JAMB UTME - Mathematics - 2011

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

Bayanin Amsa

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

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

Find the standard deviation of the above distribution.

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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)

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

-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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

Bayanin Amsa

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 $\frac{1}{8}$
= $\frac{1}{12}$

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

Bayanin Amsa

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

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

Bayanin Amsa

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

Find ${\int }_0^1$ cos4 x dx

Bayanin Amsa

${\int }_0^1$ 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

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

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

(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

Solve the inequality x2 + 2x > 15.

Bayanin Amsa

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

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

Bayanin Amsa

-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

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

Bayanin Amsa

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

Bayanin Amsa

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

Bayanin Amsa

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

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?

Bayanin Amsa

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

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

Bayanin Amsa

< 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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

Evaluate ${\int }_2^1$(3 - 2x)dx

Bayanin Amsa

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

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

Bayanin Amsa

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

If 2q35 = 778, find q

Bayanin Amsa

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

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

Find the total number of students

Bayanin Amsa

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.

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Find the mode of the above distribution.

Bayanin Amsa

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

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 $\oplus$ - $\frac{2}{3}$.

Bayanin Amsa

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

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

Bayanin Amsa

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

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

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

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

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

Bayanin Amsa

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

Factorize completely 9y2 - 16X2

Bayanin Amsa

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