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JAMB UTME - Mathematics - 2018

Ibeere 1 Ìròyìn

From a point P, R is 5km due West and 12km due South. Find the distance between P and R'.

5km

12km

13km

17km

Awọn alaye Idahun

To find the distance between P and R', we first need to determine the location of R'. R' is the reflection of R across the point P. Since R is 5km due West and 12km due South of P, we can draw a diagram to represent their positions.

           R
            |
            | 12km
            |
            P------5km------->
            |
            |

To find R', we draw a line that connects R and P, then draw a perpendicular bisector to that line at point P. The intersection of the perpendicular bisector and the line RP is the point R'.

           R'
            |
            | 
            |
            P------5km------->
            |
            |

The distance between P and R' is the length of the line segment PR'. To find this distance, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.

In this case, we have a right triangle with sides of length 5km and 12km. The hypotenuse, which is the distance between P and R', is given by:

sqrt(5^2 + 12^2) = sqrt(169) = 13km

Therefore, the distance between P and R' is 13km.

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Polynomials

Integration

Ibeere 3 Ìròyìn

How many times, correct to the nearest whole number, will a man run round a circular track of diameter 100m to cover a distance of 1000m?

3

4

5

6

Awọn alaye Idahun

In terms of distance, a circle has a total distance or perimeter of 2π or π
Where r is radius and D is the diameter
So perimeter = $\frac{22}{7}$ x 100
= 314.2857m
To cover a distance of 1000m, he is going to round the circular track for $\frac{1000}{314.2857}$ = 3.18
It means it is on the number 4 round after the third round he is going to cover up to 1000m

Ibeere 6 Ìròyìn

Integrate the expression 6x $^{2}$ - 2x + 1

3x

^{3}

- 2x

^{2}

+ x + c

2x

^{3}

- x

^{2}

+ x + c

2x

^{3}

– 3x

^{2}

+ c

x

^{3}

+ x

^{2}

– x + c

Awọn alaye Idahun

$\int 6 x^{2} - 2 x + 1 = \frac{6 x^{2 + 1}}{2 + 1} - \frac{2 x^{1 + 1}}{1 + 1} + x + c$
$\frac{6 x^{3}}{3} - \frac{2 x^{2}}{2} + x + c$
$2 x^{3} - x^{2} + x + c$

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Representation Of Data

Integration

Differentiation

Ibeere 8 Ìròyìn

Calculate 243 $_{s i x}$ – 243 $_{f i v e}$ expressing your answer in base 10.

0

1

26

46

Awọn alaye Idahun

Since they are of different base, convert to base 10
243 $_{s i x}$ = (2 x 62) + (4 x 61) + (3 x 60)
= 72 + 24 + 3 = 99 base 10
243 $_{s i x}$ = 2 x 52 + 4 x 51 +3 x 50
50 + 20 + 3 = 73 base 10
Subtracting them, 99 - 73
= 26

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Ibeere 10 Ìròyìn

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

Use the figure to answer this question

What is n(A U B)c ?

2

3

4

7

Ibeere 12 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}

Probability

Probability

Ibeere 13 Ìròyìn

Evaluate ∫ $_{1}^{2}$ $\frac{5}{x}$ dx

1.47

2.67

3.23

3.47

Awọn alaye Idahun

∫ $\frac{5}{x}$ dx = 5 ∫ $\frac{1}{x}$ = 5Inx
Since the integral of $\frac{1}{x}$ is Inx
∫ $^{2}$ $_{1}$ ∫ $\frac{5}{x}$ dx = 5
dx = 5 (In<2 – InIn1)
= 3.4657
= 3.47

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Integration

Ibeere 15 Ìròyìn

Simplify 25 $^{\frac{1}{2}}$ × 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

Ibeere 17 Ìròyìn

Add 54 $_{e i g h t}$ and 67 $_{e i g h t}$ giving your answers in base eight

111

121

123

143

Awọn alaye Idahun

54 $_{e i g h t}$ and 67 $_{e i g h t}$ = 1438
Starting with normal addition, 4 + 7 gives 11
(it is more than the base, 8) 8 goes in 11 just 1 time, remaining 3, the remainder will be written, and the 1 will be added to the sum of 5 and 6 which gives 12 altogether, 8 goes in 12 one time remaining 4, the remainder 4 was written and then the 1 that was the quotient was then written since nothing to add the 1 to.
So answer is 143 in base eight

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (JAMB)

Number Bases

Number Bases

Ibeere 18 Ìròyìn

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

8.8 x 10

^{- 1}

8.8 x 10

^{- 2}

8.8 x 10

^{- 3}

8.8 x 10

^{3}

Awọn alaye Idahun

Explanation:

