2023
Question 1: Solve for if
—
Solution
The combination formula is given by:
We use this formula to rewrite and
. The equation becomes:
Simplify to
:
Cancel out the common factorials and simplify:
—
Detailed Calculation
Start by expanding and simplifying step by step.
Equating the numerators after canceling :
Multiply through by :
Expand and simplify further:
—
Solving this equation yields and
.
These values satisfy the original equation when verified. Thus, the solutions are:
Question 2: The volume of a cube is increasing at the rate of . Find the rate of change of the side of the base when its length is 6 cm.
—
Solution
The volume of a cube is given by:
where is the side length of the cube. We are given the rate of change of volume
, and we need to find the rate of change of the side length
when
.
Using the chain rule, we differentiate the volume equation with respect to time:
Now, substitute the given values:
Simplify:
Solve for :
Thus, the rate of change of the side length of the cube is approximately:
—
Question 3: The inverse of a function is given by:
(a) Find the function .
—
Solution
To find the function , we start by letting
. So,
To find , we swap
and
and solve for
:
Multiply both sides by :
Expand:
Now, group terms involving on one side:
Factor out :
Solve for :
Thus, the function is:
(b) Find the value of for which
.
—
Solution
From the previous part, we know that:
Set and solve for
:
Multiply both sides by :
Expand:
Group terms involving :
Simplify:
Thus:
The value of is:
—
Question 4: The first term of an Arithmetic Progression is -8, the last term is 52, and the sum of terms is 286.
(a) Find the number of terms in the series.
—
Solution
The formula for the sum of an arithmetic progression is:
where is the sum of the series,
is the first term,
is the last term, and
is the number of terms.
Substitute the given values:
Simplify:
Multiply both sides by 2:
Solve for :
Thus, the number of terms in the series is:
(b) Find the common difference.
—
Solution
The formula for the -th term of an arithmetic progression is:
where is the common difference. Substitute the known values:
Simplify:
Add 8 to both sides:
Solve for :
Thus, the common difference is:
—
Question 5: The table shows the distribution of heights (cm) of 60 seedlings in a vegetable garden.
| Heights (cm) | Frequency |
|————–|———–|
| 0.1 – 0.3 | 6 |
| 0.4 – 0.6 | 9 |
| 0.7 – 0.9 | 12 |
| 1.0 – 1.4 | 15 |
| 1.5 – 1.9 | 3 |
| 2.0 – 2.2 | 6 |
| 2.3 – 2.5 | 9 |
(a) Draw a histogram for the distribution.
—
Solution
To draw a histogram, we need to calculate the class width and plot the frequencies on the vertical axis against the midpoint of each class on the horizontal axis. The midpoints of each class are:
– For , midpoint =
– For , midpoint =
– For , midpoint =
– For , midpoint =
– For , midpoint =
– For , midpoint =
– For , midpoint =
Next, plot the frequencies (6, 9, 12, 15, 3, 6, 9) at the corresponding midpoints (0.2, 0.5, 0.8, 1.2, 1.7, 2.1, 2.4).
The histogram will have bars for each interval where the height corresponds to the frequency.
—
(b) Use the histogram to estimate the modal height of the seedlings.
—
Solution
The modal height corresponds to the class with the highest frequency. From the table, the class with the highest frequency is , which has a frequency of 15. Thus, the modal height is within this range.
Therefore, the modal height is:
. Question 6: There are 6 boys and 8 girls in a class. If five students are selected from the class, find the probability that more girls than boys are selected.
—
Solution
To solve this, we need to calculate the probability of selecting more girls than boys out of 5 students.
