JAMB UTME - Mathematics - 1985

At what real value of x do the curves whose equations are y = x3 + x and y = x2 + 1 intersect?

-2

2

-1

9

1

Bayanin Amsa

y = x3 + x and y = x2 + 1
$\begin{array}{cc} x & - 2 & - 1 & 0 & 1 & 2 \\ Y = x^{3} + x & - 10 & - 2 & 0 & 2 & 10 \\ y = x^{2} + 1 & 5 & 2 & 1 & 2 & 5 \end{array}$
The curves intersect at x = 1

Karanta bayanin darasi akan Coordinate Geometry (JAMB)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Coordinate Geometry

Quadratic Equations

Duba Amsa

Tambaya 2 Rahoto

If f(x - 2) = 4x2 + x + 7, find f(1)

12

27

7

46

17

Solution Of Linear Equations

Algebraic Expressions

Duba Amsa

Tambaya 3 Rahoto

If a{ $\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}$ } = 6x. Find a in its simplest form

x2 - 1

x2 + 1

x2 + 4

x2 - 4

1

Bayanin Amsa

a{ $\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}$ } = a{ $\frac{(x + 1) (x + 2) - (x - 1) (x - 2)}{(x - 2) (x + 2)}$ }
= 6
$\frac{6 x}{x^{2} - 4}$ = 6x
a = x2 - 4

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 4 Rahoto

In the equation below, Solve for x if all the numbers are in base 2: $\frac{11}{x}$ = $\frac{1000}{x + 101}$

101

11

110

111

10

Bayanin Amsa

$\frac{11}{x}$ = $\frac{1000}{x + 101}$ = 11(x + 101)
1000x = 11x + 1111
1000x - 11x = 1111
101x = 1111
x = $\frac{1111}{101}$
x = 11

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 5 Rahoto

Find the values of p for which the equation x2 - (p - 2)x + 2p + 1 = 0

(21, 0)

(0, 12)

(1, 2)

(3, 4)

(4, 5)

Quadratic Equations

Duba Amsa

Tambaya 6 Rahoto

Arrange the following numbers in ascending order of magnitude $\frac{6}{7}$ , $\frac{13}{15}$ , 0.8650

$\frac{6}{7}$ < 0.865 < $\frac{13}{15}$

$\frac{13}{15}$ < $\frac{6}{7}$ < 0.865

$\frac{6}{7}$ < $\frac{13}{15}$ < 0.865

0.865 < $\frac{6}{7}$ < $\frac{13}{15}$

Bayanin Amsa

$\frac{6}{7}$ , $\frac{13}{15}$ , 0.8650

In ascending order, we have 0.8571, 0.8650, 0.8666

i.e. $\frac{6}{7}$ < 0.8650 < $\frac{13}{15}$

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 7 Rahoto

a sum of money was invested at 8% per annum simple interest. If after 4 years the money amounts to N330.00. Find the amount originally invested

N180.00

n165.00

N150.00

N250.00

N200.00

Bayanin Amsa

S.I = $\frac{P T R}{100}$

T = 4yrs, R = 8%, a = N330.00

330 - P = $\frac{P T R}{100}$ , A = P + I

i.e. A = P + $\frac{P T R}{100}$

330 = P + $\frac{P (4) (8)}{100}$

33000 = 32P + 100p

132P = 33000

P = N250.00

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Positive And Negative Integers, Rational Numbers

Duba Amsa

Tambaya 8 Rahoto

Two points X and Y both on latitude 60oS have longitude 147oE and 153oW respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude (Take 2 $π$ R = 4 x 104km, where R is the radius of the earth)

16667km

28850km

8333km

2233km

Bayanin Amsa

Length of an area = $\frac{θ}{360}$ × 2 $π$ r
Longitude difference = 147 + 153 = 300oN
distance between xy = $\frac{θ}{360}$ × 2 $π$ R cos60o
= $\frac{300}{360}$ × 4 × 104 × $\frac{1}{2}$
= 1.667 × 104km (1667 km)

