JAMB UTME - Mathematics - 2010

Express the product of 0.21 and 0.34 in standard form

Bayanin Amsa

0.21 x 0.34 = 0.0714
= 7.14 x 10-2

If y = (2x + 1)3, find $\frac{dy}{dx}$

Bayanin Amsa

If y = (2x + 1)3, then
Let u = 2x + 1 so that, y = u3
$\frac{dy}{du}$ = 3u2 and $\frac{dy}{dx}$ = 2
Hence by the chain rule,
$\frac{dy}{dx}$ = $\frac{dy}{du}$ x $\frac{du}{dx}$
= 3u2 x 2
= 6u2
= 6(2x + 1)2

If Cos $\theta$ = $\frac{12}{13}$. Find $\theta$ + cos2$\theta$

Bayanin Amsa

Cos $\theta$ = $\frac{12}{13}$
x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tan$\theta$ = $\frac{5}{12}$ and cos$\theta$ = $\frac{12}{13}$
If cos2$\theta$ = 1 + $\frac{1}{ta{n}^2\theta }$
= 1 + $\frac{1}{\frac{(5{)}^2}{12}}$
= 1 + $\frac{1}{\frac{25}{144}}$
= 1 + $\frac{144}{25}$
= $\frac{25+144}{25}$
= $\frac{169}{25}$

What is the locus of point that is equidistant from points P(1,3) and Q(3,5)?

Bayanin Amsa

The locus of a point that is equidistant from two given points is the perpendicular bisector of the line segment connecting those two points. In this case, the two points are P(1,3) and Q(3,5). To find the perpendicular bisector of the line segment PQ, we can first find the midpoint of PQ: Midpoint = ( (1+3)/2 , (3+5)/2 ) = (2,4) Next, we can find the slope of the line PQ: Slope of PQ = (5-3) / (3-1) = 1 The perpendicular bisector of PQ will have a slope that is the negative reciprocal of the slope of PQ, which is -1. Therefore, we can use the point-slope form of a line to find the equation of the perpendicular bisector, using the midpoint (2,4) and slope -1: y - 4 = -1(x - 2) Simplifying this equation, we get: y = -x + 6 So the equation of the locus of points that are equidistant from P and Q is y = -x + 6. Therefore, the correct option is: - y = -x + 6

A man bought a second-hand photocopying machine for N34,000. He serviced is at a cost od N2,000 and then sold it i profit of 15%. What was the selling price?

Bayanin Amsa

To find the selling price of the photocopying machine, we need to first calculate the profit the man made. He bought the machine for N34,000 and spent an additional N2,000 on servicing it, which brings his total cost to N36,000. He then sold the machine for a profit of 15%, which means he earned 15% of his cost as profit. To calculate the profit, we can use the formula: Profit = (Profit Percentage/100) * Cost Price Substituting the given values, we get: Profit = (15/100) * 36,000 Profit = 5,400 Therefore, the man made a profit of N5,400. To find the selling price, we can add the profit to the cost price: Selling Price = Cost Price + Profit Selling Price = 36,000 + 5,400 Selling Price = 41,400 Therefore, the selling price of the photocopying machine was N41,400. Therefore, the correct option is: N41,400

If y = (2x + 1)3 find dy/dx

Bayanin Amsa

y = (2x + 1)3
dy/dx = 3(2x + 1)3-1 x 2
= 3(2x + 1)2 x 2
= 6(2x + 1)2

The table above show the marks obtained in a given test.

How many student too the test

Bayanin Amsa

To find the total number of students who took the test, we need to add up the frequency of all the marks. 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (C) 20.

A cylindrical pipe 50cm long with radius 7m has one end open. What is the total surface area of the pipe?

