JAMB UTME - Mathematics - 2018

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.

Ka ìwé ẹ̀kọ́ nípa Financial Arithmetic (WAEC)

Financial Arithmetic

Wo Idahun

Ibeere 6 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 7 Ìròyìn

Express 495g as a percentage of 16.5kg

3%

3

\frac{1}{3}

%

15%

30%

Awọn alaye Idahun

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%

Ka ìwé ẹ̀kọ́ nípa Percentages (WAEC)

Percentages

Wo Idahun

Ibeere 8 Ìròyìn

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}}

Awọn alaye Idahun

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}

Wo Idahun

Ibeere 9 Ìròyìn

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

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Wo Idahun

Ibeere 10 Ìròyìn

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

20

19

17

15

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Wo Idahun

Ibeere 11 Ìròyìn

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

65

23

17

91

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Matrices And Determinants (WAEC)

Ka ìwé ẹ̀kọ́ nípa Matrices And Determinants (JAMB)

Matrices And Determinants

Wo Idahun

Ibeere 12 Ìròyìn

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

2.4

4

24

4.2

Awọn alaye Idahun

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.

Wo Idahun

Ibeere 13 Ìròyìn

Calculate the area of an equilateral triangle of side 8cm

8√3

16

4√3

16√3

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Areas

Angles

Wo Idahun

Ibeere 14 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 15 Ìròyìn

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

Wo Idahun

Ibeere 16 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Sets

Wo Idahun

Ibeere 17 Ìròyìn

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}

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Bearings (WAEC)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (WAEC)

Bearings

Coordinate Geometry

Wo Idahun

Ibeere 18 Ìròyìn

In how many ways can the letters LEADER be arranged?

72

144

360

720

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Permutation And Combination (JAMB)

Permutation And Combination

Wo Idahun

Ibeere 19 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sine, Cosine And Tangent Of An Angle (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Wo Idahun

Ibeere 20 Ìròyìn

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

1

\frac{1}{4}

2

\frac{1}{4}

6

10

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Application Of Differentiation (JAMB)

Integration

Application Of Differentiation

Wo Idahun

Ibeere 21 Ìròyìn

Approximate 0.9875 to 1 decimal place.

1.1

1.0

0.9

0.10

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Wo Idahun

Ibeere 22 Ìròyìn

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

Wo Idahun

Ibeere 23 Ìròyìn

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}

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Probability (JAMB)

Probability

Wo Idahun

Ibeere 24 Ìròyìn

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

20

30

48

60

Awọn alaye Idahun

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

\div

16 = 60

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Solution Of Linear Equations

Number Bases

Surds (radicals)

Wo Idahun

Ibeere 25 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Sets

Measures Of Dispersion

Wo Idahun

Ibeere 26 Ìròyìn

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}

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Probability (JAMB)

Probability

Wo Idahun

Ibeere 27 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 28 Ìròyìn

Convert 0.04945 to two significant figures

0.040

0.049

0.050

0.49

Awọn alaye Idahun

0.04945 to two significant figures is 0.049.

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Wo Idahun

Ibeere 29 Ìròyìn

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}

Awọn alaye Idahun

\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}

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Wo Idahun

Ibeere 30 Ìròyìn

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

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Loci (JAMB)

Ka ìwé ẹ̀kọ́ nípa Loci (WAEC)

Loci

Wo Idahun

Ibeere 31 Ìròyìn

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

Second only

Third only

Fourth only

First and third only

Awọn alaye Idahun

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

Wo Idahun

Ibeere 32 Ìròyìn

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

Wo Idahun

Ibeere 33 Ìròyìn

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}

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Probability

Linear Inequalities

Wo Idahun

Ibeere 34 Ìròyìn

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

Φ

{s}

{t, u}

{y, z}

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Sets

Wo Idahun

Ibeere 35 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Representation Of Data

Wo Idahun

Ibeere 36 Ìròyìn

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

Awọn alaye Idahun

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).

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Wo Idahun

Ibeere 37 Ìròyìn

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}

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Probability (JAMB)

Probability

Wo Idahun

Ibeere 38 Ìròyìn

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°

Awọn alaye Idahun

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