JAMB UTME - Mathematics - 1988

If $ \cos^2 \theta + \frac{1}{8} = \sin^2 \theta $, find $ \tan \theta $

3

\frac{3 \sqrt{7}}{7}

3

\sqrt{7}

\sqrt{7}

Détails de la réponse

cos2

θ

+

\frac{1}{8}

= sin2

θ

..........(i)
from trigometric ratios for an acute angle, where cos

θ

+ sin2

θ

= 1 - cos

θ

........(ii)
Substitute for equation (i) in (i) = cos2

θ

+

\frac{1}{8}

= 1 - cos2

θ

= cos2

θ

+ cos2

θ

= 1 -

\frac{1}{8}

2 cos2

θ

=

\frac{7}{8}

cos2

θ

=

\frac{7}{2 \times 3}

\frac{7}{16}

= cos

θ

\sqrt{\frac{7}{16}}

=

\frac{\sqrt{7}}{4}

but cos

θ

=

\frac{adj}{hyp}

opp2 = hyp2 - adj2
opp2 = 42 (

\sqrt{7}

)2
= 16 - 7
opp =

\sqrt{9}

= 3
than

θ

=

\frac{opp}{hyp}

=

\frac{3}{\sqrt{7}}

\frac{3}{\sqrt{7}}

x

\frac{7}{\sqrt{7}}

=

\frac{3 \sqrt{7}}{7}

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Trigonometry

Modular Arithmetic

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Question 2 Rapport

In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle

\sqrt{3}

cm

\sqrt{7}

cm

3cm

7cm

Détails de la réponse

PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o
PR2 = 12 + 22 - 2(1)(2) x - cos 60o
= 5 - 2(1)(2) x -

\frac{1}{2}

= 5 + 2 = 7
PR =

\sqrt{7}

cm

Voir la réponse

Question 3 Rapport

Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle

110o

120o

130o

140o

Détails de la réponse

Let the fifth interior angle be y: sum of interior angle of a pentagon
= (2 x 5 - 4) x 90o
= 6 x 90o
= 540o
(90 - x) + (90 + x) + (110 - 2x) + (110 + 2x) + y = 540o
400o + y = 540o
y = 540 - 400o
y = 140o

Voir la réponse

Question 4 Rapport

A tax player is allowed $ \frac{1}{8} $th of his income tax-free, and pays 20% on the remainder. If he pays ₦490.00 tax, what is his income?

₦560.00

₦2,450.00

₦2,800.00

₦3,920.00

Détails de la réponse

He pays tax on 1 - $\frac{7}{8}$ = $\frac{1}{7}$ th of his income
20% is 490, 100% is $\frac{1000}{20}$ x 490, ₦2,450.00
= $\frac{7}{8}$ of his income = ₦2,450.00

$\frac{1}{\frac{7}{8}}$ x 2450
= $\frac{8 \times 2450}{7}$
= $\frac{19600}{7}$
= ₦2800.00

Voir la réponse

Question 5 Rapport

Evaluate $ \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} $ = $ \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} $

2

5

3

3√5

3√2

Détails de la réponse

\frac{8^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{\frac{2}{10^{3}}}

=

\frac{(2^{3})^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{(2 \times 5)^{\frac{2}{3}}}

=

\frac{2 \times 5}{2^{\frac{2}{3}} \times 5^{\frac{3}{2}}}

= 21 -

\frac{2}{3}

= 2

\frac{1}{3}

= 3

\sqrt{2}

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Question 6 Rapport

The thickness of an 800 pages of book is 18mm. Calculate the thickness of one leaf of the book giving your answer in meters and in standard form

2.25 x 10-4m

50 x 10-4m

2.25 x 10-5m

4.50 x 10-5m

Détails de la réponse

Thickness of an 800 pages book = 18mm to meter
18 x 103m = 1.8 x 10-2m
One leaf =

