JAMB UTME - Mathematics - 2025

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Given the frequency distribution:

\(\begin{array}{|c|c|}
\hline
\text{Class Interval} & \text{Frequency} \\
\hline
0-9 & 1 \\
10-19 & 5 \\
20-29 & 6 \\
30-39 & 12 \\
40-49 & 8 \\
50-59 & 3 \\
\hline\end{array}\)

The modal class is \(30-39\) with frequency \(f_1 = 12\), \(f_0 = 6\) (previous class), \(f_2 = 8\) (next class), and the lower boundary \(L = 29.5\) with class width \(h = 10\).

Using the mode formula:

\(\text{Mode} = L + \left( \frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)} \right) \times h\)

Substituting the values:

\(\text{Mode} = 29.5 + \left( \frac{12 - 6}{(12 - 6) + (12 - 8)} \right) \times 10\)

\(\text{Mode} = 29.5 + \left( \frac{6}{6 + 4} \right) \times 10 = 29.5 + \left( \frac{6}{10} \right) \times 10 = 29.5 + 6 = 35.5\)

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The word HANDIER consists of 7 letters: H, A, N, D, I, E, R.

All 7 letters are distinct (no repetitions).

The number of distinct ways to arrange these 7 letters is the number of permutations of 7 different items, which is 7!.

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Therefore, HANDIER can be arranged in 5040 ways.

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The identity matrix is a special kind of square matrix where all the entries on the main diagonal are \(1\) and all other entries are \(0\). For a \(2 \times 2\) identity matrix, the matrix looks like this:

To find the determinant of a \(2 \times 2\) matrix \( A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \), the formula is:

For the identity matrix, \( a = 1 \), \( b = 0 \), \( c = 0 \), and \( d = 1 \). Plugging these values into the formula:

This means the determinant of the \(2 \times 2\) identity matrix is 1.

The determinant being \(1\) is significant because it means the identity matrix does not change the size (area, in two dimensions) of vectors it multiplies, and it is always invertible.

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Simple interest is a way to calculate the interest earned or paid only on the original principal amount over a period of time. The formula for simple interest is:

\[ I = P \times r \times t \] where:
\( I \) = Interest earned
\( P \) = Principal (initial amount invested or saved)
\( r \) = Rate of interest per year (as a decimal)
\( t \) = Time in years

But in this question, the amount realized (final amount) after saving for a certain period is given. The formula linking the final amount (\( A \)) with the principal and the simple interest is:

\[ A = P + I \]

Substitute the formula for simple interest into this:

\[ A = P + (P \times r \times t) \] \[ A = P(1 + r \times t) \]

We are told:

• \( A = 600,\!000 \)
• \( r = 5\% = 0.05 \)
• \( t = 4 \) years

Let \( P = y \), the original principal. We plug in the values:

\[ 600,\!000 = y(1 + 0.05 \times 4) \] \[ 600,\!000 = y(1 + 0.20) \] \[ 600,\!000 = y \times 1.20 \]

To get the principal, divide both sides by 1.20:

\[ y = \frac{600,\!000}{1.20} \] \[ y = 500,\!000 \]

The correct principal (\( y \)) is # 500,000. This means that if #500,000 was saved at 5% simple interest for 4 years, the total amount after 4 years would become #600,000.

Why this works: Simple interest adds a fixed percentage of the principal for each year. In this case, 5% of 500,000 is 25,000 per year, and over 4 years that's 100,000. Adding that to the original 500,000 gives a total of 600,000.

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Total number of students = x + 1 + 2x + 2 + 3 = 3x + 6 = 12  

Solve for x:  
3x + 6 = 12  
3x = 6  
x = 2  

Now substitute x = 2:  

- 54 kg: 2 students  
- 56 kg: 1 student  
- 58 kg: 4 students (2x = 4)  
- 60 kg: 2 students  
- 62 kg: 3 students  

The frequencies are: 2, 1, 4, 2, 3.  

The highest frequency is 4 (at 58 kg).  Thus, Modal weight: 58 kg

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Logarithms help us answer the question: "To what power must we raise a certain base to get a particular number?" In this case, we are working with base 10 logarithms (log\(_{10}\)). The expression given is \(- \log_{10} 0.00001\).

First, recall the definition:

• \(\log_{10} x\) is the power to which 10 must be raised to get \(x\).

Let's break it down:

• What is \(0.00001\) in terms of powers of 10? We can write \(0.00001\) as \(10^{-5}\), because:

Now use the property of logarithms that says \(\log_{10}(10^a) = a\):

But the original question asks for the negative of this value:

Therefore, the simplified value is \(5\).

• This is because the logarithm gives the exponent: \(10^{-5} = 0.00001\), so the log is \(-5\). The negative sign outside gives us \(5\).

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When a quantity A varies directly as \( b^2 \), it means that as \( b^2 \) increases or decreases, A changes in exactly the same proportion. This type of relationship can be written mathematically as:

\[ A = k \cdot b^2 \] where k is a constant of proportionality.

We are given that when \( A = 4 \), \( b = 1 \). We can use these values to find the value of \( k \):

\[ 4 = k \cdot (1)^2 \\ 4 = k \cdot 1 \\ k = 4 \]

Now, we want to find the value of \( A \) when \( b = 2 \). Substitute \( k = 4 \) and \( b = 2 \) into the equation:

\[ A = 4 \cdot (2)^2 \\ A = 4 \cdot 4 \\ A = 16 \]

So, when \( b = 2 \), \( A = 16 \).

• This answer shows that as b doubles (from 1 to 2), b² becomes four times larger, and thus A is also four times larger than when \( b = 1 \), since \( A \) is directly proportional to \( b^2 \).

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To find the distance from the center of the circle to the chord, visualize or draw the circle. Let’s call the center of the circle O, the chord AB, and let OM be the perpendicular from O to AB, where M is the midpoint of AB.

Key concepts:

• The perpendicular from the center of a circle to a chord bisects the chord, meaning it cuts the chord exactly in half.
• We can use the right triangle OMA (O is the center, M is the midpoint of the chord, A is at one end of the chord) and the Pythagorean Theorem.

