JAMB UTME - Mathematics - 2011
Question 1 Report
Solve for x and y respectively in the simultaneous equations -2x - 5y = 3, x + 3y = 0
Answer Details
-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
Question 2 Report
Answer Details
y = 5x2 + 4 comparing with
y = mx + e
m = 52
Since they are perpendicular
m1m2 = -1
m2 = −1m1 = -1
52 = -1 x 25
The equator of the line is thus
y = mn + c (4, 2)
2 = -25 (4) + c
21 + 85 = c
c = 185
10+55 = c
y = -25 x + 185
5y = -2x + 18
or 5y + 2x - 18 = 0
Question 3 Report
| Class Interval | 3−5 | 6−8 | 9−11 |
| Frequency | 2 | 2 | 2 |
.
Find the standard deviation of the above distribution.
Answer Details
¯x = ∑fx∑f
= 8+14+202+2+2
= 426
¯x = 7
S.D = √∑f(x−¯x)2∑f
= √18+0+186
= √366
= √6
Question 4 Report
If 2q35 = 778, find q
Answer Details
2q35 = 778
2 x 52 + q x 51 + 3 x 50 = 7 x 81 + 7 x 80
2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1
50 + 5q + 3 = 56 + 7
5q = 63 - 53
q = 105
q = 2
Question 5 Report
Answer Details
dydx = d(x3)dx + d(x2)dx - d(x)dx + d(1)dx
dydx = 3x2 + 2x - 1 = 0
dydx = 3x2 + 2x - 1
At the maximum point dydx = 0
3x2 + 2x - 1 = 0
(3x2 + 3x) - (x - 1) = 0
3x(x + 1) -1(x + 1) = 0
(3x - 1)(x + 1) = 0
therefore x = 13 or -1
For the maximum point
d2ydx2 < 0
d2ydx2
6x + 2
when x = 13
dx2dx2
= 6(13
) + 2
= 2 + 2 = 4
d2ydx2
> o which is the minimum point
when x = -1
d2ydx2
= 6(-1) + 2
= -6 + 2 = -4
-4 < 0
therefore, d2ydx2 < 0
the maximum point is -1
Question 6 Report
Find \( \int_{0}^{1} \cos4 x\, dx \)
Answer Details
let u = 4x
dydx = 4
dx = dy4
∫10 cos u. dy4 = 14 ∫ cos u du
= 14 sin u + k
= 14 sin4x + k
Question 7 Report
The bar chart above shows the distribution of SS2 students in a school.
Find the total number of students
Answer Details
Question 8 Report
Simplify \(\left(\sqrt{2}+\frac{1}{\sqrt{3}}\right)\left(\sqrt{2}-\frac{1}{\sqrt{3}}\right)\)
Answer Details
√4−√2√3+√2√3−1√9
= 2 - 13
= 16−13
= 53
Question 9 Report
Rationalize \( \frac{2-\sqrt{5}}{3-\sqrt{5}} \)
Answer Details
(2−√5)(3+√5)(3−√5)(3+√5) = 6+2√5−3√5−√259+3√5−3√5−√25
= 6−√5−59−5
= 1−√54
Question 10 Report
Make R the subject of the formula if \(T = \frac{KR^2 + M}{3}\)
Answer Details
3T = KR2 + M
KR2 = 3T - M
R2 = 3T−MK
R = √3T−MK
Question 11 Report
Answer Details
midpoint = (5, 8)
x + 8 = 5
y+62 = 8
x + 8 = 10
x = 10 - 8 = 2
y + 6 = 16
y + 16 - 6 = 10
therefore, P(2, 10)
Question 12 Report
Answer Details
a + a + 1 + a + 2 + a + 3 = 34
4a = 34 - 6
4a = 28
a = 284
= 7
The least of these numbers is a = 7
Question 13 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 14 Report
Answer Details
N = (43, 44, 45,..., 60)
The universal set contains 18 numbers.
The prime numbers between 43 and 60 are 4
Probability of picking a prime number = 418
= 29
Question 15 Report
Answer Details
i.e. (n - 1)!
= (5 - 1)! = 4!
= 24 ways
Question 16 Report
Answer Details
28cm = 4L
L = 284 = 7cm
Area of square = L2
= 72
= 49cm2
Question 17 Report
From the venn diagram above, the complement of the set \(P \cap Q\) is given by
Answer Details
Question 18 Report
Answer Details
(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)
6x - 15 = 8x - 3
6x - 8x = 15 - 3
-2x = 12
x = 12−2
= -6
Question 19 Report
Answer Details
a = 3, d = 3, i = 36, n = 18
Sn = n2 [2a + (n - 1)d
S18 = 182 [2 x 3 + (18 - 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513
Question 20 Report
Answer Details
θ = tan-1(1) = 45o
The bearing of x from z is ₦45oE or 135o
Question 21 Report
| No | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 1 | 4 | 3 | 8 | 2 | 5 |
From the table above, find the median and range of the data respectively.