\frac{1.8 \times 10^{- 2}}{800}

=

\frac{1.8 \times 10^{- 2}}{8 \times 10^{2}}

=

\frac{- 1.8}{8}

x 10-4
= 0.225 x 10-4
= 2.25 x 10-5m

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Question 7 Rapport

PQR is a triangle in which PQ = 10cm and QPR = 60oS is a point equidistant from P and Q. Also S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place

2.7

4.33

3.1

3.3

Détails de la réponse

△

PUS is right angled

\frac{U S}{5}

= sin60o
US = 5 x

\frac{\sqrt{3}}{2}

= 2.5

\sqrt{3}

= 4.33cm

Voir la réponse

Question 8 Rapport

Factorize completely (x2 + x)2 - (2x + 2)2

(x + 1)(x + 2)(x - 2)

(x + 1) 2(x + 2) (x - 2)

(x + 1)2 (x + 2)2

(x + 1)2 (x + 2)(x - 2)

Voir la réponse

Question 9 Rapport

If $ \cos \theta = \frac{x}{y} $, find $ \cosec \theta $

\frac{1}{y}

(x2 + y2)

\frac{x}{y}

\frac{1}{y}

\sqrt{x^{2} + y^{2}}

\frac{x - y}{y}

Détails de la réponse

Cos

θ

=

\frac{x}{y}

=

\frac{adj}{opp}

(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp =

\sqrt{x^{2} + y^{2}}

Cosec

θ

= hyp
= x2 + y2
=

\frac{1}{y}

\sqrt{x^{2} + y^{2}}

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Angles Of Elevation And Depression

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Question 10 Rapport

Simplify $ \frac{x-7}{x^2-9} \, x \, \frac{x^2-3x}{x^2-49} $

x

(x - 3) (x - 7)

x

(x + 3) (x - 7)

x

(x + 3) (x + 7)

x

(x - 3) (x + 7)

Détails de la réponse

\frac{x - 7}{x^{2} - 9}

x

\frac{x^{2} - 3 x}{x^{2} - 49}

=

\frac{x - 7}{(x - 3) (x + 3)}

x

\frac{x (x - 3)}{(x - 7) (x + 7)}

=

\frac{x}{(x + 3) (x + 7)}

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Algebraic Expressions

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Question 11 Rapport

What is the solution of the equation x2 - x - 1 + 0?

x = 1.6 and x = -0.6

x = -1.6 and x = 0.6

x = 1.6 and x = 0.6

x = -1.6 and x = -0.6

Quadratic Equations

Voir la réponse

Question 12 Rapport

If 7 and 189 are the first and fourth terms of a geometric progression respectively find the sum for the first
three terms of the progression

182

180

91

63

28

Sequence And Series

Voir la réponse

Question 13 Rapport

if a = -3, b = 2, c = 4, evaluate $ \frac{a^3-b^3-c^{\frac{1}{2}}}{b-a-c} $

37

-37

5

37

5

-37

Voir la réponse

Question 14 Rapport

A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?

20

2

1.1

0.2

Détails de la réponse

% error =

\frac{actual error}{true value}

x 100
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error =

\frac{0.1}{5.0}

x 100
=

\frac{10}{5}

= 2

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Fractions, Decimals, Approximations And Percentage

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Question 15 Rapport

If (IPO3)4 = 11510 find P

0

1

2

3

Détails de la réponse

1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = $\frac{48}{16}$
= 3

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Number Bases

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Question 16 Rapport

In the figure, PQ is a parallel to ST and QRS = $40^\circ$. Find the value of x.

55

^{\circ}

60

^{\circ}

65

^{\circ}

75

^{\circ}

Détails de la réponse

From the figure, 3x + x - 40

^{\circ}

= 180

^{\circ}

4x = 180 $^{\circ}$ + 40 $^{\circ}$

4x = 220 $^{\circ}$

x = $\frac{220}{4}$

= 55 $^{\circ}$

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Angles And Intercepts On Parallel Lines

Angles

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Question 17 Rapport

The solutions of x2 - 2x - 1 = 0 are the points of intersection of two graphs. if one of the graphs is y = 2 + x - x2, find the second graph

y = 1 - x

y = 1 + x

y = x - 1

y = 3x + 3

Algebraic Expressions

Voir la réponse

Question 18 Rapport

Two sisters, Taiwo and Keyinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later, Keyinde sells $ \frac{2}{3} $ of her share to Taiwo for ₦720.00. Find the value of the store.