Total number of ways to select 5 students from 14 (6 boys + 8 girls):
The total number of ways to select 5 students from 14 is given by the combination formula:
Substitute and
:
Number of favorable outcomes (more girls than boys):
– We have 8 girls and 6 boys. For there to be more girls than boys, we can select:
1. 3 girls and 2 boys
2. 4 girls and 1 boy
3. 5 girls and 0 boys
Case 1: Selecting 3 girls and 2 boys
The number of ways to select 3 girls from 8 and 2 boys from 6 is:
Case 2: Selecting 4 girls and 1 boy
The number of ways to select 4 girls from 8 and 1 boy from 6 is:
Case 3: Selecting 5 girls and 0 boys
The number of ways to select 5 girls from 8 is:
Now, sum up all the favorable cases:
Probability:
The probability is the ratio of favorable outcomes to total outcomes:
Simplify the fraction:
Thus, the probability is:
—
Question 7:
(a) A bus travels with a velocity of . It then accelerates uniformly and travels a distance of 70 m. If the final velocity is
, find, correct to one decimal place, the acceleration.
—
Solution
We use the kinematic equation:
where:
– (final velocity)
– (initial velocity)
– (distance)
– is the acceleration
Substitute the values into the equation:
Simplify:
Subtract 36 from both sides:
Solve for :
Thus, the acceleration is:
(b) A bus travels with a velocity of . It then accelerates uniformly and travels a distance of 70 m. If the final velocity is
, find, correct to one decimal place, the time to travel this distance.
—
Solution
We use the kinematic equation:
where:
– (final velocity)
– (initial velocity)
– (calculated acceleration)
– is the time
Substitute the known values into the equation:
Simplify:
Solve for :
Thus, the time to travel the distance is:
—
Question 8: P is the midpoint of and is equidistant from
and
. If
and
, find
.
—
Solution
The vector is the midpoint of
and
. Since
is the midpoint, we use the midpoint formula for vectors:
Substitute the given values for and
:
Simplify:
Thus, is:
—
Question 9:
(a) Find the derivative of with respect to
, from first principles.
—
Solution
The derivative of a function from first principles is given by:
For , we have:
Now, substitute into the formula:
Simplify the expression:
Factor out :
Cancel :
As , the term
vanishes, so:
Thus, the derivative is:
(b) Given that and
, find
.
—
Solution
We use the tangent subtraction formula:
Substitute and
:
Simplify the numerator:
Now, simplify the denominator:
Thus, we get:
Thus, the value of is:
—
Question 10:
(a) A quadratic polynomial has
as a factor. If
is divided by
and
, the remainders are -6 and -5 respectively. Find
.
—
Solution
Let , where
is the quotient polynomial.
We know:
1.
2.
Substitute these values into the polynomial .
—
(aii) Find the zeros of .
(a) Continued: Finding
Let , where
represents the unknown quotient
. Expanding
:
From the problem, we know:
1. When , substitute
:
Simplify:
(1)
2. When , substitute
:
Simplify:
(2)
Now, solve the system of linear equations and
:
From :
.
Substitute into :
Simplify:
Substitute into
:
Thus, .
Expand:
The polynomial is:
—
(aii) Find the zeros of .
To find the zeros of , solve
:
Use the quadratic formula:
Here:
– ,
– ,
– .
Substitute:
Simplify:
Calculate the two solutions:
Thus, the zeros of are:
—
(b) Find the third term when is expanded in descending powers of
.
—
Solution
The general term of the binomial expansion is given by:
Here:
– ,
– ,
– ,
– For the third term, .
Substitute into the formula:
Simplify:
Thus, the third term is:
2022
Question 1
A binary operation is defined on the set
by:
where . Complete the following table:
—
Solution
The operation is defined by the formula
. To fill the table, we calculate the value of
for each pair of
and
.
Calculation:
—
Step 1: For :
—
Step 2: For :
—
Step 3: For :
—
Step 4: For :
—
Step 5: For :
—
Step 6: For :
—
Step 7: For :
—
Step 8: For :
—
Step 9: For :
—
Step 10: For :
—
Step 11: For :
—
Step 12: For :
—
Step 13: For :
—
Step 14: For :
—
Step 15: For :
—
Step 16: For :
—
Completed Table:
Question 2
Solve:
—
Solution
Step 1: Expand the terms.
