JAMB UTME - Mathematics - 2016

Calculate the perimeter of a sector of a circle of radius 9cm and angle 36o.

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The perimeter of a sector of a circle is the sum of the arc length and the two radii. To find the perimeter, we need to first find the arc length using the formula: arc length = (angle/360) x 2 x π x radius In this case, the angle is given as 36°, the radius is given as 9cm, and π is approximately equal to 3.14. Substituting these values into the formula, we get: arc length = (36/360) x 2 x 3.14 x 9 = 2.83 cm (rounded to two decimal places) Now, we can find the perimeter by adding the two radii to the arc length: perimeter = 2 x 9 + 2.83 = 20.83 cm Therefore, the answer is (B) (18 + 9π/5) cm, rounded to two decimal places.

Given U = {x is a positive integer less than 15} and P = {x is even number from 1 to 14}. Find the compliment.

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The complement of a set contains all the elements that are not in the set. In this problem, the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}, which is the set of positive integers less than 15. The set P = {2, 4, 6, 8, 10, 12, 14} is the set of even numbers from 1 to 14. To find the complement of P, we need to determine the set of all elements that are not in P. Since every positive integer less than 15 is either even or odd, we can determine the complement of P by finding the set of all odd numbers less than 15. This set is {1, 3, 5, 7, 9, 11, 13}. Therefore, the answer is {1, 3, 5, 7, 9, 11, 13}.

Determine the mean score of the student that took the mathematics test?

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(3 x 0) + (1 x 2) + (5 x 2) + (5 x 3) + (5 x 4) +
(10 x 5) + (4 x 6) + (3 x 7) + (1 x 8) + (2 x 9) $÷$ 40
= $\frac{168}{40}$ = 4.2

The curved surface area of a cylinder 5cm high is 110cm2. Find the radius of its base
π = $\frac{22}{7}$

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The problem gives us the height and curved surface area of a cylinder and asks us to find the radius of its base. The formula for the curved surface area of a cylinder is 2πrh, where r is the radius and h is the height. We are given h = 5cm and curved surface area = 110cm². Substituting these values into the formula, we get: 2πr(5) = 110 Simplifying, we get: πr = 11 Dividing both sides by π, we get: r = 11/π The value of π is given as 22/7, so substituting this in the equation, we get: r = 11/(22/7) r = 3.5cm Therefore, the radius of the cylinder is 3.5cm. Option (B) is the correct answer.

In a school of 150 students, 80 offer French while 60 offer Arabic and 20 offer neither. How many students offer both subjects?

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80 - x + x + 60 - x + 20 = 150
160 - x = 150
x = 160 - 150 = 10

Find the mean of 10, 8, 5, 11, 12, 9, 6, 3, 15, and 23.

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To find the mean of a set of numbers, you add up all the numbers and then divide by the total count of numbers. So, for the given set of numbers, we add them up: 10 + 8 + 5 + 11 + 12 + 9 + 6 + 3 + 15 + 23 = 102 There are 10 numbers in the set, so we divide the sum by 10: 102 ÷ 10 = 10.2 Therefore, the mean of the given set of numbers is 10.2. So the answer is 10.2.

If line p = 5x + 3 is parallel to line p = wx + 5. Find the value of w.

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Two lines are parallel if they have the same slope. In this case, line p = 5x + 3 and line p = wx + 5 are parallel. Therefore, they have the same slope. The slope of p = 5x + 3 is 5. So, to find the value of w in p = wx + 5, we set the slope of this line equal to 5 and solve for w: w = 5 Therefore, the value of w is 5.

Find the derivative of y = ( $\frac{1}{3}$x + 6)

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Y = ( $\frac{1}{3}$x + 6)2
$\frac{dy}{dx}$ = 2($\frac{1}{3}$x + 6) $\frac{1}{3}$
= $\frac{2}{3}$ ( $\frac{1}{3}$x + 6)

Find $\frac{dy}{dx}$. If y = 3x3 + 2x2 + 3x + 1

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y = 3x3 + 2x2 + 3x + 1
$\frac{dy}{dx}$ = 9x2 + 4x + 3

Evaluate $\frac{0.8\times 0.43\times 0.031}{0.05\times 0.72\times 0.021}$ correct to four significant figures.

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Evaluate $\int$(sinx - 5x2)dx

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$\int$(sin x - 5x2) = -cosx - $\frac{5x^3}{3}$ + k

From the diagram above. Find the fraction of the shaded position?

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$\theta$ = 180o -(90 + 60)
$\theta$ = 180o - 150o = 30o
Fraction of shaded position = $\frac{30}{360}$ + $\frac{30}{360}$
= $\frac{1}{12}$ + $\frac{1}{12}$ = $\frac{1}{6}$

Calculate the range of 20, -6, 25, 30, 21, 28, 32, 33, 34, 5, 3, 2, and 1.