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 9 Rahoto

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

115o

125o

135o

145o

Volumes

Angles

Duba Amsa

Tambaya 10 Rahoto

John gives one-third of his money to Janet who has ₦105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money john has at first

₦45.00

₦48.00

₦52.00

₦60.00

₦53.00

Bayanin Amsa

Let x be John's money, Janet already had ₦105, $\frac{1}{3}$ of x was given to Janet
Janet now has $\frac{1}{3^{2}}$ x + 105 = $\frac{x + 315}{3}$
John's money left = $\frac{2}{3}$ x
= $\frac{\frac{1}{4} (x + 315)}{3}$
= $\frac{2}{3}$
24x = 3x + 945
∴ x = 45

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 11 Rahoto

The bearing of a bird on a tree from a hunter on the ground is ₦72oE. What is the bearing of the hunter from the birds?

S 18o W

S 72o W

S 72o E

S 27o E

S 27o W

Bearings

Duba Amsa

Tambaya 12 Rahoto

In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Zo at the centre of the circle, what is Z?

15

27

50

52

54

Bayanin Amsa

Total students = 120
grade = 18
Zo = $\frac{18}{120}$ x $\frac{360}{1}$
= 54o

Duba Amsa

Tambaya 13 Rahoto

If ex = 1 + $\frac{x}{2}$ + $\frac{x^{2}}{1.2}$ + $\frac{x^{3}}{1.2.3}$ + .....Find $\frac{1}{e^{x}}$

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{2.4.3}

+ ..

1 +

\frac{x}{2}

+

\frac{x^{2}}{1.2}

+

\frac{x^{3}}{1.2.3}

+

\frac{x^{4}}{1.23.4}

+ .........

1 - x +

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 +

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{1.2.6}

+ ........

Bayanin Amsa

ex = 1 + $\frac{x}{2}$ + $\frac{x^{2}}{1.2}$ + $\frac{x^{3}}{1.2.3}$ + $\frac{x^{4}}{1.23.4}$

Duba Amsa

Tambaya 14 Rahoto

The number of goals scored by a football team in 20 matches is shown below:

$\begin{array}{cc} No. of goals & 0 & 1 & 2 & 3 & 4 & 5 \\ No. of matches & 3 & 5 & 7 & 3 & 1 & 0 \end{array}$

What are the values of the mean and the mode respectively?

(1.75, 5)

(1.75, 2)

(1.75, 1)

(65, 74)

Bayanin Amsa

$\begin{array}{cc} x & f & f x \\ 0 & 3 & 0 \\ 1 & 5 & 5 \\ 2 & 7 & 14 \\ 3 & 4 & 12 \\ 4 & 1 & 4 \\ 5 & 0 & 0 \end{array}$
$\sum f$ = 20
$\sum f x$ = 35
Mean = $\frac{\sum f x}{\sum f}$
= $\frac{35}{20}$
= $\frac{7}{4}$
= 1.75
Mode = 2
= 1.75, 2

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Statistics (WAEC)

Probability

Statistics

Duba Amsa

Tambaya 15 Rahoto

If 32y + 6(3y) = 27. Find y

3

-1

2

-3

1

Solution Of Linear Equations

Duba Amsa

Tambaya 16 Rahoto

In $△$ XYZ, XKZ = 90O, XK = 15cm, XY = 35cm and YK = 8cm. Find the area of $△$ XYZ

190 sq.cm

20 sq.cm

210 sq.cm

160sq.cm

320sq.cm

Bayanin Amsa

By Pythagoras, KZ2 = 252 - 52

KZ2 = (25 + 15)(25 - 15) = 400

KZ = $\sqrt{400}$ = 20

area of XYZ = $\frac{1}{2} \times 28 \times 15$

= 210 sq. cm

Duba Amsa

Tambaya 17 Rahoto

Solve for (x, y) in the equation 2x + y = 4: x^2 + xy = -12

(6, -8): (-2, 8)

(3, -4): (-1, 4)

(8, -4): (-1, 4)

(-8, 6): (8, -2)

