JAMB UTME - Mathematics - 2018

\[ P=\left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \]

Make T the subject of the relation.

T = \frac{R + P3}{15 Q}

T = \frac{R - 15 P^{3}}{Q}

T = R - \frac{15 P^{3}}{Q}

T = \frac{15 R + Q}{P^{3}}

Akọwa Nkọwa

P = {[\frac{Q (R - T)}{15}]}^{\frac{1}{3}}

P^{3} = \frac{Q (R - T)}{15}

15 P^{3} = Q (R - T)

\frac{15 P^{3}}{Q} = R - T

T = R - \frac{15 P^{3}}{Q}

Gụọ akwụkwọ nkuzi na Change Of Subject Of A Formula/Relation (WAEC)

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Lee Azịza

Ajụjụ 2 Ripọtì

What is the loci of a distance 4cm from a given point P?

A straight line of length 4cm

a circle of radius 4cm

perpendicular to point P at 4cm

a circle of diameter 4cm

Akọwa Nkọwa

The locus of a distance 4cm from a given point P is a circle of radius 4cm centered at point P. To understand this, imagine drawing all the points that are exactly 4cm away from point P. These points form a circle, since they are equidistant from the center, which is point P. This circle has a radius of 4cm, since all points on the circle are exactly 4cm away from point P. So, the correct option is "a circle of radius 4cm". The other options are not correct because: - A straight line of length 4cm: This would only include points that are exactly 4cm away from point P in one direction. However, there are infinitely many points that are 4cm away from point P in all directions, so this cannot be the correct locus. - Perpendicular to point P at 4cm: This would be a line that is perpendicular to the line that passes through point P, but it would not include all points that are exactly 4cm away from point P. - A circle of diameter 4cm: This would only include points that are exactly 2cm away from point P in all directions, not points that are exactly 4cm away.

Gụọ akwụkwọ nkuzi na Loci (JAMB)

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Loci

Lee Azịza

Ajụjụ 3 Ripọtì

A machine valued at N20,000 depreciates by 10% every year. What will be the value of the machine at the end of two years?

N16,200

N14,200

N12,000

N8,000

Akọwa Nkọwa

Since it depreciates by 10% At the end of first year, its value = 90% of 20000
=

\frac{90}{100}

x 20000 =18000
At the end of second year, its value = 90% of 18000
=

\frac{90}{100}

x 18000 = ₦16,200

Lee Azịza

Ajụjụ 4 Ripọtì

Calculate the area of an equilateral triangle of side 8cm

8√3

16

4√3

16√3

Akọwa Nkọwa

To calculate the area of an equilateral triangle, we can use the formula: Area = (sqrt(3) / 4) * (side length)^2 Since we know that the side length is 8cm, we can substitute it into the formula: Area = (sqrt(3) / 4) * (8cm)^2 Simplifying this expression, we get: Area = (sqrt(3) / 4) * 64cm^2 Area = (16sqrt(3)) cm^2 Therefore, the area of the equilateral triangle is 16sqrt(3) square centimeters. So the answer is 16√3.

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Areas

Angles

Lee Azịza

Ajụjụ 5 Ripọtì

If $ P = \left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} $ make T the subject of the relation

T =

\frac{R + P^{3}}{15 Q}

T =

\frac{R - 15 P^{3}}{Q}

T =

\frac{R - 15 P^{3}}{Q}

T =

\frac{15 R - Q}{P^{3}}

Akọwa Nkọwa

Taking the cube of both sides of the equation give
P

^{3}

=

\frac{Q (R - T)}{15}

Cross multiplying
15P

^{3}

= Q(R - T)
Divide both sides by Q

\frac{15 P^{3}}{Q}

= R - T
Rearranging gives
T = R -

\frac{15 P^{3}}{Q}

=

\frac{R Q - 15 P^{3}}{Q}

Lee Azịza

Ajụjụ 6 Ripọtì

Evaluate $5^{-3}\log^{2} \times 2^{2\log 3}$

8

\frac{1}{8}

1

\frac{1}{8}

\frac{2}{5}

Indices, Logarithms And Surds

Logarithms

Lee Azịza

Ajụjụ 7 Ripọtì

In how many ways can the letters LEADER be arranged?