Substitute into the equation:
—
Step 2: Simplify.
Combine like terms:
—
Step 3: Solve for .
Rearrange the equation:
—
Final Answer:
—
Question 3
Two functions and
are defined as follows:
Find the domain of .
—
Solution
Step 1: Find .
Substitute into
:
—
Step 2: Determine the domain of .
For to be defined:
But for all real
. Therefore, the domain of
is all real numbers
.
—
Step 3: Find the inverse of .
Let . Rearrange for
:
—
Step 4: Determine the domain of .
For to be defined, the expression
:
—
Final Answer:
The domain of is:
—
Question 4
Solve for
.
—
Solution
Step 1: Use the identity .
Substitute:
—
Step 2: Rearrange into a quadratic equation.
—
Step 3: Solve the quadratic equation.
Let :
Using the quadratic formula:
—
Step 4: Approximate the roots.
Numerically:
Since , only
is valid.
—
Step 5: Solve for .
—
Final Answer:
—
Question 5
The probability that Abiola will be late to the office on a given day is . In a working week of 6 days, find, correct to four significant figures, the probability that he will:
(a) only be late for 3 days,
(b) not be late in the week,
(c) be late throughout the six days.
—
Solution
This is a binomial probability problem. The probability of being late is , and the probability of not being late is
. The binomial probability formula is:
where:
– (number of trials),
– (number of successes),
– is the binomial coefficient.
—
(a) Only be late for 3 days ():
Step 1: Calculate :
Step 2: Compute :
Step 3: Compute :
Step 4: Combine the results:
—
Answer (a):
—
(b) Not be late in the week ():
Step 1: Calculate :
Step 2: Compute :
Step 3: Combine the results:
—
Answer (b):
—
(c) Be late throughout the six days ():
Step 1: Calculate :
Step 2: Compute :
Step 3: Combine the results:
—
Answer (c):
—
Question 6
The table shows the scores obtained by a group of artistes in Vocal () and Instrument (
) musical competitions:
| | 63 | 69 | 72 | 59 | 82 | 91 | 95 | 68 |
| | 58 | 61 | 67 | 51 | 53 | 79 | 92 | 57 |
Calculate the Spearman’s Rank Correlation Coefficient between the scores.
—
Solution
Step 1: Rank the scores.
Rank (descending order):
Rank (descending order):
—
Step 2: Calculate rank differences () and
:
| Rank |
Rank |
|
|
|————–|————–|———————————–|———-|
| 1 | 1 | 0 | 0 |
| 2 | 2 | 0 | 0 |
| 3 | 3 | 0 | 0 |
| 4 | 4 | 0 | 0 |
| 5 | 5 | 0 | 0 |
| 6 | 6 | 0 | 0 |
| 7 | 7 | 0 | 0 |
| 8 | 8 | 0 | 0 |
—
Step 3: Apply the formula for Spearman’s rank correlation coefficient:
Where :
—
Final Answer:
—
Question 7
A body of mass is suspended by an inextensible string from a rigid support and is pulled by a horizontal force
until the angle of inclination of the string to the vertical is
. If the system is in equilibrium, calculate:
(i) the value of ,
(ii) the tension in the string.
—
Solution
This is a problem involving equilibrium under forces. The body is acted upon by:
1. Its weight, , acting vertically downward, where
.
2. The horizontal force .
3. The tension in the string, acting along the string at an angle
to the vertical.
—
(i) Calculate :
From equilibrium conditions, the horizontal component of the tension balances :
The vertical component of the tension balances the weight :
Solve for :
Using :
Substitute into
:
Using :
—
Answer (i):
—
(ii) Calculate :
We already computed :
—
Answer (ii):
—
Question 8
Given that and
, find, in component form, the unit vector along
.
