A carregar...






Pressione e Mantenha para Arrastar
Clique aqui para fechar

JAMB UTME - Mathematics - 2018

Pergunta 1 Relatório

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

Detalhes da Resposta

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.

Ler a nota de aula em Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Ver Resposta

Pergunta 2 Relatório

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}

Detalhes da Resposta

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}

Ler a nota de aula em Probability (WAEC)

Ler a nota de aula em Probability (JAMB)

Probability

Probability

Ver Resposta

Pergunta 8 Relatório

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

1

\frac{1}{4}

2

\frac{1}{4}

6

10

Detalhes da Resposta

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}

Ler a nota de aula em Integration (JAMB)

Ler a nota de aula em Application Of Differentiation (JAMB)

Integration

Application Of Differentiation

Ver Resposta

Pergunta 10 Relatório

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

8.8 x 10

^{- 1}

8.8 x 10

^{- 2}

8.8 x 10

^{- 3}

8.8 x 10

^{3}

Detalhes da Resposta

Explanation:

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

= \frac{0.000016384}{0.002}

= 0.008192

Ler a nota de aula em Number Bases (WAEC)

Ler a nota de aula em Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Ver Resposta

Pergunta 11 Relatório

In how many ways can the letters LEADER be arranged?

72

144

360

720

Detalhes da Resposta

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

Ler a nota de aula em Permutation And Combination (JAMB)

Permutation And Combination

Ver Resposta

Pergunta 12 Relatório

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°

Detalhes da Resposta

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

Ler a nota de aula em Bearings (WAEC)

Ver Resposta

Pergunta 15 Relatório

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

Detalhes da Resposta

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

Ler a nota de aula em Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Ver Resposta

Pergunta 18 Relatório

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.

\frac{1}{10}

\frac{2}{10}

\frac{9}{20}

\frac{11}{20}

Detalhes da Resposta

Pr(both John and James passed)
=

\frac{3}{4}

x

\frac{3}{5}

=

\frac{9}{20}

Pr(John and James failed)= 1 - Pr(John and James passed)
1 –

\frac{9}{20}

=

\frac{11}{20}

Ler a nota de aula em Probability (WAEC)

Ler a nota de aula em Probability (JAMB)

Probability

Probability

Ver Resposta

Pergunta 19 Relatório

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}

Detalhes da Resposta

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.

Ler a nota de aula em Probability (WAEC)

Ler a nota de aula em Probability (JAMB)

Probability

Probability

Ver Resposta

Pergunta 22 Relatório

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

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}}

Detalhes da Resposta

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}

Ler a nota de aula em Change Of Subject Of A Formula/Relation (WAEC)

Ler a nota de aula em Solution Of Linear Equations (WAEC)

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Ver Resposta

Pergunta 23 Relatório

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}}

Detalhes da Resposta

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}

Ver Resposta

Pergunta 24 Relatório

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

Detalhes da Resposta

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

Ler a nota de aula em Sets (JAMB)

Ler a nota de aula em Measures Of Dispersion (JAMB)

Measures Of Dispersion

Ver Resposta

Pergunta 25 Relatório

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

Ver Resposta

Pergunta 26 Relatório

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

Detalhes da Resposta

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

Ver Resposta

Pergunta 28 Relatório

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}

Detalhes da Resposta

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

Ler a nota de aula em Bearings (WAEC)

Ler a nota de aula em Coordinate Geometry (WAEC)

Coordinate Geometry

Ver Resposta

Pergunta 29 Relatório

Find the values of x for which

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

x < 8

x > -6

x < 4

x > -3

Detalhes da Resposta

\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

Ler a nota de aula em Solution Of Linear Equations (WAEC)

Ler a nota de aula em Inequalities (JAMB)

Solution Of Linear Equations

Inequalities

Ver Resposta

Pergunta 30 Relatório

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

Detalhes da Resposta

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

Ler a nota de aula em Sine, Cosine And Tangent Of An Angle (WAEC)

Ler a nota de aula em Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Ver Resposta

Pergunta 31 Relatório

Express 495g as a percentage of 16.5kg

3%

3

\frac{1}{3}

%

15%

30%

Detalhes da Resposta

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%

Ler a nota de aula em Percentages (WAEC)

Percentages

Ver Resposta

Pergunta 33 Relatório

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

20

30

48

60

Detalhes da Resposta

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

\div

16 = 60

Ler a nota de aula em Solution Of Linear Equations (WAEC)

Ler a nota de aula em Number Bases (WAEC)

Ler a nota de aula em Surds (radicals) (WAEC)

Solution Of Linear Equations

Number Bases

Surds (radicals)

Ver Resposta

Pergunta 34 Relatório

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}

Detalhes da Resposta

\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}

Ler a nota de aula em Solution Of Linear Equations (WAEC)

Ler a nota de aula em Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Ver Resposta

Pergunta 37 Relatório

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

1110

10111

11101

111100

Detalhes da Resposta

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}

Ler a nota de aula em Number Bases (WAEC)

Ler a nota de aula em Number Bases (JAMB)

Number Bases

Number Bases

Ver Resposta

Pergunta 38 Relatório

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}

Detalhes da Resposta

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.

Ler a nota de aula em Probability (WAEC)

Ler a nota de aula em Linear Inequalities (WAEC)

Probability

Linear Inequalities

Ver Resposta

Pergunta 40 Relatório

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

32 cm

24 cm

16 cm

12 cm

Detalhes da Resposta

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

Ler a nota de aula em Circles (WAEC)

Ler a nota de aula em Angles Of Elevation And Depression (WAEC)

Ler a nota de aula em Angles And Intercepts On Parallel Lines (WAEC)

Ler a nota de aula em Triangles And Polygons (WAEC)

Angles Of Elevation And Depression

Angles And Intercepts On Parallel Lines

Triangles And Polygons

Ver Resposta

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