72

144

360

720

Akọwa Nkọwa

The word LEADER has 1L 2E 1A 1D and 1R making total of 6!

\frac{6}{1! 2! 1! 1! 1!}

=

\frac{6!}{2!}

=

\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}

= 360

Gụọ akwụkwọ nkuzi na Permutation And Combination (JAMB)

Permutation And Combination

Lee Azịza

Ajụjụ 8 Ripọtì

A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room.

\frac{15}{17}

\frac{9}{17}

\frac{8}{15}

\frac{12}{17}

Akọwa Nkọwa

$a^{2} = 12^{2} + 9^{2}$

Gụọ akwụkwọ nkuzi na Bearings (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Bearings

Coordinate Geometry

Lee Azịza

Ajụjụ 9 Ripọtì

A car uses one litre of petrol for every 14km. If one litre of petrol cost ₦63.00, how far can the car go with ₦900.00 worth of petrol?

420km

405km

210km

200km

Akọwa Nkọwa

To solve this problem, we need to first find out how many litres of petrol we can buy with ₦900.00, and then use that to calculate how far the car can go. We know that one litre of petrol costs ₦63.00, so we can divide ₦900.00 by ₦63.00 to find out how many litres we can buy: ₦900.00 ÷ ₦63.00 = 14.29 litres (rounded to two decimal places) So we can buy 14.29 litres of petrol with ₦900.00. Next, we need to use the information given to calculate how far the car can go with 14.29 litres of petrol. We are told that the car uses one litre of petrol for every 14km, so we can multiply 14.29 litres by 14km to find out how far the car can go: 14.29 litres x 14km = 200.06km (rounded to two decimal places) Therefore, the car can go 200.06km with ₦900.00 worth of petrol. So the correct answer is: 200km (rounded to the nearest kilometre).

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 10 Ripọtì

Find the equation of the line through (5,7) parallel to the line 7x + 5y = 12.

5x + 7y = 20

7x + 5y = 70

xy = 7

15x + 17y = 90

Akọwa Nkọwa

To find the equation of a line parallel to another line, we need to use the fact that parallel lines have the same slope. The given line 7x + 5y = 12 can be rearranged into slope-intercept form, which is y = (-7/5)x + (12/5), where the slope is -7/5. Since the line we want to find is parallel to this line, it must also have a slope of -7/5. We also know that the line passes through the point (5,7). To find the equation of this line, we can use the point-slope form, which is y - y1 = m(x - x1), where (x1,y1) is the given point and m is the slope. Substituting in the values we know, we get: y - 7 = (-7/5)(x - 5) Simplifying this equation, we get: y = (-7/5)x + (49/5) So the equation of the line through (5,7) parallel to the line 7x + 5y = 12 is y = (-7/5)x + (49/5), which is option (A) 5x + 7y = 20.

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 11 Ripọtì

If X, Y can take values from the set (1, 2, 3 ,4), find the probability that the product of X and Y is not greater than 6.

\frac{5}{8}

\frac{5}{16}

\frac{1}{2}

\frac{3}{8}

Akọwa Nkọwa

Each multiplication of elements gives 16 results out of which the ones that are not greater than 6 are asterisked (*) and they are 8 in numbers
Pr(product of x and y NOT > 6) =

\frac{8}{16}

=

\frac{1}{2}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Probability (JAMB)

Probability

Lee Azịza

Ajụjụ 12 Ripọtì

if $y = 23_{\mathit{five}} + 101_{\mathit{three}}$ find y leaving your answer in base two

