JAMB UTME - Mathematics - 2011

Find the remainder when X3 - 2X2 + 3X - 3 is divided by X2 + 1

Maelezo ya Majibu

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

T varies inversely as the cube of R. When R = 3, T = $\frac{2}{81}$, find T when R = 2

Maelezo ya Majibu

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 $\frac{1}{8}$
= $\frac{1}{12}$

The bar chart above shows the distribution of SS2 students in a school.

Find the total number of students

Maelezo ya Majibu

Evaluate $\left|\begin{array}{ccc}4& 2& -1\\ 2& 3& -1\\ -1& 1& 3\end{array}\right|$

Maelezo ya Majibu

$\left|\begin{array}{ccc}4& 2& -1\\ 2& 3& -1\\ -1& 1& 3\end{array}\right|$
4 $\left|\begin{array}{cc}3& -1\\ 1& 3\end{array}\right|$ -2 $\left|\begin{array}{cc}2& -1\\ -1& 3\end{array}\right|$ -1 $\left|\begin{array}{cc}2& 3\\ -1& 1\end{array}\right|$
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

$\begin{array}{ccccccc}\text{No}& 0& 1& 2& 3& 4& 5\\ \text{Frequency}& 1& 4& 3& 8& 2& 5\end{array}$.

From the table above, find the median and range of the data respectively.

Maelezo ya Majibu

Find the sum of the first 18 terms of the series 3, 6, 9,..., 36.

Maelezo ya Majibu

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

A man invested ₦5,000 for 9 months at 4%. What is the simple interest?

Maelezo ya Majibu

S.I. = $\frac{P\times R\times T}{100}$
If T = 9 months, it is equivalent to $\frac{9}{12}$ years
S.I. = $\frac{5000\times 4\times 9}{100\times 12}$
S.I. = ₦150

Find the probability that a number picked at random from the set(43, 44, 45, ..., 60) is a prime number.

Maelezo ya Majibu

If log318 + log33 - log3x = 3, Find x.

Maelezo ya Majibu

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

Rationalize $\frac{2-\sqrt{5}}{3-\sqrt{5}}$

Maelezo ya Majibu

$\frac{2-\sqrt{5}}{3-\sqrt{5}}$ x $\frac{3+\sqrt{5}}{3+\sqrt{5}}$
$\frac{(2-\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$ = $\frac{6+2\sqrt{5}-3\sqrt{5}-\sqrt{2}5}{9+3\sqrt{5}-3\sqrt{5}-\sqrt{2}5}$
= $\frac{6-\sqrt{5}-5}{9-5}$
= $\frac{1-\sqrt{5}}{4}$

$\begin{array}{cccccc}\text{Class Intervals}& 0-2& 3-5& 6-8& 9-11& \\ \text{Frequency}& 3& 2& 5& 3& \end{array}$

Find the mode of the above distribution.

Maelezo ya Majibu

Mode = L1 + ($\frac{D_1}{D_1+D_2}$)C
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 + ($\frac{D_1}{D_1+D_2}$)C
= 5.5 + $\frac{3}{2+3}$C
= 5.5 + $\frac{3}{5}$ x 3
= 5.5 + $\frac{9}{5}$
= 5.5 + 1.8
= 7.3 $\approx$ = 7

Factorize completely 9y2 - 16X2

Maelezo ya Majibu

9y2 - 16x2
= 32y2 - 42x2
= (3y - 4x)(3y +4x)

Evaluate ${\int }_2^1$(3 - 2x)dx

Maelezo ya Majibu

${\int }_0^1$(3 - 2x)dx
[3x - x2]o
[3(1) - (1)2] - [3(0) - (0)2]
(3 - 1) - (0 - 0) = 2 - 0
= 2

The pie chart shows the distribution of courses offered by students. What percentage of the students offer English?

Maelezo ya Majibu

$\frac{90}{360}\times 100=\frac{1}{4}\times 100$
$=25\%$

Find the value of x at the minimum point of the curve y = x3 + x2 - x + 1

Maelezo ya Majibu

y = x3 + x2 - x + 1
$\frac{dy}{dx}$ = $\frac{d\left(x^3\right)}{dx}$ + $\frac{d\left(x^2\right)}{dx}$ - $\frac{d\left(x\right)}{dx}$ + $\frac{d\left(1\right)}{dx}$
$\frac{dy}{dx}$ = 3x2 + 2x - 1 = 0
$\frac{dy}{dx}$ = 3x2 + 2x - 1
At the maximum point $\frac{dy}{dx}$ = 0
3x2 + 2x - 1 = 0
(3x2 + 3x) - (x - 1) = 0
3x(x + 1) -1(x + 1) = 0
(3x - 1)(x + 1) = 0
therefore x = $\frac{1}{3}$ or -1
For the maximum point
$\frac{d^{2}y}{d{x}^2}$ < 0

