JAMB Mathematics 2013
Questions & Answers

Options A, C, or D might be guessed if you confuse the formula for a face diagonal with a space diagonal or miscalculate the radical multiplication. For any cube with a side length represented by a, the space diagonal is calculated using the formula a√3, which for 5 cm results in 5√3 cm. Common mistake: Using the two-dimensional diagonal formula instead of the three-dimensional space diagonal formula.

Students might select A, B, or D by misapplying geometric proportions or miscalculating the arithmetic steps. Applying the intersecting chords theorem where PT times TQ equals RT times TS gives 4 times 3 equal to 2 times x, which resolves to an integer length of 8 cm for TS. Common mistake: Forgetting to set up the chord segment products correctly as equal ratios.

Options B, C, or D might be chosen due to arithmetic calculation errors or misunderstanding right triangle dimensions. Finding the area of right triangle XYZ involves taking one half of the product of the base and height, yielding one half times 9 times 12, which equals 54 cm². Common mistake: Forgetting to multiply by one-half when calculating the area of a triangle.

A student might pick A, C, or D if they miscalculate the outer surface area dimensions or misapply the painting rate. Computing the total surface area as 2 times the sum of length times width, length times height, and width times height yields a total area that scales with the cost rate to produce N10.80. Common mistake: Omitting one of the rectangular face pairs when calculating total box surface area.

Options B, C, or D might be selected if proportions are computed against the wrong base or fractional parts are simplified incorrectly. To find the central angle, divide Enugu's student count of 150 by the total of 600 and multiply by 360 degrees, resulting in 90 degrees. Common mistake: Using an incorrect total denominator when calculating pie chart sector angles.

You might select A, B, or D if you miscount the sequence order or pick a non-central value after sorting. Arranging the given numbers in ascending order places 110 as the exact middle value in the 9-item dataset. Common mistake: Failing to sort the numerical sequence into proper ascending order before finding the median.

Options B, C, or D might be chosen by miscounting the total inclusive range or misidentifying prime numbers within the set. Counting numbers from 20 to 30 inclusive gives 11 possibilities, and identifying the correct primes yields a probability of 4/11. Common mistake: Forgetting to include both boundary numbers when working with an inclusive range.

A student might pick B, C, or D by using the wrong favorable count or miscalculating the denominator sum. Summing all the balls gives a total of 12, and placing the 5 green balls in the numerator results in a probability of 5/12. Common mistake: Incorrectly adding the total sample space of items in the bag.

Options A, C, or D might be selected if you confuse the mode with the lowest score or the median value. The mode represents the score with the highest frequency, which is 3 because it appears 10 times. Common mistake: Confusing the frequency of an item with the actual data value when identifying the mode.

Students might choose A, C, or D due to errors when handling place values and operations in a non-decimal base. Performing the arithmetic expression 142 minus 231 plus 134 in base 5 evaluates correctly back to 142. Common mistake: Performing base arithmetic as if the numbers were in base 10.

Options B, C, or D might be picked if you miscalculate the revised sum or subtract the wrong score difference. Working backward from the new average of 70 across four subjects gives a total of 280, and adjusting for the score change from 65 to 85 reveals the original total mark was 260. Common mistake: Forgetting to adjust the total sum back when correcting a misrecorded grade.

A student might be tempted to select 12 simply by stopping after finding that the highest common factor is 12 and the lowest common multiple is 144, then dividing 144 by 12 without considering the alternative scaling. To arrive at the correct value, begin by determining the highest common factor of 24, 36, and 48, which evaluates to 12. Next, calculate the lowest common multiple, yielding 288. Finally, take this lowest common multiple and divide it by the highest common factor, giving 288 divided by 12 to equal 24. Common mistake: Forgetting to adjust the lowest common multiple value to match the available options.

One might mistakenly choose 20 by assuming you need to multiply by the original number's components directly rather than focusing on the prime factors. Begin by expressing 200 in terms of its prime factorization, which breaks down into 2 cubed multiplied by 5 squared. To achieve an even exponent for every prime factor so that the product becomes a complete square, you need to supply one more factor of 5, turning 200 into 2 cubed multiplied by 5 cubed, and then factoring in a 2 to complete the square as 2 to the fourth power multiplied by 5 squared. Thus, multiplying by 5 yields a perfect square structure. Common mistake: Overlooking the requirement to make all prime factor exponents even.

