2014 UTME Mathematics Full Simulation

A number is chosen at random from 10 to 30, both inclusive. This means **all whole numbers from 10 up to 30** are possible. First, count the total number of possible numbers. Total numbers = 30 − 10 + 1 The +1 is added because both 10 and 30 are included. 30 − 10 = 20 So: Total numbers = 21 Next, list the numbers between 10 and 30 that are divisible by 3. A number is divisible by 3 if it can be written as 3 × an integer. The numbers divisible by 3 in this range are: 12, 15, 18, 21, 24, 27, 30 Now count them. There are **7** such numbers. Probability = number of favourable outcomes ÷ total number of possible outcomes Probability = 7 ÷ 21 Now simplify the fraction. 7 ÷ 21 = 1 ÷ 3 1/3

The outcomes that are **at least 4** are the numbers 4, 5, and 6. “At least 4” means **4 or more**, so we include 4, 5, and 6. From the table, their frequencies are: For 4 → 16 For 5 → 10 For 6 → 14 Now add these frequencies to get the total number of favourable outcomes. 16 + 10 + 14 = 40 The die was thrown **100 times**, so the total number of outcomes is 100. Probability is calculated as: Probability = (number of favourable outcomes) ÷ (total number of outcomes) So we write: Probability = 40 ÷ 100 Now simplify the fraction. 40 ÷ 100 = 4 ÷ 10 = 2 ÷ 5 So the probability is: 2/5 C

A team of 3 girls is to be selected from 7 girls. This is a **combination** problem because the **order of selection does not matter**. The formula for combinations is: ⁿCᵣ = n! / r!(n − r)! Here: n = 7 (total girls) r = 3 (girls to be selected) Substitute into the formula. Number of ways = 7! / 3!(7 − 3)! First simplify the bracket. 7 − 3 = 4 So we have: 7! / 3!4! This expression already matches one of the options given. C

5, 4, 3, 2, 1 First, find the mean of the numbers. The mean is the sum of the values divided by the number of values. Sum = 5 + 4 + 3 + 2 + 1 Sum = 15 Number of values = 5 Mean = 15 ÷ 5 Mean = 3 Next, find the deviation of each value from the mean. This is done by subtracting the mean from each value. For 5: 5 − 3 = 2 For 4: 4 − 3 = 1 For 3: 3 − 3 = 0 For 2: 2 − 3 = −1 For 1: 1 − 3 = −2 Now square each deviation. This removes negative signs and prepares for averaging. 2² = 4 1² = 1 0² = 0 (−1)² = 1 (−2)² = 4 Now add the squared deviations. 4 + 1 + 0 + 1 + 4 = 10 Next, divide by the number of values to get the variance. (Standard deviation in UTME uses division by n.) Variance = 10 ÷ 5 Variance = 2 Finally, take the square root to get the standard deviation. Standard deviation = √2 A

The median is the **middle value** when the numbers are arranged in **ascending order**. First, list all the numbers given: 5, 9, 6, 10, 3, 8, 9, 2, 4, 5, 5, 7, 3, 6 Next, arrange them in ascending order (from smallest to largest): 2, 3, 3, 4, 5, 5, 5, 6, 6, 7, 8, 9, 9, 10 Now count the total number of values. There are 14 numbers. Since 14 is an **even number**, the median is the **average of the 7th and 8th terms**. Identify the 7th term: 7th term = 5 Identify the 8th term: 8th term = 6 Now find their average. Median = (5 + 6) ÷ 2 Median = 11 ÷ 2 Median = 5.5 From the given options, the closest correct choice is: 5

The mode of a distribution is the **number that occurs most frequently**. So we look for the number with the **highest frequency**. From the table: Numbers: 0, 1, 2, 3, 4 Frequencies: 1, 2, 2, 1, 9 Now compare the frequencies one by one. The number 0 occurs 1 time. The number 1 occurs 2 times. The number 2 occurs 2 times. The number 3 occurs 1 time. The number 4 occurs 9 times. The highest frequency is **9**, and it corresponds to the number **4**. Therefore, the mode of the distribution is: 4

The mean of several quantities is found by **adding all the quantities together and dividing by the number of quantities**. The given quantities are: 2 − t , 4 + t , 3 − 2t , 2 + t , t − 1 There are **5** quantities in total. First, add all the expressions together. (2 − t) + (4 + t) + (3 − 2t) + (2 + t) + (t − 1) Now remove the brackets and group like terms. = 2 − t + 4 + t + 3 − 2t + 2 + t + t − 1 Next, collect all the numbers (constants). 2 + 4 + 3 + 2 − 1 = 10 Now collect all the t terms. −t + t − 2t + t + t The +t and −t cancel out. So we have: −2t + t + t = 0t This means all the t terms cancel completely. So the total sum becomes: 10 Now divide by the number of terms, which is 5. Mean = 10 ÷ 5 Mean = 2 C

The pie chart shows angles representing how the salary is shared among food items. From the chart: Rice = 80° Beans = 50° Garri = 70° First, find the angle for yam. The total angle in a circle is 360°. So we subtract the known angles from 360°. Yam angle = 360° − (80° + 50° + 70°) First add the given angles. 80° + 50° = 130° 130° + 70° = 200° Now subtract from 360°. Yam angle = 360° − 200° = 160° Now compare yam and rice using their angles. Rice corresponds to 80° and costs ₦8,000. Yam corresponds to 160°. Set up a ratio using angles and money. Rice : Yam 80° : 160° Simplify the ratio. 80 : 160 = 1 : 2 This means yam costs **twice** as much as rice. So, cost of yam = 2 × ₦8,000 = ₦16,000 C