Answer Details
No012345F143825fx04624825
∑fx = 0 + 4 + 6 + 24 + 8 + 25 = 67
∑f = 23
Median = ∑fx∑f = 6723 = 2.913
= ≈ 3
Range = 5 - 0 = 5
(3, 5)
Question 22 Report
In a right angled triangle, if \(\tan \theta = \frac{3}{4}\). What is \(\cos \theta - \sin \theta\)?
Answer Details
from Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin θ = 35 and cosθ = 43
cosθ - sinθ
45 - 35 = 15
Question 23 Report
Answer Details
= 32y2 - 42x2
= (3y - 4x)(3y +4x)
Question 24 Report
In the diagram, STUV is a straight line. < TSY = < UXY = 40o and < VUW = 110o. Calculate < TYW
Answer Details
< TUW = 180∘ - 110∘ = 70∘
In △ XTU, < XUT + < TXU = 180∘
i.e. < YTS + 70∘ = 180
< XTU = 180 - 110∘ = 70∘
Also < YTS + < XTU = 180 (< s on a straight line)
i.e. < YTS + < XTU - 180(< s on straight line)
i.e. < YTS + 70∘ = 180
< YTS = 180 - 70 = 110∘
in △ SYT + < YST + < YTS = 180∘ (Sum of interior < s)
SYT + 40 + 110 = 180
< SYT = 180 - 150 = 30
< SYT = < XYW (vertically opposite < s)
Also < SYX = < TYW (vertically opposite < s)
but < SYT + < XYW + < SYX + < TYW = 360
i.e. 30 + 30 + < SYX + TYW = 360
but < SYX = < TYW
60 + 2(< TYW) = 360
2(< TYW) = 360∘ - 60
2(< TYW) = 300∘
TYW = 3002 = 150∘
< SYT
Question 25 Report
Evaluate \(\begin{vmatrix}4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3\end{vmatrix}\)
Answer Details
4 ∣∣∣3−113∣∣∣ -2 ∣∣∣2−1−13∣∣∣ -1 ∣∣∣23−11∣∣∣
4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)]
= 4[9 + 1] -2 [6 - 1] -1 [2 + 3]
= 4(10) - 2(5) - 1(5)
= 40 - 10 - 5
= 25
Question 26 Report
Answer Details
2x + 1 d(3x+1)d + (3x + 1) d(2x+1)d
2x + 1 (3) + (3x + 1) (2)
6x + 3 + 6x + 2 = 12x + 5
Question 27 Report
Answer Details
= −6X3+n−2X2+2X−3
= (−2X2−2)2X−1
Remainder is 2X - 1
Question 28 Report
Evaluate \( \int_{2}^{1} (3 - 2x)\,dx \)
Answer Details
[3x - x2]o
[3(1) - (1)2] - [3(0) - (0)2]
(3 - 1) - (0 - 0) = 2 - 0
= 2
Question 29 Report
Answer Details
x2 + 2x - 15 > 0
(x2 + 5x) - (3x - 15) > 0
x(x + 5) - 3(x + 5) >0
(x - 3)(x + 5) > 0
therefore, x = 3 or -5
then x < -5 or x > 3
i.e. x< 3 or x < -5
Question 30 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
Answer Details
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, T4Tr = ar3ar = 164
r2 = 4 and r = 2
but ar = 4
a = 4r = 42
a = 2
Sn = a(rn−1)r−1
S5 = 2(25−1)2−1
= 2(32−1)2−1
= 2(31)
= 62
Question 32 Report
Answer Details
Question 33 Report
Answer Details
Question 34 Report
Simplify \(\frac{3\frac{2}{3}\times\frac{5}{6}\times\frac{2}{3}}{\frac{11}{15}\times\frac{3}{4}\times\frac{2}{27}}\)
Answer Details
113×56×231115×34×227
11054÷661620
50
Question 35 Report
Answer Details
|OA|2 = |AN|2 + |ON|2
72 = |AN|2 + (5)2
49 = |AN|2 + 25
|AN|2 = 49 - 25 = 24
|AN| = √24
= √4×6
= 2√6 cm
|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)
|AN| + |NB| = |AB|
2√6 + 2√6 = |AB|
|AB| = 4√6 cm
Question 36 Report
\(T\) varies inversely as the cube of \(R\). When \(R = 3\), \(T = \frac{2}{81}\), find \(T\) when \(R = 2\)
Answer Details
T = kR3
k = TR3
= 281 x 33
= 281 x 27
dividing 81 by 27
k = 22
therefore, T = 23 x 1R3
When R = 2
T = 23 x 123 = 23 x 18