1110

10111

11101

111100

Akọwa Nkọwa

First we convert the numbers to base ten
23

_{f i v e}

= 2 x 51 + 3 x 50
= 10 + 3 = 13
101

_{f i v e}

= (1 x 32) + (0 x 31) + (1 x 30)
= 9 + 0 + 1 = 10
So, y = 13 + 10 = 23
To convert 23 to base 2 (as in the diagram above)
y = 23
= 10111

_{f i v e}

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Number Bases

Lee Azịza

Ajụjụ 13 Ripọtì

Find the average of the first four prime numbers greater than 10

20

19

17

15

Akọwa Nkọwa

The first four prime numbers greater than 10 are 11, 13, 17, and 19. To find the average of these numbers, we add them all up and then divide by the number of items in the set. So, (11 + 13 + 17 + 19) / 4 = 60 / 4 = 15. Therefore, the average of the first four prime numbers greater than 10 is 15.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Lee Azịza

Ajụjụ 14 Ripọtì

In the figure below, /MX/ = 8cm, /XN/ = 12cm, /NZ/ = 4cm and ∠ XMN = ∠ XZY. Calculate /YM/

32 cm

24 cm

16 cm

12 cm

Akọwa Nkọwa

From the figure,
∠ XMN = ∠ XZY
Angle X is common
So, ∠ XNM = ∠ XYZ
Then from the angle relationship

\frac{X M}{X Z}

=

\frac{X N}{X Y}

=

\frac{M N}{Z Y}

XM = 8, XZ = 12 + 4 = 16,
XN = 12, XY = 8 + YM

\frac{8}{16}

=

\frac{12}{(8 + Y M)}

Cross multiply
8(8 + YM) = 192
64 + 8YM = 192
8YM = 128
YM =

\frac{128}{8}

= 16cm

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Angles Of Elevation And Depression (WAEC)

Gụọ akwụkwọ nkuzi na Angles And Intercepts On Parallel Lines (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Angles Of Elevation And Depression

Angles And Intercepts On Parallel Lines

Triangles And Polygons

Lee Azịza

Ajụjụ 15 Ripọtì

A boy walks 800m in 20 minutes. Calculate his average speed in km/h.

2.4

4

24

4.2

Akọwa Nkọwa

The average speed of the boy can be calculated by dividing the distance he traveled by the time it took him to travel that distance. To convert the distance from meters to kilometers, divide 800 by 1000, which gives 0.8 km. To convert the time from minutes to hours, divide 20 by 60, which gives 0.3333 hours. Now, we can calculate the average speed by dividing the distance by the time: 0.8 km ÷ 0.3333 hours = 2.4 km/h. So, the average speed of the boy is 2.4 km/h.

Lee Azịza

Ajụjụ 16 Ripọtì

If the simple interest on a sum of money invested at 3% per annum for $2\frac{1}{2}$ years is ₦123, find the principal.

₦676.50

₦820

₦1,640

₦4,920

Akọwa Nkọwa

To find the principal, we need to use the formula for simple interest: Simple Interest = (Principal * Rate * Time) / 100 where Rate is the interest rate, Time is the time period in years, and Principal is the amount of money invested. In this case, the rate is 3%, the time is 2 years, and the simple interest is ₦123. We can plug these values into the formula and solve for the Principal: ₦123 = (Principal * 3 * 2) / 100 ₦123 = (6 * Principal) / 100 ₦123 * 100 = 6 * Principal Principal = (₦123 * 100) / 6 Principal = ₦2050 Therefore, the principal is ₦2050. Answer (E) is not among the options provided. Among the given options, the closest answer to ₦2050 is option (C) ₦1,640.

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Financial Arithmetic

Lee Azịza

Ajụjụ 17 Ripọtì

The pie chart shows the monthly expenditure of a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is ₦7200?

