Nkojọpọ....






Tẹ mọ́ & Dì mú láti fà yíká.
Tẹ ibi lati pa

JAMB UTME - Mathematics - 1995

Ibeere 1 Ìròyìn

Age in years	13	14	15	16	17
No. of students	3	10	30	42	15

The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is

27o

30o

54o

108o

Awọn alaye Idahun

To find the angle corresponding to the 15-year-olds in the pie chart, we need to first calculate the total number of students in the school. From the frequency distribution, we can see that there are: 3 + 10 + 30 + 42 + 15 = 100 students in total. Next, we need to find the fraction of students that are 15 years old. From the frequency distribution, we can see that there are 30 students who are 15 years old. Therefore, the fraction of students that are 15 years old is: 30/100 = 0.3 To convert this fraction to an angle, we need to multiply it by 360 degrees (the total angle in a pie chart). Therefore, the angle corresponding to the 15-year-olds in the pie chart is: 0.3 x 360 = 108 degrees Therefore, the answer is (D) 108o.

Ibeere 2 Ìròyìn

P is on the locus of points equidiatant from two given points X and Y. UV is a straight line throuh Y parallel to the locus. If < PYU is 40°, find < XPY.

100o

80 o

50 o

40 o

Awọn alaye Idahun

< PYU = < YPQ = 40°(Alternative) = PQ

∥

UV

< XPQ = < YPQ = 40°, Since QX = QY

therefore < XPY = 80°

Ka ìwé ẹ̀kọ́ nípa Loci (WAEC)

Ibeere 3 Ìròyìn

Find T in terms of K, Q and S if S = 2r\( \pi \)QT + K)

\frac{S^{2}}{2 π r^{2} Q} - \frac{k}{Q}

\frac{S^{2}}{2 π r^{2} Q}

- k

\frac{S^{2}}{4 π r^{2} Q} - \frac{k}{Q}

\frac{s^{2}}{4 π r^{2} Q}

Awọn alaye Idahun

\frac{s^{2}}{4 r^{2}}

= QT

π

+ KT

\frac{s^{2}}{4 r^{2}}

- k

π

= QT

π

T = $\frac{s^{2}}{4 Q π r^{2}}$ - k

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Ibeere 5 Ìròyìn

PT is a tangent to the circle TYZX. YT = YX and < PTX = 50o. Calculate < TZY

50o

65o

85o

130o

Awọn alaye Idahun

XTY = 50o = TXY =

\frac{130^{o}}{2}

= 65o

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ibeere 7 Ìròyìn

Class Interval	Frequency	Class boundaries	Class Mid−point
\(1.5-1.9\)	\(2\)	\(1.45-1.95\)	\(1.7\)
\(2.0-2.4\)	\(21\)	\(1.95-2.45\)	\(2.2\)
\(2.5-2.9\)	\(4\)	\(2.45-2.95\)	\(2.7\)
\(3.0-2.9\)	\(15\)	\(2.95-3.45\)	\(3.2\)
\(3.5-3.9\)	\(10\)	\(3.45-3.95\)	\(3.7\)
\(4.0-4.4\)	\(5\)	\(3.95-4.45\)	\(4.2\)
\(4.5-4.9\)	\(3\)	\(4.45-4.95\)	\(4.7\)

Find the mode of the distribution above to find the mode of the distribution.

3.2

3.3

3.7

4.2

Awọn alaye Idahun

Mode = a + (b - a)(fm - Fb)

2Fm - Fa - Fb

= 3.0 + $\frac{(3.4 - 3) (15 - 4)}{2 (15) - 4 - 10}$

= 3 + $\frac{(6.4) (11)}{30 - 14}$

= 3 + $\frac{4.4}{16}$

= 3 + 0.275

= 3.275

= 3.3cm

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Representation Of Data

Measures Of Dispersion

Ibeere 9 Ìròyìn

Express \( \frac{5x-12}{(x-2)(x-3)} \) in partial fractions

\frac{2}{x + 2} - \frac{3}{x - 3}

\frac{2}{x - 2} + \frac{3}{x - 3}

\frac{2}{x - 3} - \frac{3}{x - 2}

\frac{5}{x - 3} - \frac{4}{x - 2}

Awọn alaye Idahun

$\frac{5 x - 12}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$

= $\frac{A (x - 3) + B (x - 2)}{(x - 2) (x - 3)}$

⟹5x−12=Ax−3A+Bx−2B

$A + B = 5 . . . (i)$

−(3A+2B)=−12⟹3A+2B=12...(ii)