∫ (2x + 3)¹ᐟ² dx This is an integration involving a power of a linear expression. First, rewrite the expression clearly. (2x + 3)¹ᐟ² means the square root of (2x + 3). To integrate this, we use **substitution**, because the inside of the bracket is a linear function. Let: u = 2x + 3 Now differentiate u with respect to x. du/dx = 2 So: du = 2 dx This means: dx = du / 2 Now rewrite the integral in terms of u. ∫ (2x + 3)¹ᐟ² dx = ∫ u¹ᐟ² × (1/2) du Take the constant 1/2 outside the integral. = 1/2 ∫ u¹ᐟ² du Now apply the power rule for integration. ∫ uⁿ du = uⁿ⁺¹ / (n + 1) Here, n = 1/2. So: ∫ u¹ᐟ² du = u³ᐟ² / (3/2) Dividing by 3/2 is the same as multiplying by 2/3. So: = 2/3 u³ᐟ² Now multiply by the outside factor 1/2. 1/2 × 2/3 u³ᐟ² = 1/3 u³ᐟ² Now substitute back u = 2x + 3. = 1/3 (2x + 3)³ᐟ² + k B

∫ sin 2x dx This is an integration problem involving a trigonometric function with a coefficient inside the angle. First, recall the basic integral: ∫ sin u du = −cos u + k where k is the constant of integration. Here, the angle is not just x, but 2x. So we must account for the derivative of 2x. Let: u = 2x Now differentiate u with respect to x. du/dx = 2 This means dx = du / 2. Now rewrite the integral in terms of u. ∫ sin 2x dx = ∫ sin u × (1/2) du The factor 1/2 comes from replacing dx with du/2. So we have: = 1/2 ∫ sin u du Now integrate sin u. ∫ sin u du = −cos u So: 1/2 × (−cos u) = −1/2 cos u Now substitute back u = 2x. = −1/2 cos 2x + k C

y = x² − 2x − 3 This is a quadratic function of the form y = ax² + bx + c, where a = 1, b = −2, and c = −3. Since the coefficient of x² is positive, the graph of this function opens upwards. This means the function has a **minimum value** at its turning point (vertex). The x-coordinate of the vertex of a quadratic function is given by: x = −b / (2a) Substitute the values of a and b. x = −(−2) / (2 × 1) x = 2 / 2 x = 1 Now substitute x = 1 into the original equation to find the minimum value of y. y = (1)² − 2(1) − 3 First evaluate each term. 1² = 1 −2(1) = −2 So: y = 1 − 2 − 3 y = −4 D

y = cos 3x We are required to find dy/dx, which means we differentiate y with respect to x. This is a **composite function** because the angle of the cosine is not just x, but 3x. So we must use the **chain rule**. First, recall the basic differentiation rule: If y = cos u, then dy/du = −sin u Here, u = 3x. So we proceed step by step. Differentiate y = cos 3x with respect to 3x first. dy/d(3x) = −sin 3x This comes from the derivative of cosine. Now differentiate 3x with respect to x. d(3x)/dx = 3 This is because the derivative of ax is a. Now multiply the two results together, according to the chain rule. dy/dx = −sin 3x × 3 dy/dx = −3 sin 3x D

y = 4x³ − 2x² + x We are asked to find dy/dx, which means we differentiate y with respect to x. Differentiate **each term separately**, using the basic rule: If y = axⁿ, then dy/dx = a × n × xⁿ⁻¹ First term: 4x³ Differentiate 4x³. The power of x is 3, so multiply 4 by 3 and reduce the power by 1. dy/dx of 4x³ = 4 × 3 × x² = 12x² Second term: −2x² Differentiate −2x². The power of x is 2, so multiply −2 by 2 and reduce the power by 1. dy/dx of −2x² = −2 × 2 × x = −4x Third term: +x Differentiate x. The power of x is 1, so multiply by 1 and reduce the power to 0. dy/dx of x = 1 Now combine all the differentiated terms. dy/dx = 12x² − 4x + 1 D

sin θ = 12/13 This means the sine of angle θ is the ratio of the opposite side to the hypotenuse in a right-angled triangle. So we take: Opposite side = 12 Hypotenuse = 13 Now find the adjacent side using Pythagoras’ theorem. This is because sin, cos, and tan are all related in a right-angled triangle. Hypotenuse² = Opposite² + Adjacent² 13² = 12² + Adjacent² 169 = 144 + Adjacent² Subtract 144 from both sides. Adjacent² = 25 Take the square root. Adjacent = 5 Now find cos θ. cos θ is defined as: Adjacent ÷ Hypotenuse cos θ = 5/13 Now evaluate the given expression: 1 + cos θ Substitute cos θ = 5/13. 1 + 5/13 Write 1 as a fraction with denominator 13. 1 = 13/13 So: 13/13 + 5/13 Add the numerators. 18/13 B

