JAMB UTME - Mathematics - 2018

Evaluate ∫${}_1^2$ $\frac{5}{x}$ dx

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

What is the modal age?

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

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.

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

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Explanation:

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

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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

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

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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

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.

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The probability of drawing a red ball on the first draw is 2/6 (since there are two red balls out of a total of six balls), and the probability of drawing a red ball on the second draw is also 2/6 (since the first ball is replaced before the second draw, and the number of balls remains the same). To find the probability of drawing two red balls, we need to multiply the probability of the first draw by the probability of the second draw, since the two events are independent of each other. So the probability of drawing two red balls is: 2/6 * 2/6 = 4/36 = 1/9 Therefore, the answer is 1/9.

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?

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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

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.

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${\mathrm{a2=122+92\; a\; 2\; =\; 12\; 2\; +\; 9\; 2\; a=\surd 122+92\; a\; =\; 12\; 2\; +\; 9\; 2\; a=15 m\; a\; =\; 15\; m\; b2=152+82\; b\; 2\; =\; 15\; 2\; +\; 8\; 2\; b=\surd 152+82\; b\; =\; 15\; 2\; +\; 8\; 2\; b=17 m\; b\; =\; 17\; m\; cos\theta =ab=1517}}^{}$

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

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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.

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

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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.

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

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

Use the quadratic equation curve to answer this question.

What is the 80th percentile?

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

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

Use the table to answer the question below:

How many children are in the hospital

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To answer this question, we need to add up the frequency of children in each age category. Looking at the table, we can see that there are 3 children aged 1, 4 children aged 2, 5 children aged 3, 6 children aged 4, 7 children aged 5, 6 children aged 6, 5 children aged 7, and 4 children aged 8. We can add up these numbers to find the total number of children in the hospital: 3 + 4 + 5 + 6 + 7 + 6 + 5 + 4 = 40 Therefore, there are 40 children in the hospital. The correct answer is.

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

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In how many ways can the letters LEADER be arranged?

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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

Express 495g as a percentage of 16.5kg

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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%

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?

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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.

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.

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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

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

Find the value of $x$

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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${}^o$

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

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First we convert the numbers to base ten
23${}_{five}$= 2 x 51 + 3 x 50
= 10 + 3 = 13
101${}_{five}$ = (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${}_{five}$

Convert 0.04945 to two significant figures

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0.04945 to 2s.f is 0.05

Evaluate (2√3 - 4) (2√3 + 4) 

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2√3 - 4) ( 2√3 + 4)
= 12 + 8√3 - 8√3 – 16
= 12 – 16
= -4
The two expressions in the bracket are conjugate of each other

Approximate 0.9875 to 1 decimal place.

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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

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 ?

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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

Calculate the area of an equilateral triangle of side 8cm

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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.

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?

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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

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

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To form words having 3 consonants and 2 vowels out of 7 consonants and 4 vowels, the number of such words is 7/3C x 4/2C = 35 x 6
= 210

Calculate 243${}_{six}$ – 243${}_{five}$ expressing your answer in base 10.

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

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

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LCM of 80, 64, 48 = 960
HCF of 80, 64, 48 = 16
960 $÷$ 16 = 60

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

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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.

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

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

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

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The problem can be solved using algebra. Let's start by assuming that Taiwo's share is 11x and Kehinde's share is 9x. Then the total value of the store is 20x. Kehinde sells 2/3 of her share to Taiwo for ₦720.00. This means that Taiwo paid ₦720.00 for 2/3 of 9x, which is (2/3) * 9x = 6x. So we have: 6x = 720 Solving for x, we get: x = 120 Now we can find the total value of the store: 20x = 20 * 120 = ₦2,400.00 Therefore, the answer is option (B), ₦2,400.00. In summary, we used algebra to solve the problem by assuming values for Taiwo's and Kehinde's shares and finding the total value of the store. Then we used the information about Kehinde's sale to find the value of x and ultimately the total value of the store.

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

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Using law of indices:
25$\frac{1}{2}$ × 8$\frac{-2}{3}$
= √25 x ($\sqrt[3]{38}$) -2
= 5 x 2-2
= 5 x $\frac{1}{2^2}$ =
$\frac{5}{4}$ = $\frac{11}{4}$

Evaluate log${}_2$ 8 – log${}_3$ $\frac{1}{9}$

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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

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

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From the figure,
∠ XMN = ∠ XZY
Angle X is common
So, ∠ XNM = ∠ XYZ
Then from the angle relationship
$\frac{XM}{XZ}$ = $\frac{XN}{XY}$ = $\frac{MN}{ZY}$
XM = 8, XZ = 12 + 4 = 16,
XN = 12, XY = 8 + YM
$\frac{8}{16}$ = $\frac{12}{(8+YM)}$
Cross multiply
8(8 + YM) = 192
64 + 8YM = 192
8YM = 128
YM = $\frac{128}{8}$
= 16cm

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

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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.

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

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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
$\frac{15P^3}{Q}$ = R - T
Rearranging gives
T = R - $\frac{15P^3}{Q}$
= $\frac{RQ-15P^3}{Q}$

Add 54 ${}_{eight}$ and 67${}_{eight}$ giving your answers in base eight

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54 ${}_{eight}$ and 67${}_{eight}$ = 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

Use the cumulative frequency curve to answer the question.
What is the 80th percentile?

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