$\frac{d^{2}y}{d{x}^2}$ 6x + 2
when x = $\frac{1}{3}$
$\frac{d{x}^2}{d{x}^2}$ = 6($\frac{1}{3}$) + 2
= 2 + 2 = 4
$\frac{d^{2}y}{d{x}^2}$ > o which is the minimum point
when x = -1
$\frac{d^{2}y}{d{x}^2}$ = 6(-1) + 2
= -6 + 2 = -4
-4 < 0

therefore, $\frac{d^{2}y}{d{x}^2}$ < 0

the maximum point is -1

From the venn diagram above, the complement of the set P$\cap$Q is given by

Maelezo ya Majibu

In the diagram, STUV is a straight line. < TSY = < UXY = 40o and < VUW = 110o. Calculate < TYW

Maelezo ya Majibu

< TUW = 110${}^{\circ }$ = 180${}^{\circ }$ (< s on a straight line)
< TUW = 180${}^{\circ }$ - 110${}^{\circ }$ = 70${}^{\circ }$
In $△$ XTU, < XUT + < TXU = 180${}^{\circ }$
i.e. < YTS + 70${}^{\circ }$ = 180
< XTU = 180 - 110${}^{\circ }$ = 70${}^{\circ }$
Also < YTS + < XTU = 180 (< s on a straight line)
i.e. < YTS + < XTU - 180(< s on straight line)
i.e. < YTS + 70${}^{\circ }$ = 180
< YTS = 180 - 70 = 110${}^{\circ }$
in $△$ SYT + < YST + < YTS = 180${}^{\circ }$(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${}^{\circ }$ - 60
2(< TYW) = 300${}^{\circ }$
TYW = $\frac{300}{2}$ = 150${}^{\circ }$
< SYT

A chord of circle of radius 7cm is 5cm from the centre of the maximum possible area of the square?

Maelezo ya Majibu

From Pythagoras theorem
|OA|2 = |AN|2 + |ON|2
72 = |AN|2 + (5)2
49 = |AN|2 + 25
|AN|2 = 49 - 25 = 24
|AN| = $\sqrt{24}$
= $\sqrt{4\times 6}$
= 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

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 $\oplus$ - $\frac{2}{3}$.

Maelezo ya Majibu

N + Y = XY + X + Y
3 + -$\frac{2}{3}$ = 3(- $\frac{2}{3}$) + 3 + (- $\frac{2}{3}$)
= -2 + 3 -$\frac{2}{3}$
= $\frac{1-2}{1-3}$
= $\frac{1}{3}$

What is the size of each interior angle of a 12-sided regular polygon?

Maelezo ya Majibu

Interior angle = (n - 2)180
but, n = 12
= (12 -2)180
= 10 x 180
= 1800
let each interior angle = x
x = $\frac{(n-2)180}{n}$
x = $\frac{1800}{12}$
= 150o

The inverse of matrix N = $\left|\begin{array}{cc}2& 3\\ 1& 4\end{array}\right|$ is

Maelezo ya Majibu

N = [2 3]
N-1 = $\frac{adjN}{|N|}$
adj N = $\left|\begin{array}{cc}4& -3\\ -1& 2\end{array}\right|$
|N| = (2 x4) - (1 x 3)
= 8 - 3
=5
N-1 = $\frac{1}{5}$ $\left|\begin{array}{cc}4& -3\\ -1& 2\end{array}\right|$

Solve for x and y respectively in the simultaneous equations -2x - 5y = 3, x + 3y = 0

Maelezo ya Majibu

-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

In a right angled triangle, if tan $\theta$ = $\frac{3}{4}$. What is cos$\theta$ - sin$\theta$?

Maelezo ya Majibu

tan$\theta$ = $\frac{3}{4}$
from Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin $\theta$ = $\frac{3}{5}$ and cos$\theta$ = $\frac{4}{3}$
cos$\theta$ - sin$\theta$
$\frac{4}{5}$ - $\frac{3}{5}$ = $\frac{1}{5}$

If 2q35 = 778, find q

Maelezo ya Majibu

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 = $\frac{10}{5}$
q = 2

Find the derivative of $\frac{\sin \theta }{\cos \theta }$

Maelezo ya Majibu

$\frac{\sin \theta }{\cos \theta }$
$\frac{\cos \theta \frac{d\left(\sin \theta \right)}{d\theta }-\sin \theta \frac{d\left(\cos \theta \right)}{d\theta }}{{\cos }^2\theta }$
$\frac{\cos \theta .\cos \theta -\sin \theta (-\sin \theta )}{co{s}^2\theta }$
$\frac{co{s}^2\theta +{\sin }^2\theta }{co{s}^2\theta }$
Recall that sin2 $\theta$ + cos2 $\theta$ = 1
$\frac{1}{{\cos }^2\theta }$ = sec2 $\theta$