The given line is: 4x + 3y − 5 = 0 First, find the gradient of this line. Rewrite the equation to make y the subject. 3y = −4x + 5 This is done by moving 4x to the right side and keeping the constant. y = (−4/3)x + 5/3 So, the gradient of this line is −4/3. For two lines to be perpendicular, the product of their gradients must be −1. That is: m₁ × m₂ = −1 Here, m₁ = −4/3. Let m₂ be the gradient of the required line. (−4/3) × m₂ = −1 Now solve for m₂. m₂ = (−1) ÷ (−4/3) Dividing by a fraction means multiplying by its reciprocal. m₂ = (−1) × (−3/4) m₂ = 3/4 So, the gradient of the required line is 3/4. The line passes through the point (−2, 3). Use the point–slope form of a straight line: y − y₁ = m(x − x₁) Substitute the values. y − 3 = (3/4)(x − (−2)) x − (−2) becomes x + 2. So: y − 3 = (3/4)(x + 2) Now expand the bracket. y − 3 = (3/4)x + 3/2 Add 3 to both sides to make y the subject. y = (3/4)x + 3/2 + 3 Write 3 as a fraction with denominator 2. 3 = 6/2 So: y = (3/4)x + 9/2 Now remove fractions by multiplying every term by 4. 4y = 3x + 18 Rearrange into the general form. 3x − 4y + 18 = 0 A

The line is given as: y − 4x + 3 = 0 First, rewrite the equation to make y the subject. This is done so we can easily find the intercepts. y = 4x − 3 Here, y depends on x with gradient 4 and y-intercept −3. Now find the point where the line meets the **y-axis**. On the y-axis, x = 0. Substitute x = 0 into the equation. y = 4(0) − 3 y = −3 So, the y-intercept is the point: (0, −3) Next, find the point where the line meets the **x-axis**. On the x-axis, y = 0. Substitute y = 0 into the equation y = 4x − 3. 0 = 4x − 3 Add 3 to both sides. 3 = 4x Divide both sides by 4. x = 3/4 So, the x-intercept is the point: (3/4, 0) Now find the midpoint of the line segment joining (0, −3) and (3/4, 0). The midpoint formula is: Midpoint = ((x₁ + x₂)/2 , (y₁ + y₂)/2) Substitute the values. Midpoint x-coordinate: (0 + 3/4) ÷ 2 = 3/4 ÷ 2 = 3/8 Midpoint y-coordinate: (−3 + 0) ÷ 2 = −3/2 So, the midpoint is: (3/8, −3/2) A

The line passes through the points (0, 5) and (5, 0). We use the formula for the gradient of a straight line. Gradient = (y₂ − y₁) / (x₂ − x₁) This formula measures how much y changes compared to x. Let the first point be (0, 5). Let the second point be (5, 0). Substitute the values. Gradient = (0 − 5) / (5 − 0) First simplify the numerator. 0 − 5 = −5 Then simplify the denominator. 5 − 0 = 5 So the gradient becomes: Gradient = −5 / 5 Gradient = −1 Now use the equation of a straight line in the form: y = mx + c Here, m is the gradient and c is the y-intercept. From the question: Gradient m = −1 The line cuts the y-axis at (0, 5), so the y-intercept c = 5. Substitute these values into the equation. y = −1x + 5 This simplifies to: y = −x + 5 B

The gradient of a straight line joining two points is given by the formula gradient = (y₂ − y₁) / (x₂ − x₁) This formula compares the change in y-coordinates to the change in x-coordinates. The two points are: (x, 4) and (1, 2) Let: x₁ = x , y₁ = 4 x₂ = 1 , y₂ = 2 Substitute these values into the gradient formula. Gradient = (2 − 4) / (1 − x) First simplify the numerator. 2 − 4 = −2 So the gradient becomes: −2 / (1 − x) We are told that the gradient is equal to 1/2. So we equate the expressions. −2 / (1 − x) = 1/2 Now cross-multiply to remove the fractions. This means multiply −2 by 2, and 1/2 by (1 − x). −2 × 2 = 1 × (1 − x) −4 = 1 − x Now make x the subject of the equation. Subtract 1 from both sides. −4 − 1 = −x −5 = −x Multiply both sides by −1. x = 5 A

Points S(−5, 4) and T(−3, −2) are given. The midpoint of two points is found by taking the average of their x-coordinates and the average of their y-coordinates. First, write the midpoint formula. Midpoint = ((x₁ + x₂) / 2 , (y₁ + y₂) / 2) Here: x₁ = −5, y₁ = 4 x₂ = −3, y₂ = −2 Now substitute the values. Midpoint = ((−5 + (−3)) / 2 , (4 + (−2)) / 2) Simplify the x-coordinate. −5 + (−3) = −8 −8 ÷ 2 = −4 Now simplify the y-coordinate. 4 + (−2) = 2 2 ÷ 2 = 1 So the midpoint is: (−4, 1) C

A dog is tethered to a pole with a rope of length 4 m. This means the dog can move freely in all directions **as long as the distance from the pole does not exceed 4 m**. The pole acts as a fixed point. The rope acts as a fixed distance from that point. The locus of a point that moves such that it is always at a fixed distance from a fixed point is a **circle**. The fixed point is the **centre** of the circle. The fixed distance is the **radius** of the circle. Here, the fixed distance (length of the rope) is 4 m. So, the radius of the circle is 4 m. The dog can move all around the pole, not just on one side, so the shape is a **full circle**, not a semi-circle. Therefore, the locus is a **circle with radius 4 m**. B