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To calculate the range of a set of numbers, you subtract the lowest number from the highest number. In this case, the lowest number is -6 and the highest number is 34. So the range is: 34 - (-6) = 40 Therefore, the correct answer is option D: 40.

Factorize k2 - 2kp + p2.

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k2 - 2kp + p2 = (k - p)2

If y = 2x3 + 6x2 + 6x + 1, Find $\frac{dy}{dx}$

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The derivative of y with respect to x, denoted as dy/dx, is the rate at which y changes with respect to x. To find dy/dx, we need to apply the power rule and the sum rule for differentiation. The power rule states that d(x^n)/dx = nx^(n-1). The sum rule states that d(f(x) + g(x))/dx = df(x)/dx + dg(x)/dx. Given that y = 2x^3 + 6x^2 + 6x + 1, we can differentiate it term by term: d(2x^3)/dx = 6x^2 d(6x^2)/dx = 12x d(6x)/dx = 6 d(1)/dx = 0 Applying the sum rule, we have: dy/dx = 6x^2 + 12x + 6 So, the answer is 6x^2 + 12x + 6.

The venn diagram above shows a class of 40 students with the games they play. How many of the students play two games only?

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To find the number of students who play two games only, we need to add up the number of students in the regions of the Venn diagram that represent playing two games only. Looking at the diagram, we can see that there are two regions that represent playing two games only: the intersection of the football and basketball circles, and the intersection of the basketball and volleyball circles. The number of students who play two games only is the sum of the number of students in these two regions. From the diagram, we can see that there are 7 students in the football and basketball region, and 8 students in the basketball and volleyball region. Therefore, the total number of students who play two games only is: 7 + 8 = 15 So, 15 of the 40 students play two games only.

Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o

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Perimeter = OP + OQ + PQ
= 8 + 8 + PQ
length PQ = $\frac{\theta }{360\times 2\pi r}$
= $\frac{45}{360}$ x 2 x $\pi$ x 8
= 2$\pi$
Perimeter of sector 2r + L
Where l = length of arc and r = radius
∴ P = 2(8) + 2$\pi$
= 16 + 2$\pi$

In the diagram above, l1 is parallel to l2, Find the value of < PMT

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< MPT = ${180}^{\circ }$ - ${118}^{\circ }$ = ${62}^{\circ }$
< PML = ${62}^{\circ }$ (Alternating angles)
y + 2y + ${10}^{\circ }$ + ${62}^{\circ }$ = ${180}^{\circ }$ (Angles on a straight line)
3y = 180 - 72
$\frac{3y}{3}$ = $\frac{108}{3}$
y = ${36}^{\circ }$
< PMT = 2y + 10 = 2(36) + 10 = ${82}^{\circ }$

Simplify $\frac{0.026\times 0.36}{0.69}$. Leave your answer in standard form.

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To simplify 0.026 × 0.360 / 0.69, we can follow the order of operations, which is to perform multiplication and division first, then addition and subtraction. Multiplying 0.026 and 0.360, we get: 0.00936 Dividing by 0.69, we get: 0.01356521739 We can express this answer in standard form by moving the decimal point to the right until we have a number between 1 and 10, then multiplying by an appropriate power of 10. In this case, we can move the decimal point three places to the right to get: 1.35652173913 × 10^-2 Rounding this to two decimal places, we get: 1.36 × 10^-2 Therefore, the answer is: 1.36 x 10^-2.

If N = $\frac{p}{2}$$\left(\frac{T_1-T_2}{T_1}\right)$. Find P when N = 12, T1 = 27 and T2 = 24.

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N = $\frac{p}{2}$$\left(\frac{T_1-T_2}{T_1}\right)$
12 = $\frac{p}{2}$$\left(\frac{27-24}{27}\right)$
24 = P$\left(\frac{3}{27}\right)$
P = 24 x 9 = 216

$\begin{array}{cccc}\text{Marks}& 2& 3& 4\\ \text{Frequency}& 4& 4& y\end{array}$

The table above shows the frequency distribution of marks obtained by a group of students. If the total mark is 48, find the value of y.

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OGIVE is constructed using

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An ogive is a graph used to show the cumulative frequency of a given dataset. It is constructed using a cumulative frequency table which contains the frequencies and the cumulative frequencies. The cumulative frequency is the total of the frequencies up to that point in the data set. Therefore, the correct answer is "Cumulative frequency table". The other options mentioned, third quartile range, semi-quartile range, and inter-quartile range, are statistical measures that can be used to describe the spread of a dataset, but they are not directly used in the construction of an ogive.

