JAMB UTME - Mathematics - 2010
Question 1 Report
Answer Details
Using the binomial prob. distribution,
(H + T)3 = H3 + 3H2T1 + 3HT2 + T3
Hence the probability that three heads show in a toss of the three coins is H3
= (12 )3
= 18
Question 2 Report
Make Q the subject of formula if \( p = \frac{M}{5}(X + Q) + 1 \)
Answer Details
Question 3 Report
| Marks | 1 | 2 | 3 | 4 | 5 |
| Frequency | 2 | 2 | 8 | 4 | 4 |
The table above show the marks obtained in a given test.
Find the mean mark
Answer Details
Question 4 Report
Solve the inequality \( (x - 3)(x - 4) \le 0 \)
Answer Details
Case 1 (+, -) = x - 3 ≥ 0, X - 4 ≥ 0
= X ≤ 3, x ≥ 4
= 3 < x ≥ 4 (solution)
Case 2 = (-, +) = x - 3 ≤
0, x - 4 ≥
0
= x ≤
3, x ≥
4
therefore = 3 ≤
x ≤
4
Question 5 Report
If x is inversely proportional to y and x = \( \frac{1}{2} \) when y = 2, find x if y = 4
Answer Details
x = k x 1y .........(2)
When x = 212
= 52 , y = 2
(2) becomes 52 = k x 12
giving k = 5
from (2), x = 5y
so when y =4, x = 5y = 114
Question 6 Report
Answer Details
Question 7 Report
If the area of \( \triangle PQR \) above is \(12\sqrt{3}\text{ cm}^2\), find the value of q?
Answer Details
Using Sin 60∘ = hq
h = q sin 60∘
So A = 12b(qsin60o)
12√3=12×8×q×√33
12√3 - 2q√3
q = 122=6 cm
Question 8 Report
For what range of values of x is \( \frac{1}{2}x + \frac{1}{4} > \frac{1}{3}x + \frac{1}{2} \)?
Answer Details
Multiply through by through by the LCM of 2, 3 and 4
12 x 12 x + 12 x 14 > 12 x 13 x + 12 x 12
6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3
2x2 > 32
x > 32
Question 9 Report
Answer Details
6 x 82 + r x 81 + 7 x 8o = 5 x 92 + 1 x 91 + 1 x 9o
6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1
384 + 8r + 7 = 405 + 9 + 1
391 + 8r = 24
r = 248
= 3
Question 10 Report
If \(x \ast y = x + y^2\), find the value of \((2 \ast 3) \ast 5\)
Answer Details
X ∗ y = X + y2
(2 ∗ 3) ∗ 5 = (2 + 32)∗ 5
= (2 + 9)∗ 5 = 11 ∗ 5
Hence 11 ∗ 5 = 11 + 52
= 11 + 25 = 36
Question 11 Report
Evaluate \( \int_{1}^{3} (X^2 - 1)\,dx \)
Answer Details
Question 12 Report
Answer Details
x2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x = 124
= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1
Question 13 Report
Answer Details
Let the gradient of the line parallel to the given line be m2,
then, m2 = m1 = -4 (condition for parallelism)
Using: y - y1 = m2(x - x1)
Hence the equation of the parallel line is
y - 3 = -4(x-2)
y - 3 = -4 x + 8
y + 4x = 8 + 3
y + 4x = 11
y + 4x - 11 = 0
Question 14 Report
| Rationalise | 2√3+√5 |
| √5-√3 |
Answer Details
Question 15 Report
If \(p : q = \frac{2}{3} : \frac{5}{6}\) and \(q : r = \frac{3}{4} : \frac{1}{2}\), find \(p : q : r\)
Answer Details
If q : r = 34 : 12 , then the sum S2 of ratio = 34 + 12 = 54
Let p + q = T1, then
q = (56÷96 )T1 = (56×69 )T1 = 59 T1
Again, let q + r = T2, then
q = (34÷54 )T2 = (34×45 )T2 = 35 T2
Using q = q
59 T1 = 35 T2
5 x 5T1 = 9 x 3T2
T1T2 = 9×35x5 = 275
Giving that, T1 = 27 and T2 = 25
P = (23÷S1 )T1 = (23÷96 )T1
= (23×69 )27 = 12
q = (56÷S1 )T1 = (56÷96 )T1
= (56×69 )27 = 15
and r = (12÷S2 )T2 = (12÷54 )T2
= (12×45 )25 = 10
Hence p : q : r = 12: 15 : 10
Question 16 Report
If the area of ΔPQR above is 12√3 cm2, find the value of q?
Answer Details
12√3 = 1/2 x 8 x q sin 60
12√3 = 4 x q x √3/2
12√3 = 2q√3
| q = | 12√3 |
| 2√3 |
q = 6
Question 17 Report
| Marks | 1 | 2 | 3 | 4 | 5 |
| Frequency | 2 | 2 | 8 | 4 | 4 |
The table above show the marks obtained in a given test.