Capacity of a cylindrical tank is the same as its volume. The formula for the volume of a cylinder is: Volume = πr²h Here, r is the radius of the base and h is the depth (height) of the tank. We are given: Volume = 6160 m³ Radius r = 28 m π = 22/7 Substitute these values into the formula. 6160 = (22/7) × 28² × h First, square the radius. 28² = 28 × 28 = 784 Now substitute this value. 6160 = (22/7) × 784 × h Next, simplify (22/7) × 784. Since 784 ÷ 7 = 112, we rewrite: (22/7) × 784 = 22 × 112 22 × 112 = 2464 So the equation becomes: 6160 = 2464h Now divide both sides by 2464 to make h the subject. h = 6160 ÷ 2464 h = 2.5 Therefore, the depth of the tank is: 2.5 m D

Radius of the circle = 10.5 cm Angle of the sector = 120° The perimeter of a sector consists of two straight radii and the curved arc. First, find the length of the arc. Arc length = (θ / 360°) × 2πr This formula is used because the arc is only a fraction of the full circle. Substitute the given values. Arc length = (120 / 360) × 2π × 10.5 Simplify the fraction. 120 / 360 = 1 / 3 So we have: Arc length = (1 / 3) × 2π × 10.5 Multiply 2 × 10.5. 2 × 10.5 = 21 So: Arc length = (1 / 3) × 21π Arc length = 7π Using π = 22 / 7, 7π = 7 × 22 / 7 = 22 cm Next, find the length of the two radii. Two radii = 2 × 10.5 = 21 cm Now add the arc length and the two radii to get the perimeter of the sector. Perimeter = 22 + 21 Perimeter = 43 cm B

The figure is a triangle with angles 30°, 60°, and the third angle at the top. First, find the third angle of the triangle. The sum of angles in a triangle is 180°. 30° + 60° + third angle = 180° Add the given angles. 30° + 60° = 90° So the third angle is: 180° − 90° = 90° This means the triangle is a right-angled triangle. Now identify the sides opposite each angle. The side marked **10 cm** is opposite the **30°** angle. The side marked **x** is opposite the **60°** angle. We now use the sine rule. The sine rule states that in any triangle: side₁ / sin(angle₁) = side₂ / sin(angle₂) So we write: x / sin 60° = 10 / sin 30° Now substitute the known sine values. sin 60° = √3 / 2 sin 30° = 1 / 2 So the equation becomes: x / (√3 / 2) = 10 / (1 / 2) Dividing by a fraction is the same as multiplying by its reciprocal. x × (2 / √3) = 10 × 2 Simplify the right-hand side. 10 × 2 = 20 So we have: x × (2 / √3) = 20 Now multiply both sides by √3 / 2 to make x the subject. x = 20 × (√3 / 2) Simplify. 20 ÷ 2 = 10 So: x = 10√3 10√3 cm

Two straight lines intersect, forming vertical and adjacent angles. From the figure, the angle marked q° and the angle marked 30° are **vertically opposite angles**. Vertically opposite angles are equal. So we write: q = 30 Next, look at the angle marked (p + 2q)°. This angle is **adjacent** to the 30° angle, and together they form a straight line. Angles on a straight line add up to 180°. So we write: (p + 2q) + 30 = 180 Now substitute q = 30 into the equation. This is done because we already found the value of q. p + 2(30) + 30 = 180 First multiply: 2(30) = 60 So the equation becomes: p + 60 + 30 = 180 Add the constants: 60 + 30 = 90 So: p + 90 = 180 Now subtract 90 from both sides to isolate p. p = 180 − 90 p = 90 B

KL ∥ NM This means any angle made with KL will have a corresponding equal angle made with NM, because alternate interior angles between parallel lines are equal. Angle KLN = 54° This angle is formed between KL and LN. Since KL is parallel to NM, the angle between LN and NM is also 54°. So, ∠LNM = 54° This is because LN acts as a transversal cutting the two parallel lines. LN bisects ∠KNM To bisect an angle means to divide it into two equal parts. So, ∠KNM is split into two equal angles by LN. Each half equals ∠LNM. Therefore, ∠KNM = 2 × 54° ∠KNM = 108° Now consider triangle KMN. We are given: ∠MKN = 35° This is the angle at K in triangle KMN. We have also found: ∠KNM = 108° This is the angle at N in triangle KMN. The sum of interior angles of a triangle is 180°. This means all three angles in triangle KMN must add up to 180°. So we write: ∠KMN + ∠MKN + ∠KNM = 180° Substitute the known values: ∠KMN + 35° + 108° = 180° First add the known angles. 35° + 108° = 143° So: ∠KMN + 143° = 180° Now subtract 143° from both sides. ∠KMN = 180° − 143° ∠KMN = 37° 37°

