JAMB UTME - Mathematics - 2020

A sector of circle of radius 7.2cm which substends an angle of 300o ${}^o$ at the centre is used to form a cone. What s the radius of the base of the cone?

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A cylindrical tank has a capacity of 3080m3 ${}^3$. What is the depth of the tank if the diameter of its base is 14m?

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V = 2080cm3 ${}^3$, h = ? 

r = 7cm

V = Vπr2h $\pi r^{2}h$

h = Vπr2=3080227×49 $\frac{V}{\pi r^2}=\frac{3080}{\frac{22}{7}\times 49}$ 

308054 $\frac{3080}{54}$ = 20cm

If x is positive real number, find the range of values for which 13x $\frac{1}{3x}$ + 12 $\frac{1}{2}$x = 2+3x6x $\frac{2+3x}{6x}$ > 14x $\frac{1}{4x}$ 

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13x $\frac{1}{3x}$ + 12 $\frac{1}{2}$x = 2+3x6x $\frac{2+3x}{6x}$ > 14x $\frac{1}{4x}$ 

= 4(2 + 3x) > 6x = 12x2 ${}^2$ - 2x = 0

= 2x(6x - 1) > 0 = x(6x - 1) > 0

Case 1 (-, -) = x < 0, 6x - 1 > 0

= x < 0, x < 16 $\frac{1}{6}$ (solution) 

Case 2 (+, +) = x > 0, 6x - 1 > 0 = x > 0

x > 16 $\frac{1}{6}$

Combining solutions in cases (1) and (2) 

= x > 0, x < 16 $\frac{1}{6}$ = 0 < x < 16 $\frac{1}{6}$ 

Calculate the standard deviation of the following data: 7, 8, 9, 10, 11, 12, 13 

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S.D = ∑√(x−x)2N=∑d2N=28√7 $\frac{\sqrt{\sum }(x-x{)}^2}{N}=\frac{\sum d^2}{N}=\frac{\sqrt{28}}{7}$ 

= 4–√ $\sqrt{4}$ 

= 2

Find the non-zero positive value of x which satisfies the equation 

⎡⎣⎢x101x101x⎤⎦⎥ = 0

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In a class of 150 students, the sector in a pie chart representing the students offering physics has angle 12o ${}^o$. How many students are offering physics? 

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No of students offering physics are 

12360x 150 = 5

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h = 30 tan 60 

= 303–√ $\sqrt{3}$ 

The equation of the line in the graph is 

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Gradient of line = Change in yChange in x $\frac{\text{Change in y}}{\text{Change in x}}$ = y2−Yx2−x $\frac{y_2-Y}{x_2-x}$ 

y2 ${}_2$  = 01 ${}_1$ 

Y1 ${}_1$  = 4 

x2 ${}_2$  = 3 and x1 ${}_1$  = 0

y2−y1x2−x1=0−43−0=−43 $\frac{y_2-y_1}{x_2-x_1}=\frac{0-4}{3-0}=\frac{-4}{3}$

Equation of straight line y = mx + c

Where m = gradient and c = y 

intercept = 4 

y = 4x + 43 $\frac{4}{3}$ multiply through y 

3y = 4x + 23 

What is the product of 275 $\frac{27}{5}$, (3)−3 $(3{)}^{-3}$ and (15)−1 $\frac{1}{5}{)}^{-1}$?

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275×3−3×(1)−15 $\frac{27}{5}\times 3^{-3}\times \frac{(1{)}^{-1}}{5}$

= 275×133×115 $\frac{27}{5}\times \frac{1}{3^3}\times \frac{1}{\frac{1}{5}}$

275 $\frac{27}{5}$ x 127 $\frac{1}{27}$ x 51 $\frac{5}{1}$ = 1

Express the product of 0.0014 and 0.011 in standard form 

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A crate of soft drinks contains 10 bottle of Coca-Cola 8 of Fanta and 6 of Sprite. If one bottle is selected at random, what is the probability that it is Not a Coca-Cola bottle?

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Coca-Cola = 10 bottles, Fanta = 8 bottles, Spirite = 6 bottles 

Total = 24

P(Coca-Cola) = 1024 $\frac{10}{24}$; P(not Coca-Cola) 

1 - 1024 $\frac{10}{24}$ 

24−1024=1424=712 $\frac{24-10}{24}=\frac{14}{24}=\frac{7}{12}$ 

The mean age group of some students is 15years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages become 18 years. Find the number of students in the group. 