How many student too the test
Answer Details
Question 18 Report
Answer Details
___________________________________
¯x = ∑xN
= 164
= 4
S = √(x−¯x)2N
= √(10)4
= √(5)2
Question 19 Report
Answer Details
Question 20 Report
If \( \cos \theta = \frac{12}{13} \). Find \( \theta + \cos^2 \theta \)
Answer Details
x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tanθ = 512 and cosθ = 1213
If cos2θ = 1 + 1tan2θ
= 1 + 1(5)212
= 1 + 125144
= 1 + 14425
= 25+14425
= 16925
Question 21 Report
Answer Details
dy/dx = 3(2x + 1)3-1 x 2
= 3(2x + 1)2 x 2
= 6(2x + 1)2
Question 22 Report
| 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?
Answer Details
Question 23 Report
If \( y = (2x + 1)^3 \), find \( \frac{dy}{dx} \)
Answer Details
Let u = 2x + 1 so that, y = u3
dydu = 3u2 and dydx = 2
Hence by the chain rule,
dydx = dydu x dudx
= 3u2 x 2
= 6u2
= 6(2x + 1)2
Question 24 Report
Find the distance between the points \( \left(\frac{1}{2}, -\frac{1}{2}\right) \).
Answer Details
D = √(x2−x1)2+(y2−y1)2
= √(−12−12)2+(−12−12)2
= √(−1)2+(−1)2
= √1+1
= √2
Question 25 Report
Evaluate \( \left(\frac{81}{16}\right)^{-\frac{1}{4}} \times 2^{-1} \)
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Question 26 Report
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Question 27 Report
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Question 28 Report
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Question 29 Report
Simplify \( \left(\frac{16}{81}\right)^{\frac{1}{4}} \div \left(\frac{9}{16}\right)^{-\frac{1}{2}} \)
Answer Details
(1681)14÷(916)-12
(1681)14÷(169)12
(2434)14÷(4232)12
24×1434×14÷42×1232×12
23÷43
23×34
24
12
Question 31 Report
If \( P = \begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix} \)
Answer Details
|P| = 2 - 1 x -3 = 5
P-1 = 15 (13?12)
= (1535?1525)
Question 32 Report
If \(y = x \sin x\), Find \(\frac{d^2y}{d^2x}\)
Answer Details
Question 33 Report
Answer Details
2 * 3 = 2 + 32
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 52
= 11 + 25
= 36
Question 34 Report
| Marks | 1 | 2 | 3 | 4 | 5 |
| Frequency | 2 | 2 | 8 | 4 | 4 |
The table above shows the marks obtained in a given test. How many students took the test?
Answer Details
Question 35 Report
In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find < RPS
Answer Details
< PRS = 90? (Angle in a semicircle)
< PRS = 55? (Angle between a chord and a tangent = Angle in the alternate segment)
< PSR + < PRS + < RSP = 180?
90v + 55? + < RSP = 180?
< RSP = 180? - 145?
= 35?
Question 36 Report
Answer Details
But 9x2 + 6xy + 4y2 = (3x +2y)2
So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2
Hence the other factor is 3x - 2y
Question 37 Report
Answer Details
when n = 4
x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o
5x - 50o = (8 - 4)90o
5x - 50o = 4 x 90o = 360o
5x = 360o + 50o
5x = 410o
x = 410o5
= 82o
Hence, the value of the least interior angle is (x - 30o)
= (82 - 30)o
= 52o
Question 38 Report
Simplify \( \frac{3}{5} \div \left(\frac{2}{7} x \frac{4}{3} \div \frac{4}{9}\right) \)
Answer Details
= 35 ÷ 67
= 35 x 76
= 710
Question 39 Report
From the cyclic quadrilateral TUVW above, find the value of x
Answer Details
3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3χ + 108 = 180
3χ = 180 - 108
3χ = 72
χ = 72/3χ = 24∘
Question 40 Report
actorize completely \( \frac{x^3+3x^2-10x}{2x^2-8} \)
Answer Details
= x(x2+5x−2x−10)2(x+2)(x−2)
= x(x−2)(x+5)2(x+2)(x−2)
= x(x+5)2(x+2)
Question 41 Report
Answer Details
r = 0.050.5 = 110
S∞ = 0.5110
= 0.5(910)
= 0.5×109
= 59
Question 42 Report
An arc subtends an angle of \(50^\circ\) at the center of circle of radius 6cm. Calculate the area of the sector formed
Answer Details
| Area of a sector = | θ | x πr2 |
| 360 |
= 50/360 x 22/7 x 36/1
= 110/7
Question 43 Report
Evaluate \(\begin{vmatrix}2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1\end{vmatrix}\)
Answer Details
= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)
= 2(-21) - 0 + 5(-12)
= -42 + 5(-12)
= -42 - 60
= -102
Question 44 Report
If \( \begin{vmatrix} x & 3 \\ 2 & 7 \end{vmatrix} = 15 \), find the value of x
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Question 45 Report
Solve the inequality \( -6(x + 3) \le 4(x - 2) \)
Answer Details
-6(x +3) ≤ 4(x - 2)
-6x -18 ≤ 4x - 8
-18 + 8 ≤ 4x +6x
-10x ≤ 10x
10x ≤ -10
x ≤ 1
Question 46 Report
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Question 47 Report
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Question 49 Report
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Question 50 Report
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