Bayanin Amsa

To calculate the total surface area of the cylindrical pipe, we need to add the surface area of the curved part and the surface area of the two circular ends. The surface area of the curved part can be calculated by multiplying the circumference of the circle (2πr) by the length of the pipe (50cm), which gives us: 2πr x h = 2π x 7m x 50cm = 7π m^2 The surface area of one circular end can be calculated by multiplying the area of the circle (πr^2) by 1, since one end of the pipe is open and has no surface area. Thus, the total surface area of both circular ends is: 2πr^2 = 2π x 7m^2 = 14π m^2 Finally, we add the surface area of the curved part and the surface area of the two circular ends to get the total surface area of the pipe: 7π m^2 + 14π m^2 = 21π m^2 Therefore, the total surface area of the pipe is 21π square meters. The closest option to this answer is 749π, but it is not the correct answer.

If y varies directly as the square root of x and y = 3 when x = 16. Calculate y when x = 64

Bayanin Amsa

If y varies directly as the square root of x, this means that the ratio between y and the square root of x remains constant. We can write this as: y/sqrt(x) = k where k is the constant of proportionality. We are given that y = 3 when x = 16. Using this information and the equation above, we can solve for k: 3/sqrt(16) = k 3/4 = k Now that we know k, we can use the same equation to find y when x = 64: y/sqrt(64) = 3/4 y/8 = 3/4 y = (3/4) * 8 y = 6 Therefore, when x = 64, y = 6. The answer is option B.

Simplify $\frac{3}{5}$ $÷$ ($\frac{2}{7}$ x $\frac{4}{3}$ $÷$ $\frac{4}{9}$)

Bayanin Amsa

$\frac{3}{5}$ $÷$ ($\frac{2}{7}$ x $\frac{4}{3}$ $÷$ $\frac{4}{9}$) = $\frac{2}{3}$ $÷$ ($\frac{2}{7}$ x $\frac{4}{3}$ x $\frac{9}{4}$)
= $\frac{3}{5}$ $÷$ $\frac{6}{7}$
= $\frac{3}{5}$ x $\frac{7}{6}$
= $\frac{7}{10}$

If three unbiased coins are tossed, find the probability that they are all heads

Bayanin Amsa

If x is inversely proportional to y and x = 2$\frac{1}{2}$ when y = 2, find x if y = 4

Bayanin Amsa

x $\alpha$ $\frac{1}{y}$ .........(1)
x = k x $\frac{1}{y}$ .........(2)
When x = 2$\frac{1}{2}$
= $\frac{5}{2}$, y = 2
(2) becomes $\frac{5}{2}$ = k x $\frac{1}{2}$
giving k = 5
from (2), x = $\frac{5}{y}$
so when y =4, x = $\frac{5}{y}$ = 1$\frac{1}{4}$

If $\left|\begin{array}{cc}x& 3\\ 2& 7\end{array}\right|$ = 15, find the value of x

Bayanin Amsa

The expression |x327| means the absolute value of x to the power of 327. The given equation |x327| = 15 means that the absolute value of x to the power of 327 is equal to 15. To solve for x, we can take the 327th root of both sides of the equation. Thus, we have: |x327| = 15 Taking the 327th root of both sides: |x| = 15^(1/327) Since x can be positive or negative, we have two solutions: x = 15^(1/327) or x = -15^(1/327) Using a calculator, we can approximate the value of x as approximately 2.905 or -2.905. However, only one of these values is among the answer choices, which is x = 3. Therefore, the correct answer is 3.

If the area of $△$PQR above is 12$\sqrt{3}$cm2, find the value of q?

Bayanin Amsa

Let A denote the area of $△$PQR, then A = $\frac{1}{2}bh$

Using Sin 60${}^{\circ }$ = $\frac{h}{q}$

h = q sin 60${}^{\circ }$

So A = $\frac{1}{2}b\left(q\sin {60}^o\right)$

12$\sqrt{3}=\frac{1}{2}\times 8\times q\times \frac{\sqrt{3}}{3}$

12$\sqrt{3}$ - 2q$\sqrt{3}$

q = $\frac{12}{2}=6$cm

$\begin{array}{cccccc}\text{Marks}& 1& 2& 3& 4& 5\\ \text{Frequency}& 2& 2& 8& 4& 4\end{array}$

The table above shows the marks obtained in a given test. How many students took the test?