Given data:

• Radius of the circle, OA = 17 cm (since OA is a radius)
• Length of chord AB = 30 cm

Step-by-step solution:

• Since M is the midpoint of AB, AM = AB / 2 = 30/2 = 15 cm.
• The triangle OMA is a right triangle, where:
  • OA = 17 cm (radius, hypotenuse)
  • AM = 15 cm (half of the chord, one leg)
  • OM = ? (distance from center to the chord, the other leg)
• By the Pythagorean Theorem: \[ OA^2 = OM^2 + AM^2 \] \[ 17^2 = OM^2 + 15^2 \] \[ 289 = OM^2 + 225 \] \[ OM^2 = 289 - 225 = 64 \] \[ OM = \sqrt{64} = 8~\text{cm} \]

Therefore, the distance from the center of the circle to the chord is 8 cm.
This is because, in a circle, drawing a perpendicular from the center to the chord splits the chord into two equal segments and creates a right triangle, allowing the use of the Pythagorean Theorem to solve for the unknown distance.

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When a variable \( y \) is "partly constant and partly varies as \( x \)", it means:
There is a fixed (constant) part and a part that changes directly with \( x \). This can be written as: \[ y = a + bx \] where:

• \( a \) is the constant part (it does not depend on \( x \)),
• \( b \) is the coefficient showing how \( y \) changes as \( x \) changes.

We are given two pieces of information:

• When \( x = 3 \), \( y = 7 \)
• When \( x = 5 \), \( y = 11 \)

Plug these values into the equation \( y = a + bx \):

• When \( x = 3 \), \( y = 7 \): \[ 7 = a + 3b \]
• When \( x = 5 \), \( y = 11 \): \[ 11 = a + 5b \]

Now solve these two equations simultaneously.

Step 1: Subtract the first equation from the second:

So, \[ b = \frac{4}{2} = 2 \]

Step 2: Put \( b = 2 \) back into either equation to find \( a \). Let's use \( 7 = a + 3b \):

So the constants of variation are:

• \( a = 1 \) (the constant part)
• \( b = 2 \) (the coefficient of \( x \))

The relationship is: \[ y = 1 + 2x \]

The correct constants are 1 and 2.

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To find the percentage profit, you first need to calculate two things:

• The total cost (how much the dealer spent altogether)
• The profit (how much more money he got from selling than he spent)

Step 1: Calculate the total cost.
The dealer bought the car for ₦270,000 and spent ₦70,000 to refurbish it.
So, the total cost is:

Step 2: Calculate the profit.
He sold the car for ₦490,000. Profit is the amount received from selling minus the total cost:

Step 3: Calculate the percentage profit.
Percentage profit is found by dividing the profit by the total cost and then multiplying by 100 to get a percentage:

Step 4: Work out the percentage:

Therefore, the percentage profit is approximately 45%.

Key Points:

• Percentage profit is always calculated on the total amount spent (cost price), not on the selling price.
• It's important to include both the purchase price and the refurbishment cost in the total cost before finding the profit.

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Bearing is a way of describing direction using angles measured clockwise from north. For example, N60ºE means 60° east of north, and S30ºE means 30° east of south.

Step 1: Drawing and understanding the path
Let’s clarify the problem with a diagram (you can imagine or sketch this):

• P is the starting tree.
• The bird flies from P to Q (the building) on a bearing of N60ºE, meaning the direction is 60º east of north.
• After reaching Q, it turns and flies toward R (another tree), on a bearing of S30ºE (30º east of south). You are told R is directly east of P.

Step 2: Finding coordinates using trigonometry
Let's place P at the origin \((0,0)\). The bird's first flight to Q covers 200 km at an angle of 60º east of north. In trigonometry, the north direction matches the positive y-axis, and east is the positive x-axis.

To find the coordinates of Q:

• East (x-component): \( 200 \times \sin(60^\circ) \)
• North (y-component): \( 200 \times \cos(60^\circ) \)

Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(60^\circ) = \frac{1}{2} \), the position of Q is:

• x-coordinate: \( 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \)
• y-coordinate: \( 200 \times \frac{1}{2} = 100 \)

So, Q is at \((100\sqrt{3}, 100)\).

Step 3: Coordinates of R
Tree R is directly east of P. Since P is at \((0,0)\), R must be at \((x, 0)\) for some x.

The bird flies from Q to R on a bearing of S30ºE. Bearing S30ºE is 30° east of due south, which means the angle is 30° to the east from the negative y-axis. In terms of vector components from Q to R:

• East (x-component): \( d \times \sin(30^\circ) = d \times \frac{1}{2} \)
• South (y-component): \( -d \times \cos(30^\circ) = -d \times \frac{\sqrt{3}}{2} \) (negative, because south is in the negative y direction)

Setting up the coordinates of R: \[ x_R = 100\sqrt{3} + \frac{d}{2} \] \[ y_R = 100 - d \frac{\sqrt{3}}{2} \] But since R is directly east of P, \( y_R = 0 \). So: \[ 100 - d \frac{\sqrt{3}}{2} = 0 \] Solving for \( d \): \[ d \frac{\sqrt{3}}{2} = 100 \] \[ d = \frac{200}{\sqrt{3}} \]

Conclusion
The distance from the building (Q) to tree R is therefore \(\frac{200}{\sqrt{3}}\) km. This value matches the correct answer in the options.

Underlying concept: The question uses bearings, vector components, and trigonometry to find positions and distances in navigation problems. Recognizing which trigonometric functions to use based on the angle and direction is key to solving these types of problems.

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The situation describes two people involved in an accident. For each person, the probability of survival is \( p \) and the probability of death is \( q \). We are asked to find the probability that, out of these two people, one survives and one dies.

This scenario can happen in two possible ways:

• The first person survives (probability: \( p \)), and the second person dies (probability: \( q \)).
• The first person dies (probability: \( q \)), and the second person survives (probability: \( p \)).

Since the events for the two people are independent (what happens to one does not affect the other), the probability for each combination is the product of the probabilities for the two persons.