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

= \frac{0.000016384}{0.002}

= 0.008192

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Ibeere 19 Ìròyìn

Two sisters, Taiwo and Kehinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later Kehinde sells $\frac{2}{3}$ of her share to Taiwo for ₦720.00. Find the value of the store

₦1,080.00

₦2,400.00

₦3,000.00

₦3,600.00

Sine, Cosine And Tangent Of An Angle

Ibeere 20 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

Ibeere 21 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)

Ibeere 23 Ìròyìn

Given that sin (5 $x$ − 28) $^{o}$ = cos(3 $x$ − 50) $^{o}$ , O $x$ < 90 $^{o}$

Find the value of $x$

14

^{o}

21

^{o}

32

^{o}

39

^{o}

Awọn alaye Idahun

Sin(5x - 28) = Cos(3x - 50)………..i
But Sinα = Cos(90 - α)
So Sin(5x - 28) = Cos(90 - [5x - 28])
Sin(5x - 28) = Cos(90 - 5x + 28)
Sin(5x - 28) = Cos(118 - 5x)………ii
Combining i and ii
Cos(3x - 50) = Cos(118 - 5x)
3x - 50 = 118 - 5x
Collecting the like terms
3x + 5x = 118 + 50
8x = 168
x = $\frac{168}{8}$
x = 21 $^{o}$

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

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Sine, Cosine And Tangent Of An Angle

Ibeere 24 Ìròyìn

Express 495g as a percentage of 16.5kg

3%

3

\frac{1}{3}

%

15%

30%

Percentages

Ibeere 25 Ìròyìn

Which one of the following gives the members of the set Ac $\cap$ B $\cap$ C?

Φ

{s}

{t, u}

{y, z}

Ibeere 26 Ìròyìn

A box contains two red balls and four blue balls. A ball is drawn at random from the box and then replaced before a second ball is drawn. Find the probability of drawing two red balls.

\frac{2}{3}

\frac{1}{3}

\frac{1}{4}

\frac{1}{9}

Probability

Surds (radicals)

Ibeere 27 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)

Coordinate Geometry

Ibeere 30 Ìròyìn

Find the equation of the tangent at the point (2, 0) to the curve y = x $^{2}$ - 2x

y = 2x - 4

y = 2x + 4

y = 2x - 2

y = 2x + 2

Awọn alaye Idahun

The gradient to the curve is found by differentiating the curve equation with respect to x
So $\frac{d y}{d x}$ 2x - 2
The gradient of the curve is the same with that of the tangent.
At point (2, 0) $\frac{d y}{d x}$ = 2(2) - 2
= 4 – 2 = 2
The equation of the tangent is given by (y - y1) $\frac{d y}{d x}$ (x – x1)
At point (x1, y1) = (2, 0)
y - 0 = 2(x - 2)
y = 2x - 4

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (JAMB)

Coordinate Geometry

Ibeere 31 Ìròyìn

if y = 23 $_{f i v e}$ + 101 $_{t h r e e}$ find y leaving your answer in base two

1110

10111

11101

111100

Awọn alaye Idahun

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 $_{f i v e}$

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (JAMB)

Number Bases

Number Bases

Ibeere 33 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

Ibeere 34 Ìròyìn

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

Ibeere 35 Ìròyìn

Evaluate log $_{2}$ 8 – log $_{3}$ $\frac{1}{9}$

-1 1

\frac{1}{2}

-1

1

5

Awọn alaye Idahun

log $_{2}$ 8 – log $_{3}$ $\frac{1}{9}$
= log $_{2}$ 2 $^{3}$ – log $_{3}$ 9 $^{- 1}$
= log $_{2}$ 2 $^{3}$ – log $_{3}$ 3 $^{- 2}$
Based on law of logarithm
= 3 log $_{2}$ 2 – (-2 log $_{3}$ 3)
But log $_{2}$ 2 = 1,
log $_{3}$ 3 = 1
So, = 3 + 2
= 5

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals, Approximations And Percentage (JAMB)

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Fractions, Decimals, Approximations And Percentage

Ibeere 38 Ìròyìn

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

32 cm

24 cm

16 cm

12 cm

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Angles Of Elevation And Depression (WAEC)

Ka ìwé ẹ̀kọ́ nípa Angles And Intercepts On Parallel Lines (WAEC)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Angles Of Elevation And Depression

Angles And Intercepts On Parallel Lines

Triangles And Polygons

Ibeere 40 Ìròyìn

A man stands on a tree 150cm high and sees a boat at an angle of depression of 74°. Find the distance of the boat from the base of the tree.

52cm

43cm

40cm

15cm

Awọn alaye Idahun

$\tan 74 = \frac{150}{x}$
$x = \frac{150}{\tan 74}$
= 43.01cm

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

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