(-4, 3): (4, -1)

Solution Of Linear Equations

Duba Amsa

Tambaya 18 Rahoto

Make c the subject of formula v = 1 - $\frac{a}{5}$ (b + $\frac{3 c}{7}$ )

[

\frac{7}{3}

+

\frac{5}{a}

(v - 1)] + b

[

\frac{7 b}{3}

+

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5 v - 1}{a}

]

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Duba Amsa

Tambaya 19 Rahoto

Solve the simultaneous equations 2x - 3y + 10 = 10x - 6y = 5

x = 2

\frac{1}{2}

, y = 3

\frac{1}{2}

x =

\frac{1}{2}

, y =

\frac{3}{2}

x = 2

\frac{1}{4}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 2

\frac{1}{2}

Bayanin Amsa

2x - 3y = -10; 10x - 6y = -5
2x - 3y = -10 x 2
10x - 6y = 5
4x - 6y = -20 .......(i)
10x - 6y = 5.......(ii)
eqn(ii) - eqn(1)
6x = 15
x = $\frac{15}{6}$
= $\frac{5}{2}$
x = 2 $\frac{1}{2}$
Sub. for x in equ.(ii) 10( $\frac{5}{2}$ ) - 6y = 5
y = 3 $\frac{1}{2}$

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 20 Rahoto

( $\sqrt{2} + 4 \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) is equal to

1

(

\sqrt{2} + 4 \sqrt{2}

)

(6

\sqrt{2}

8

Bayanin Amsa

( $\sqrt{2} + 4 \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ )
( $\sqrt{3} + \sqrt{2}$ + ( $\sqrt{3} + \sqrt{2}$ ) = 3 + ( $\sqrt{6} - \sqrt{6}$ ) - 2
= 1

Duba Amsa

Tambaya 21 Rahoto

In the figure, GHIJKLMN is a cube of side a. Find the length of HN.

\sqrt{3 a}

3a

3a2

a

\sqrt{2}

a

\sqrt{3}

Bayanin Amsa

HJ2 = a2 + a2 = 2a2

HJ = $\sqrt{2 a^{2}} = a \sqrt{2}$

HN2 = a2 + (a $\sqrt{2}$ )2 = a2 + 2a2 = 3a2

HN = $\sqrt{3 a^{2}}$

= a $\sqrt{3}$ cm

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Mensuration (JAMB)

Karanta bayanin darasi akan Inequalities (JAMB)

Polynomials

Mensuration

Inequalities

Duba Amsa

Tambaya 22 Rahoto

Find x if log9x = 1.5

72.0

27.0

36.0

3.5

24.5

Logarithms

Duba Amsa

Tambaya 23 Rahoto

Solve the following equation $\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$

(

\frac{5}{2}

, 1)

(5, -4)

(2, 1)

(1,

\frac{- 5}{2}

)

(

\frac{- 5}{2}

, 1)

Bayanin Amsa

$\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$
$\frac{2}{2 r - 1}$ - $\frac{1}{r + 2}$ = $\frac{5}{3}$
$\frac{2 r + 4 - 2 r + 1}{2 r - 1 (r + 2)}$ = $\frac{5}{3}$
$\frac{5}{(2 r + 1) (r + 2)}$ = $\frac{5}{3}$
5(2r - 1)(r + 2) = 15
(10r - 5)(r + 2) = 15
10r2 + 20r - 5r - 10 = 15
10r2 + 15r = 25
10r2 + 15r - 25 = 0
2r2 + 3r - 5 = 0
(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)
(r - 1)(2r + 5) = 0
r = 1 or $\frac{- 5}{2}$

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 24 Rahoto

22 $\frac{1}{2}$ % of the Nigerian Naira is equal to 17 $\frac{1}{10}$ % of a foreign currency M. What is the conversion rate of the M to the Naira?

