JAMB UTME - Mathematics - 2020

Arrange in ascending order

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

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

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 $\frac{24-10}{24}=\frac{14}{24}=\frac{7}{12}$ 

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 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.

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.

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%.

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 

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.

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.

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

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 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.

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 $\frac{60}{360}\times 2\pi \times r=\frac{\pi }{3}$ 

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.

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 

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

40 = 32 - x + x + 24 + 4

40 = 60 - x

x = 60 - 40 

x = 20

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 

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.

0.021741×1.20470.023789 = 0.0255×1.20.024(to 216) 

= 0.02640.024= 1.1

[(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) 

No of students offering physics are 

12360x 150 = 5

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.

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 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.

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 $\frac{3}{2}$ 

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.

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

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 $\frac{1}{6}$ 

x	x - x	(x - x)2 ${}^2$
7 8 9 10 11 12 13	-3 -2 -1 0 1 2 3	9 4 1 1 0 4 9 -------------- 28

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

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.

h = 30 tan 60 

= 303–√ $\sqrt{3}$ 

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 

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

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

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