₦1,080.00

₦2,400.00

₦3,000.00

₦3,600.00

Détails de la réponse

Let value of store = X
Ratio of Taiwo's share to kehine's is 11:9 Keyinde sells

\frac{2}{3}

of her share to Taiwo for ₦720

\frac{2}{3}

of 9 = 6
∴ Sum of the ratio = 11 + 9 = 20

\frac{6}{20}

of x = ₦720

\frac{6 x}{20}

= 720
∴ x =

\frac{720 \times 20}{6}

x = ₦24,000

Voir la réponse

Question 19 Rapport

In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid

36

π

cm 3

54

π

cm 2

18

π

cm 2

108

π

cm 2

Détails de la réponse

The volume of the solid = vol. of cone + vol. of hemisphere

volume of cone = $\frac{1}{2} π^{2} h$

= $\frac{1 π}{3} \times (3)^{2} x 6 = 18 π c m^{2}$

vol. of hemisphere = $\frac{4 π r^{3}}{6} = \frac{2 π r^{3}}{3}$

= $\frac{2 π}{3} \times (3)^{3} = 18 π c m^{3}$

vol. of solid = 18 $π$ + 18 $π$

= 36 $π$ cm3

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Euclidean Geometry

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Question 20 Rapport

If two dice are thrown together, what is the probability of obtaining at least a score of 10?

1

6

1

12

5

6

11

12

Détails de la réponse

The total sample space when two dice are thrown together is 6 x 6 = 36

\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6 \\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6 \\ 3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6 \\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6 \\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6 \\ 6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{matrix}

At least 10 means 10 and above
P(at least 10) =

\frac{6}{36}

=

\frac{1}{6}

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Probability

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Question 21 Rapport

If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3

1

3

2

3

8

27

4

9

Variation

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Question 22 Rapport

If a metal pipe 10cm long has an external diameter of 12cm and a thickness of 1cm find the volume of the metal used in making the pipe

120

π

cm3

110

π

cm3

60

π

cm3

50

π

cm3

Détails de la réponse

The volume of the pipe is equal to the area of the cross section and length.
let outer and inner radii be R and r respectively.
Area of the cross section = (R2 - r2)
where R = 6 and r = 6 - 1
= 5cm
Area of the cross section = (62 - 52)

π

= (36 - 25)

π

cm sq
vol. of the pipe =

π

(R2 - r2)L where length (L) = 10
volume = 11

π

x 10
= 110

π

cm3

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Indices, Logarithms And Surds

Areas

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Question 23 Rapport

In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS

56cm2

112cm2

120cm2

176cm2

Détails de la réponse

From the figure, PS = QR = YT = 7cm

Area of parallelogram PQRS = 56cm

56 = base x height, where base = 7

7 x h = 56cm,

h = $\frac{56}{7}$

= 8cm

Area of trapezium $\frac{1}{2}$ (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm

Area of trapezium PQTS = $\frac{1}{2}$ (23 + 7) x 8

$\frac{1}{2}$ x 30 x 8 = 120cmsq

Voir la réponse

Question 24 Rapport

If cos 60o = 1/2, which of the following angle has cosine of -1/2?