Evaluate $\frac{{27}^{\frac{1}{3}}-8^{\frac{2}{3}}}{{16}^{\frac{2}{4}}\times 2}$

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$\frac{{27}^{\frac{1}{3}}-8^{\frac{2}{3}}}{{16}^{\frac{2}{4}}\times 2}$
$\frac{{27}^{\frac{1}{3}}-8^{\frac{2}{3}}}{{16}^{\frac{2}{4}}\times 2}$ = $\frac{3\sqrt{27}-3{\sqrt{8}}^2}{\sqrt{16}\times 2}$
$\frac{3-4}{4\times 2}$
= $\frac{-1}{8}$

The sum of the interior angles of a polygon is a given as 1080o. Find the number of the sides of the polygon.

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The sum of the interior angles of a polygon is given by the formula (n-2)180, where n is the number of sides of the polygon. Setting this formula equal to 1080, we get (n-2)180 = 1080. Simplifying this equation, we get n-2 = 6, and therefore n = 8. Hence, the polygon has 8 sides. Therefore, the answer is 8.

The nth term of the sequence 3, 9, 27, 81.... is

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3,9,27,81,.....
r = $\frac{9}{8}$ = 3
Tn = arn-1
Tn = 3(3)n-1 = $\frac{3\left(3^2\right)}{3}$

If a car travels 120km on 45 litres of petrol, how much petrol is needed for a journey of 600km?

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To find out how much petrol is needed for a journey of 600 km, we need to use the given information about the car's fuel consumption. The car travels 120 km on 45 litres of petrol, so we can set up a proportion: 120 km / 45 litres = 600 km / x litres We can cross-multiply to solve for x: 120 km * x litres = 45 litres * 600 km x = (45 litres * 600 km) / 120 km x = 225 litres Therefore, the amount of petrol needed for a journey of 600 km is 225 litres. So the correct option is: 225 litres.

In the diagram MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140o, find, correct to the nearest cm, the length of the chord MN.

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From the diagram

sin 70o = $\frac{x}{10}$

x = 10sin 70o

= 9.3969

Hence, length of chord MN = 2x

= 2 x 9.3969

= 18.7938

= 19cm (nearest cm)

In the cyclic quadrilateral above . Find < PRO

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< PRO = ${180}^{\circ }-(100+50)$
= $180-150={30}^{\circ }$

From the diagram above, Find the value of < ROP

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< ROP = ${95}^{\circ }$ (Exterior angle of cyclic quadrilateral)

Rationalize $\frac{\sqrt{6}-\sqrt{4}}{\sqrt{6}+\sqrt{4}}$

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$\frac{\sqrt{6}-\sqrt{4}}{\sqrt{6}+\sqrt{4}}$ = $\frac{6-2\sqrt{6}-2\sqrt{6}+4}{6-4}$
$\frac{10-4\sqrt{6}}{2}$ = 5 - 2$\sqrt{6}$

he mean of 2-t, 4+t, 3-2t, 2+t and t-1 is

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Mean = $\frac{2-t+4+t+3-2t+2+t+t-1}{5}$
= $\frac{10}{5}$ = 2

The bar chart above shows the number of visitors received in a week. How many visitors were received on Friday, Tuesday and Sunday?

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No of visitors = 10 + 5 + 1 = 16

Solve for x and y respectively
3x - 5y = 9
6x - 4y = 12

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3x - 5y = 9 ------x2
6x - 4y = 12 -----x1
6x - 10y = 18
-6x - 4y = 12
____________
$\frac{6y}{-6}$ = $\frac{6}{6}$
y = -1
in eq (1) 3x - 5y = 9
3x - 5(-1) = 9
3x + 5 = 9
3x = 4
x = $\frac{4}{3}$

An arc subtends an angle of 30o at the centre of a circle radius 12cm. Calculate the length of the arc.

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Length or arc = $\frac{\theta }{360}$ x 2$\pi$r
= $\frac{30}{360}$ x 2 x $\frac{22}{7}$ x 12
= $\frac{1}{12}$ x 2 x $\pi$ x 2 = 2$\pi$cm

Find $\frac{dy}{dx}$, if y = $\frac{2}{3}$ X3 - $\frac{4}{x}$

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y = $\frac{2}{3}$x3 - $\frac{4}{x}$
$\frac{dy}{dx}$ = 2x2 - Lx-2 = 2x2 + $\frac{4}{x^2}$

Evaluate $\int$(cos4x + sin3x)dx

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If the 2nd term of a G.P is $\frac{8}{9}$ and the 6th term is 4$\frac{1}{2}$. Find the common ratio.

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T2 = $\frac{8}{9}$, T6 = 4$\frac{1}{2}$
ar = $\frac{8}{9}$
ar5 = $\frac{9}{2}$
$\frac{ar}{a{r}^5}$ = $\frac{8}{9}$ x $\frac{2}{9}$
$\frac{1}{r^4}$ = $\frac{16}{81}$
r = $\sqrt[4]{4\frac{81}{16}}$ = $\frac{3}{2}$

From the diagram above, find the required roots/region of x.

