JAMB UTME - Mathematics - 2011

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

Detalhes da Resposta

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

If 2q35 = 778, find q

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

-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

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Solve the inequality x2 + 2x > 15.

Detalhes da Resposta

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

Find ${\int }_0^1$ cos4 x dx

Detalhes da Resposta

${\int }_0^1$ 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

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

Detalhes da Resposta

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

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?

Detalhes da Resposta

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

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.

Detalhes da Resposta

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

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

Detalhes da Resposta

< 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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

-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

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

Detalhes da Resposta

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 $\frac{1}{8}$
= $\frac{1}{12}$

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Find the standard deviation of the above distribution.

Detalhes da Resposta

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

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

Find the total number of students

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Evaluate ${\int }_2^1$(3 - 2x)dx

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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?

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Find the mode of the above distribution.

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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 $\oplus$ - $\frac{2}{3}$.

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Factorize completely 9y2 - 16X2

Detalhes da Resposta

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

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

Detalhes da Resposta

(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