₦1000

₦2000

₦3000

₦4000

Representation Of Data

Areas

Lee Azịza

Ajụjụ 18 Ripọtì

Express 495g as a percentage of 16.5kg

3%

3

\frac{1}{3}

%

15%

30%

Akọwa Nkọwa

To express 495g as a percentage of 16.5kg, we need to convert both values to the same units of measurement, such as grams or kilograms. We can convert 16.5kg to grams by multiplying by 1000: 16.5kg x 1000 = 16500g Now we can express 495g as a percentage of 16500g: 495g / 16500g x 100% = 0.03 x 100% = 3% Therefore, 495g is 3% of 16.5kg. So the answer is: 3%

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Percentages

Lee Azịza

Ajụjụ 19 Ripọtì

Which one of the following gives the members of the set $A^c \cap B \cap C$?

Φ

{s}

{t, u}

{y, z}

Akọwa Nkọwa

A1 = Elements in the universal set but not in A = {s, w, x, y, z}
B = {r, s. t, u}
C = {t, u, v, w, x}
A1 n B n C = elements common to the three sets = none = empty set = Φ

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 20 Ripọtì

Divide the L.C.M of 48, 64 and 80 by their H.C.F.

20

30

48

60

Akọwa Nkọwa

LCM of 80, 64, 48 = 960
HCF of 80, 64, 48 = 16
960

\div

16 = 60

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Solution Of Linear Equations

Number Bases

Surds (radicals)

Lee Azịza

Ajụjụ 21 Ripọtì

A trader realises 10x - x$^{2}$ Naira profit from the sale of x bags of corn. How many bags will give him the maximum profit?

7

6

5

4

Akọwa Nkọwa

Profit (P) =

10 x - x^{2}

Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0
i.e.

\frac{δ p}{δ x}

= 0

\frac{δ p}{δ x}

= 10 - 2x = 0
10 = 2x
Then x =

\frac{10}{2}

= 5

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 22 Ripọtì

A man's initial salary is ₦540.00 a month and increases after each period of six months by ₦36.00 a month. Find his salary in the eight month of the third year.

₦828.00

₦756.00

₦720.00

₦684.00

Akọwa Nkọwa

Since the salary increases by 36 after every 6 months
Every 6 months that can be counted on the eight month of the third year is 5
(i.e. 2 times in the first year, 2 times in the second year and once in the third year)
His salary then = initial salary + increment
= 540 + 5(36)
= 540 + 180
= ₦720.00
It can also be solves using a sequence in form of an AP

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 23 Ripọtì

Use the cumulative frequency curve to answer the question.
Estimate the median of the date represented on the graph.

35.5

36.5

37.5

38.5

Representation Of Data

Measures Of Location

Lee Azịza

Ajụjụ 24 Ripọtì

The pie chart shows the monthly expenditure of a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is ₦7200?

₦1000

₦2000

₦3000

₦4000

Akọwa Nkọwa

Let the monthly expenditure angle for school fees is x, then that of housing will be 2x. Since the total angle in the circle is 360.
Using 90 as the angle for transport
So, x + 2x + 120 + 90 = 360
3x + 210 = 3605
3x = 360 - 210
= 150
x =

\frac{150}{3}

= 50
So the angle for housing is 2x = 2 × 50
= 100
Amount spent on housing =

\frac{100}{360}

× 7200
= ₦2000

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Lee Azịza

Ajụjụ 25 Ripọtì

\[ \frac{0.00256 \times 0.0064}{0.025 \times 0.08} \]

8.8 x 10

^{- 1}

8.8 x 10

^{- 2}

8.8 x 10

^{- 3}

8.8 x 10

^{3}

Akọwa Nkọwa

Explanation:

\frac{0.00256 \times 0.0064}{0.025 \times 0.08}

= \frac{0.000016384}{0.002}

= 0.008192

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Lee Azịza

Ajụjụ 26 Ripọtì

Solve for t in the equation $ \frac{3}{4}t + \frac{1}{3}(21 - t) = 11 $

\frac{9}{13}

\frac{7}{13}

5

9

\frac{3}{5}

Akọwa Nkọwa

\frac{3}{4}

t +

\frac{1}{3}

(21 - t) = 11
Multiply through by the LCM of 4 and 3 which is 12
12 x(

\frac{3}{4}

t) + 12 x (

\frac{1}{3}

(21 - t)) = (11 x 12)
9t + 4(21 - t) = 132
9t + 84 - 4t = 132
5t + 84 = 132
5t = 132 - 84 = 48
t =

\frac{48}{5}

t = 9

\frac{3}{5}

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Lee Azịza

Ajụjụ 27 Ripọtì

The figure below is a Venn diagram showing the elements arranged within sets A, B, C, $ \epsilon $.