The interior angle of a regular polygon is given by the formula Interior angle = (n − 2) × 180° ÷ n Here, n represents the number of sides of the polygon. We are told that the interior angle is 135°, so substitute this value into the formula. 135 = (n − 2) × 180 ÷ n To remove the denominator n, multiply both sides by n. This is done to clear the fraction. 135n = (n − 2) × 180 Now expand the bracket on the right-hand side. (n − 2) × 180 = 180n − 360 So the equation becomes: 135n = 180n − 360 Next, collect like terms by bringing all n-terms to one side. Subtract 135n from both sides. 0 = 180n − 135n − 360 0 = 45n − 360 Now add 360 to both sides. 360 = 45n Divide both sides by 45 to make n the subject. n = 360 ÷ 45 n = 8 D

| 0 3 2 | | 1 7 8 | | 0 5 4 | We evaluate this 3 × 3 determinant using expansion along the **first column**. This column is chosen because it contains two zeros, which makes the calculation easier. The first column entries are 0, 1, and 0. Using cofactor expansion along the first column: Determinant = 0 × (minor of 0) − 1 × (minor of 1) + 0 × (minor of 0) The first and third terms are zero because they are multiplied by 0. So we focus only on the middle term. Determinant = −1 × | 3 2 |       | 5 4 | The minus sign comes from the position of the element 1, which is in row 2, column 1, and (−1)²⁺¹ = −1. Now evaluate the 2 × 2 determinant. | 3 2 | | 5 4 | For a 2 × 2 determinant, we use: ad − bc So we calculate: 3 × 4 − 2 × 5 3 × 4 = 12 2 × 5 = 10 12 − 10 = 2 Now return to the earlier step. Determinant = −1 × 2 = −2 D

The determinant of a 2 × 2 matrix | a b | | c d | is calculated using the rule: ad − bc This rule means multiply the numbers on the main diagonal, then subtract the product of the other diagonal. So we evaluate the determinant step by step. The matrix is: | −x 12 | | −1 4 | First, multiply the main diagonal terms. (−x) × 4 = −4x This is because −x multiplied by 4 gives −4x. Next, multiply the other diagonal terms. 12 × (−1) = −12 This is because 12 multiplied by −1 gives −12. Now subtract the second product from the first product. −4x − (−12) Subtracting a negative number is the same as adding the positive value. So we get: −4x + 12 The determinant is given as −12, so we equate: −4x + 12 = −12 Now subtract 12 from both sides to isolate the term containing x. −4x = −12 − 12 −4x = −24 Now divide both sides by −4. This step makes x the subject. x = (−24) ÷ (−4) x = 6 D

5x − 6y = 7 This equation comes from multiplying the first row of the matrix by the column vector, meaning 5 multiplied by x minus 6 multiplied by y gives 7. 2x − 7y = −11 This equation comes from multiplying the second row of the matrix by the column vector, meaning 2 multiplied by x minus 7 multiplied by y gives −11. We now solve the two equations simultaneously. First equation: 5x − 6y = 7 Second equation: 2x − 7y = −11 The aim is to eliminate one variable. We will eliminate x. Multiply the first equation by 2. This is done so that the coefficient of x becomes the same as in the second equation. 2(5x − 6y) = 2(7) 10x − 12y = 14 Multiply the second equation by 5. This is done so that the coefficient of x matches 10x. 5(2x − 7y) = 5(−11) 10x − 35y = −55 Now subtract the first new equation from the second new equation. This removes x because their coefficients are equal. (10x − 35y) − (10x − 12y) = −55 − 14 10x − 35y − 10x + 12y = −69 −23y = −69 Now divide both sides by −23. This isolates y. y = −69 ÷ −23 y = 3 C

x * y = xy This definition means the operation * is ordinary multiplication. So, x² means x * x, which equals x × x. We are given: x² = 12 − x First, bring all terms to one side to form a quadratic equation. This is done so that the equation equals zero. x² + x − 12 = 0 Now factorize the quadratic expression. We look for two numbers that multiply to −12 and add up to +1. These numbers are +4 and −3. So we write: (x + 4)(x − 3) = 0 Now solve each factor separately. x + 4 = 0 Subtract 4 from both sides. x = −4 x − 3 = 0 Add 3 to both sides. x = 3 So the possible values of x are −4 and 3. B

(√10 + √5), √10 + 2√5, … These are the first two terms of a geometric progression. In a geometric progression, the common ratio r is found by dividing the second term by the first term. So we write: r = (second term) ÷ (first term) r = (√10 + 2√5) ÷ (√10 + √5) To simplify this surd fraction, we factor where possible. First, factor √5 from the numerator. √10 = √5 × √2 So: √10 + 2√5 = √5√2 + 2√5 = √5(√2 + 2) Now factor √5 from the denominator as well. √10 + √5 = √5√2 + √5 = √5(√2 + 1) So the ratio becomes: r = [√5(√2 + 2)] ÷ [√5(√2 + 1)] Cancel √5 from the numerator and denominator, because the same non-zero factor appears in both. r = (√2 + 2) ÷ (√2 + 1) Now simplify the fraction by rationalizing the denominator. Multiply the numerator and denominator by the conjugate of (√2 + 1), which is (√2 − 1). r = (√2 + 2)(√2 − 1) ÷ [(√2 + 1)(√2 − 1)] First expand the denominator. (√2 + 1)(√2 − 1) is a difference of two squares. (√2)² − (1)² = 2 − 1 = 1 So the denominator becomes 1. Now expand the numerator. √2 × √2 = 2 √2 × (−1) = −√2 2 × √2 = 2√2 2 × (−1) = −2 Add all the terms: 2 − √2 + 2√2 − 2 Combine like terms. 2 − 2 = 0 −√2 + 2√2 = √2 So the numerator becomes √2. Since the denominator is 1: r = √2 A