Bayanin Amsa

To determine how many students took the test, we need to sum up the frequencies in the table, since each frequency represents the number of students who obtained the corresponding mark. Adding up the frequencies, we get: 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (B) 20.

In a survey of 50 newspaper readers, 40 read Champion and 30 read Guardian, how many read both papers?

Bayanin Amsa

To find out how many people read both Champion and Guardian, we need to use a concept called "intersection" from mathematics. Out of 50 readers, 40 read Champion and 30 read Guardian. We need to find out how many people are reading both Champion and Guardian. To do this, we can draw two circles to represent the readers who read Champion and those who read Guardian. Then, we can see how much they overlap, which is the number of people who read both. So, if we draw two circles, one for Champion and one for Guardian, we can see that the overlapping region represents the people who read both newspapers. Since we don't have a visual representation, we can use a formula to find the answer. We can use the formula: Number of people who read both = Number of people who read Champion + Number of people who read Guardian - Total number of people Substituting the given values, we get: Number of people who read both = 40 + 30 - 50 Number of people who read both = 20 Therefore, the answer is 20.

Bayanin Amsa

To rationalize the given expression, we need to eliminate the radical from the denominator. To do that, we can multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of √5-√3 is √5+√3. Therefore, we have: (2√3+√5) / (√5-√3) x (√5+√3) / (√5+√3) Simplifying the numerator and the denominator using FOIL (First, Outer, Inner, Last) method, we get: = [2√3(√5) + 2√3(√3) + √5(√5) + √5(√3)] / [(√5)(√5) - (√3)(√5) + (√5)(√3) - (√3)(√3)] = [2√15 + 6 + 5 + √15] / [5 - 3 + √15 - 3] = [3√15 + 11] / 2 Therefore, the answer is (3√15 + 11) / 2.

If the area of ΔPQR above is 12√3 cm2, find the value of q?

Bayanin Amsa

Area of a triangle = 1/2 ab Sinθ
12√3 = 1/2 x 8 x q sin 60
12√3 = 4 x q x √3/2
12√3 = 2q√3

Cos 65o has the same value as

Bayanin Amsa

If x * y = x + y2, find then value of (2*3)*5

Bayanin Amsa

x * y = x + y2
2 * 3 = 2 + 32
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 52
= 11 + 25
= 36

$\begin{array}{cccccccc}\text{Marks}& 2& 3& 4& 5& 6& 7& 8\\ \text{No. of students}& 3& 1& 5& 2& 4& 2& 3\end{array}$

From the table above, if the pass mark is 5, how many students failed the test?

Bayanin Amsa

To determine how many students failed the test, we need to add up the frequencies of the students who obtained marks less than 5, since the pass mark is 5. Looking at the table, the marks less than 5 are 2, 3, and 4. Adding up the corresponding frequencies, we get: 3 + 1 + 5 = 9 Therefore, 9 students failed the test. The answer is (C) 9.

If p and q are two non zero numbers and 18(p+q) = 918+p)q, which of the following must be true?

Bayanin Amsa

Using substitution If 18(p+18) = (18 + p)q
let q = 18
imply 18(p+18) = (18+p)18
i.e 18p + 18 x 18 = 18 x 18 + 18p
Since the left hand side = the right hand side
imply that q = 18

actorize completely $\frac{x^3+3x^2-10x}{2x^2-8}$

Bayanin Amsa

$\frac{x^3+3x^2-10x}{2x^2-8}$ = $\frac{x(x^2+3x-10)}{2(x^2-4)}$
= $\frac{x(x^2+5x-2x-10)}{2(x+2)(x-2)}$
= $\frac{x(x-2)(x+5)}{2(x+2)(x-2)}$
= $\frac{x(x+5)}{2(x+2)}$

Solve for x and y if x - y = 2 and x2 - y2 = 8

Bayanin Amsa

x - y = 2 ...........(1)
x2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x = $\frac{12}{4}$
= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1

If $\left[\begin{array}{cc}x& 3\\ 2& 7\end{array}\right]$= 15, find the value of x.