1M = 1

\frac{15}{57}

N

1M = 38

\frac{1}{4}

N

1M = 1

\frac{18}{57}

N

1M = 384

\frac{3}{4}

N

Bayanin Amsa

N = 22 $\frac{1}{2}$ %, M = 17 $\frac{1}{10}$ %
M = $\frac{171}{10}$ %, N = $\frac{45}{2}$
$\frac{45}{2}$ x $\frac{10}{171}$
= $\frac{225}{171}$
= 1 $\frac{54}{171}$
= 1 $\frac{18}{57}$

Karanta bayanin darasi akan Percentages (WAEC)

Karanta bayanin darasi akan Financial Arithmetic (WAEC)

Percentages

Financial Arithmetic

Duba Amsa

Tambaya 25 Rahoto

List all integers satisfying the inequality -2 $\leq$ 2 x -6 < 4

2, 3, 4, 5

2, 3, 4

2, 5

3, 4, 5

4, 5

Bayanin Amsa

-2 $\leq$ 2x - 6 < 4 = 2x - 6 < 4

= 2x < 10

= x < 5

2x $\geq$ -2 + 6 $\geq$

= x $\geq$ 2

∴ 2 $\leq$ x < 5 [2, 3, 4]

Karanta bayanin darasi akan Inequalities (JAMB)

Inequalities

Duba Amsa

Tambaya 26 Rahoto

If the quadratic function 3x2 - 7x + R is a perfect square, find R

\frac{49}{24}

\frac{49}{12}

\frac{49}{13}

\frac{49}{3}

\frac{49}{36}

Bayanin Amsa

3x2 - 7x + R. Computing the square, we have
x2 - $\frac{7}{3}$ = - $\frac{R}{3}$
( $\frac{x}{1} - \frac{7}{6}$ )2 = - $\frac{R}{3}$ + $\frac{49}{36}$
$\frac{- R}{3}$ + $\frac{49}{36}$ = 0
R = $\frac{49}{36}$ x $\frac{3}{1}$
= $\frac{49}{12}$

Karanta bayanin darasi akan Fractions, Decimals, Approximations And Percentage (JAMB)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Fractions, Decimals, Approximations And Percentage

Quadratic Equations

Duba Amsa

Tambaya 27 Rahoto

The factors of 9 - (x2 - 3x - 1)2 are

-(x - 4)(x + 1) (x - 1)(x - 2)

(x - 4)(x - 2) (x - 1)(x + 1)

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

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

Algebraic Expressions

Duba Amsa

Tambaya 28 Rahoto

Find the missing value in the table below

x	-4	-3	-2	-1	0	1	2	3
y=4?3x?x^3	80		18	8	4	0	-10	-32

-32

-14

40

22

37

Duba Amsa

Tambaya 29 Rahoto

What is the probability that a number chosen at random from the intergers between 1 and 10 inclusive is either a prime or a multiple of 3?

\frac{7}{10}

\frac{3}{5}

\frac{4}{5}

\frac{3}{10}

Bayanin Amsa

Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10)
Prime: (2, 3, 5, 7)
multiples of 3: (3, 6, 9)
Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6)
Probability = $\frac{6}{10}$
= $\frac{3}{5}$

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 30 Rahoto

In the figure, PQ is the tangent from P to the circle QRS with SR as its diameter. If QRS = $θ$ $^{\circ}$ and RQP = $ϕ$ $^{\circ}$ , which of the following relationships between $θ$ $^{\circ}$ and $ϕ$ $^{\circ}$ is correct

θ

^{\circ}

+

ϕ

^{\circ}

= 902

ϕ

^{\circ}

= 902 - 2

θ

^{\circ}

θ

^{\circ}

=

ϕ

^{\circ}

ϕ

^{\circ}

= 2

θ

^{\circ}

θ

^{\circ}

+ 2

ϕ

^{\circ}

Bayanin Amsa

180 - $ϕ$ $^{\circ}$ = $θ$ $^{\circ}$ + $ϕ$ $^{\circ}$ (Sum of opposite interior angle equal to its exterior angle)

180 = 2 $ϕ$ + $θ$ $^{\circ}$

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Karanta bayanin darasi akan Circles (WAEC)

