JAMB UTME - Mathematics - 2011

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 $\oplus$ - $\frac{2}{3}$.

Détails de la réponse

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?

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

Factorize completely 9y2 - 16X2

Détails de la réponse

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

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

Détails de la réponse

-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

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

Détails de la réponse

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

Find the total number of students

Détails de la réponse

If 2q35 = 778, find q

Détails de la réponse

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

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

Détails de la réponse

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

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

Find the standard deviation of the above distribution.

Détails de la réponse

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

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?

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

Find ${\int }_0^1$ cos4 x dx

Détails de la réponse

${\int }_0^1$ 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

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

Détails de la réponse

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

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

Détails de la réponse

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)

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

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

Détails de la réponse

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

< 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?

Détails de la réponse

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

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

Détails de la réponse

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

Evaluate ${\int }_2^1$(3 - 2x)dx

Détails de la réponse

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

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.

Détails de la réponse

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

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

Détails de la réponse

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

Détails de la réponse

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

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

Détails de la réponse

(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

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

Détails de la réponse

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$\theta$
$\frac{4}{5}$ - $\frac{3}{5}$ = $\frac{1}{5}$

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

Détails de la réponse

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

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

Détails de la réponse

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

Détails de la réponse

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

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

Détails de la réponse

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

Détails de la réponse

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

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

Détails de la réponse

-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

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

Détails de la réponse

$\left|\begin{array}{cc}2& 3\\ 5& 3x\end{array}\right|$ = $\left|\begin{array}{cc}4& 1\\ 3& 2x\end{array}\right|$
(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

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

Détails de la réponse

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 $\frac{1}{8}$
= $\frac{1}{12}$

Solve the inequality x2 + 2x > 15.

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Find the mode of the above distribution.

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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

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

Détails de la réponse

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 = $\frac{1}{5}$ $\left|\begin{array}{cc}4& -3\\ -1& 2\end{array}\right|$