30o

120o

150o

210o

300o

Détails de la réponse

cos60o = 1/2, cos(180o/60o) = -1/2

cos120o = -1/2

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Logarithms

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Question 25 Rapport

If log102 = 0.3010 and log103 = 0.4771, evaluate; without using logarithm tables, log104.5

0.3010

0.4771

0.6532

0.9542

Détails de la réponse

If log102 = 0.3010 and log103 = 0.4771,
log104.5 = log10

\frac{(3 \times 3)}{2}

log103 + log103 - log102 = 0.4771 + 0.4771 - 0.0310
= 0.6532

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Logarithms

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Question 26 Rapport

The solution of the quadratic equation px2 + qx + b = 0 is

\sqrt{\frac{- b \pm b^{2} - 4 a c}{2 a}}

\frac{- b \pm \sqrt{p^{2} - 4 p b}}{2 a}

\frac{- q \pm \sqrt{q^{2} - 4 b p}}{2 p}

\frac{- q \pm \sqrt{p^{2} - 4 b p}}{2 p}

Détails de la réponse

px2 + qx + b = 0
Using almighty formula

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

.........(i)
Where a = p, b = q and c = b
substitute for this value in equation (i)
=

\frac{- q \pm \sqrt{q^{2} - 4 b p}}{2 p}

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Quadratic Equations

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Question 27 Rapport

If $ \frac{1}{p} = \frac{a^2+2ab+b^2}{a-b} $ and $ \frac{l}{q} = \frac{a+b}{a^2-2ab+b^2} $ Find $ \frac{p}{q} $

a + b

a - b

1

a² - b²

a - b

a + b

a2 - b2

Algebraic Expressions

Algebraic Fractions

Simple Operations On Algebraic Expressions

Quadratic Equations

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Question 28 Rapport

If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the $ \frac{x}{y} $ correct to one decimal place

18

15

1.0

5

Détails de la réponse

Mean

\bar{x}

=

\frac{156}{10}

= 15.6
Median =

\bar{y}

=

\frac{15 + 16}{2}

\frac{31}{2}

= 15.5

\frac{x}{y}

=

\frac{15.6}{15.5}

= 1.0065
1.0(1 d.p)

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Surds (radicals)

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Question 29 Rapport

Solve the following equation equation for $x$

\[ x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0 \]

r2

1

r⁴

-

\frac{1}{r^{2}}

1 - r

Détails de la réponse

x2 +

\frac{2 x}{r^{2}}

+

\frac{1}{r^{4}}

= 0
(x +

\frac{1}{r^{2}}

) = 0
x +

\frac{1}{r^{2}}

= 0
x =

\frac{- 1}{r^{2}}

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Number Bases

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Question 30 Rapport

Tope bought X oranges at N5.00 each and some mangoes at N4.00 each. if she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of values of X.

4≤X≤5

5≤X≤8

5≤X≤10

8≤X≤10

Number Bases

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Question 31 Rapport

Simplify $ \frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2} $

a - b

2a + b

2a + 7b

a - b

2a - 7b

a + b

2a - 7b

a - b

Détails de la réponse

\frac{4 a^{2} - 49 b^{2}}{2 a^{2} - 5 a b - 7 b^{2}}

=

\frac{(2 a)^{2} - (7 b)^{2}}{(a - b) (2 a + 7 b)}

=

\frac{(2 a + 7 b) (2 a - 7 b)}{(a - b) (2 a + 7 b)}

=

\frac{2 a - 7 b}{a - b}

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Solution Of Linear Equations

Algebraic Expressions

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Question 32 Rapport

A basket contain green, black and blue balls in the ratio 5 : 2 : 1. If there are 10 blue balls. Find the corresponding new ratio when 10 green and 10 black balls are removed from the basket

1 : 1 : 1

4 : 2 : 1

5 : 1 : 1

4 : 1 : 1

Détails de la réponse

Let x represent total number of balls in the basket.
If there are 10 blue balls,

\frac{1}{8}

of x = 10
x = 10 x 8 = 80 balls
Green balls will be

\frac{5}{8}

x 80 = 50 and black balls =

\frac{2}{8}

x 80 = 20
Ratio = Green : black : blue
50 : 20 : 10
-10 : 10 : -
------------------
New Ratio 40 : 10 : 10
4 : 1 : 1

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Positive And Negative Integers, Rational Numbers