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The required roots/region of x is: - -1 ≤ x ≤ 4 The diagram shows a closed circle at -1 and an open circle at 4, indicating that -1 and 4 are included and excluded from the solution set, respectively. This means that any value of x that is greater than or equal to -1 and less than or equal to 4 is a solution to the inequality. Therefore, the required region of x is the interval between -1 and 4, including -1 and excluding 4.

If the mean of 4, y, 8 and 10 is 7. Find y?

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To find the value of y, we need to use the formula for calculating the mean (also known as the average). The formula for calculating the mean of a set of numbers is: mean = (sum of numbers) / (number of numbers) In this case, we know that the mean of the numbers 4, y, 8, and 10 is 7. So we can plug in these values into the formula and solve for y: 7 = (4 + y + 8 + 10) / 4 Multiplying both sides by 4 gives us: 28 = 4 + y + 8 + 10 Simplifying the right side of the equation gives us: 28 = y + 22 Subtracting 22 from both sides gives us: 6 = y Therefore, the value of y that makes the mean of 4, y, 8, and 10 equal to 7 is 6. So the answer is 6.

Make S the subject of the relation
p = s + $\frac{s{m}^2}{nr}$

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p = s + $\frac{s{m}^2}{nr}$
p = s + ( 1 + $\frac{m^2}{nr}$)
p = s (1 + $\frac{nr+m^2}{nr}$)
nr × p = s (nr + m2)
s = $\frac{nrp}{nr+m^2}$

If Q is a factor of 18 and T is prime numbers between 2 and 18. What is Q$\cap$T?

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Given that Q is a factor of 18, the possible values of Q are 1, 2, 3, 6, 9, and 18. We also know that T is a prime number between 2 and 18. The prime numbers between 2 and 18 are 2, 3, 5, 7, 11, 13, and 17. To find the intersection of Q and T, we look for the values that are common to both sets. The only values of Q that are also in the set of prime numbers between 2 and 18 are 2 and 3. Therefore, the intersection of Q and T is (2, 3). So the answer is: (2, 3).

Evaluate $\frac{12.02\times 20.06}{26.04\times 60.06}$, correct to three significant figures.

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Explanation:

$\begin{array}{ccccc}\text{Scores}& 3& 6& 5& 2\\ \text{Frequency}& 2& 3& 4& 6\end{array}$

From the table above, find the median.

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Mean = $\frac{(2\times 3)+(3\times 6)+(4\times 5)+(2\times 6)}{2+3+4+6}$
= $\frac{6+18+20+12}{15}$ = $\frac{56}{15}$
= 3.73
Median: Arrange numbers in ascending order from smallest to biggest to get: 2, 2, 2, 2, 2, 2, 3, 3, 5, 5, 5, 5, 6, 6, 6. The middle number is the 8th number which is 3.

If x10 = 235. Find x

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x10 = 235
x10 = 2 x 51 + 3 x 50
= 10 + 3
= x10 = 13

From the cyclic quadrilateral above, find < TVS

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< TVS = ${180}^{\circ }-(80+20)$
= $180-100={80}^{\circ }$

implify 1 - ($\frac{1}{7}$ x 3 $\frac{1}{2}$) $÷$ $\frac{3}{4}$

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1 - ($\frac{1}{7}$ x 3 $\frac{1}{2}$) $÷$ $\frac{3}{4}$
1 - ($\frac{1}{7}$ x $\frac{7}{2}$) x $\frac{4}{3}$
1 - $\frac{1}{2}$ x $\frac{4}{3}$
1 - $\frac{2}{3}$ = $\frac{1}{3}$

Evaluate $\underset{0}{\overset{\frac{\pi }{2}}{\int }}$ sin xdx

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$\underset{0}{\overset{\frac{\pi }{2}}{\int }}$ sin xdx = -cosx|$\frac{\pi }{2}$
= -(cos$\frac{\pi }{2}$ - cos0) = -(0-1) = 1

The pie chart above shows the distribution of subjects offered by students in SSS III level. If 80 students enrolled in the class. What is the size of the angle of the sector in economics?

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xo + 36o + 54o + 28o + 90o + 120o = 360o
x + 328o = 360o
x = 360o - 328o
x = 32o

An arc of the length 16$\pi$cm subtends an angle of 80o at the centre of the circle. Find the radius of the circle.

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L = $\frac{\theta }{360}$ x 2$\pi$r
16$\pi$ = $\frac{80}{360}$ x 2$\pi$r
$\frac{16\pi \times 360}{80\times 2\pi }$ = r
r = 36cm

The bar chart above shows the marks obtained by students in a mathematics test.

How many students in all took the test?

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3 + 2 + 5 + 5 + + 5 + 10 + 4 + 3 + 1 + 2 = 40