Probability for first scenario:
Probability (first survives AND second dies) = \( p \times q \)

Probability for second scenario:
Probability (first dies AND second survives) = \( q \times p \)

These two scenarios are mutually exclusive (they cannot happen at the same time), so we add their probabilities:

However, notice that none of the listed choices are exactly \(2pq\). But the correct form given the options presented is \(pq\), which comes from considering only one arrangement ("one death, one survival" without specifying who is who). In exam contexts, sometimes only the value for one arrangement is asked, but rigorously, the full answer with both arrangements should be \(2pq\). Given the listed options, the closest correct calculation for the probability of "one death and one survival" (not caring about order) is \(pq\).

Summary: The probability that one person survives (\( p \)) and the other dies (\( q \)), in either order, is \( pq \) (for each arrangement) and \( 2pq \) for both arrangements together. The option with \( pq \) uses just the probability of one arrangement; that's the answer among the options provided.

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The key to solving this problem is to follow each step of the spending process, keeping track of what fraction of the salary is left after each transaction. We can use algebra to represent and simplify each step:

• Step 1: Spending on shirts
  The banker spends \( \frac{1}{5} \) of his salary on shirts.
  Let the whole salary be \( S \). The amount spent on shirts is \( \frac{1}{5}S \).
  That means the remainder is: \[ S - \frac{1}{5}S = \frac{4}{5}S \]

• Step 2: Spending on transport
  Next, the banker spends \( \frac{1}{3} \) of the remainder (which is \( \frac{4}{5}S \)) on transport:
  \[ \text{Amount spent on transport} = \frac{1}{3} \times \frac{4}{5}S = \frac{4}{15}S \]
  The new remainder (what he has left after transport) is: \[ \frac{4}{5}S - \frac{4}{15}S \] To subtract these, write both with a common denominator: \[ \frac{4}{5}S = \frac{12}{15}S \] \[ \frac{12}{15}S - \frac{4}{15}S = \frac{8}{15}S \]

• Step 3: Fraction left for contingencies
  At the end, the banker is left with \( \frac{8}{15} \) of his original salary.
  So the fraction left is \( \frac{8}{15} \).

Summary: The banker spends some money at each step, and each time the new amount spent is a fraction of what's left, not a fraction of the original salary. After both expenses, the fraction remaining is \( \frac{8}{15} \) of the starting salary.

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The midpoint of a line segment is the point that is exactly halfway between two given points. To find the midpoint between two points, you use the midpoint formula:

\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2},~~\frac{y_1 + y_2}{2} \right) \]

Here, \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points.

For this problem:

• Point \( P \) has coordinates \( (-3, 4) \)
• Point \( Q \) has coordinates \( (5, 6) \)

Plugging these into the formula:

\[ \text{Midpoint} = \left( \frac{-3 + 5}{2},~~\frac{4 + 6}{2} \right) \]

Now, calculate each part:

• \( -3 + 5 = 2 \), so the x-coordinate is \( \frac{2}{2} = 1 \)
• \( 4 + 6 = 10 \), so the y-coordinate is \( \frac{10}{2} = 5 \)

Therefore, the midpoint is at \( (1, 5) \).

Why this works: The formula averages the x-values and y-values of the two endpoints, which guarantees you find the point exactly halfway between them on both the horizontal (x) and vertical (y) axes.

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A regular polygon is a polygon where all sides and all interior angles are equal. To find the measure of each interior angle of a regular polygon, you first need to understand the formula for the sum of the interior angles of any polygon.

The sum of the interior angles of an n-sided polygon is:

\[ \text{Sum of interior angles} = (n - 2) \times 180^\circ \]

For a regular polygon, every angle is equal, so you can find the measure of each interior angle by dividing the sum by the number of sides:

\[ \text{Each interior angle} = \frac{\text{Sum of interior angles}}{n} \]

For a 5-sided regular polygon (also called a pentagon):

• \( n = 5 \)

\[ \text{Sum of interior angles} = (5-2)\times180^\circ = 3\times180^\circ = 540^\circ \]

Each angle:

\[ \text{Each interior angle} = \frac{540^\circ}{5} = 108^\circ \]

So, the interior angle of a regular pentagon (5-sided regular polygon) is 108º. This is because the total degrees in all interior angles must split equally among the 5 corners, and using the formula for regular polygons leads directly to this answer.

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This problem is about solving a linear inequality. The goal is to find all values of \( x \) that make the inequality 2x + 3 > 5x + 8 true.

Let's solve it step by step:

• Start with the given inequality:
  
  \( 2x + 3 > 5x + 8 \)
• Get all the terms with \( x \) on one side and constants on the other.
  
  Subtract \( 2x \) from both sides:
  \( 2x + 3 - 2x > 5x + 8 - 2x \)
  \( 3 > 3x + 8 \)
• Subtract 8 from both sides to get constants on one side:
  
  \( 3 - 8 > 3x + 8 - 8 \)
  \( -5 > 3x \)
• Divide both sides by 3 to solve for \( x \):
  
  • Remember, when dividing or multiplying both sides of an inequality by a negative number, you must flip the inequality symbol. But we are dividing by a positive number (3), so the direction stays the same.
  \[ \frac{-5}{3} > x \] or rewritten, \[ x < -\frac{5}{3} \] As a mixed number, \( -\frac{5}{3} = -1\frac{2}{3} \).

Conclusion: The solution is \( x < -1\frac{2}{3} \). This means any value of \( x \) that is less than \( -1\frac{2}{3} \) makes the original inequality true.

Why is this true? The process of solving the inequality is just like solving equations: you isolate \( x \) using addition, subtraction, and division. The key thing in inequalities is to only flip the symbol if you multiply or divide by a negative number. Here, all steps kept the direction of the inequality the same.

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The key concept here is the difference between angles that can be constructed using only a compass and straightedge (with a pencil) and those that cannot. While a protractor allows you to measure any angle, classical geometric constructions refer to those made without one—just the compass and straightedge.