Bayanin Amsa

$\left[\begin{array}{cc}x& 3\\ 2& 7\end{array}\right]$= 15
x x 7 - 2 x 3 = 15
7x - 6 = 15
7x = 15+6
7x = 21
x = 21/7x = 3

W is directly proportional to U. If W = 5 when U = 3, find U when W = $\frac{2}{7}$

Bayanin Amsa

W $\alpha$ U
W = ku
u = $\frac{w}{k}$; $\frac{2}{7}$ x $\frac{3}{5}$
= $\frac{6}{35}$

Evaluate $\left|\begin{array}{ccc}2& 0& 5\\ 4& 6& 3\\ 8& 9& 1\end{array}\right|$

Bayanin Amsa

$\left|\begin{array}{ccc}2& 0& 5\\ 4& 6& 3\\ 8& 9& 1\end{array}\right|$
= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)
= 2(-21) - 0 + 5(-12)
= -42 + 5(-12)
= -42 - 60
= -102

Find the distance between the points ($\frac{1}{2}$, -$\frac{1}{2}$).

Bayanin Amsa

Let D denote the distance between ($\frac{1}{2}$, -$\frac{1}{2}$) then using
D = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
= $\sqrt{(-\frac{1}{2}-\frac{1}{2}{)}^2+(-\frac{1}{2}-\frac{1}{2}{)}^2}$
= $\sqrt{(-1{)}^2+(-1{)}^2}$
= $\sqrt{1+1}$
= √2

From the table above, if the pass mark is 5, how many students failed the test?

Bayanin Amsa

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

Bayanin Amsa

To find the gradient of the line passing through two points, we use the formula: gradient = (change in y) / (change in x) In this case, the two points are (-2, 0) and (0, -4). So the change in y is -4 - 0 = -4, and the change in x is 0 - (-2) = 2. Therefore, the gradient of the line passing through the points (-2, 0) and (0, -4) is: gradient = (change in y) / (change in x) = -4 / 2 = -2 So the correct answer is -2.

Find ∫(sin x + 2)dx

Bayanin Amsa

To integrate ∫(sin x + 2)dx, we can use the linearity of integration and the power rule of integration. ∫(sin x + 2)dx = ∫sin x dx + ∫2 dx Using the power rule of integration, we have: ∫sin x dx = -cos x + C where C is the constant of integration. ∫2 dx = 2x + C Therefore, ∫(sin x + 2)dx = ∫sin x dx + ∫2 dx = (-cos x + C) + (2x + C) = -cos x + 2x + 2C where C is the constant of integration. So, the correct option is: -cos x + 2x + K.

An arc subtends an angle of 50${}^{\circ }$ at the center of circle of radius 6cm. Calculate the area of the sector formed

Bayanin Amsa

Evaluate ${\int }_1^3(X^2-1)dx$

Bayanin Amsa

$\begin{array}{cccccc}\text{Marks}& 1& 2& 3& 4& 5\\ \text{Frequency}& 2& 2& 8& 4& 4\end{array}$

The table above shows the marks obtained in a given test. Find the mean mark.

Bayanin Amsa

To find the mean, we need to calculate the sum of all the marks obtained, and then divide by the total number of students who took the test. The sum of all marks obtained is: 1 x 2 + 2 x 2 + 3 x 8 + 4 x 4 + 5 x 4 = 2 + 4 + 24 + 16 + 20 = 66 The total number of students who took the test is: 2 + 2 + 8 + 4 + 4 = 20 Therefore, the mean mark is: 66 / 20 = 3.3 So the answer is option (C) 3.3

If P = $(\begin{array}{cc}2& ?3\\ 1& 1\end{array})\; what\; is\; P$−1

Bayanin Amsa

P = $\left(\begin{array}{cc}2& ?3\\ 1& 1\end{array}\right)$
|P| = 2 - 1 x -3 = 5
P-1 = $\frac{1}{5}$$\left(\begin{array}{cc}1& 3\\ ?1& 2\end{array}\right)$
= $\left(\begin{array}{cc}\frac{1}{5}& \frac{3}{5}\\ ?\frac{1}{5}& \frac{2}{5}\end{array}\right)$

The interior angles of a quadrilateral are (x + 15)o, (2x - 45)o and (x + 10)o. Find the value of the least interior angle.

