JAMB UTME - Mathematics - 2020

Answer Details

h = 30 tan 60 

= 303–√ $\sqrt{3}$ 

The equation of the line in the graph is 

Answer Details

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 

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? 

Answer Details

40 = 32 - x + x + 24 + 4

40 = 60 - x

x = 60 - 40 

x = 20

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?

Answer Details

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

Answer Details

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.

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

Answer Details

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

What is the circumference of latitude 0o ${}^o$S if R is the radius of the earth? 

Answer Details

Circumstances of latitude 0o ${}^o$s where R is the radius of the earth 2π $\pi$r cos θ

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

Answer Details

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

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?

Answer Details

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

Answer Details

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

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

⎡⎣⎢x101x101x⎤⎦⎥ = 0

Answer Details

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. 

Answer Details

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.

Simplify 324−4x22x+18

Answer Details

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 goals scored by 40 football teams from three league  divisions are recorded below

What is the total number of goals scored by all the teams?

Answer Details

The total number of goals scored by all the teams can be calculated by multiplying the number of goals by its frequency and then adding all of them up. For example, the number of goals scored by teams with 0 goals is 4 * 0 = 0. The number of goals scored by teams with 1 goal is 3 * 1 = 3. The number of goals scored by teams with 2 goals is 15 * 2 = 30. And so on. So, the total number of goals scored by all the teams is 0 + 3 + 30 + 48 + 4 + 0 + 6 = 91. Therefore, the answer is 91.

The acres for rice, pineapple, cassava, cocoa and palm oil in a certain district are given respectively as 2, 5, 3, 11 and 9. What is the angle of the sector of cassava in a pie chart? 

Answer Details

The size of each sector in a pie chart represents the proportion of the data it represents compared to the whole. In this case, the total number of acres for all the crops is 2 + 5 + 3 + 11 + 9 = 30. To find the angle of the sector for cassava, we need to find the proportion of 3 acres to 30 acres, and then convert this proportion to degrees. The proportion of 3 acres to 30 acres is 3/30 = 1/10. In a full circle, there are 360 degrees, so the angle of the sector representing 1/10 of the data would be 360 x (1/10) = 36 degrees. Therefore, the angle of the sector for cassava in a pie chart is 36 degrees.

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

Answer Details

0.021741×1.20470.023789 = 0.0255×1.20.024(to 216) 

= 0.02640.024= 1.1

In preparing rice cutlets, a cook used 75g of rice, 40g of margarine, 105g of meat and 20g of bread crumbs. Find the angle of the sector which represent meat in a pie chart? 

Answer Details

Rice = 75g, Margarine = 40g, Meat = 105g

Bread = 20g

Total = 240

Angle of sector represented by meat 

= 105240×360o1 $\frac{105}{240}\times \frac{{360}^o}{1}$ 

= 157.5 

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

Answer Details

[(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) 

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

Answer Details

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.

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 $\frac{1}{4x}$ 

Answer Details

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 

Answer Details

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

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

Answer Details

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.

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

Answer Details

Arrange in ascending order

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

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

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? 

Answer Details

No of students offering physics are 

12360x 150 = 5

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

Answer Details

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.

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

Answer Details

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.

Express the product of 0.0014 and 0.011 in standard form 

Answer Details

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

Answer Details

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.

If N225.00 yields N27.00 in x years simple interest at the rate of 4% per annum, find x. 

Answer Details

Principal = N225.00, interest = N27.00

Year = x, Rate = 4% 

1 = PRT100 $\frac{PRT}{100}$

27 = 225×4×x100 $\frac{225\times 4\times x}{100}$ = 2700 = 900T

T = 2700900 $\frac{2700}{900}$ 

= 3 years

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

Answer Details

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

Answer Details

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

Answer Details

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

Answer Details

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

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

Answer Details

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

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

Answer Details

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 

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

Answer Details

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.

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?