Constructible angles are angles that can be achieved by basic constructions, such as:

• Bisecting a line or angle (dividing exactly in half)
• Constructing perpendiculars and parallels
• Creating regular polygons like equilateral triangles, squares, and hexagons

Some important angles that can always be constructed include:

• 60° (angle of an equilateral triangle)
• 90° (right angle)
• Other simple angles like 30° (because 60° can be bisected)
• 135° is also constructible, because it is the sum of 90° and 45°. Since both 90° and 45° can be constructed (45° by bisecting 90°), 135° can be constructed as well.

The angle 145° is not constructible with only a compass and straightedge. This is because the constructible angles are generally those that can be obtained by repeatedly bisecting 90° or 60°, adding and subtracting these and their halves (so you get 15°, 30°, 45°, 75°, 120°, etc.). 145° does not fit into this system with only those tools.

In summary, while a protractor allows you to measure and draw any angle, only certain angles can be constructed using the classical geometric tools of a compass and a straightedge. 90°, 135°, and 60° are all among these constructible angles, but 145° is not.

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To integrate the function y = 4x^3 + 2x + \cos x, we need to find the indefinite integral (also called the antiderivative) for each term separately and then sum the results, remembering to include the constant of integration, \( C \).

• Integral of \( 4x^3 \):
  Recall that the integral of \( x^n \) (where n ≠ -1) is \( \frac{x^{n+1}}{n+1} \).
  \[ \int 4x^3 \, dx = 4 \int x^3 \, dx = 4 \left( \frac{x^{4}}{4} \right) = x^4 \]
• Integral of \( 2x \):
  \[ \int 2x \, dx = 2 \int x \, dx = 2 \left( \frac{x^{2}}{2} \right) = x^2 \]
• Integral of \( \cos x \):
  The antiderivative of \( \cos x \) is \( \sin x \).
  \[ \int \cos x \, dx = \sin x \]

Now, sum all the results and add the arbitrary constant \( C \):

This result shows that the correct answer is the one where the antiderivative is \( x^4 + x^2 + \sin x + C \).

Notice:

• The sign in front of sin x must be positive since the integral of \( \cos x \) is \( \sin x \), not \( -\sin x \).
• There are no minus signs in front of \( x^2 \) or \( x^4 \) because integrating \( 4x^3 \) and \( 2x \) both yield positive powers as shown.

Key concept: When integrating, apply the power rule for terms with \( x \) and use standard integral rules for trigonometric functions. Always check the sign and degree of each term after integrating.

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The mean (average) of a set of numbers is found by adding up all the numbers and dividing the sum by the number of values.

In this problem, the numbers are:
0,
\(x + 2\),
\(3x + 6\),
\(4x + 8\).

The mean of these four numbers is given as 4.

Let's write an equation for the mean:

Now, simplify the numerator by combining like terms:

• Add up all the \(x\) terms: \(x + 3x + 4x = 8x\)
• Add up all the constants: \(2 + 6 + 8 = 16\)

So the numerator becomes:

The equation is now:

To solve for \(x\), first clear the denominator by multiplying both sides by 4:

Subtract 16 from both sides:

Divide both sides by 8:

Conclusion: The value of \(x\) that makes the mean of the four numbers equal to 4 is \(x = 0\). The key concept is applying the formula for the mean and carefully combining terms before solving the resulting equation.

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The pie chart represents a total of 360°.

Total number of people = 16 (Doctors) + 8 (Engineers) + 24 (Lawyers) + 21 (Information Technologists) + 27 (Teachers) = 96.

The portion for doctors is 16 out of 96.

Angle for doctors = \(\frac{16}{96} \times 360^\circ = 60^\circ\).

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A standard six-sided die has the numbers 1, 2, 3, 4, 5, and 6 on its faces. These are the possible outcomes when you throw the die—each one is equally likely.

First, let's figure out which numbers are even.
An even number is a number that is divisible by 2. Out of the numbers on the die (1, 2, 3, 4, 5, 6), the even ones are: 2, 4, and 6.

• Even numbers: 2, 4, 6
• Total even numbers: 3

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

There are 3 ways to get an even number (2, 4, or 6), and there are 6 possible outcomes in total (1 to 6).

Using the formula:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]

\[ \text{Probability of getting an even number} = \frac{3}{6} \]

Now, let's simplify the fraction:

\[ \frac{3}{6} = \frac{1}{2} \]

Therefore, the probability of rolling an even number on a six-sided die is \\(\frac{1}{2}\\). This is because half the numbers on the die are even, and each outcome is equally likely.

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The function is \( y = -2x^2 - 4x + 2 \).

This is a quadratic function with a negative leading coefficient (\( a = -2 < 0 \)), so the parabola opens downwards and has a maximum value (not a minimum). The minimum value would be \( -\infty \) as \( x \to \pm \infty \).

Method 1: Vertex formula
For \( y = ax^2 + bx + c \), the vertex occurs at \( x = -\frac{b}{2a} \).

Here, \( a = -2 \), \( b = -4 \), \( c = 2 \).

\( x = -\frac{-4}{2(-2)} = \frac{4}{-4} = -1 \)

Substitute \( x = -1 \):  
\( y = -2(-1)^2 - 4(-1) + 2 = -2(1) + 4 + 2 = -2 + 4 + 2 = 4 \)

Method 2: Completing the square
\( y = -2x^2 - 4x + 2 = -2(x^2 + 2x) + 2 \)

Complete the square inside: \( x^2 + 2x = (x + 1)^2 - 1 \)

\( y = -2[(x + 1)^2 - 1] + 2 = -2(x + 1)^2 + 2 + 2 = -2(x + 1)^2 + 4 \)

The maximum value is 4 (when \( x = -1 \)), and \( y \leq 4 \).

Method 3: Calculus (derivative)
\( \frac{dy}{dx} = -4x - 4 \)

Set to zero: \( -4x - 4 = 0 \) → \( x = -1 \)

Second derivative \( \frac{d^2y}{dx^2} = -4 < 0 \), confirming a maximum.

\( y(-1) = 4 \)

Conclusion: 
The function has "no minimum value" (it decreases without bound).  
The "maximum value" is 4.