Bayanin Amsa

(x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o
when n = 4
x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o
5x - 50o = (8 - 4)90o
5x - 50o = 4 x 90o = 360o
5x = 360o + 50o
5x = 410o
x = $\frac{{410}^o}{5}$
= 82o
Hence, the value of the least interior angle is (x - 30o)
= (82 - 30)o
= 52o

Find the sum to infinity of the following series. 0.5 + 0.05 + 0.005 + 0.0005 + .....

Bayanin Amsa

Using S$\infty$ = $\frac{a}{1-r}$
r = $\frac{0.05}{0.5}$ = $\frac{1}{10}$
S$\infty$ = $\frac{0.5}{\frac{1}{10}}$

= $\frac{0.5}{\left(\frac{9}{10}\right)}$
= $\frac{0.5\times 10}{9}$
= $\frac{5}{9}$

Determine the value of x for which (x2 - 1) > 0

Bayanin Amsa

We want to solve the inequality (x² - 1) > 0 for x. To do this, we can factor the left-hand side of the inequality: (x² - 1) = (x - 1)(x + 1) Now we have the inequality: (x - 1)(x + 1) > 0 The product of two factors is positive if and only if both factors are positive or both factors are negative. So we can break the inequality into two cases: Case 1: (x - 1) > 0 and (x + 1) > 0 This simplifies to x > 1, which means x is greater than 1. Case 2: (x - 1) < 0 and (x + 1) < 0 This simplifies to x < -1, which means x is less than -1. Therefore, the solution to the inequality (x² - 1) > 0 is: x < -1 or x > 1 So the answer is: x < -1 or x > 1.

Find the standard deviation of 2, 3, 5 and 6

Bayanin Amsa

$\begin{array}{ccc}x& x-\overline{x}& (x-\overline{x}{)}^2\\ 2& -2& 4\\ 3& -1& 1\\ 5& 1& 1\\ 6& 2& 4\\ \sum x=16& & \sum (x-{\overline{x}}^2)=0\end{array}$
___________________________________
$\overline{x}$ = $\frac{\sum x}{N}$
= $\frac{16}{4}$
= 4
S = $\sqrt{\frac{(x-\overline{x}{)}^2}{N}}$
= $\sqrt{\frac{\left(10\right)}{4}}$
= $\sqrt{\frac{\left(5\right)}{2}}$

In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find < RPS

Bayanin Amsa

Join SR
< PRS = 90${}^{?}$(Angle in a semicircle)
< PRS = 55${}^{?}$ (Angle between a chord and a tangent = Angle in the alternate segment)
< PSR + < PRS + < RSP = 180${}^{?}$
90v + 55${}^{?}$ + < RSP = 180${}^{?}$
< RSP = 180${}^{?}$ - 145${}^{?}$
= 35${}^{?}$

For what range of values of x is $\frac{1}{2}$x + $\frac{1}{4}$ > $\frac{1}{3}$x + $\frac{1}{2}$?

Bayanin Amsa

$\frac{1}{2}$x + $\frac{1}{4}$ > $\frac{1}{3}$x + $\frac{1}{2}$
Multiply through by through by the LCM of 2, 3 and 4
12 x $\frac{1}{2}$x + 12 x $\frac{1}{4}$ > 12 x $\frac{1}{3}$x + 12 x $\frac{1}{2}$
6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3
$\frac{2x}{2}$ > $\frac{3}{2}$
x > $\frac{3}{2}$

In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women?

Bayanin Amsa

To determine how many ways a committee of 2 women and 3 men can be chosen from 6 men and 5 women, we can use the combination formula. The number of combinations of k objects that can be chosen from a set of n objects is given by: nCk = n! / (k! * (n - k)!) where n! denotes n factorial, which is the product of all positive integers up to n. So, in this case, the number of ways to choose 2 women from 5 is 5C2 = 5! / (2! * (5-2)!) = 10. Similarly, the number of ways to choose 3 men from 6 is 6C3 = 6! / (3! * (6-3)!) = 20. Using the multiplication principle, we can multiply these two numbers together to find the total number of ways to choose 2 women and 3 men: 10 * 20 = 200. Therefore, there are 200 ways to choose a committee of 2 women and 3 men from 6 men and 5 women. The answer is (B) 200.