From (i), $A = 5 - B$

$3 (5 - B) + 2 B = 12$

15−3B+2B=12⟹B=3

A+3=5⟹A=2

$\frac{5 x - 12}{(x - 2) (x - 3)} = \frac{2}{x - 2} + \frac{3}{x - 3}$

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Ibeere 12 Ìròyìn

Use the graph of the curve \(y = f(x)\)to solve the inequality \(f(x) \le 0\)

-1

\geq

x

\geq

1, x

\geq

2

x

\leq

-1, 1

\geq

x

\geq

2

x

\geq

-1, 1

\geq

x

\geq

2

x

\geq

2, -1

\geq

x

\geq

1

Awọn alaye Idahun

-1

\geq

x

\geq

1, 1 < x

\geq

2

Combining solutions

= x $\leq$ 1; 1 $\geq$ x $\geq$ 2

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Linear Inequalities

Ibeere 17 Ìròyìn

Solve the inequality \( (x - 3)(x - 4) \le 0 \)

3

\leq

x

\leq

4

3 < x < 4

3

\leq

x < 4

3 < x

\leq

4

Awọn alaye Idahun

To solve this inequality, we need to find the values of x that make the expression (x - 3)(x - 4) less than or equal to 0. To do this, we can use a method called the "sign chart" or "interval notation". First, we can find the critical values of x by setting the expression equal to 0: (x - 3)(x - 4) = 0. This gives us x = 3 and x = 4. Next, we can create a sign chart by dividing the number line into intervals using the critical values of x. We then choose a test point from each interval and evaluate the expression (x - 3)(x - 4) using that test point. If the result is positive, the expression is positive in that interval. If the result is negative, the expression is negative in that interval. Intervals: (-∞, 3), (3, 4), (4, ∞) Test point: -1, 3.5, 5 Expression: (-)(-)=+ (+)(-)= - (+)(+)=+ From the sign chart, we can see that the expression is less than or equal to 0 when x is between 3 and 4, inclusive. Therefore, the solution is: 3 ≤ x ≤ 4 So the correct option is: - 3 ≤ x ≤ 4

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Ka ìwé ẹ̀kọ́ nípa Inequalities (JAMB)

Linear Inequalities

Inequalities

Ibeere 22 Ìròyìn

Class Interval	Frequency	Class boundaries	Class Mid-point
\(1.5 - 1.9\)	\(2\)	\(1.45 - 1.95\)	\(1.7\)
\(2.0 - 2.4\)	\(21\)	\(1.95 - 2.45\)	\(2.2\)
\(2.5 - 2.9\)	\(4\)	\(2.45 - 2.95\)	\(2.7\)
\(3.0 - 2.9\)	\(15\)	\(2.95 - 3.45\)	\(3.2\)
\(3.5 - 3.9\)	\(10\)	\(3.45 - 3.95\)	\(3.7\)
\(4.0 - 4.4\)	\(5\)	\(3.95 - 4.45\)	\(4.2\)
\(4.5 - 4.9\)	\(3\)	\(4.45 - 4.95\)	\(4.7\)

The median of the distribution above is

4.0

3.4

3.2

3.0

Awọn alaye Idahun

Median =

\frac{a + b}{f m}

(

\frac{1}{2} \sum f - C F_{b}

)
= 2.95 +

\frac{0.5}{15}

(2.-7)
= 2.95 +

\frac{0.5}{15}

x 13
= 2.95 + 0. 43
= 3.38
= 3.4

Ibeere 24 Ìròyìn

Find the distance between two towns p(45\(^{o}\)N, 30\(^{o}\)W) and Q(15\(^{o}\)S, 30\(^{o}\)W) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)]