= 112
Question 37 Report
Find the derivative of \( \frac{\sin \theta}{\cos \theta} \)
Answer Details
cosθd(sinθ)dθ−sinθd(cosθ)dθcos2θ
cosθ.cosθ−sinθ(−sinθ)cos2θ
cos2θ+sin2θcos2θ
Recall that sin2 θ + cos2 θ = 1
1cos2θ = sec2 θ
Question 38 Report
Answer Details
log183 + log33 - logx3 = 3log33
log183 + log33 - logx3 = log333
log3(18×3X ) = log333
18×3X = 33
18 x 3 = 27 x X
x = 18×327
= 2
Question 39 Report
Answer Details
33 = 27cm3
volume of rectangular tank = L x B X h
= 3 x 4 x 5
= 60cm3
volume of H2O the tank can now hold
= volume of rectangular tank - volume of cube
= 60 - 27
= 33cm3
Question 40 Report
Answer Details
= 7 x 6 x 5
= 210 ways
or 7P3 = 7!(7−3)! = 7!4!
= 7×6×5×4!4!
= 210 ways
Question 41 Report
If the numbers M, N, Q are in the ratio 5:4:3, find the value of \( \frac{2N-Q}{M} \)
Answer Details
i.e M = 5, N = 4, Q = 3
Substituting values into equation, we have...
2N−QM
= 2(4)−35
= 8−35
= 55
= 1
Question 42 Report
If x varies directly as square root of y and x = 81 when y = 9, Find x when y = 1 \( \frac{7}{9} \)
Answer Details
x = k√y
81 = k√9
k = 813
= 27
therefore, x = 27√y
y = 179 = 169
x = 27 x √169
= 27 x 43
dividing 27 by 3
= 9 x 4
= 36
Question 43 Report
A binary operation \( \oplus \) on real numbers is defined by \( x \oplus y = xy + x + y \) for two real numbers \( x \) and \( y \). Find the value of \( 3 \oplus -\frac{2}{3} \).
Answer Details
3 + -23 = 3(- 23 ) + 3 + (- 23 )
= -2 + 3 -23
= 1−21−3
= 13
Question 44 Report
The pie chart shows the distribution of courses offered by students. What percentage of the students offer English?
Answer Details
=25%
Question 45 Report
In a right angled triangle, if \( \tan \theta = \frac{3}{4} \). What is \( \cos \theta - \sin \theta \)?
Answer Details
from Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin θ = 35 and cosθ = 43
cosθ - sinθ
45 - 35 = 15
Question 46 Report
The inverse of matrix \(N = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix}\) is
Answer Details
N-1 = adjN|N|
adj N = ∣∣∣4−3−12∣∣∣
|N| = (2 x4) - (1 x 3)
= 8 - 3
=5
N-1 = 15 ∣∣∣4−3−12∣∣∣
Question 47 Report
Answer Details
but, n = 12
= (12 -2)180
= 10 x 180
= 1800
let each interior angle = x
x = (n−2)180n
x = 180012
= 150o
Question 48 Report
| Class Intervals | 0 − 2 | 3 − 5 | 6 − 8 | 9 − 11 |
| Frequency | 3 | 2 | 5 | 3 |
Find the mode of the above distribution.
Answer Details
D1 = frequency of modal class - frequency of the class before it
D1 = 5 - 2 = 3
D2 = frequency of modal class - frequency of the class that offers it
D2 = 5 - 3 = 2
L1 = lower class boundary of the modal class
L1 = 5 - 5
C is the class width = 8 - 5.5 = 3
Mode = L1 + (D1D1+D2 )C
= 5.5 + 32+3 C
= 5.5 + 35 x 3
= 5.5 + 95
= 5.5 + 1.8
= 7.3 ≈ = 7
Question 49 Report
Answer Details
Since there are 7 letters. The first letter can be arranged in 7 ways, , the second letter in 6 ways, the third letter in 5 ways, the 4th letter in four ways, the 3rd letter in three ways, the 2nd letter in 2 ways and the last in one way.
therefore, 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7! ways