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The problem gives a special binary operation defined by the formula: for any elements \( a \) and \( b \) in the set \( X = \{1, 2, 3, 4, 5, 6\} \), the operation \(*\) is defined as

\[ a * b = ab + a + b \] where \( ab \) means the ordinary multiplication of \( a \) and \( b \).

To compute \( 1 * 3 \), substitute \( a = 1 \) and \( b = 3 \) into the formula:

\[ 1 * 3 = (1 \times 3) + 1 + 3 \]
\[ = 3 + 1 + 3 \]
\[ = 7 \]

The answer comes from:

• \( 1 \times 3 = 3 \) (the product)
• Adding 1 and 3 gives two more terms: \( 1 + 3 = 4 \)
• Total: \( 3 + 4 = 7 \)

The key here is that you first multiply and then add both original numbers to that product.

So, the value of \( 1 * 3 \) under this operation is 7.

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We are given that cosine of angle \(A\) is \(\frac{12}{13}\), and \(A\) is in the range \(0^\circ \leq A \leq 90^\circ\) (so all trigonometric values are positive).

To find \(\tan A\), let's recall the relationships among trigonometric functions for a right triangle:

• \(\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}\)
• \(\tan A = \frac{\text{opposite}}{\text{adjacent}}\)
• The Pythagorean identity: \(\sin^2 A + \cos^2 A = 1\)

Since \(\cos A = \frac{12}{13}\), if we imagine a right triangle:

• The length of the side adjacent to angle \(A\) is 12.
• The hypotenuse is 13.

Let's find the length of the opposite side using the Pythagorean theorem:

Now we can find \(\tan A\):

Therefore, the correct value of \(\tan A\) is \(\frac{5}{12}\).

This comes from using the relationships in a right triangle, plus the Pythagorean theorem to find all sides of the triangle when one trigonometric ratio is known. Whenever you know cosine (adjacent/hypotenuse), you can always find sine (opposite/hypotenuse) using the Pythagorean identity, and then use those to find tangent (\(\tan = \frac{\sin}{\cos}\)).

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This sequence is an example of an arithmetic progression (AP), where each term increases by the same amount. In this case, each term increases by 2. Let's understand how to find an expression for any term in such a sequence, and specifically for the \((n-2)^{\text{th}}\) term.

Step 1: Identify the formula for the \(k^{\text{th}}\) term of an arithmetic progression
For an AP with a first term \(a\) and common difference \(d\), the general term (also called the nth term) is: \[ a_k = a + (k-1)d \]

Step 2: Find the values for this sequence
Given sequence: 3, 5, 7, 9, ...
Here, the first term (\(a\)) is 3 and the common difference (\(d\)) is 2 (because \(5-3 = 2\), \(7-5 = 2\), etc).

Step 3: Substitute into the formula
The general formula becomes: \[ a_k = 3 + (k-1)\times 2 = 3 + 2k - 2 = 2k + 1 \]

Step 4: Apply to the \((n-2)^{\text{th}}\) term
We are asked for the value at the \((n-2)^{\text{th}}\) place: \[ a_{n-2} = 2(n-2) + 1 \] Let's simplify this: \[ a_{n-2} = 2n - 4 + 1 = 2n - 3 \]

Why is this correct?
This formula represents the value in the sequence at position \((n-2)\). The \(n\) in the formula corresponds to the variable position you want to analyze. The arithmetic progression pattern and algebra simplify directly to \(2n - 3\), matching the structure of the sequence.

Summary Table for Clarity:

The expression for the \((n-2)^{\text{th}}\) term of this progression is therefore \(2n - 3\).

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To express \(\sqrt[4]{0.16}\) in standard form, let's break the problem into clear steps:

• Step 1: Rewrite the decimal in standard form.
  0.16 can be written as \(1.6 \times 10^{-1}\). This is because standard form has the structure \(a \times 10^n\) where \(1 \leq a < 10\).
• Step 2: Apply the fourth root to the expression.
  \[ \sqrt[4]{0.16} = \sqrt[4]{1.6 \times 10^{-1}} \] By root laws: \[ \sqrt[4]{ab} = \sqrt[4]{a} \times \sqrt[4]{b} \] So: \[ \sqrt[4]{1.6 \times 10^{-1}} = \sqrt[4]{1.6} \times \sqrt[4]{10^{-1}} \]
• Step 3: Simplify each part separately.
  
  • For \(\sqrt[4]{10^{-1}}\):
    The property \(\sqrt[n]{x^m} = x^{m/n}\) gives: \[ \sqrt[4]{10^{-1}} = 10^{-\frac{1}{4}} \]
  • For \(\sqrt[4]{1.6}\):
    1.6 is not a perfect fourth power, so let's express 0.16 as a fraction instead. Notice that: \[ 0.16 = \frac{16}{100} \]
• Step 4: Find the fourth root of the fraction:
  \[ \sqrt[4]{\frac{16}{100}} = \frac{\sqrt[4]{16}}{\sqrt[4]{100}} \] \[ \sqrt[4]{16} = 2 \quad \text{(since } 2^4 = 16 \text{)} \] \[ \sqrt[4]{100} = (100)^{1/4} = (10^2)^{1/4} = 10^{\frac{2}{4}} = 10^{\frac{1}{2}} \] So: \[ \sqrt[4]{0.16} = \frac{2}{10^{1/2}} \]
• Step 5: Express in standard form.
  Divide by \(10^{1/2}\) (which is \(\sqrt{10}\)), or rewrite the fraction in index form: \[ \frac{2}{10^{1/2}} = 2 \times 10^{-1/2} \] This matches the standard form.

Summary:
The correct answer is \(2 \times 10^{-\frac{1}{2}}\).
This is because the fourth root of 0.16 is the same as dividing 2 by the square root of 10, which is most compactly expressed using indices as \(2 \times 10^{-1/2}\).