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Let's call the number of students in the group "n". The total of the ages of all students is 15n. When the teacher's age is added, the total becomes 15n + 45. The mean of the ages becomes 18, which means the total of the ages is divided by the number of people, which is "n + 1". We can set up an equation to represent the situation: (15n + 45) / (n + 1) = 18 Expanding the equation: 15n + 45 = 18 * (n + 1) 15n + 45 = 18n + 18 Subtracting 18n from both sides: -3n + 45 = 18 Subtracting 45 from both sides: -3n = -27 Dividing both sides by -3: n = 9 So, there are 9 students in the group.

A binary operation x is defined by a x b = ab ${}^b$. If a x 2 = 2 - a, find the possible values of a?

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The equation a x 2 = 2 - a can be solved for the possible values of a. To do this, we can simplify the equation by adding a to both sides: a x 2 + a = 2 Combining like terms: a x 2 + a = 2 a(x + 1) = 2 Dividing both sides by x + 1: a = 2 / (x + 1) Since the binary operation x is defined as a x b = a, we can substitute this definition into the equation: a = 2 / (a x 1 + 1) a = 2 / (a + 1) Since a x b = a, we know that a x 1 = a, so we can substitute this into the equation: a = 2 / (a + 1) a = 2 / (a + 1) Solving for a, we get: a = 2 / (a + 1) = 2 / (2) = 1 So, the possible values of a are 1.

The pie chart shows the income of a civil servant in month. If his monthly income is N6,000. Find his monthly basic salary. 

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360o ${}^o$ - (60o ${}^o$ + 60o ${}^o$ + 67 + 50 = 237o ${}^o$) 

360o ${}^o$ - 237 = 130o ${}^o$ 

B. Salary = 123360XN60001 $\frac{123}{360}X\frac{N6000}{1}$ 

= N2,050

A surveyor walks 500m up a hill which slopes at an angle of 30o ${}^o$. Calculate the vertical height through which he rises

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The vertical height through which the surveyor rises can be calculated using trigonometry. When a person walks up a hill, they are moving in two dimensions: horizontally and vertically. We can use a right triangle to represent this movement, where the horizontal distance is one side, the vertical distance is another side, and the slope of the hill is the angle between these sides. In this case, the horizontal distance is 500m and the angle is 30 degrees, so we can use the trigonometric function sine to find the vertical distance: sin(30) = vertical distance / 500 Multiplying both sides by 500, we get: 500 * sin(30) = vertical distance Using a calculator, we can find that sin(30) = 0.5, so: 500 * 0.5 = 250 Therefore, the vertical height through which the surveyor rises is 250m.

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

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The equation of a line can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept. To find the equation of a line parallel to another line, we need to use the same slope. The line 7x + 5y = 12 can be rewritten in the form y = -(7/5)x + 12/5. The slope of this line is -(7/5). To find a line through the point (5, 7) with the same slope, we can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope. Plugging in the values for (x1, y1) and m, we get: y - 7 = -(7/5)(x - 5) Expanding and simplifying, we get: y = -(7/5)x + 7 + (35/5) = -(7/5)x + 70/5 = 14x + 7. So the equation of the line through the point (5, 7) parallel to the line 7x + 5y = 12 is 7x + 5y = 70.

The sum of the interior angle of pentagon is 6x + 6y. Find y in terms of x.

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The sum of the interior angles of a pentagon is equal to (5-2) * 180 = 540 degrees. So, the equation 6x + 6y = 540 can be used to find the value of y in terms of x. Solving for y, we get y = 90 - x. This means that if we know the value of x, we can find the value of y by subtracting x from 90. So, the correct answer is y = 90 - x.

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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ABCD is the floor. By pathagoras 

AC2 ${}^2$ = 144 + 81 = 225−−−√ $\sqrt{225}$ 

AC = 15cm

Height of room 8m, diagonal of floor is 15m

Therefore, the cosine of the angle which a diagonal of the room makes with the floor is 

EC2 ${}^2$ = 152 ${}^2$ + 82 ${}^2$ cosine

adjHyp=1517 $\frac{adj}{Hyp}=\frac{15}{17}$ 

EC2 ${}^2$ = 225+64−−−−−−−√ $\sqrt{225+64}$

EC = 289−−−√ $\sqrt{289}$

= 17 

Four boys and ten girls can cut a field first in 5 hours. If the boys work at 54 $\frac{5}{4}$ the rate at which the girls work, how many boys will be needed to cut the field in 3 hours? 