If the numbers M, N, Q are in the ratio 5:4:3, find the value of $\frac{2N-Q}{M}$

Maelezo ya Majibu

M:N:Q == 5:4:3
i.e M = 5, N = 4, Q = 3
Substituting values into equation, we have...
$\frac{2N-Q}{M}$
= $\frac{2\left(4\right)-3}{5}$
= $\frac{8-3}{5}$
= $\frac{5}{5}$
= 1

The midpoint of P(x, y) and Q(8, 6). Find x and y. midpoint = (5, 8)

Maelezo ya Majibu

P(x, y) Q(8, 6)
midpoint = (5, 8)
x + 8 = 5
$\frac{y+6}{2}$ = 8
x + 8 = 10
x = 10 - 8 = 2
y + 6 = 16
y + 16 - 6 = 10
therefore, P(2, 10)

A solid metal cube of side 3 cm is placed in a rectangular tank of dimension 3, 4 and 5 cm. What volume of water can the tank now hold

Maelezo ya Majibu

Volume of cube = L3
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

If x varies directly as square root of y and x = 81 when y = 9, Find x when y = 1$\frac{7}{9}$

Maelezo ya Majibu

x $\alpha \sqrt{y}$
x = k$\sqrt{y}$
81 = k$\sqrt{9}$
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

$\begin{array}{cccc}\text{Class Interval}& 3-5& 6-8& 9-11\\ \text{Frequency}& 2& 2& 2\end{array}$.

Find the standard deviation of the above distribution.

Maelezo ya Majibu

$\begin{array}{cccc}\text{Class Interval}& 3-3& 6-8& 9-11\\ x& 4& 7& 10\\ f& 2& 2& 2\\ f-x& 8& 14& 20\\ |x-\overline{x}{|}^2& 9& 0& 9\\ |x-\overline{x}{|}^2& 180& 18& \end{array}$
$\overline{x}$ = $\frac{\sum fx}{\sum f}$
= $\frac{8+14+20}{2+2+2}$
= $\frac{42}{6}$
$\overline{x}$ = 7
S.D = $\sqrt{\frac{\sum f(x-\overline{x}{)}^2}{\sum f}}$
= $\sqrt{\frac{18+0+18}{6}}$
= $\sqrt{\frac{36}{6}}$
= $\sqrt{6}$

A man walks 100 m due West from a point X to Y, he then walks 100 m due North to a point Z. Find the bearing of X from Z.

Maelezo ya Majibu

tan$\theta$ = $\frac{100}{100}$ = 1
$\theta$ = tan-1(1) = 45o
The bearing of x from z is ₦45oE or 135o

If | 2 3 | = | 4 1 |. find the value of y. 7

Maelezo ya Majibu

$\left|\begin{array}{cc}2& 3\\ 5& 3x\end{array}\right|$ = $\left|\begin{array}{cc}4& 1\\ 3& 2x\end{array}\right|$
(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)
6x - 15 = 8x - 3
6x - 8x = 15 - 3
-2x = 12
x = $\frac{12}{-2}$
= -6

The derivatives of (2x + 1)(3x + 1) is

Maelezo ya Majibu

(2x + 1)(3x + 1) IS
2x + 1 $\frac{d(3x+1)}{d}$ + (3x + 1) $\frac{d(2x+1)}{d}$
2x + 1 (3) + (3x + 1) (2)
6x + 3 + 6x + 2 = 12x + 5

I how many was can the letters of the word ELATION be arranged?

Maelezo ya Majibu

ELATION
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

Simplify ($\sqrt{2}+\frac{1}{\sqrt{3}}\left)\right(\sqrt{2}-\frac{1}{\sqrt{3}}$)

Maelezo ya Majibu

($\sqrt{2}+\frac{1}{\sqrt{3}}\left)\right(\sqrt{2}-\frac{1}{\sqrt{3}}$)
$\sqrt{4}-\frac{\sqrt{2}}{\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{3}}-\frac{1}{\sqrt{9}}$
= 2 - $\frac{1}{3}$
= $\frac{16-1}{3}$
= $\frac{5}{3}$

The perpendicular bisector of a line XY is the locus of a point B. whose distance from Y is always twice its distance from X. C

Maelezo ya Majibu

Solve the inequality x2 + 2x > 15.