The 4th term of an arithmetic progression is 13, and the 10th term is 31. In an arithmetic progression, the nth term is given by the formula: Tₙ = a + (n − 1)d where a is the first term and d is the common difference. Start by writing the expression for the 4th term. T₄ = a + (4 − 1)d T₄ = a + 3d We are told that the 4th term is 13, so: a + 3d = 13 Next, write the expression for the 10th term. T₁₀ = a + (10 − 1)d T₁₀ = a + 9d We are told that the 10th term is 31, so: a + 9d = 31 Now we have two equations: a + 3d = 13 a + 9d = 31 Subtract the first equation from the second to eliminate a. (a + 9d) − (a + 3d) = 31 − 13 a − a + 9d − 3d = 18 6d = 18 Now divide both sides by 6 to find d. d = 3 Substitute d = 3 into one of the earlier equations to find a. a + 3d = 13 a + 3(3) = 13 a + 9 = 13 Subtract 9 from both sides. a = 4 Now find the 24th term using the formula. T₂₄ = a + (24 − 1)d T₂₄ = 4 + 23 × 3 23 × 3 = 69 T₂₄ = 4 + 69 T₂₄ = 73 C

x/2 + 3/4 ≤ 5x/6 − 7/12 This is a linear inequality involving fractions, so the first step is to remove all fractions by using a common denominator. Look at the denominators: 2, 4, 6, and 12. The lowest common multiple of these numbers is 12. Multiply **every term** on both sides of the inequality by 12. This is done to clear all denominators while keeping the inequality balanced. 12 × (x/2) + 12 × (3/4) ≤ 12 × (5x/6) − 12 × (7/12) Now simplify each term carefully. 12 × (x/2) = 6x This is because 12 ÷ 2 = 6. 12 × (3/4) = 9 This is because 12 ÷ 4 = 3, and 3 × 3 = 9. 12 × (5x/6) = 10x This is because 12 ÷ 6 = 2, and 2 × 5x = 10x. 12 × (7/12) = 7 This is because 12 ÷ 12 = 1. So the inequality becomes: 6x + 9 ≤ 10x − 7 Next, collect like terms. Subtract 6x from both sides. This is done to bring all x-terms to one side. 9 ≤ 4x − 7 Now add 7 to both sides. This removes the constant on the right-hand side. 16 ≤ 4x Now divide both sides by 4. This isolates x. 4 ≤ x This can be written as: x ≥ 4 A

(x − 5)/(x + 3) < −1 This is an inequality involving a rational expression, so we must first note where the expression is undefined. The denominator is x + 3. Set x + 3 = 0 to find the restriction. x + 3 = 0 x = −3 So, x ≠ −3 because division by zero is not allowed. Next, remove the fraction by bringing all terms to one side. Add 1 to both sides of the inequality. (x − 5)/(x + 3) + 1 < 0 Write 1 with the same denominator (x + 3). 1 = (x + 3)/(x + 3) Now combine the fractions. (x − 5 + x + 3)/(x + 3) < 0 Simplify the numerator. x − 5 + x + 3 = 2x − 2 So the inequality becomes: (2x − 2)/(x + 3) < 0 Factor the numerator. 2x − 2 = 2(x − 1) So we have: 2(x − 1)/(x + 3) < 0 The number 2 is positive, so it does not affect whether the fraction is positive or negative. We can focus on: (x − 1)/(x + 3) < 0 A fraction is negative when the numerator and denominator have opposite signs. Find the critical values. From the numerator: x − 1 = 0 gives x = 1 From the denominator: x + 3 = 0 gives x = −3 These values divide the number line into three intervals. Check each interval. For x < −3: x − 1 is negative x + 3 is negative Negative ÷ negative = positive This does not satisfy the inequality. For −3 < x < 1: x − 1 is negative x + 3 is positive Negative ÷ positive = negative This satisfies the inequality. For x > 1: x − 1 is positive x + 3 is positive Positive ÷ positive = positive This does not satisfy the inequality. So the solution is: −3 < x < 1 A

P varies directly as Q and inversely as R. This means P is proportional to Q and also proportional to 1/R. So we write the relationship as: P = kQ / R Here, k is the constant of variation. We are given that when Q = 36, R = 16, P = 27. Substitute these values into the formula to find k. 27 = k × 36 / 16 First simplify the fraction. 36 / 16 = 9 / 4 This is done by dividing both numerator and denominator by 4. So the equation becomes: 27 = k × 9 / 4 Now remove the denominator by multiplying both sides by 4. This is done to isolate k. 27 × 4 = 9k 27 × 4 = 108 So: 108 = 9k Now divide both sides by 9. This step makes k the subject. k = 108 ÷ 9 k = 12 Now substitute k back into the general formula. P = 12Q / R B

y varies directly as w² This means y is proportional to the square of w, so we write the relationship as: y = kw² Here, k is the constant of variation, which does not change. We are told that when y = 8, w = 2. Substitute these values into the formula to find k. 8 = k(2)² This is done by replacing y with 8 and w with 2. 2² = 4 So the equation becomes: 8 = 4k Now divide both sides by 4 to make k the subject. k = 8 ÷ 4 k = 2 Now that the value of k is known, write the full formula. y = 2w² Next, find y when w = 3. Substitute w = 3 into the formula. y = 2(3)² 3² = 9 So the equation becomes: y = 2 × 9 2 × 9 = 18 A

y − 1 is a factor This means that when y = 1 is substituted into the expression, the result must be zero. This rule is called the factor theorem, and it is used to test factors of polynomials. Start with the given expression: y³ + 4y² + ky − 6 Substitute y = 1 into every term. This is done because y − 1 = 0 when y = 1. 1³ + 4(1)² + k(1) − 6 Now evaluate each part step by step. 1³ = 1 This is because 1 multiplied by itself three times is 1. 4(1)² = 4 × 1 = 4 This is because 1 squared is still 1. k(1) = k Multiplying by 1 does not change the value. Now substitute these values back: 1 + 4 + k − 6 Combine the constant numbers first. 1 + 4 = 5 5 − 6 = −1 So the expression becomes: k − 1 Because y − 1 is a factor, the result must equal zero. So we set: k − 1 = 0 Now solve for k by adding 1 to both sides. k = 1 D