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Let x rep. numbers of boys that can work at 54 $\frac{5}{4}$ the rate at

 which the 10 girls work 

For 1 hrs, x boys will work for 15×104 $\frac{\frac{1}{5}\times 10}{4}$ 

x = 54 $\frac{5}{4}$ x 10 

= 8 boys 

8 boys will do the work of ten girls at the same rate 

4 + 8 = 12 bous cut the field in 5 hrs for 3 hrs, 

12×53 $\frac{12\times 5}{3}$ boys will be needed = 20 boys.

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hn−2=n+3n $\frac{h}{n-2}=\frac{n+3}{n}$

n2 ${}^2$ = (n + 3) (n - 2) 

n2 ${}^2$ = n2 ${}^2$ + n - 6

n2 ${}^2$ + n - 6 - n2 ${}^2$ = 0

n - 6 = 0

n = 6

Common ratio: nn−2=66−2=64 $\frac{n}{n-2}=\frac{6}{6-2}=\frac{6}{4}$ = 32 $\frac{3}{2}$ 

The chord ST of a circle is equal to the radius r of the circle. Find the length of arc ST

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r2r $\frac{\frac{r}{2}}{r}$ Sin θ $\theta$ = 12 $\frac{1}{2}$ 

θ $\theta$ = sin−1 ${}^{-1}$ (12 $\frac{1}{2}$) = 30o ${}^o$ = 60o ${}^o$ 

Length of arc (minor)

ST = θ360 $\frac{\theta }{360}$ x 2πr $\pi r$ 

60360×2π×r=π3 $\frac{60}{360}\times 2\pi \times r=\frac{\pi }{3}$ 

Find all median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119 

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Arrange in ascending order

89, 100, 108, 119, 120,130, 131, 131, 141, 161

Median = 120+1302 $\frac{120+130}{2}$ = 125

Reach each number to two significant figures and then evaluate 0.02174×1.20470.023789

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0.021741×1.20470.023789 = 0.0255×1.20.024(to 216) 

= 0.02640.024= 1.1

Find all real number x which satisfy the inequality 13 $\frac{1}{3}$ (x + 1) - 1 > 15 $\frac{1}{5}$(x + 4) 

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13 $\frac{1}{3}$ (x + 1) - 1 > 15 $\frac{1}{5}$(x + 4)  = x+13−1 $\frac{x+1}{3}-1$ > x+45 $\frac{x+4}{5}$ 

x+13−x+45−1 $\frac{x+1}{3}-\frac{x+4}{5}-1$ > 0 

= 5x+5−3x−1215 $\frac{5x+5-3x-12}{15}$ 

2x - 7 > 15

2x > 22 = x > 11

Find the gradient of the line passing through the points (-2, 0) and (0, -4) 

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The gradient of a line describes how steep the line is, and it can be calculated as the change in the y-coordinate divided by the change in the x-coordinate between two points on the line. In this case, the two points are (-2, 0) and (0, -4), so the change in x is 0 - (-2) = 2, and the change in y is -4 - 0 = -4. Therefore, the gradient of the line passing through these two points is -4 ÷ 2 = -2. So, the answer is -2.

A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime. 

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Probability is the measure of how likely an event is to occur. In this question, we want to find the probability of selecting a prime number between 20 and 30, both numbers inclusive. The prime numbers between 20 and 30 are 23 and 29. The total number of possible outcomes (or numbers between 20 and 30) is 11 (20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30). So, the probability of selecting a prime number is the number of favorable outcomes (prime numbers) divided by the total number of possible outcomes. The probability of selecting a prime number is 2/11. So, the answer is 2/11.

Find the simple interest rate percent annum at which N1000 accumulates to N1240 in 3 years.

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Simple interest is the interest calculated on the original amount of a loan or deposit, without taking into account any compound interest. To find the simple interest rate at which N1000 accumulates to N1240 in 3 years, we can use the formula for simple interest: I = Prt where I is the interest, P is the principal (original amount), r is the interest rate, and t is the time in years. We know that I = N1240 - N1000 = N240, and t = 3 years. So, we can plug in these values into the formula and solve for r: N240 = N1000 x r x 3 Dividing both sides by N1000 x 3, we get: r = N240 / (N1000 x 3) = 0.08 So, the interest rate is 8%. So, the answer is 8%.

Simplify 324−4x22x+18

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234 - 4x2 ${}^2$ = 182 ${}^2$ - (2x)2 ${}^2$ = (18 - 2x)(18 + 2x)

2x + 18 = 2x + 18 = (2x + 18)

18 - 2x = 2(a - x) or -2(x - a) 

The angle of a sector of a circle radius 10.5 cm is 48o ${}^o$. Calculate the perimeter of the sector.