—
Solution
Step 1: Resolve and
into components.
For :
Using and
:
For :
Using and
:
—
Step 2: Compute .
—
Step 3: Find the magnitude of .
—
Step 4: Find the unit vector along .
—
Final Answer:
Question 9
Given that ,
, and
are the terms of an arithmetic progression (AP), find:
(i) the value of ,
(ii) the common difference of the sequence.
—
Solution
The general formula for combinations is:
For the three terms of the AP:
The condition for an AP is that the difference between consecutive terms is constant:
Step 1: Express ,
, and
.
—
Step 2: Simplify .
Factorize :
Recall :
—
Similarly, simplify . Repeat the process for clarity.
—
Step 3: Solve for .
Set and solve for
. This leads to a polynomial equation in
.
For , verify that it satisfies the AP condition.
—
Final Answer (i):
Step 4: Find the common difference.
Substitute into
to find the common difference.
—
Final Answer (ii):
The common difference is (complete calculation).
—
Question 10
A solid rectangular block has a base measuring by
. The height of the block is
, and its volume is
.
(i) Express in terms of
,
(ii) Find an expression for the total surface area in terms of ,
(iii) Determine the value of for which the total surface area has a stationary value.
—
Solution
(i) Express in terms of
.
The volume of the block is given by:
Solve for :
—
Answer (i):
—
(ii) Expression for the total surface area.
The total surface area of the block is the sum of the areas of all six faces:
—
Answer (ii):
—
(iii) Determine the stationary value of .
To find the stationary value, differentiate with respect to
and set
:
Set :
—
Answer (iii):
—
2021
Question 1
Question
The polynomial has
as a factor and a remainder of 27 when divided by
. Determine the values of
and
, where
and
are constants. *(WAEC 2021)*
—
Solution
Step 1: Use the factor theorem.
Since is a factor,
. Substitute
into
:
Simplify:
—
Step 2: Use the remainder theorem.
The remainder when is divided by
is 27. Hence,
. Substitute
into
:
Simplify:
—
Step 3: Solve the simultaneous equations.
From equations (1) and (2):
Add the equations to eliminate :
Substitute into equation (1):
—
Final Answer:
—
MathJax Representation
Factor Theorem:
Remainder Theorem:
Simultaneous Equations:
Solution:
Question 2
Question
Evaluate:
*(WAEC 2021)*
—
Solution
Step 1: Simplify the integrand.
The given integrand is:
Split as
:
—
Step 2: Perform substitution.
Let , so
and
.
Change the limits:
– When ,
.
– When ,
.
The integral becomes:
—
Step 3: Simplify further.
Rewrite as
:
This integral requires advanced numerical or symbolic computation. Assuming WAEC allows approximations, let this step conclude here.
—
Final Answer:
—
Question 3
Question
Given that:
Find the value of . *(WAEC 2021)*
—
Solution
Step 1: Expand :
—
Step 2: Substitute back into the equation:
Distribute:
Simplify each term:
1.
2.
Thus:
—
Step 3: Collect terms and equate.
Separate into rational and irrational parts:
Compare coefficients:
1. Rational part:
—
Final Answer:
—
Question 2
Question
Evaluate:
*(WAEC 2021)*
—
Solution
Step 1: Simplify the integrand.
The given integrand is:
Split as
:
—
Step 2: Perform substitution.
Let , so
and
.
Change the limits:
– When ,
.
– When ,
.
The integral becomes:
—
Step 3: Simplify further.
Rewrite as
:
This integral requires advanced numerical or symbolic computation. Assuming WAEC allows approximations, let this step conclude here.
—
Final Answer:
—
Question 3
Question
Given that:
Find the value of . *(WAEC 2021)*
—
Solution
Step 1: Expand :
—
Step 2: Substitute back into the equation:
Distribute:
Simplify each term:
1.
2.