\frac{1100}{3}

km

\frac{2200}{3}

km

\frac{22000}{3}

km

\frac{1100}{3}

km

Coordinate Geometry

Ibeere 25 Ìròyìn

class	1−3	4−6	7−9
Frequency	5	8	5

Find the standard deviation of the data using the table above

5

\(\sqrt{6}\0

\frac{5}{3}

\sqrt{5}

Awọn alaye Idahun

\begin{matrix} class intervals & F r e (F) & class-marks(x) & F x & (x - x) & (x - x)^{2} & F (x - x)^{2} \\ 1 - 3 & 5 & 10 & 10 & - 3 & 9 & 90 \\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \\ 18 & 90 & 450 \end{matrix}

x =

\frac{\sum f x}{\sum f}

=

\frac{90}{18}

= 5
S.D =

\frac{\sum f (x - x)^{2}}{\sum f}

=

\frac{450}{18}

=

\sqrt{25}

= 5

Ibeere 33 Ìròyìn

Let \(p\) be a probability function on set \(S\), where \(S = (a_1, a_2, a_3, a_4)\). Find \(P(a_1)\) if \(P(a_2) = \frac{1}{3}\), \(p(a_3) = \frac{1}{6}\) and \(p(a_4) = \frac{1}{5}\)

\frac{7}{3}

\frac{2}{3}

\frac{1}{3}

\frac{3}{10}

Probability

Ibeere 34 Ìròyìn

If S = (x : x2 = 9, x > 4), then S is equal to

4

{0}

\oplus

{

\oplus

}

Algebraic Fractions

Ibeere 35 Ìròyìn

A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank.

60m

30

\sqrt{3}

m

20

\sqrt{3}

m

10

\sqrt{3}

m

Awọn alaye Idahun

To solve this problem, we can use trigonometry. Let h be the height of the water tank, and d be the horizontal distance between the foot of the water tank tower and the position of the boy. Then, we have: tan(60°) = h / d We know that d = 30m, and tan(60°) = √3, so we can solve for h: h = d * tan(60°) h = 30m * √3 h = 30√3 m Therefore, the height of the water tank is 30√3 meters.

Ibeere 38 Ìròyìn

A die has four of it's faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive time, the top face will be white in both cases?

\frac{2}{3}

\frac{1}{9}

\frac{4}{9}

\frac{1}{36}

Awọn alaye Idahun

The probability of getting a white face on a single roll of the die is 4/6 or 2/3, since four of the six faces are white. Similarly, the probability of getting a white face on a second roll of the die is also 2/3. To find the probability of both events occurring (getting a white face on both rolls), we multiply the probabilities of each event together. So, the probability of getting a white face on the first roll AND getting a white face on the second roll is (2/3) x (2/3) = 4/9. Therefore, the correct answer is not provided in the options given.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Probability (JAMB)

Probability

Probability

Ibeere 39 Ìròyìn

Determine the value of x in the figure.

134

^{\circ}

81

^{\circ}

53

^{\circ}

46

^{\circ}

Ibeere 45 Ìròyìn

Which of the following binary operations is cumulative in the set of integers?

a

*

b = a + 2b

a

*

b = a + b - ab

a

*

b = a2 + b

a

*

b =

\frac{a (b + 1)}{2}

Binary Operations

Ibeere 46 Ìròyìn

Simplify \( \frac{x^2 - 1}{x^3 + 2x^2 - x - 2} \)

\frac{1}{x + 2}

\frac{x - 1}{x + 1}

\frac{x - 1}{x + 2}

x−1x+2

\frac{1}{x - 2}

Awọn alaye Idahun

\frac{x^{2} - 1}{(x - 1) (x + 2) (x + 1)}

=

\frac{(x - 1) (x + 1)}{(x - 1) (x + 2) (x + 1)}

=

\frac{1}{x + 2}

Tẹ́lẹ̀

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47