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.