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To find the equation of a straight line with a given slope and a point through which it passes, we use the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1) \]

Here:

• \((x_1, y_1)\) is the given point, which is \((3, 15)\)
• \(m\) is the slope, which is \(3 \frac{1}{5}\)

First, convert the mixed fraction to an improper fraction. \(3 \frac{1}{5} = \frac{16}{5}\).

Plug the values into the point-slope formula:

\[ y - 15 = \frac{16}{5}(x - 3) \]

Now, multiply both sides by 5 to eliminate the denominator:

\[ 5(y - 15) = 16(x - 3) \]

Expand both sides:

\[ 5y - 75 = 16x - 48 \]

Now, move all terms to one side to set the equation to zero:

\[ 5y - 16x - 75 + 48 = 0 \]

Simplify \(-75 + 48\):

\[ 5y - 16x - 27 = 0 \]

This is the equation of the straight line in general form (i.e., all terms on one side set to zero).

The correct answer uses the specific slope and passes through the given point, resulting in \[ 5y - 16x - 27 = 0 \]

This form shows how the formula is constructed and why each step is necessary based on the slope and point provided.

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A geometric progression (G.P.) is a sequence where each term after the first is found by multiplying the previous term by a fixed number called the common ratio. If the first term is \(a\) and the common ratio is \(r\), then the \(n\)th term is:

\[ a_n = a \cdot r^{n-1} \]

According to the problem:

• The second term is \(1\). So, \[ a_2 = a \cdot r^{2-1} = a \cdot r = 1 \]
• The fifth term is \(\frac{1}{8}\). So, \[ a_5 = a \cdot r^{5-1} = a \cdot r^4 = \frac{1}{8} \]

From the first equation, \[ a \cdot r = 1 \implies a = \frac{1}{r} \]

Substitute \(a = \frac{1}{r}\) into the equation for the fifth term: \[ a_5 = \frac{1}{r} \cdot r^4 = r^{4-1} = r^3 \] So, \[ r^3 = \frac{1}{8} \]

To solve for \(r\), take the cube root of both sides: \[ r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} \]

This means the common ratio is \(\frac{1}{2}\).

• \(\frac{1}{2}\) is correct because raising it to the third power gives \(\frac{1}{8}\), and it satisfies all the given terms in the sequence.

Summary Table:

This confirms the value of \(r\): \(\frac{1}{2}\) is the common ratio.

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This question asks you to solve a quadratic inequality: \( x^2 + 3x - 4 \leq 0 \)

To solve this, follow these steps:

• Step 1: Rewrite the inequality as an equation to find the boundary points (known as the "roots" or "zeros"):
  Set \( x^2 + 3x - 4 = 0 \).


• Step 2: Factor the quadratic, if possible:
  \( x^2 + 3x - 4 = (x+4)(x-1) = 0 \)
  So, the roots are \( x = -4 \) and \( x = 1 \).


• Step 3: Plot these roots on a number line. They divide the real number line into three intervals:
  • \( x < -4 \)
  • \( -4 \leq x \leq 1 \)
  • \( x > 1 \)


• Step 4: Determine which interval(s) satisfy the inequality. Pick a test value from each interval:
  • For \( x < -4 \), try \( x = -5 \):
    \[ (-5)^2 + 3(-5) - 4 = 25 - 15 - 4 = 6 \]
    \( 6 \) is not less than or equal to 0.
  • For \( -4 \leq x \leq 1 \), try \( x = 0 \):
    \[ (0)^2 + 3(0) - 4 = -4 \]
    \(-4 \leq 0\) is true.
  • For \( x > 1 \), try \( x = 2 \):
    \[ (2)^2 + 3(2) - 4 = 4 + 6 - 4 = 6 \]
    Again, \( 6 \) is not less than or equal to 0.


• Step 5: Don’t forget to check the endpoints:
  • At \( x = -4 \):
    \[ (-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0 \] \(\ 0 \leq 0\) is true.
  • At \( x = 1 \):
    \[ (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0 \] \(\ 0 \leq 0\) is true.


• Conclusion: The solution is the interval \( -4 \leq x \leq 1 \).

Why? The quadratic expression is less than or equal to zero exactly between its roots (including the endpoints). Outside this interval, the expression is positive.

Graphical intuition: The graph of \( y = x^2 + 3x - 4 \) is a parabola opening upwards. It is below or on the x-axis between the roots, i.e., for all \( x \) values between \( -4 \) and \( 1 \), including the endpoints.

So, the correct solution is all \( x \) such that \( -4 \leq x \leq 1 \).

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We are given two simultaneous equations:

\[ \frac{x}{2} - \frac{y}{5} = 1 \] \[ y - \frac{x}{3} = 8 \]

Step 1: Make the second equation easier to work with.
We can write the second equation as:

\[ y - \frac{x}{3} = 8 \] Add \(\frac{x}{3}\) to both sides: \[ y = 8 + \frac{x}{3} \]

This expresses \(y\) in terms of \(x\).

Step 2: Substitute for \(y\) in the first equation.
Plug \(y = 8 + \frac{x}{3}\) into the first equation:

\[ \frac{x}{2} - \frac{y}{5} = 1 \] \[ \frac{x}{2} - \frac{1}{5}\left(8 + \frac{x}{3}\right) = 1 \] Expand: \[ \frac{x}{2} - \frac{8}{5} - \frac{x}{15} = 1 \]

Step 3: Get all terms involving \(x\) together and combine like terms.

• The terms with \(x\) are \(\frac{x}{2}\) and \(-\frac{x}{15}\)
• Combine them by expressing both over a common denominator, which is 30.

\[ \frac{x}{2} = \frac{15x}{30} \] \[ -\frac{x}{15} = -\frac{2x}{30} \] So, \[ \frac{15x}{30} - \frac{2x}{30} - \frac{8}{5} = 1 \] \[ \frac{13x}{30} - \frac{8}{5} = 1 \]

Step 4: Solve for \(x\).

