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JAMB UTME - Mathematics - 2018

Question 1 Rapport

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}

Détails de la réponse

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.

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Probability

Linear Inequalities

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Question 2 Rapport

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

Détails de la réponse

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

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Measures Of Dispersion

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Question 3 Rapport

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

8.8 x 10

^{- 1}

8.8 x 10

^{- 2}

8.8 x 10

^{- 3}

8.8 x 10

^{3}

Détails de la réponse

Explanation:

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

= \frac{0.000016384}{0.002}

= 0.008192

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Number Bases

Modular Arithmetic

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Question 4 Rapport

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}

Détails de la réponse

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

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Coordinate Geometry

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Question 5 Rapport

In how many ways can the letters LEADER be arranged?

72

144

360

720

Détails de la réponse

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

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Permutation And Combination

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Question 6 Rapport

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

1

\frac{1}{4}

2

\frac{1}{4}

6

10

Détails de la réponse

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}

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Integration

Application Of Differentiation

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Question 9 Rapport

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

32 cm

24 cm

16 cm

12 cm

Détails de la réponse

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

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Angles Of Elevation And Depression

Angles And Intercepts On Parallel Lines

Triangles And Polygons

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Question 11 Rapport

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

Détails de la réponse

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

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Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

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Question 12 Rapport

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°

Détails de la réponse

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

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Voir la réponse

Question 14 Rapport

Express 495g as a percentage of 16.5kg

3%

3

\frac{1}{3}

%

15%

30%

Détails de la réponse

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%

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Percentages

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Question 15 Rapport

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

20

30

48

60

Détails de la réponse

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

\div

16 = 60

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Solution Of Linear Equations

Number Bases

Surds (radicals)

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Question 16 Rapport

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}

Détails de la réponse

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}

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Probability

Probability

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Question 18 Rapport

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

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}}

Détails de la réponse

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}

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Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

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Question 19 Rapport

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}

Détails de la réponse

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.

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Probability

Probability

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Question 21 Rapport

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}}

Détails de la réponse

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}

Voir la réponse

Question 25 Rapport

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}

Détails de la réponse

\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}

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Solution Of Linear Equations

Quadratic Equations

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Question 26 Rapport

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

1110

10111

11101

111100

Détails de la réponse

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}

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Number Bases

Number Bases

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Question 28 Rapport

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

Détails de la réponse

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

Voir la réponse

Question 30 Rapport

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

Voir la réponse

Question 33 Rapport

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

Détails de la réponse

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

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Solution Of Linear Equations

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Question 34 Rapport

Find the values of x for which

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

x < 8

x > -6

x < 4

x > -3

Détails de la réponse

\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

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Solution Of Linear Equations

Inequalities

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Question 40 Rapport

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}

Détails de la réponse

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}

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Probability

Probability

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