Use the figure to answer this question

What is $ n((A \cup B)^c) $?

2

3

4

7

Akọwa Nkọwa

A = (p, q, r, t, u, v)
B = (r, s, t, u)
A U B = Elements in both A and B = (p, q, r, s, t, u, v)
(A U B)1 = elements in the universal set E but not in (A U B)= (w, x, y, z)
n(A U B) 1 = number of the elements in (A U B)1 = 4

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 28 Ripọtì

If \[ N=\begin{pmatrix}3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1\end{pmatrix} \] , find $|N|$.

65

23

17

91

Akọwa Nkọwa

3[(-3 × 1) - (2 × -5)] - 5[(6 × 1) - (-5 × -2)] + (-4)[(6 × 2) - (-3 × -2)]
3(-3 + 10) - 5(6 - 10) - 4(12 - 6)
3(7) - 5(-4) - 4(6) = 17

Gụọ akwụkwọ nkuzi na Matrices And Determinants (WAEC)

Gụọ akwụkwọ nkuzi na Matrices And Determinants (JAMB)

Matrices And Determinants

Lee Azịza

Ajụjụ 29 Ripọtì

om a point R, 300m north of P, a man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degree

026°

045°

210°

240°

Akọwa Nkọwa

$\sin y = \frac{300 m}{600 m} = \frac{1}{2}$
$y = \sin^{- 1} 0.5$
$y = 30^{\circ}$
$x = 180^{\circ} + (90^{\circ} - 30^{\circ})$
$x = 180^{\circ} + 60^{\circ}$
$x = 240^{\circ}$

Gụọ akwụkwọ nkuzi na Bearings (WAEC)

Bearings

Lee Azịza

Ajụjụ 30 Ripọtì

Tossing a coin and rolling a die are two separate events. What is the probability of obtaining a tail on the coin and an even number on the die?

\frac{1}{16}

\frac{1}{6}

\frac{1}{4}

\frac{3}{8}

Akọwa Nkọwa

The probability of obtaining a tail on a coin is 1/2 since there are two possible outcomes (either a head or a tail) and they are equally likely. The probability of obtaining an even number on a die is 3/6 or 1/2 since there are three even numbers (2, 4, 6) and six possible outcomes (1, 2, 3, 4, 5, 6) and they are equally likely. Since the events of obtaining a tail on the coin and an even number on the die are independent (i.e., the outcome of one event does not affect the outcome of the other), we can multiply their probabilities to find the probability of obtaining both: P(tail and even) = P(tail) x P(even) = (1/2) x (1/2) = 1/4 Therefore, the probability of obtaining a tail on the coin and an even number on the die is 1/4 or 0.25.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Probability (JAMB)

Probability

Lee Azịza

Ajụjụ 31 Ripọtì

The table shown gives the marks scored by a group of student in a test. Use the table to answer the question given.

Mark	0	1	2	3	4	5
Frequency	1	2	7	5	4	3

What is the median mark?

1

2

3

4

Akọwa Nkọwa

Total frequency = 1 + 2 + 7 + 5 + 4 + 3 = 22
Median is the middle number
=

\frac{N t h}{2}

Term =

\frac{22 t h}{2}

= 11th term
Going in ascending order, 11th term is 3, going in descending order 11th term is 3
Median = 3 +

\frac{3}{2}

=

\frac{6}{2}

= 3

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Sets

Measures Of Dispersion

Lee Azịza

Ajụjụ 32 Ripọtì

Use the quadratic equation curve to answer this question.