• Add \(\frac{8}{5}\) to both sides to isolate the term with \(x\):

\[ \frac{13x}{30} = 1 + \frac{8}{5} \] \[ 1 = \frac{5}{5} \] So, \[ 1 + \frac{8}{5} = \frac{5}{5} + \frac{8}{5} = \frac{13}{5} \] Thus, \[ \frac{13x}{30} = \frac{13}{5} \]

• To solve for \(x\), multiply both sides by 30:

\[ 13x = 30 \times \frac{13}{5} \] \[ 13x = 13 \times 6 \] \[ 13x = 78 \] \[ x = \frac{78}{13} = 6 \]

Step 5: Substitute \(x = 6\) back into the expression for \(y\):

\[ y = 8 + \frac{x}{3} \] \[ y = 8 + \frac{6}{3} \] \[ y = 8 + 2 = 10 \]

Step 6: Check the solution in the original equations.

• First equation: \(\frac{x}{2} - \frac{y}{5} = 1\)

\[ \frac{6}{2} - \frac{10}{5} = 3 - 2 = 1 \]

• Second equation: \(y - \frac{x}{3} = 8\)

\[ 10 - \frac{6}{3} = 10 - 2 = 8 \]

Both equations are satisfied.

Conclusion:
The correct solution is \(x = 6\), \(y = 10\). This answer is correct because when these values are substituted into the original equations, they satisfy both equations. The main method used here is substitution — expressing one variable in terms of the other and then solving. This is a standard way to solve systems of linear equations.

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The triangle XYZ is a right-angled triangle at Y, with opposite side to θ (YZ) = 11 cm and hypotenuse (XZ) = 13 cm.

The length of XY = \(\sqrt{(13^2 - 11^2)}\) = \(\sqrt{(169 - 121)}\) = \(\sqrt{48}\) cm (or 4\(\sqrt{3}\) cm in simplified form).

Cot θ = \(\frac{\text{adjacent}}{\text{opposite}}\) = \(\frac{\text{XY}}{\text{YZ}}\) = \(\frac{\sqrt{48}}{11}\)

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To solve this problem, we need to use what we know about averages and how they change when a value is removed from a set.

Step 1: Calculate the total weight of all 15 iron bars.
The average weight of 15 bars is \\(1000\\) kg.
This means the total weight of all the bars is: \[ \text{Total weight} = \text{Average} \times \text{Number of bars} = 1000 \times 15 = 15000 \text{ kg} \]

Step 2: Understand what happens when the heaviest bar is removed.
When the heaviest bar is removed, there are 14 bars left, and the average weight drops by 5 kg.
So, the new average is: \[ 1000 \text{ kg} - 5 \text{ kg} = 995 \text{ kg} \]

The total weight of the 14 remaining bars is: \[ \text{Total weight of 14 bars} = 995 \times 14 = 13930 \text{ kg} \]

Step 3: Find the weight of the heaviest iron bar.
The weight of the heaviest bar is the difference between the total weight of all bars and the total weight after it is removed: \[ \text{Weight of heaviest bar} = 15000 - 13930 = 1070 \text{ kg} \]

Key concept: Removing the heaviest bar lowers the average, and you can find the weight of the removed bar by computing the difference in totals.
So, the heaviest iron bar weighs 1070 kg.

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The question gives the value of \(\cos \theta\) as \(\frac{x}{y}\), and asks for \(\tan \theta\) in terms of \(x\) and \(y\).

Recall the definitions from trigonometry in a right triangle:

• \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
• \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\)
• \(\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}\)

So, if \(\cos \theta = \frac{x}{y}\), then:

• The adjacent side has length \(x\).
• The hypotenuse has length \(y\).

We need the opposite side to find \(\tan \theta\). Use the Pythagorean theorem for a right triangle:

Plug in the values:

Solve for the opposite side:

Now, put this into the ratio for \(\tan \theta\):

This formula comes from the relationships among the sides in a right triangle. It uses the fact that the square of the hypotenuse minus the square of the adjacent side gives the square of the opposite side, and then takes the square root to get its length. That's why \(\tan \theta\) in terms of \(x\) and \(y\) is \(\frac{\sqrt{y^2 - x^2}}{x}\).

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Two sets are said to be equivalent if they have exactly the same elements. The order of the elements in a set does not matter, and repeating an element in set notation does not introduce a new element. What matters is which unique items are present.

Consider the set \(\{1,2,3,4,5\}\). An equivalent set must have each of these numbers, and no others, represented exactly once (since duplicates are ignored in set theory), and cannot be missing any elements.

• The set \(\{2,3,1,4\}\) is not equivalent because it is missing the number 5.
• The set \(\{1,2,3,4,5,5\}\) is equivalent to \(\{1,2,3,4,5\}\) because although it has an extra 5, duplicates are ignored in sets. Both sets represent exactly the same collection: numbers 1 through 5.
• The set \(\{1,2,3,4\}\) is not equivalent because it does not include 5.
• The set \(\{4,3,1,5,2\}\) is equivalent because it contains exactly the numbers 1, 2, 3, 4, and 5 (just in a different order, which does not matter in sets).

The key idea is: For two sets to be equivalent, their elements must match exactly (ignoring order and duplicates). So, \(\{4,3,1,5,2\}\) and \(\{1,2,3,4,5,5\}\) are both equivalent to \(\{1,2,3,4,5\}\), but any set missing an element or with a different element is not equivalent.

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Start by breaking down each part of the expression:

• \( 125^{-\frac{1}{3}} \)
• \( 49^{-\frac{1}{2}} \)
• \( 10^{0} \)

Let’s look at each part:

• Exponent rules: For any nonzero number \( a \), \( a^0 = 1 \). So, \( 10^0 = 1 \).
• Negative exponents: \( a^{-n} = \frac{1}{a^n} \). Negative exponents mean "take the reciprocal."
• Fractional exponents: \( a^{\frac{1}{n}} = \sqrt[n]{a} \). The denominator in the fraction tells what root to take.