Introductory Calculus

Trigonometry

Circles

Duba Amsa

Tambaya 31 Rahoto

Find correct to two decimals places 100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$

100.02

1000.02

100.22

100.01

100.51

Bayanin Amsa

100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$
$\frac{1000, 000 + 100 + 30 + 27}{10000}$ = $\frac{1, 000.157}{10000}$
= 100.02

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Number Bases

Areas

Duba Amsa

Tambaya 32 Rahoto

Write h in terms of a, b, c, d if a = $\frac{b (1 - c h)}{a - d h}$

h = $\frac{a - b}{a d}$

h = $\frac{1 - b}{a d - b c}$

h = $\frac{(a - b)^{2}}{a d - b c}$

h =

\frac{a - b}{a d - b c}

h = $\frac{b - a}{a b - d c}$

Bayanin Amsa

a = $\frac{b (1 - c h)}{a - d h}$
a = $\frac{b - b c h}{1 - d h}$
= a - adh
= b - bch
a - b = bch + adn
a - b = adh
a - b = h(ad - bc)
h = $\frac{a - b}{a d - b c}$

Duba Amsa

Tambaya 33 Rahoto

In a restaurant, the cost of providing a particular type of food is partly constant and partially inversely proportional to the number of people. If cost per head for 100 people is 30k and the cost for 40 people is 60k, Find the cost for 50 people?

15k

20k

50k

40k

45k

Bayanin Amsa

C = a + k
$\frac{1}{N}$ = c
= $\frac{a N + k}{N}$
CN = aN + K
30(100) = a(100) + k
3000 = 100a + k.......(i)
60(40) = a(40) + k
2400 = 40a + k.......(ii)
eqn (i) - eqn (ii)
600 = 60a
a = 10
subt. for a in eqn (i) 3000 = 100(10) + K
3000 - 1000 = k
k = 2000
CN = 10N + 2000. when N = 50,
50C = 10(50) + 2000
50C = 500 + 2000
C = $\frac{2500}{50}$
= 50k

Karanta bayanin darasi akan Variation (WAEC)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Variation

Linear Inequalities

Duba Amsa

Tambaya 34 Rahoto

Find the area of a regular hexagon inscribed in a circle of radius 8cm

16

\sqrt{3}

cm2

96

\sqrt{3}

cm2

192

\sqrt{3}

cm2

16

\sqrt{3}

cm2

33cm2

Bayanin Amsa

Area of a regular hexagon = 8 x 8 x sin 60o
= 32 x $\frac{\sqrt{3}}{2}$ =
Area = 16 $\sqrt{3}$ x 6 = 96 $\sqrt{3}$ cm2

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 35 Rahoto

PRSQ is a trapezium of area 14cm2 in which PQ||RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm2

7.0

6.0

5.2

5.0

Bayanin Amsa

Area of trapezium = 14cm2
Area of trapezium = $\frac{1}{2}$ (a + b)h
14 = $\frac{1}{2}$ (4 + 3)h
14 = $\frac{7}{2}$ h
h = $\frac{14 \times 2}{7}$
= 4
Area of SQR = $\frac{1}{2}$ (3 x 4)
= $\frac{12}{2}$
= 6.0

Duba Amsa

Tambaya 36 Rahoto

If the hypotenuse of right angled isosceles triangle is 2, what is the length of each of the other sides?

\sqrt{2}

\frac{1}{2}

22

1

2 - 1

Bayanin Amsa

45o = $\frac{x}{2}$ , Since 45o = $\frac{1}{\sqrt{2}}$
x = 2 x $\frac{1}{\sqrt{2}}$
= $\frac{2 \sqrt{2}}{2}$
= $\sqrt{2}$

Karanta bayanin darasi akan Angles And Intercepts On Parallel Lines (WAEC)

Karanta bayanin darasi akan Statistics (WAEC)

Karanta bayanin darasi akan Angles (WAEC)