Ratio, Proportions And Rates

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Question 33 Rapport

Find $m$ such that $(m + \sqrt{3})(1 - \sqrt{3})^2 = 6 - 2\sqrt{2}$

1

2

3

4

Détails de la réponse

(m +

\sqrt{3}

)(1 -

\sqrt{3}

)2 = 6 - 2

\sqrt{2}

(m +

\sqrt{3}

)(4 - 2

\sqrt{3}

) = 6 - 2

\sqrt{2}

= 6 - 2

\sqrt{3}

4m - 6 + 4 - 2m

\sqrt{3}

= 6 - 2

\sqrt{3}

comparing co-efficients,
4m - 6 = 6.......(i)
4 - 2m = -2.......(ii)
in both equations, m = 3

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Surds (radicals)

Lengths And Perimeters

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Question 34 Rapport

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

3

x + 1

3x + 1

(x + 1) (x + 2)

5x + 1

(x + 1) (x + 2)

2x²+ 5x + 2

(x + 1) (x + 2)

Détails de la réponse

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

-

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

=

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

=

\frac{(x^{2} + 4 x + 4) - (x^{2} - x - 2)}{(x + 1) (x + 2)}

=

\frac{x^{2} + 4 x + 4 - x^{2} + x + 2}{(x + 1) (x + 2)}

=

\frac{5 x + 6}{(x + 1) (x + 2)}

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Algebraic Expressions

Quadratic Equations

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Question 35 Rapport

if x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y

27

30

33

90

Détails de la réponse

Prime numbers between 1 and 6 are 2, 3 and 5
x = 2 + 3
= 5 = 10
H.C.F. of 6, 9, 15 = 3
∴ y = 3
X x y = 10 x 3
= 30

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Probability

Number Bases

Sets

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Question 36 Rapport

Find correct to one decimal place, $0.24633 \div 0.0306$

0.8

1.8

8.0

8.1

Détails de la réponse

\frac{0.24633}{0.03060}

multiplying throughout by 100,000
=

\frac{24633}{3060}

= 8.05
= 8.1

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Fractions, Decimals And Approximations

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Question 37 Rapport

For what values of x is the curve $y = \frac{x^2 + 3}{x + 4}$ decreasing?

-3 < x

\leq

0

-3

\geq

x < 0

0 < x < 3

0

\leq

x

\leq

3

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Voir la réponse

Question 38 Rapport

Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is

(

\frac{- 1}{8}, \frac{19}{24}

)

8,

\frac{24}{19}

-8,

\frac{24}{19}

(

\frac{19}{24}, \frac{- 1}{8}

)

Détails de la réponse

3x - 5y = 3, 2y - 6x = -5
-5y + 3x = 3........{i} x 2
2y - 6x = -5.........{ii} x 5
Substituting for x in equation (i)
-5y + 3(

\frac{19}{24}

) = 3
-5y + 3 x

\frac{19}{24}

= 3
-5y =

\frac{3 - 19}{8}

-5 =

\frac{24 - 19}{8}

=

\frac{5}{8}

y =

\frac{5}{8 \times 5}

y =

\frac{- 1}{8}

(x, y) = (

\frac{19}{24}, \frac{- 1}{8}

)

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Algebraic Expressions

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Question 39 Rapport

Simplify $ \frac{1\frac{1}{2}}{2+\frac{1}{4}\text{ of }32} $

3

256

3

32

6

85

Détails de la réponse

\frac{1 \frac{1}{2}}{2 + \frac{1}{4} of 32}

=

\frac{\frac{3}{2}}{2 + \frac{1}{4} \times 32}

\frac{3}{2}

x 4 = 6

Lisez la note de cours sur Number Bases (WAEC)

Lisez la note de cours sur Number Bases (JAMB)

Number Bases

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Question 40 Rapport

In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR

47

5

32

5

22

5

Détails de la réponse

From similar triangle,

\frac{Q S}{Q P} = \frac{T Q}{Q R} = \frac{5}{12} = \frac{6}{Q R}

QT = $\frac{6 \times 12}{5} = \frac{72}{5} = S R = Q R - Q S$

= $\frac{72}{5} - 5 = \frac{72 - 25}{5}$

= $\frac{47}{5}$

Voir la réponse

Question 41 Rapport

If $ g(y) = \frac{y - 3}{11} + \frac{11}{y^2 - 9} $. what is $ g(y + 3) $?