2y² − 15xy + 18x² This is a quadratic expression involving y², xy, and x², and the goal is to write it as a product of two brackets. First, identify the coefficient of y² and the constant term in terms of x. The coefficient of y² is 2. The constant term is 18x². Next, multiply the coefficient of y² by the constant term. 2 × 18x² = 36x² This value helps us split the middle term. Now look for two terms that multiply to give +36x² and add to give −15x. These two terms are −12x and −3x. This is because −12x × −3x = +36x² and −12x + (−3x) = −15x. Now rewrite the middle term using these two terms. 2y² − 12xy − 3xy + 18x² Next, group the terms into two pairs. (2y² − 12xy) − (3xy − 18x²) Factor each group separately. From the first group: 2y² − 12xy = 2y(y − 6x) From the second group: 3xy − 18x² = 3x(y − 6x) Since the group is subtracted, it becomes −3x(y − 6x). Now write the expression with the common factor shown. 2y(y − 6x) − 3x(y − 6x) Factor out the common bracket (y − 6x). (2y − 3x)(y − 6x) B

gt² − k − w = 0 This equation shows that g multiplied by t², then minus k and minus w, gives zero. First, move the terms that do not contain g to the other side of the equation. This is done to isolate the term containing g. gt² = k + w Here, k and w are added together because they were subtracted on the left side and are moved to the right side as positives. Now g is multiplied by t², so we divide both sides by t². This step is done to make g stand alone as the subject of the formula. g = (k + w) / t² A

The shaded region shows parts that lie **inside P and Q together**, and also parts that lie **inside P and R together**. There is no shading outside P, and there is no shading that belongs to only Q and R without P. First, identify the shaded areas carefully. One shaded part is the overlap between P and Q. This region contains elements that are in P and in Q at the same time. That region is written as P ∩ Q. Another shaded part is the overlap between P and R. This region contains elements that are in P and in R at the same time. That region is written as P ∩ R. Both of these shaded regions are included together. When we combine two regions using “or”, we use the union symbol ∪. So, combining the two shaded regions gives: (P ∩ Q) ∪ (P ∩ R) A

P = {1, 2, 3, 4, 5} P ∪ Q = {1, 2, 3, 4, 5, 6, 7} The union of two sets contains **all elements that are in P, or in Q, or in both**. First, list all the elements already in set P. P contains the elements 1, 2, 3, 4, and 5. Now look at the elements in the union set P ∪ Q. The union contains 1, 2, 3, 4, 5, 6, and 7. Compare the two sets carefully. Elements 1, 2, 3, 4, and 5 are already present in P. These do not need to come from Q because they are already included. The elements that appear in the union but are **not** in P are: 6 and 7 These missing elements must have come from set Q, because the union includes everything from both sets. So, the elements in Q are: {6, 7} C

log₂8 + log₂16 − log₂4 These are logarithms to base 2, which means each logarithm tells us the power to which 2 must be raised to give the number. First, evaluate log₂8. 8 can be written as a power of 2. 8 = 2 × 2 × 2 = 2³ So, log₂8 = 3 This means 2 raised to the power 3 gives 8. Next, evaluate log₂16. 16 can also be written as a power of 2. 16 = 2 × 2 × 2 × 2 = 2⁴ So, log₂16 = 4 This means 2 raised to the power 4 gives 16. Next, evaluate log₂4. 4 = 2 × 2 = 2² So, log₂4 = 2 This means 2 raised to the power 2 gives 4. Now substitute all the evaluated values into the original expression. log₂8 + log₂16 − log₂4 = 3 + 4 − 2 First add 3 and 4. 3 + 4 = 7 Then subtract 2. 7 − 2 = 5 C

(2√2 − √3) ÷ (√2 + √3) This is a fraction involving surds in the denominator, so we must remove the surds from the denominator by rationalizing. To rationalize, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of √2 + √3 is √2 − √3. This is done because the product of a surd pair of the form (a + b)(a − b) gives a difference of squares. So we multiply: (2√2 − √3)(√2 − √3) ÷ (√2 + √3)(√2 − √3) First, expand the numerator step by step. 2√2 × √2 = 2 × 2 = 4 This is because √2 × √2 = 2. 2√2 × (−√3) = −2√6 This comes from multiplying the numbers and the surds. −√3 × √2 = −√6 This is another surd multiplication. −√3 × (−√3) = +3 This is because √3 × √3 = 3 and minus times minus is plus. Now add all the numerator terms together: 4 − 2√6 − √6 + 3 Combine the surd terms: −2√6 − √6 = −3√6 Combine the constant terms: 4 + 3 = 7 So the numerator becomes: 7 − 3√6 Next, expand the denominator. (√2 + √3)(√2 − √3) This is a difference of squares. √2 × √2 = 2 √3 × √3 = 3 So the denominator becomes: 2 − 3 = −1 Now divide the numerator by −1. (7 − 3√6) ÷ (−1) = −7 + 3√6 Rewriting in standard form gives: 3√6 − 7 A