Maelezo ya Majibu

x2 + 2x > 15
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

Simplify $(\frac{16}{81}{)}^{\frac{1}{4}}÷(\frac{9}{16}{)}^{-\frac{1}{2}}$

Maelezo ya Majibu

$(\frac{16}{81}{)}^{\frac{1}{4}}÷(\frac{9}{16}{)}^{-\frac{1}{2}}$
$(\frac{16}{81}{)}^{\frac{1}{4}}÷(\frac{16}{9}{)}^{\frac{1}{2}}$
$(\frac{2^4}{3^4}{)}^{\frac{1}{4}}÷(\frac{4^2}{3^2}{)}^{\frac{1}{2}}$
$\frac{2^{4\times \frac{1}{4}}}{3^{4\times \frac{1}{4}}}÷\frac{4^{2\times \frac{1}{2}}}{3^{2\times \frac{1}{2}}}$
$\frac{2}{3}÷\frac{4}{3}$
$\frac{2}{3}\times \frac{3}{4}$
$\frac{2}{4}$
$\frac{1}{2}$

The sum of four consecutive integers is 34. Find the least of these numbers

Maelezo ya Majibu

Let the numbers be a, a + 1, a + 2, a + 3
a + a + 1 + a + 2 + a + 3 = 34
4a = 34 - 6
4a = 28
a = $\frac{28}{4}$
= 7
The least of these numbers is a = 7

Make R the subject of the formula if T = $\frac{K{R}^2+M}{3}$

Maelezo ya Majibu

A circle of perimeter 28cm is opened to form a square. What is the maximum possible area of the square?

Maelezo ya Majibu

Perimeter of circle = Perimeter of square
28cm = 4L
L = $\frac{28}{4}$ = 7cm
Area of square = L2
= 72
= 49cm2

Find the equation of a line perpendicular to line 2y = 5x + 4 which passes through (4, 2).

Maelezo ya Majibu

2y = 5x + 4 (4, 2)
y = $\frac{5x}{2}$ + 4 comparing with
y = mx + e
m = $\frac{5}{2}$
Since they are perpendicular
m1m2 = -1
m2 = $\frac{-1}{m_1}$ = -1
$\frac{5}{2}$ = -1 x $\frac{2}{5}$
The equator of the line is thus
y = mn + c (4, 2)
2 = -$\frac{2}{5}$(4) + c
$\frac{2}{1}$ + $\frac{8}{5}$ = c
c = $\frac{18}{5}$
$\frac{10+5}{5}$ = c
y = -$\frac{2}{5}$x + $\frac{18}{5}$
5y = -2x + 18
or 5y + 2x - 18 = 0

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?

Maelezo ya Majibu

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

Which of these angles can be constructed using ruler and a pair of compasses only?

Maelezo ya Majibu

Simplify $\frac{3\frac{2}{3}\times \frac{5}{6}\times \frac{2}{3}}{\frac{11}{15}\times \frac{3}{4}\times \frac{2}{27}}$

Maelezo ya Majibu

$\frac{3\frac{2}{3}\times \frac{5}{6}\times \frac{2}{3}}{\frac{11}{15}\times \frac{3}{4}\times \frac{2}{27}}$
$\frac{\frac{11}{3}\times \frac{5}{6}\times \frac{2}{3}}{\frac{11}{15}\times \frac{3}{4}\times \frac{2}{27}}$
$\frac{110}{54}÷\frac{66}{1620}$
50

Find ${\int }_0^1$ cos4 x dx

Maelezo ya Majibu

${\int }_0^1$ cos4 x dx
let u = 4x
$\frac{dy}{dx}$ = 4
dx = $\frac{dy}{4}$
${\int }_0^1$cos u. $\frac{dy}{4}$ = $\frac{1}{4}$$\int$cos u du
= $\frac{1}{4}$ sin u + k
= $\frac{1}{4}$ sin4x + k

In how many ways can five people sit round a circular table?

Maelezo ya Majibu

The first person will sit down and the remaining will join.
i.e. (n - 1)!
= (5 - 1)! = 4!
= 24 ways

The seconds term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms

Maelezo ya Majibu

T2 = 4, T4 = 16
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, $\frac{T_4}{T_r}$ = $\frac{a{r}^3}{ar}$ = $\frac{16}{4}$
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

Solve the inequality -6(x + 3) $\le$ 4(x - 2)

Maelezo ya Majibu

-6(x + 3) $\le$ 4(x - 2)
-6(x +3) $\le$ 4(x - 2)
-6x -18 $\le$ 4x - 8
-18 + 8 $\le$ 4x +6x
-10x $\le$ 10x
10x $\le$ -10
x $\le$ 1