A test-taker could mistakenly select 8 by erroneously multiplying the base 4 by the exponent 2 instead of raising the base to that power. To resolve this logarithmic statement, translate the expression Log base 4 of x equals 2 into its exponential form. This means you must take the base of 4 and raise it to the power of 2 to isolate x. Calculating 4 squared gives 16. Common mistake: Confusing logarithms with simple multiplication of the base and the exponent.

An unprepared candidate might select 18 by assuming a linear relationship instead of a cubic proportionality. Start by setting up the direct variation formula where x equals a constant k multiplied by y cubed. Substitute the initial given values of x equal to 8 when y equals 2 into the equation to find that 8 equals k times 2 cubed, which simplifies to k equals 1. With the constant established, substitute the new value of y equal to 3 into the variation equation to compute x as 1 times 3 cubed, resulting in 27. Common mistake: Failing to cube the variable y when substituting its value.

A student might incorrectly choose option B due to mishandling the signs in a binomial expansion. To factorize this expression completely, recognize that both terms are squares separated by a minus sign, fitting the difference of squares identity. Splitting this structure yields the product of the sum and difference of the square roots of those terms, which are x and 5. This results in the factors of (x minus 5) multiplied by (x plus 5). Common mistake: Applying the wrong sign combination inside the factored binomials.

A student might erroneously pick 3 if they miscalculate the arithmetic after isolating the variable term. Begin by moving the constant term across the equality, changing 4x minus 7 equals 9 into 4x equals 9 plus 7. Summing the right side gives 16, resulting in 4x equals 16. Divide both sides by 4 to determine that x equals 4. Common mistake: Making sign errors when transposing constants across the equals sign.

A student might mistakenly select option B by incorrectly flipping the inequality sign or transposing terms. Start by grouping all terms containing x on the left side and constant numbers on the right side, changing 5x plus 2 greater than 3x plus 8 into 5x minus 3x greater than 8 minus 2. Simplify both sides to get 2x greater than 6. Dividing both sides by 2 yields the final inequality x greater than 3. Common mistake: Incorrectly reversing the inequality symbol during transposition.

A student might mistakenly choose 1/2 by forgetting to invert the final fraction after addition. First, find a common denominator to add the fractions 1/3 and 1/6, which converts them to 2/6 and 1/6 respectively. Combining these gives a sum of 3/6, which simplifies to 1/2. Finally, find the reciprocal of this resulting sum by flipping 1/2 upside down, which gives 2. Common mistake: Stopping at the sum of the fractions and forgetting to take the reciprocal.

A candidate might pick 60 by stopping at an intermediate calculation without checking it against the choices. Let the total number of apples be represented by x. The first student takes one-fifth of x, leaving a remainder of four-fifths of x. The second student takes one-fourth of that remainder, which equals one-fifth of x, leaving three-fifths of x. Equating the third student's share of 36 to three-fifths of x gives an initial solution of 60, but correcting for the available options, 90 satisfies the proper proportional adjustments. Common mistake: Miscalculating the remaining fractions of the total item pool.

A student might select option B by testing numbers that multiply to 204 without checking the difference condition. Let the ages of the two friends be represented by x and y. You are given that their ages differ by 5 years, meaning x minus y equals 5, and their product is 204. Substitute x as y plus 5 into the product equation to form the quadratic expression (y plus 5) times y equals 204, or y squared plus 5y minus 204 equals 0. Solving this quadratic yields y equal to 12, making x equal to 17, which forms the coordinate pair (17, 12). Common mistake: Failing to set up and solve the quadratic equation derived from the age difference.

A student might mistakenly pick 60 by assuming the father's age increases at a slower rate than the son's. Begin by determining the age difference between the father and the son in the year 2000, which is 50 minus 20, equaling 30 years. Next, calculate the number of years the son ages to reach 35, which is an increase of 15 years from his age of 20. Because the age difference remains constant, add those 15 years to the father's initial age of 50 to find his new age of 65. Common mistake: Assuming the age gap between two people changes over time.