Angles And Intercepts On Parallel Lines

Statistics

Angles

Duba Amsa

Tambaya 37 Rahoto

In the figure, MNQP is a cyclic quadrilateral. MN and Pq are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = 86 $^{\circ}$ and NQP = 122 $^{\circ}$ find (x $^{\circ}$ , y $^{\circ}$ )

28

^{\circ}

, 36

^{\circ}

36

^{\circ}

, 28

^{\circ}

43

^{\circ}

, 61

^{\circ}

61

^{\circ}

, 43

^{\circ}

36

^{\circ}

, 43

^{\circ}

Bayanin Amsa

y $^{\circ}$ = 180 $^{\circ}$ - (86 $^{\circ}$ + 58 $^{\circ}$ )

180 - 144 = 36 $^{\circ}$

x $^{\circ}$ = 180 - (94 + 58)

180 -152 = 28

(x $^{\circ}$ , y $^{\circ}$ ) = (28 $^{\circ}$ , 36 $^{\circ}$ )

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Differentiation (JAMB)

Karanta bayanin darasi akan Inequalities (JAMB)

Polynomials

Differentiation

Inequalities

Duba Amsa

Tambaya 38 Rahoto

A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?

5kg

16kg

19kg

6kg

Bayanin Amsa

$\frac{1 \sqrt{3}}{(\frac{1}{2})^{2}}$
= $\frac{4}{\sqrt{3}}$
= $\frac{\sqrt{3}}{\sqrt{3}}$
= $\frac{4 \sqrt{3}}{\sqrt{3}}$
m = 64kg, V = $\frac{4 π r^{3}}{3}$
= $\frac{4 π (4)^{3}}{3}$
= $\frac{256 π}{3}$ x 10-6m3
density(P) = $\frac{Mass}{Volume}$
= $\frac{64}{\frac{256 π}{3 \times 10^{- 6}}}$
= $\frac{64 \times 3 \times 10^{- 6}}{256}$
= $\frac{3}{4 \times 10^{- 6}}$
m = PV = $\frac{3}{4 π \times 10^{- 6}}$ x $\frac{4}{3}$ $π$ [32 - 22] x 10-6
$\frac{3}{4 \times 10^{- 6}}$ x $\frac{4}{3}$ x 5 x 10-6
= 5kg

Duba Amsa

Tambaya 39 Rahoto

If two fair coins are tossed, what is the probability of getting at least one head?

\frac{1}{4}

\frac{1}{2}

1

\frac{43}{78}

\frac{3}{4}

Bayanin Amsa

Prob. of getting at least one head
Prob. of getting one head + prob. of getting 2 heads
= $\frac{1}{4}$ + $\frac{2}{4}$
= $\frac{3}{4}$

Duba Amsa

Tambaya 40 Rahoto

A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?

\frac{2}{3}

\frac{2}{15}

\frac{1}{2}

\frac{1}{3}

Bayanin Amsa

P(R1) = $\frac{6}{10}$
= $\frac{2}{3}$
P(R1 n R11) = P(both red)
$\frac{3}{5}$ x $\frac{5}{9}$
= $\frac{1}{3}$

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Probability (JAMB)

Probability

Duba Amsa

Tambaya 41 Rahoto

In $△$ XYZ, XY = 13cm, YZ = 9cm, XZ = 11cm and XYZ = $θ$ . Find cos $θ$ o

\frac{4}{39}

\frac{43}{39}

\frac{209}{3}

\frac{43}{78}

Bayanin Amsa

cos $θ$ = $\frac{13^{2} + 9^{2} - 11^{2}}{2 (13) (9)}$
= $\frac{169 + 81 - 21}{26 \times 9}$
cos $θ$ = $\frac{129}{26 \times 9}$
= $\frac{43}{78}$

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Introductory Calculus

Trigonometry

Duba Amsa

Tambaya 42 Rahoto

If cos $θ$ = $\frac{\sqrt{3}}{2}$ and $θ$ is less than 90o. Calculate cos $\frac{90 - θ}{s i n^{2} θ}$

