2}{3}\).
A
- \(\frac{1}{2}\)
B
\(\frac{1}{3}\)
C
-1
D
2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
N + Y = XY + X + Y
3 + -\(\frac{2}{3}\) = 3(- \(\frac{2}{3}\)) + 3 + (- \(\frac{2}{3}\))
= -2 + 3 -\(\frac{2}{3}\)
= \(\frac{1 - 2}{1 - 3}\)
= \(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #29421)