\frac{y}{11} + \frac{11}{y (y + 6)}

\frac{y}{11} + \frac{11}{y (y + 6)}

\frac{y}{11} + \frac{11}{y (y + 6)}

\frac{y}{11} + \frac{11}{y (y + 6)}

Solution Of Linear Equations

Algebraic Expressions

Number Bases

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Question 42 Rapport

If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference

4

8

6

\frac{2}{3}

9

\frac{1}{3}

Surds (radicals)

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Question 43 Rapport

Scores($x$)	0	1	2	3	4	5	6	7
Frequency($f$)	7	11	6	7	7	5	3

In the distribution above, the mode and median respectively are

1,3

1, 2

3, 3

0, 2

Détails de la réponse

From the distribution, Mode = 1 and
Median =

\frac{2 + 2}{2}

= 2
= 1, 2

Lisez la note de cours sur Representation Of Data (JAMB)

Lisez la note de cours sur Linear Inequalities (WAEC)

Representation Of Data

Linear Inequalities

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Question 44 Rapport

Simplify $ \frac{x-y}{x^{\frac{1}{3}}-x^{\frac{1}{3}}} $

x2 + xy + y2

x

\frac{2}{3}

+ x

\frac{1}{3}

+ y

\frac{2}{3}

x

\frac{2}{3}

- x

\frac{1}{3}

y

\frac{2}{3}

y

\frac{2}{3}

Voir la réponse

Question 45 Rapport

In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.

80

^{\circ}

60

^{\circ}

45

^{\circ}

40

^{\circ}

Détails de la réponse

SRT is a straight line, where QRT = 120

SRQ = 180 $^{\circ}$ - 120 $^{\circ}$ = 60 $^{\circ}$ - (angle on a straight line)

also angle QRS = 180 $^{\circ}$ - 100 $^{\circ}$ (angle on a straight line) . In angles where QR = SR and angle SRQ = 60 $^{\circ}$

x = 100 - 60 = 40 $^{\circ}$

Lisez la note de cours sur Circles (WAEC)

Circles

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Question 46 Rapport

For which of the following exterior angles is a regular polygon possible? i. 36o ii. 18o iii. 15o

i and ii

ii only

ii and iii

iii only

Détails de la réponse

for a regular polygon to be possible, it must have all sides angles equal.

\frac{360}{18}

= 20 sides and

\frac{360}{15}

= 24 sides
(ii) and (iii) are right

Lisez la note de cours sur Polynomials (JAMB)

Polynomials

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Question 47 Rapport

In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ

135o

125o

100o

90o

Détails de la réponse

< RXZ = < ZYX = 45O(Alternate segment)

< ZYQ = 90 + 45

= 135O

Lisez la note de cours sur Circles (WAEC)

Circles

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Question 48 Rapport

In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12o. How many students are offering Physics?

18

15

10

5

Détails de la réponse

No of students offering Physics are

\frac{12}{360}

x 150
= 5

Voir la réponse

Question 49 Rapport

In the figure, PS = RS = QS and QRS = 50o. Find QPR

25o

40o

50o

65o

Détails de la réponse

In the figure PS = RS = QS, they will have equal base QR = RP

In angle SQR, angle S = 50O

In angle QRP, 65 + 65 = 130O

Since RQP = angle RPQ = $\frac{180 - 130}{2}$

= $\frac{50}{2} = 25^{o}$

QPR = 25O

Lisez la note de cours sur Polynomials (JAMB)

Polynomials

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Scores(\(x\))	0	1	2	3	4	5	6	7
Frequency(\(f\))	7	11	6	7	7	5	3