\frac{4}{\sqrt{3}}

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

\sqrt{\frac{76}{2}}

4

\sqrt{\frac{76}{2}}

Bayanin Amsa

cos $\frac{90 - θ}{s i n^{2} θ}$
= $\frac{\tan θ}{s i n^{2} θ}$
Sin2 $θ$ = $\frac{1}{4}$
$\frac{c o s (90 - θ)}{s i n^{2} θ}$
= $\frac{1}{\sqrt{3}}$
= $\frac{4}{\sqrt{3}}$

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Duba Amsa

Tambaya 43 Rahoto

The ratio of the length of two similar rectangular blocks is 2 : 3. If the volume of the larger block is 351cm3, then the volume of the other block is

234.00 cm3

526.50 cm3

166.00 cm3

687cm3

Bayanin Amsa

Let x represent total vol. 2 : 3 = 2 + 3 = 5
$\frac{3}{5}$ x = 351
x = $\frac{351 \times 5}{3}$
= 585
Volume of smaller block = $\frac{2}{3}$ x 585
= 234.00

Karanta bayanin darasi akan Polynomials (JAMB)

Polynomials

Duba Amsa

Tambaya 44 Rahoto

Without using table, calculate the value of 1 + sec2 30o

2

\frac{1}{3}

\frac{2}{15}

\frac{5}{3}

3

\frac{1}{2}

Bayanin Amsa

1 + sec2 30o = sec 30o
= $\frac{2}{\sqrt{3}}$
$\frac{(2)^{2}}{3}$
= $\frac{4}{3}$
1 + sec2 30o = sec 30o
= 1 + $\frac{4}{\sqrt{3}}$
= 2 $\frac{1}{3}$

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Loci (JAMB)

Karanta bayanin darasi akan Matrices And Determinants (JAMB)

Sine, Cosine And Tangent Of An Angle

Introductory Calculus

Loci

Matrices And Determinants

Duba Amsa

Tambaya 45 Rahoto

In the figure, the area of the shaded segment is

3

π

9

\frac{\sqrt{3}}{4}

3

π - 3 \frac{\sqrt{3}}{4}

\frac{(\sqrt{3 - π)}}{4}

π + \frac{9 \sqrt{3}}{4}

Bayanin Amsa

Area of sector = $\frac{120}{360} \times π \times (3)^{2} = 3 π$

Area of triangle = $\frac{1}{2} \times 3 \times 3 \times \sin 120^{o}$

= $\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9 \sqrt{3}}{4}$

Area of shaded portion = 3 $π - \frac{9 \sqrt{3}}{4}$

= 3 $π - 3 \frac{\sqrt{3}}{4}$

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Loci (WAEC)

Volumes

Loci

Duba Amsa

Tambaya 46 Rahoto

If three numbers P, Q, R are in ratio 6 : 4 : 5, find the value of $\frac{3 p - q}{4 q + r}$

3

2

3

2

3

18

Bayanin Amsa

P : Q : r = 6 : 4 : 5
5 = 6 + 4 + 5
= 15
P = $\frac{6}{15}$ , q = $\frac{4}{15}$ , r = $\frac{5}{15}$ = $\frac{1}{3}$
To find $\frac{3 p - q}{4 q + r}$
3p - q = 3 x $\frac{6}{15}$ - $\frac{4}{15}$
$\frac{18}{15}$ - $\frac{4}{15}$ = $\frac{14}{15}$
∴ 4q + r = 4 x $\frac{4}{15}$ + $\frac{5}{15}$
$\frac{16}{15}$ = $\frac{16}{15}$ + $\frac{5}{15}$
= $\frac{21}{15}$
$\frac{14}{15}$ x $\frac{15}{21}$ = $\frac{14}{21}$
= $\frac{2}{3}$

Duba Amsa

Tambaya 47 Rahoto

Factorize abx2 + 8y - 4bx - 2axy

(ax - 4)(bx - 2y)

(ax + b)(x - 8y)

(ax - 2y)(bx - 4)

(bx - 4)(ax - 2y)

(abx - 4)(x - 2y)

Algebraic Expressions

Duba Amsa






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