JAMB UTME - Mathematics - 2018

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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

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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.

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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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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.

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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

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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.

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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${}_{five}$

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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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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).

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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

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Profit (P) = $10x-x^2$
Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0
i.e. $\frac{\delta p}{\delta x}$ = 0
$\frac{\delta p}{\delta x}$ = 10 - 2x = 0
10 = 2x
Then x = $\frac{10}{2}$ = 5

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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

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

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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

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

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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.

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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

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

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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

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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

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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

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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.

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

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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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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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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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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.

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0.04945 to two significant figures is 0.049.

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

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$\frac{3}{4}$ 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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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

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Total frequency = 1 + 2 + 7 + 5 + 4 + 3 = 22
Median is the middle number
= $\frac{Nth}{2}$
Term = $\frac{22th}{2}$ = 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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Using law of indices:
25$\frac{1}{2}$ × 8$\frac{-2}{3}$
= √25 x ($\sqrt[3]{38}$) -2
= 5 x 2-2
= 5 x $\frac{1}{2^2}$ =
$\frac{5}{4}$ = $\frac{11}{4}$

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Explanation:

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

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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 - $\frac{15P^3}{Q}$
= $\frac{RQ-15P^3}{Q}$