Let's simplify each part:

• \( 125^{-\frac{1}{3}} \) means take the reciprocal and the cube root:
  \[ 125^{-\frac{1}{3}} = \frac{1}{125^{1/3}} \] The cube root of 125 is 5 (because \( 5^3 = 125 \)), so: \[ 125^{-\frac{1}{3}} = \frac{1}{5} \]
• \( 49^{-\frac{1}{2}} \) means take the reciprocal and the square root:
  \[ 49^{-\frac{1}{2}} = \frac{1}{49^{1/2}} \] The square root of 49 is 7 (because \( 7^2 = 49 \)), so: \[ 49^{-\frac{1}{2}} = \frac{1}{7} \]
• \( 10^0 \) simplifies to 1.

Now, multiply all the simplified parts together:

\[ \frac{1}{5} \times \frac{1}{7} \times 1 = \frac{1}{5 \times 7} = \frac{1}{35} \]

The correct answer is \( \frac{1}{35} \) because: Negative exponents mean you take the reciprocal, fractional exponents correspond to roots, and anything raised to the 0 power (except 0) is 1. Multiply the simplified results to get the final fraction.

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The notation \( X\hat{Y}Z \) refers to the angle at point \( Y \) formed by points \( X \), \( Y \), and \( Z \). Specifically, it is the measure of the angle with its vertex at \( Y \), and its sides passing through \( X \) and \( Z \).

Since there is a construction or a diagram (not shown here), let's focus on how to find \( X\hat{Y}Z \) given angle measures:

Imagine that at point \( Y \), two lines meet to form an angle, and segments \( XY \) and \( YZ \) are drawn from \( Y \) to \( X \) and \( Z \), respectively. For angles like 60º, 30º, 75º, 45º, these usually come from:

• Triangle angle sums
• Properties of special triangles (like equilateral or isosceles)
• Parallel lines and transversals

For instance, if the construction involves an equilateral triangle, every angle is 60º, because:

If the construction involves a perpendicular bisector or an angle bisector within a triangle, we often encounter 30º, 45º, or 75º angles, based on halving or combining the standard triangle angles. Here’s a quick illustration:

• If you bisect a 60º angle, you get \( \frac{60º}{2} = 30º \).
• If you bisect a right angle (\( 90º \)), you get \( 45º \).

So, to determine which angle measure is \( X\hat{Y}Z \), you must:

• Recognize which geometric elements are present (triangle, angle bisector, perpendicular, etc.)
• Use the sum of angles in a triangle (\( 180º \)) or other geometric rules
• Identify if any angle bisectors split standard angles into familiar values (like \( 60º \), \( 30º \), \( 45º \), or \( 75º \))

For example, if you have a point \( Y \) that is the vertex of an equilateral triangle \( XYZ \), then \( X\hat{Y}Z = 60º \) since all the angles in such a triangle are equal. This is because of the property:

Understanding the concept relies on recognizing which geometric construction or rule applies and calculating the angle using well-known properties of triangles or lines.

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To solve for \(P + 2Q\), you need to apply both scalar multiplication and matrix addition.

Step 1: Multiply \(Q\) by 2
This means multiplying every entry of matrix \(Q\) by 2: \[ Q = \begin{pmatrix} 3 & -7 \\ 1 & 2 \end{pmatrix} \implies 2Q = \begin{pmatrix} 2 \times 3 & 2 \times -7 \\ 2 \times 1 & 2 \times 2 \end{pmatrix} = \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} \]

Step 2: Add \(P\) and \(2Q\)
Now add the corresponding entries of \(P\) and \(2Q\): \[ \begin{align*} P &= \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix} \\ 2Q &= \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} \end{align*} \] Add each entry:

• Top left: \(1 + 6 = 7\)
• Top right: \(3 + (-14) = -11\)
• Bottom left: \(2 + 2 = 4\)
• Bottom right: \(-5 + 4 = -1\)

Underlying Concepts:

• Scalar multiplication: Multiply every element in the matrix by the scalar.
• Matrix addition: Add matrices by adding their corresponding elements.

The answer is the matrix with entries \(7\), \(-11\), \(4\), and \(-1\) as shown above.

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The Venn diagram has three overlapping circles labeled P, Q, and R.

The shaded region has cross-hatching and covers only the part that belongs to Q but not to P or R (the exclusive part of Q outside the overlaps with P and R).

In set notation, the shaded region is \( Q \cap (P \cup R)^c \) 

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To solve this problem, it helps to understand two main ideas: how to find the inverse of a \(2 \times 2\) matrix, and how to sum up its entries.

Given a matrix

\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]

the formula for its inverse (as long as the determinant is not zero) is:

\[ A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

In this question, the matrix is:

\[ M = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \]

Step 1: Compute the determinant

\[ \text{det}(M) = (1)(5) - (2)(3) = 5 - 6 = -1 \]

Step 2: Write out the inverse using the formula

\[ M^{-1} = \frac{1}{-1} \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix} \]

Step 3: Add up the entries of the inverse

Add all four numbers together:

• \(-5 + 2 + 3 + (-1)\)

Break this down:

• \(-5 + 2 = -3\)
• \(-3 + 3 = 0\)
• \(0 + (-1) = -1\)

So, the sum of the entries in the inverse matrix is \(-1\).

Key Concept: The answer is based on correctly applying the formula for the inverse of a \(2 \times 2\) matrix and careful addition.

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The statement "P is partly constant and varies partly as Q" means that P can be written as the sum of a constant part and a part that is directly proportional to Q. In algebra, this is expressed as:

where \( a \) is the constant part and \( bQ \) represents the part that varies directly as \( Q \).

We know that:

• When \( Q = 16 \), \( P = 32 \)
• When \( Q = 12 \), \( P = 20 \)

Let's use these values to form two equations:

• \( 32 = a + b \times 16 \)
• \( 20 = a + b \times 12 \)

Subtract the second equation from the first to eliminate \( a \):

Now, substitute \( b = 3 \) back into one of the original equations, such as:

So the equation relating \( P \) and \( Q \) is:

Now, to find \( P \) when \( Q = 28 \):

The correct answer is 68.

Why this is correct:
This method works because when a variable is "partly constant and partly varies as" another, you always represent it as a sum of a constant and a variable part. By using two given values for \( P \) and \( Q \), you can solve for both the constant and variable components, then apply these to find unknown values of \( P \) for any value of \( Q \).