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Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If y varies directly as the square root of x and y = 3 when x = 16. Calculate y when x = 64
A
12
B
6
C
3
D
5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
y ∝ \sqrtx
y = K\sqrtx
K = y/\sqrtx = 3/\sqrt16
= 3/4
y = 3/4\sqrtx
= 3/4\sqrt64 when x = 64
= 3/4 x 8/1
= 6
Source: Myschool Past Questions Archive (Record #9796)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If x * y = x + y\(^2\), find then value of (2*3)*5
A
36
B
25
C
11
D
55
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x * y = x + y\(^2\)
2 * 3 = 2 + 3\(^2\)
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 5\(^2\)
= 11 + 25
= 36
Source: Myschool Past Questions Archive (Record #9802)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If p and q are two non zero numbers and 18(p+q) = (18+p)q, which of the following must be true?
A
q = 18
B
p <1
C
p = 18
D
q < 1
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
If 18(p + q) = (18 + p)q
then 18p + 18q = 18q + pq
18p = pq \(\implies\) q = 18.
Source: Myschool Past Questions Archive (Record #9803)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If two smaller sides of a right angled triangle are 4cm and 5cm, find its area
A
10 cm2
B
6 cm2
C
24 cm2
D
8 cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of a triangle = 1/2 base x height
= 1/2 x 4 x 5
= 10 cm2
Source: Myschool Past Questions Archive (Record #9951)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If y = (2x + 1)\(^3\) find \(\frac{\text{d}}{\text{dx}}\)
A
3(2x+1)2
B
6(2x+1)
C
3(2x+1)
D
6(2x+1)2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = (2x + 1)\(^3\)
\(\frac{dy}{dx}\) = 3(2x + 1)\(^{3-1}\) x 2
= 3(2x + 1)\(^2\)x 2
= 6(2x + 1)\(^2\)
Source: Myschool Past Questions Archive (Record #9959)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
Marks
2
3
4
5
6
7
8
No. of students
3
1
5
2
4
2
3
from the table above, if the pass mark is 5, how many students failed the test?
A
6
B
2
C
7
D
9
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Marks
Number of students
2
3
3
1
4
5
Total
9
Source: Myschool Past Questions Archive (Record #10316)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
Find r, if 6r7\(_8\) = 511\(_9\)
A
3
B
2
C
6
D
5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
6r7\(_8\) = 511\(_9\)
6 x 8\(^2\) + r x 8\(^1\) + 7 x 8\(^0\) = 5 x 9\(^2\) + 1 x 9\(^1\) + 1 x 9\(^0\)
6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1
384 + 8r + 7 = 405 + 9 + 1
8r = 415 - 391
8r = 24
r = \(\frac{24}{8}\)
= 3
Source: WAEC / JAMB / NECO Classroom Archive (Record #29544)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
Simplify (\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
A
\(\frac{1}{5}\)
B
\(\frac{1}{4}\)
C
\(\frac{1}{36}\)
D
\(\frac{1}{25}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
Applying the rule of BODMAS, we have:
(\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\)
(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)
\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)
= \(\frac{1}{36}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #29546)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
A student measures a piece of rope and found that it was 1.26m long. If the actual length of the rope was 1.25m, what was the percentage error in the measurement?
A
0.25%
B
0.01%
C
0.80%
D
0.40%
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Actual length of rope = 1.25m
Measured length of rope = 1.26m
error = (1.26 - 1.25)m - 0.01m
Percentage error = \(\frac{error}{\text{actual length}}\) x 100%
= \(\frac{0.01}{1.25}\) x 100%
= 0.8%
Source: WAEC / JAMB / NECO Classroom Archive (Record #29552)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\) and q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), find p : q : r
A
12 : 15 : 10
B
12 : 15 : 16
C
10 : 15 : 24
D
9 : 10 : 15
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\)
If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\)
Let p + q = T1, then
q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1
Again, let q + r = T2, then
q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2
Using q = q
\(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2
5 x 5T1 = 9 x 3T2
\(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{25}\)
Giving that, T1 = 27 and T2 = 25
P = (\(\frac{2}{3} \div S_1\))T1 = (\(\frac{2}{3} \div \frac{9}{6}\))T1
= (\(\frac{2}{3} \times \frac{6}{9}\))27 = 12
q = (\(\frac{5}{6} \div S_1\))T1 = (\(\frac{5}{6} \div \frac{9}{6}\))T1
= (\(\frac{5}{6} \times \frac{6}{9}\))27 = 15
and r = (\(\frac{1}{2} \div S_2\))T2 = (\(\frac{1}{2} \div \frac{5}{4}\))T2
= (\(\frac{1}{2} \times \frac{4}{5}\))25 = 10
Hence p : q : r = 12: 15 : 10
Source: WAEC / JAMB / NECO Classroom Archive (Record #29554)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If x is inversely proportional to y and x = 2\(\frac{1}{2}\) when y = 2, find x if y = 4
A
4
B
5
C
1\(\frac{1}{4}\)
D
2\(\frac{1}{4}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x \(\alpha\) \(\frac{1}{y}\) .........(1)
x = k x \(\frac{1}{y}\) .........(2)
When x = 2\(\frac{1}{2}\)
= \(\frac{5}{2}\), y = 2
(2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\)
giving k = 5
from (2), x = \(\frac{5}{y}\)
so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #29570)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
W is directly proportional to U. If W = 5 when U = 3, find U when W = \(\frac{2}{7}\)
A
\(\frac{6}{35}\)
B
\(\frac{10}{21}\)
C
\(\frac{21}{10}\)
D
\(\frac{35}{6}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
W \(\alpha\) U
W = ku
u = \(\frac{w}{k}\); \(\frac{2}{7}\) x \(\frac{3}{5}\)
= \(\frac{6}{35}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #29652)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
Make Q the subject of formula if p = \(\frac{M}{5}\)(X + Q) + 1
A
\(\frac{5P - MX + 5}{M}\)
B
\(\frac{5P - MX - 5}{M}\)
C
\(\frac{5P + MX + 5}{M}\)
D
\(\frac{5P + MX - 5}{M}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
p = \(\frac{M}{5}\)(X + Q) + 1
P - 1 = \(\frac{M}{5}\)(X + Q)
\(\frac{5}{M}\)(p - 1) = X + Q
\(\frac{5}{M}\)(p - 1)- x = Q
Q = \(\frac{5(p -1) - Mx}{M}\)
= \(\frac{5p - 5 - Mx}{M}\)
= \(\frac{5p - Mx - 5}{M}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #29565)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
If 9x\(^2\) + 6xy + 4y\(^2\) is a factor of 27x\(^3\) - 8y\(^3\), find the other factor.
A
2y + 3x
B
2y - 3x
C
3x + 2y
D
3x - 2y
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Given that \( 9x^2 + 6xy + 4y^2 \) is a factor of \( 27x^3 - 8y^3 \), we can express:
\(27x^3 - 8y^3 = (3x - 2y)(9x^2 + 6xy + 4y^2)\)
Thus, the other factor is: 3x - 2y
Source: WAEC / JAMB / NECO Classroom Archive (Record #29566)
Mathematics
UTME / JAMB 2010
Number and Numeration
Solution
For what range of values of x is \(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)?
A
x < \(\frac{3}{2}\)
B
x > \(\frac{3}{2}\)
C
x < -\(\frac{3}{2}\)
D
x > -\(\frac{3}{2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)
Multiply through by through by the LCM of 2, 3 and 4
12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\)
6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3
\(\frac{2x}{2}\) > \(\frac{3}{2}\)
x > \(\frac{3}{2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #29571)