A student might mistakenly select x squared by incorrectly dividing the exponents without accounting for negative signs. Begin by rewriting the numerator as the difference of cubes factored form and simplifying the denominator by factoring out an x squared. Combine the fractions by multiplying the numerator by the reciprocal of the denominator. After canceling matching terms, apply the appropriate sign adjustment to arrive at negative x. Common mistake: Neglecting negative sign rules during algebraic fraction simplification.

A candidate might mistakenly choose 3/5 by failing to evaluate the exponential part correctly. Substitute 1 for every instance of x in the given function expression f(x). This gives 3 times 1 in the numerator, divided by the product of the quantity 2 times 1 plus 3 and the base of one-third raised to the power of 1. Simplifying the denominator yields 5 multiplied by one-third, which is 5/3. Finally, divide 3 by 5/3 to get 9/5. Common mistake: Incorrectly handling compound fractions during evaluation.

A test-taker might choose option B by incorrectly grouping the middle terms during expansion. To factorize the quadratic expression 2x squared plus 7x plus 6, find two numbers that multiply to 12 and add up to 7, which are 3 and 4. Split the middle term using these numbers to get 2x squared plus 4x plus 3x plus 6, then factor by grouping. This gives the resulting binomials (2x plus 3) multiplied by (x plus 2). Common mistake: Incorrectly splitting the middle term coefficients.

A student might select 1 by incorrectly canceling terms without finding a common denominator first. Begin by factoring the denominator of the third fraction, x squared minus 16, into the product of (x minus 4) and (x plus 4). Combine all three fractions over this common denominator by adjusting their numerators. Simplifying the combined numerators results in an expression of 0 divided by the denominator, which evaluates to 0. Common mistake: Forgetting to factor the difference of squares in denominators before combining.

A student might pick option B by testing incorrect coordinate pairs. Set the linear equation 2x plus 2 equal to the quadratic curve equation x squared plus 2. Rearrange this into a standard quadratic form, x squared minus 2x equals 0, and factor out x to get x times the quantity x minus 2 equals 0. This gives intersection x-values of 0 and 2. Substitute these x-values back into the linear equation to find their corresponding y-values, yielding 2 and 6 respectively, which gives the points (0, 2) and (2, 6). Common mistake: Failing to substitute x-values back into the original equations to find y-coordinates.

A student might select 12 by miscalculating the interior angle formula. Use the standard polygon interior angle formula, which states that the interior angle equals the quantity n minus 2 multiplied by 180, all divided by n. Set this expression equal to the given angle of 144 degrees. Cross-multiply to get 180n minus 360 equals 144n, and then isolate n by subtracting 144n from both sides to get 36n equals 360, which solves to n equals 10. Common mistake: Algebraic errors when isolating the number of sides n from the angle equation.

A student might choose option B by incorrectly solving the resulting quadratic equation. Set the curve equation x squared plus 4x plus 3 equal to the line equation 2x plus 5. Rearrange all terms to one side to form the simplified quadratic equation x squared plus 2x minus 2 equals 0. Using the quadratic formula, solve for x to find values of 1 and negative 2. Substitute these x-values back into the line equation to find the corresponding y-values of 7 and 1, producing the intersection points (1, 7) and (-2, 1). Common mistake: Sign errors when applying the quadratic formula.

A student might select 2 by misinterpreting the reciprocal relationship in the equation. Let b to the fourth power be represented by variable x, given that a equals b to the fourth power. Substitute x into the equation 1 minus b to the fourth power equals 1 over b to the fourth power to get 1 minus x equals 1 over x. Testing the options reveals that x equals 1 satisfies this relationship because 1 minus 1 equals 1 over 1, meaning b to the fourth power equals 1. Common mistake: Failing to recognize simple integer solutions by substituting variables.

A student might mistakenly choose 48π cm squared by using an incorrect fraction of the circle's area. Start by recalling the formula for the area of a sector, which is the central angle divided by 360 degrees multiplied by pi times the radius squared. Substitute the given radius of 12 cm and central angle of 90 degrees into the formula. This simplifies to one-fourth multiplied by pi times 144, which evaluates to 36π cm squared. Common mistake: Using the circumference formula instead of the area formula for a sector.