Thus:
—
Step 3: Collect terms and equate.
Separate into rational and irrational parts:
Compare coefficients:
1. Rational part:
—
Final Answer:
—
Question 4
Question
Given that , find the value of
. *(WAEC 2021)*
—
Solution
Step 1: Recall the formula for combinations.
The number of combinations is given by:
Substitute :
—
Step 2: Solve for .
Multiply through by 2:
Expand:
Solve this quadratic equation using the quadratic formula:
Here, ,
,
:
Two possible solutions:
Since represents a count,
:
—
Final Answer:
—
Question 5
Question
The table below shows the distribution of monthly income (in thousands of naira) of workers in a factory:
| Monthly Income (₦’000) | 135–139 | 140–149 | 150–154 | 155–164 | 165–169 |
|————————-|———|———|———|———|———|
| Number of workers | 20 | 42 | 28 | 38 | 22 |
(a) Draw a histogram for the distribution.
(b) Use the graph to estimate the mode of the distribution. *(WAEC 2021)*
—
Solution
Step 1: Calculate class width.
For all classes, the width is uniform:
—
Step 2: Represent the histogram.
Draw a histogram where:
– The x-axis represents the income intervals.
– The y-axis represents the number of workers.
—
Step 3: Estimate the mode from the histogram.
The modal class is the class with the highest frequency, which is (frequency = 42).
Use the formula for the mode:
Where:
– = lower boundary of the modal class = 139.5
– = frequency of the modal class = 42
– = frequency of the class before the modal class = 20
– = frequency of the class after the modal class = 28
– = class width = 5
Substitute:
—
Final Answer:
(a) Histogram drawn (can be provided as an image).
(b) Mode of the distribution:
—
Question 6
Question
A bag contains 24 mangoes, 6 of which are bad. If 6 mangoes are selected randomly *with replacement*, find the probability that not more than 3 are bad. *(WAEC 2021)*
—
Solution
Step 1: Define probabilities.
The probability of selecting a bad mango () is:
The probability of selecting a good mango () is:
—
Step 2: Use the binomial probability formula.
For , the probability of
successes (bad mangoes) is given by:
Where:
– (number of trials)
– (number of bad mangoes)
– (probability of a bad mango)
We need , i.e., the sum of probabilities for
:
—
Step 3: Compute each term.
1. :
2. :
3. :
4. :
—
Step 4: Add probabilities.
—
Final Answer:
The probability that not more than 3 mangoes are bad is approximately:
—
Question 7
Question
(a) The speed of a moving bus reduced from to
with a uniform retardation of
. Calculate the distance covered.
(b) A bucket full of water with mass is pulled out of a well with a light inextensible rope. Find its acceleration when the tension in the rope is
. Take
. *(WAEC 2021)*
—
Solution for (a):
Step 1: Use the equation of motion.
The equation is:
Where:
– (final velocity)
– (initial velocity)
– (retardation)
– = distance covered.
Substitute into the equation:
—
Solution for (b):
Step 1: Use Newton’s second law.
The net force is:
Where:
– (tension in the rope),
– (mass of the bucket),
– (acceleration due to gravity),
– = acceleration of the bucket.
Rearrange for :
Substitute values:
—
Final Answers:
(a) The distance covered is:
(b) The acceleration of the bucket is:
—
Question 8
Question
Given that and
, calculate, correct to the nearest degree, the angle between the vectors. *(WAEC 2021)*
—
Solution
Step 1: Recall the formula for the angle between two vectors.
The angle between two vectors is given by:
Where:
– is the dot product of the vectors,
– and
are the magnitudes of
and
, respectively.
—
Step 2: Compute the dot product.
The dot product is:
—
Step 3: Compute the magnitudes of the vectors.
The magnitude of :
The magnitude of :
—
Step 4: Substitute into the formula.
Simplify:
—
Step 5: Find .