8x² = 2/25 This equation shows that 8 multiplied by the square of x gives the fraction 2 over 25. First, isolate x² by dividing both sides by 8. This is done to remove the coefficient of x². x² = (2/25) ÷ 8 Dividing by 8 is the same as multiplying by 1/8. x² = 2/25 × 1/8 Now multiply the fractions. Multiply the numerators together and the denominators together. x² = 2 / (25 × 8) 25 × 8 = 200 So the denominator becomes 200. x² = 2/200 Now simplify the fraction by dividing both numerator and denominator by 2. x² = 1/100 Next, take the square root of both sides to remove the square on x. The square root is used because x² means x multiplied by itself. x = √(1/100) The square root of 1 is 1, and the square root of 100 is 10. x = 1/10 From the given options, this corresponds to **10**. D

2 log 75 + log 750 This expression involves logarithms to base 10, which is the standard meaning of “log” in UTME mathematics. First, express 75 and 750 in terms of 7.5, because the value of log 7.5 is given. 75 = 7.5 × 10 This is done by dividing 75 by 10 to get 7.5, then multiplying back by 10. 750 = 7.5 × 100 This is done by dividing 750 by 100 to get 7.5, then multiplying back by 100. Now apply the law of logarithms: log(ab) = log a + log b This law allows us to separate the product inside the logarithm into a sum. So, log 75 = log(7.5 × 10) log 75 = log 7.5 + log 10 log 10 = 1 This is because 10¹ = 10. Therefore, log 75 = 0.8751 + 1 log 75 = 1.8751 Now multiply by 2 as required. 2 log 75 = 2 × 1.8751 2 × 1.8751 = 3.7502 Next, evaluate log 750. log 750 = log(7.5 × 100) Using the same law: log 750 = log 7.5 + log 100 log 100 = 2 This is because 10² = 100. So, log 750 = 0.8751 + 2 log 750 = 2.8751 Now add the two results. 2 log 75 + log 750 = 3.7502 + 2.8751 3.7502 + 2.8751 = 6.6253 C

10% of the man’s monthly net earnings is given as ₦4,500. This means ₦4,500 represents 10 parts out of every 100 equal parts of his total income. First, write 10% as a fraction. 10% means 10 ÷ 100, which equals 1 ÷ 10. Now represent the situation mathematically. 1 ÷ 10 of the man’s net monthly income equals ₦4,500. This shows that when his total income is divided into 10 equal parts, one part is ₦4,500. To find the full income, multiply ₦4,500 by 10. This is because the total income has 10 equal parts. ₦4,500 × 10 = ₦45,000 B

0.00043 × 2000 This expression means we are multiplying a very small decimal number by a large whole number. First, write each number in standard (scientific) form. 0.00043 is written as 4.3 × 10⁻⁴ This is because the decimal point is moved 4 places to the right to get 4.3, and moving right gives a negative power of 10. 2000 is written as 2 × 10³ This is because 2000 = 2 multiplied by 1000, and 1000 equals 10³. Now multiply the numbers in standard form. (4.3 × 10⁻⁴) × (2 × 10³) First, multiply the ordinary numbers (called coefficients). 4.3 × 2 = 8.6 This step combines the numerical parts only. Next, multiply the powers of 10. 10⁻⁴ × 10³ = 10⁻¹ This is done by adding the indices: −4 + 3 = −1. Now combine both results. 8.6 × 10⁻¹ C

The original cost price is ₦60,000. This is the amount the woman paid to buy the grinder before selling it. A loss of 15% means she lost 15 out of every 100 of the cost price. Percentage loss is always calculated using the cost price as the base. First, convert 15% to a fraction of the cost price. 15% means 15 ÷ 100, which equals 0.15. Now calculate the amount of money lost. Loss = 0.15 × 60,000 This multiplication means we are finding 15% of ₦60,000. 0.15 × 60,000 = 9,000 So, the woman lost ₦9,000. To find the selling price, subtract the loss from the cost price. Selling price = 60,000 − 9,000 This subtraction removes the amount lost from the original cost. 60,000 − 9,000 = 51,000 C

110111₂ * 010100₂ This alignment is done by placing the two binary numbers one under the other so that digits with the same place value (units, twos, fours, etc.) are in the same column. Starting from the rightmost column (the units column): 1 + 0 = 1 This is because in base 2, adding zero does not change the value, so we write 1 and carry nothing. Next column (2¹ place): 1 + 0 = 1 Again, adding zero gives the same digit, so we write 1 and carry nothing. Next column (2² place): 1 + 1 = 10₂ This equals 2 in base 10, which is written as 0 in this column and 1 carried to the next column because base 2 allows only 0 or 1 in a single place. Next column (2³ place): 0 + 0 + 1 = 1 The two digits give 0, but the carried 1 from the previous step makes the total 1, so we write 1 and carry nothing. Next column (2⁴ place): 1 + 1 = 10₂ This again equals 2 in base 10, so we write 0 and carry 1 to the next column. Next column (2⁵ place): 1 + 0 + 1 = 10₂ The digits sum to 1, plus the carried 1 gives 2, so we write 0 and carry 1. Final carried value (2⁶ place): 1 This is written as the leftmost digit because there are no more digits to add. Putting all the written digits together from left to right gives: 1001011₂ C