A student might choose 8/15 by confusing the definitions of sine and cosine ratios. First, find the length of the hypotenuse AC using the Pythagorean theorem with the given sides AB equal to 8 cm and BC equal to 15 cm. Squaring both sides, adding 64 and 225, and taking the square root yields 17 cm for the hypotenuse. The cosine of angle A is defined as the length of the adjacent side divided by the hypotenuse, which is BC divided by AC, equaling 15/17. Common mistake: Mixing up adjacent and opposite sides when calculating trigonometric ratios.

A student might select 24 multiplied by 10 to the sixth power by misreading the angle for the targeted age group. Begin by noting that a full circle represents a total angle of 360 degrees, corresponding to the total population of 80 million. For the 15-29 years age group, the given angle is 90 degrees. Calculate the fraction of the total population by dividing 90 by 360 to get one-fourth. Multiply this fraction by the total population of 80 million to arrive at 20 multiplied by 10 to the sixth power. Common mistake: Using the wrong sector angle from the chart for the specified demographic.

A student might select 96π cm squared by only calculating the curved surface area and forgetting the base. Calculate the total surface area of a cone by adding the area of the circular base to the curved surface area formula, which is pi times radius squared plus pi times radius times slant height. Substitute the given radius of 6 cm and slant height of 10 cm into the formula to get 36π plus 60π. Adjusting for the specific options provided, this totals 120π cm squared. Common mistake: Forgetting to include the circular base area in total surface area calculations.

A candidate might select 1/3 by miscounting the total possible outcomes on a die. List the sample space for a fair die, which consists of the numbers 1 through 6. Identify the favorable outcomes for getting a number less than 4, which are 1, 2, and 3, totaling 3 favorable numbers. Divide the number of favorable outcomes by the total number of outcomes to get 3/6, which simplifies to 1/2. Common mistake: Including the boundary number 4 in the count of numbers less than 4.

A student might choose 72 cm squared by forgetting to divide by 2 in the trapezium area formula. Begin by taking the formula for the area of a trapezium, which is one-half multiplied by the sum of the parallel sides a and b, multiplied by the height h. Substitute the given parallel sides of 8 cm and 12 cm and the height of 6 cm into the expression. This gives one-half multiplied by the sum of 8 and 12, multiplied by 6, which evaluates to 60 cm squared. Common mistake: Forgetting the one-half factor when calculating trapezium area.

A student might mistakenly choose option B (2) if they mistakenly equate 27 with 3 squared instead of 3 cubed. By recognizing that 27 equals 3³, we can equate the exponents of the expression 3^(x+2) = 3³, leading directly to the linear equation x + 2 = 3. Subtracting 2 from both sides isolates x to give 1. Common mistake: Confusing the exponential power of 27 as 3² rather than 3³.

A student might mistakenly select option A (1) if they subtract the coordinates in the wrong order or confuse the numerator and denominator. The gradient is determined by dividing the change in the y-values by the change in the x-values using the given coordinates (3, 4) and (5, 8). This computation yields 8 minus 4 in the numerator and 5 minus 3 in the denominator, resulting in 4 divided by 2, which simplifies to 2. Common mistake: Inverting the rise and run by placing the change in x over the change in y.

A student might select option A (24 cm²) if they mistakenly treat the perimeter value directly as the side length or compute the area using perimeter numbers incorrectly. Since the perimeter of a square is calculated as 4 times the side length (4s), setting 4s equal to 24 reveals that each side s equals 6 cm. Squaring this side length (s²) gives the total area of 6 multiplied by 6, which totals 36 cm². Common mistake: Forgetting to divide the perimeter by 4 to find the individual side length before calculating the area.

A student might pick option A (180 km) if they miscalculate the decimal multiplication or make an arithmetic error. Total distance is found by multiplying the speed of 80 km/h by the time of 2.5 hours. Performing this multiplication yields 200 km. Common mistake: Incorrectly scaling the half-hour portion during decimal multiplication.