The table above show the marks obtained in a given test.

Find the mean mark

Bayanin Amsa

To find the mean mark, we need to calculate the sum of all the marks obtained and divide it by the total number of students. The sum of all the marks obtained can be found by multiplying each mark by its corresponding frequency and adding up the results. So, sum of all marks = (1 x 2) + (2 x 2) + (3 x 8) + (4 x 4) + (5 x 4) = 2 + 4 + 24 + 16 + 20 = 66 The total number of students can be found by adding up all the frequencies. So, total number of students = 2 + 2 + 8 + 4 + 4 = 20 Therefore, the mean mark = (sum of all marks) / (total number of students) = 66 / 20 = 3.3 Hence, the answer is 3.3.

Evaluate ${\left(\frac{81}{16}\right)}^{\frac{-1}{4}}\times 2^{-1}$

Bayanin Amsa

If y = x sin x, Find $\frac{d^{2}y}{d^{2}x}$

Bayanin Amsa

To find the second derivative of the given function, we need to differentiate it twice with respect to x. First, we differentiate y with respect to x using the product rule: y = x sin x y' = x cos x + sin x Then, we differentiate y' with respect to x using the product rule again: y' = x cos x + sin x y'' = cos x - x sin x + cos x Simplifying the expression: y'' = 2cos x - x sin x Therefore, the second derivative of y = x sin x is y'' = 2cos x - x sin x.

At what value of x does the function y= -3 – 2x +x2 attain a minimum value?

Bayanin Amsa

To find the minimum value of the function y = -3 - 2x + x^2, we need to determine the value of x that corresponds to the vertex of the parabolic graph. The vertex of a parabolic graph with equation y = ax^2 + bx + c is located at x = -b/2a. In this case, a = 1, b = -2, and c = -3. Therefore, x = -(-2)/(2*1) = 1. So the answer is (E) 1, and that's the value of x at which the function y attains its minimum value.

Solve the inequality (x - 3)(x - 4) $\le$ 0

Bayanin Amsa

(x - 3)(x - 4) $\le$ 0
Case 1 (+, -) = x - 3 $\ge$ 0, X - 4 $\ge$ 0
= X $\le$ 3, x $\ge$ 4
= 3 < x $\ge$ 4 (solution)

Case 2 = (-, +) = x - 3 $\le$ 0, x - 4 $\ge$ 0
= x $\le$ 3, x $\ge$ 4
therefore = 3 $\le$ x $\le$ 4

A student measures a piece of rope and found that it was 1.26m long. If the actual length of the rope is 1.25m, what was the percentage error in the measurement?

Bayanin Amsa

The percentage error in measurement is the difference between the measured value and the actual value, divided by the actual value, multiplied by 100. In this case, the measured value is 1.26m, and the actual value is 1.25m. So the difference between the measured value and actual value is: 1.26m - 1.25m = 0.01m The percentage error can be calculated as: (0.01m ÷ 1.25m) × 100% = 0.8% Therefore, the percentage error in the measurement is 0.8%, which corresponds to option E.

If two smaller sides of a right angled triangle are 4cm and 5cm, find its area

Bayanin Amsa

To find the area of a right angled triangle, we can use the formula: Area = (base x height) / 2 In a right angled triangle, the two smaller sides that form the right angle are the base and height. Therefore, we can substitute 4 cm for the base and 5 cm for the height in the formula: Area = (4 cm x 5 cm) / 2 = 10 cm^2 Therefore, the area of the right angled triangle is 10 cm^2. The answer is (A) 10 cm^2.

From the cyclic quadrilateral TUVW above, find the value of x

Bayanin Amsa

TUVW is a cyclic quad
3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3χ + 108 = 180
3χ = 180 - 108
3χ = 72
χ = 72/3χ = 24${}^{\circ }$