What is the 80th percentile?

-5.3

0.5

3

8

Akọwa Nkọwa

The minimum value is the lowest value of the curve on y axis which gives a value of -5.3.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Representation Of Data

Quadratic Equations

Lee Azịza

Ajụjụ 33 Ripọtì

The table below shows the frequency of children of age x years in a hospital:

x	1	2	3	4	5	6	7	8
f	3	4	5	6	7	6	5	4

Use the table to answer the question below:

What is the modal age?

4

5

6

7

Akọwa Nkọwa

The modal age is the age with the highest frequency, and that is age 5 years with f of 7

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Representation Of Data

Lee Azịza

Ajụjụ 34 Ripọtì

Simplify $25^{\frac{1}{2}} \times 8^{-\frac{2}{3}}$

1

\frac{1}{4}

2

\frac{1}{4}

6

10

Akọwa Nkọwa

Using law of indices:
25

\frac{1}{2}

× 8

\frac{- 2}{3}

= √25 x (

\sqrt[3]{8}

) -2
= 5 x 2-2
= 5 x

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

=

\frac{5}{4}

=

\frac{11}{4}

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Application Of Differentiation (JAMB)

Integration

Application Of Differentiation

Lee Azịza

Ajụjụ 35 Ripọtì

Approximate 0.9875 to 1 decimal place.

1.1

1.0

0.9

0.10

Akọwa Nkọwa

9 is on one decimal place, the next number to it is 8 which will be rounded up to 1 because it is greater than 5 and then added to 9 to give 10, 10 cannot be written, it will then be rounded up to 1 and added to 0.
So the answer is 1.0

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 36 Ripọtì

Tanθ is positive and Sinθ is negative. In which quadrant does θ lies

Second only

Third only

Fourth only

First and third only

Akọwa Nkọwa

First quadrant: Sin, Cos and Tan are all positive
Second quadrant: Sin is positive, Cos is negative and Tan is negative
Third quadrant: Tan is positive, Sin is negative and Cos is negative
Fourth quadrant: Cos is positive, Sin is negative and Tan is negative
The correct option is the third quadrant only where Tanθ is positive and Sinθ is negative

Lee Azịza

Ajụjụ 37 Ripọtì

The table shown gives the marks scored by a group of student in a test. Use the table to answer the question given.

Mark	0	1	2	3	4	5
Frequency	1	2	7	5	4	3

What is the probability of selecting a student from the group that scored 2 or 3?

\frac{1}{11}

\frac{5}{22}

\frac{7}{22}

\frac{6}{11}

Akọwa Nkọwa

To find the probability of selecting a student from the group who scored 2 or 3, we need to add the number of students who scored 2 and 3, and then divide that by the total number of students in the group. The number of students who scored 2 is 7, and the number of students who scored 3 is 5. So, 7 + 5 = 12 students scored either 2 or 3. The total number of students in the group is 1 + 2 + 7 + 5 + 4 + 3 = 22. Therefore, the probability of selecting a student from the group who scored 2 or 3 is 12/22 = 6/11.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Probability

Linear Inequalities

Lee Azịza

Ajụjụ 38 Ripọtì

Find the values of x for which

$ \frac{x+2}{4} - \frac{2x-3}{3} < 4 $

x < 8

x > -6

x < 4

x > -3

Akọwa Nkọwa

\frac{x + 2}{4} - \frac{2 x - 3}{3} < 4

\frac{3 x + 6 - (8 x - 12)}{12} < 4

\frac{- 5 x + 18}{12} < 4

- 5 x + 18 < 48

- 5 x < 30

x > - 6

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Solution Of Linear Equations

Inequalities

Lee Azịza

Ajụjụ 39 Ripọtì

The probabilities that John and James pass an examination are $ \frac{3}{4} $ and $ \frac{3}{5} $ respectively. Find the probability of both boys failing the examination.