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The perimeter of a sector of a circle is equal to the circumference of the circle that is part of the sector plus the length of the arc of the sector. To calculate the perimeter of a sector, we need to find the length of the arc and the circumference of the circle first. The length of the arc is given by the formula: Arc length = (Angle of sector / 360) x Circumference of the circle The circumference of a circle is given by the formula: Circumference = 2 * pi * radius Plugging in the values we have, the circumference of the circle is 2 * pi * 10.5 = 65.97 cm. The length of the arc is (48 / 360) x 65.97 = 12.19 cm. Adding the length of the arc to the circumference of the circle, we get 65.97 + 12.19 = 78.16 cm, which is the perimeter of the sector. Therefore, the answer is 78.16 cm, which is closest to 29.8cm.

In a class of 40 students, 32 offer mathematics, 24 offer physics and 4 offer neither mathematics nor physics. How many offer both mathematics and physics? 

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40 = 32 - x + x + 24 + 4

40 = 60 - x

x = 60 - 40 

x = 20

If log10 ${}_{10}$2 = 0.3010 and log10 ${}_{10}$3 = 0.4771, eventually without using the logarithm tables, log10 ${}_{10}$4.5

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log10 ${}_{10}$2 = 0.3010 and log10 ${}_{10}$3 = 04771

log104.5=log10 ${}_{10}4.5=lo{g}_{10}$ (3×32 $\frac{3\times 3}{2}$)

log10 ${}_{10}$ 3 + log10 ${}_{10}$ 3 - log10 ${}_{10}$2 = 0.4471 + 0.771 - 0.3010

= 0.6532

Evaluate (212)3 ${}_3$ - (121)3 ${}_3$ + (222)3

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The answer is (1020)3, which is equal to (1 * (3^3)) + (0 * (3^2)) + (2 * (3^1)) + (0 * (3^0)). In this question, each number is given in base-3 format, which means that each digit in the number represents a power of 3. For example, the number (212)3 is equal to (2 * (3^2)) + (1 * (3^1)) + (2 * (3^0)). To evaluate the expression, we simply add and subtract the numbers as written, and convert the result back to base-3 format.

In this fiqure, PQ = PR = PS and SRT = 68o ${}^o$. Find QPS 

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Since PQRS is quadrilateral

2y + 2x + QPS = 360o ${}^o$

i.e. (y + x) + QPS = 360o ${}^o$ 

QPS = 360o ${}^o$ - 2 (y + x)

But x + y + 68o ${}^o$ = 180o ${}^o$

There; x + y = 180o ${}^o$ - 68o ${}^o$ = 112o ${}^o$

QPS = 360 - 2(112o ${}^o$)

= 360o ${}^o$ - 224 = 136o

Find the value of x if 2√x+2√ $\frac{\sqrt{2}}{x+\sqrt{2}}$ = 1x−2√ $\frac{1}{x-\sqrt{2}}$ 

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2√x+2 $\frac{\sqrt{2}}{x+2}$ = x - 12√ $\frac{1}{\sqrt{2}}$

x2–√ $\sqrt{2}$ (x - 2–√ $\sqrt{2}$) = x + 2–√ $\sqrt{2}$ (cross multiply)

x2–√ $\sqrt{2}$ - 2 = x + 2–√ $\sqrt{2}$ 

= x2–√ $\sqrt{2}$ - x 

= 2 + 2–√ $\sqrt{2}$

x (2–√ $\sqrt{2}$ - 1) = 2 + 2–√ $\sqrt{2}$

= 2+2√2√−1×2√+12√+1 $\frac{2+\sqrt{2}}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

x = 22√+2+2+2√2−1 $\frac{2\sqrt{2}+2+2+\sqrt{2}}{2-1}$ 

= 32–√ $\sqrt{2}$ + 4 

Factorise (4a + 3) 2 ${}^2$ - (3a - 2)2

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[(4a + 3) 2 ${}^2$ - (3a - 2)2 ${}^2$ = a2 ${}^2$ - b2 ${}^2$ = (a + b) (a - b) 

= [(4a + 3) + (3a - 2)] [(4a + 3) - (3a - 2)]

= [4a + 3 + 3a - 2] [4a + 3 - 3a + 2]

= (7a + 1)(a + 5)

(a + 5) (7a + 1)