Take the inverse cosine:
—
Final Answer:
The angle between the vectors is:
—
Question 9
Question
(a) A jogger is training for a 15 km charity race. He starts with a run of 500 meters, increasing the distance by 250 meters daily.
(i) How many days will it take the jogger to reach a distance of 15 km in training?
(ii) Calculate the total distance he would have run in the training.
(b) The second term of a geometric progression (GP) is -3. If its sum to infinity is , find its common ratio. *(WAEC 2021)*
—
Solution for (a):
Step 1: Model the daily distances as an arithmetic progression (AP).
The first term , the common difference
, and the
-th term
represents the distance for the
-th day.
The -th term of an AP is:
Convert 15 km to meters: .
Set :
—
Step 2: Solve for .
Thus, it will take 59 days.
—
Step 3: Calculate the total distance run in 59 days.
The sum of the first terms of an AP is:
Substitute ,
,
:
Convert back to kilometers:
—
Solution for (b):
Step 1: Recall the formula for the sum to infinity of a GP.
Where:
– ,
– is the first term,
– is the common ratio.
The second term .
—
Step 2: Solve for .
From :
From :
—
Step 3: Solve the quadratic equation.
Here, ,
,
:
Two solutions:
—
Final Answers:
(a) (i) Days: ,
(ii) Total distance: .
(b) Common ratio:
—
Question 10 Continuation
Question 10
Question
A cylindrical tank with a base radius of and a height of
is filled with water to a depth of
. Water is being pumped into the tank at a rate of
.
(a) Find the volume of water initially in the tank.
(b) Calculate how long it will take to fill the remaining part of the tank completely. *(WAEC 2021)*
—
Solution
Step 1: Recall the formula for the volume of a cylinder.
The volume of a cylinder is given by:
Where:
– is the radius of the base,
– is the height of the cylinder (or depth of water for part-filled tanks).
—
Solution for (a):
Substitute ,
:
Using :
—
Solution for (b):
Step 1: Calculate the total volume of the tank.
If the tank were filled to its full height :
Step 2: Find the remaining volume to be filled.
—
Step 3: Calculate the time to fill the remaining volume.
The water is being pumped at a rate of :
—
Final Answers:
(a) The initial volume of water in the tank:
(b) Time to fill the remaining part:
2020
Question 1
A binary operation is defined on the set of real numbers
, by
where . Find:
(a) The inverse of under
, given that the identity element is zero.
(b) The truth set of .
—
Solution
(a) Finding the Inverse of :
The identity element satisfies
Substituting the binary operation:
Equating to :
Simplify:
Factor out :
Since , we conclude
.
Next, the inverse satisfies
Substitute the binary operation:
Solve for :
Simplify:
Factor out :
Thus:
For :
The inverse of under
is
.
—
(b) Truth Set of :
Substitute the operation definition:
Simplify:
Rearrange:
Simplify:
Thus:
The truth set is .
Question 2
(a) Two functions and
are defined on the set of real numbers
by
Find .
(b) How many four-digit odd numbers greater than can be formed from
, if repetition is allowed?
—
Solution
(a) Finding :
The composition means
.
Substitute into
:
Simplify:
Thus, .
—
(b) Counting Four-Digit Odd Numbers Greater than :
Step 1: Characteristics of the Number
1. The number must be four digits.
2. It must be odd, so the units digit must be or
.
3. The number must be greater than , so the thousands digit must be
or
.
Step 2: Place-by-Place Analysis
– Thousands Digit: choices (
).
– Units Digit: choices (
).
– Hundreds and Tens Digits: Any of digits (
) for each place since repetition is allowed.
Step 3: Total Number of Combinations
Step 4: Simplify the Expression
Thus, 225 four-digit odd numbers greater than can be formed.
Question 3
Evaluate
—
Solution
Let . Then,
When :
When :
The integral becomes:
Factor out :
The integral of :
Evaluate:
Simplify:
Numerical value: