JAMB UTME - Mathematics - 1995
Tambaya 1 Rahoto
Find the distance between two towns p(45\(^{o}\)N, 30\(^{o}\)W) and Q(15\(^{o}\)S, 30\(^{o}\)W) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)]
Tambaya 2 Rahoto
The length of a notebook 15cm, was measured as 16.8cm. Calculate the percentage error to 2 significant figures.
Bayanin Amsa
Tambaya 3 Rahoto
If S = (x : x2 = 9, x > 4), then S is equal to
Tambaya 5 Rahoto
Find the area bounded by the curve \(y = 3x^2 - 2x + 1\), the coordinates \(x = 1\) and \(x = 3\) and the x-axis
Bayanin Amsa
Tambaya 6 Rahoto
In a triangle XYZ, < YXZ = 44° and < XYZ = 112°. Calculate the acute angle between the internal bisectors of < XYZ and < XZY.
Bayanin Amsa
Tambaya 7 Rahoto
Find the sum to infinity of the following sequence 1. \( \frac{9}{10} \), 2. (\( \frac{9}{10} \)), 3. (\( \frac{9}{10} \))
Tambaya 8 Rahoto
In the diagram, find PQ if the area of triangle PQR is 35cm2
Bayanin Amsa
To find the length of PQ, we need to use the formula for the area of a triangle, which is: Area = 1/2 * base * height In this case, the area is given as 35cm^2, and we know that the height is PQ. We also know that the base is QR, since the height is perpendicular to the base. So, we can write: 35 = 1/2 * QR * PQ Multiplying both sides by 2 and dividing by PQ, we get: 70/PQ = QR We don't have the value of QR, but we can use the Pythagorean theorem to find it. In triangle PQR, we have: PR^2 = PQ^2 + QR^2 We don't know the value of PR, but we can express it in terms of PQ and QR using the fact that PQR is a right-angled triangle. We have: PR = sqrt(PQ^2 + QR^2) Substituting this expression into the Pythagorean theorem, we get: PQ^2 + QR^2 = (sqrt(PQ^2 + QR^2))^2 Simplifying, we get: PQ^2 + QR^2 = PQ^2 + QR^2 This is an identity, which means it is true for all values of PQ and QR. But we can use it to eliminate QR from the equation we got earlier: 70/PQ = QR Multiplying both sides by QR, we get: 70 = PQ^2 + QR^2 Substituting QR^2 = 70 - PQ^2, we get: 70/PQ = sqrt(70 - PQ^2) Squaring both sides, we get: 4900/PQ^2 = 70 - PQ^2 Rearranging, we get: PQ^4 - 70PQ^2 + 4900 = 0 This is a quadratic equation in PQ^2, which we can solve using the quadratic formula: PQ^2 = (70 ± sqrt(70^2 - 4*1*4900))/2 PQ^2 = (70 ± 20sqrt(3))/2 We can discard the negative solution, since PQ is a length and can't be negative. So, we have: PQ^2 = 35 + 10sqrt(3) Taking the square root, we get: PQ ≈ 7.98 So, the closest option is 8cm, which corresponds to the first option, "7cm".
Tambaya 9 Rahoto
Express the product of 0.0014 and 0.011 in standard form
Bayanin Amsa
To express the product of 0.0014 and 0.011 in standard form, we need to multiply these two numbers together and then express the result in the form of a x 10^n, where 1 ≤ a < 10 and n is an integer. 0.0014 multiplied by 0.011 gives us 0.0000154. We can then express this in standard form by moving the decimal point to the right until there is only one non-zero digit to the left of the decimal point. This gives us 1.54 x 10^-5, which is option D. Therefore, the answer is: - 1.54 x 10-5
Tambaya 11 Rahoto
The variance of the scores 1, 2, 3, 4, 5 is
Bayanin Amsa
To calculate the variance of a set of numbers, you first need to find the mean (average) of the set. The mean of 1, 2, 3, 4, 5 is: (1 + 2 + 3 + 4 + 5) ÷ 5 = 3 Next, subtract the mean from each number in the set and square the differences. This gives you the squared deviations from the mean: (1 - 3)^2 = 4 (2 - 3)^2 = 1 (3 - 3)^2 = 0 (4 - 3)^2 = 1 (5 - 3)^2 = 4 The sum of the squared deviations is: 4 + 1 + 0 + 1 + 4 = 10 Finally, divide the sum of the squared deviations by the number of items in the set: 10 ÷ 5 = 2 Therefore, the variance of the scores 1, 2, 3, 4, 5 is 2. Therefore, the option that represents the variance of the scores 1, 2, 3, 4, 5 is 2.05.
Tambaya 12 Rahoto
The derivative of cosec x is
Bayanin Amsa
The derivative of cosec x is -cot x cosec x. Cosec x is the reciprocal of sin x, so we can rewrite it as (1/sin x). Using the quotient rule of differentiation, we can find the derivative of cosec x as follows: d/dx (cosec x) = d/dx (1/sin x) = (-1/sin^2x) * d/dx (sin x) Now, using the chain rule, we can find the derivative of sin x: d/dx (sin x) = cos x Substituting this back into our original expression, we get: d/dx (cosec x) = (-1/sin^2x) * cos x Simplifying this, we get: d/dx (cosec x) = -cot x cosec x
Tambaya 13 Rahoto
Express \( \frac{5x-12}{(x-2)(x-3)} \) in partial fractions
Bayanin Amsa
5x−12(x−2)(x−3)=Ax−2+Bx−3
= A(x−3)+B(x−2)(x−2)(x−3)
⟹5x−12=Ax−3A+Bx−2B
A+B=5...(i)
−(3A+2B)=−12⟹3A+2B=12...(ii)
From (i), A=5−B
3(5−B)+2B=12
15−3B+2B=12⟹B=3
A+3=5⟹A=2
5x−12(x−2)(x−3)=2x−2+3x−3
Tambaya 14 Rahoto
Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR
Tambaya 15 Rahoto
Four members of a social first eleven cricket team are also members of the first fourteen rugby team. How many boys play for at least one of the two teams?
Bayanin Amsa
If there are 11 members in the cricket team and 4 of them also play rugby, that means there are 7 cricket players who do not play rugby. Similarly, there are 14 members in the rugby team and only 4 of them play cricket, which means there are 10 rugby players who do not play cricket. To find the total number of boys who play at least one of the two sports, we need to add the number of cricket players who don't play rugby to the number of rugby players who don't play cricket, and then subtract the overlap (i.e., the 4 boys who play both sports) since we don't want to count them twice. So, total number of boys who play at least one of the two sports = (number of cricket players who don't play rugby) + (number of rugby players who don't play cricket) - (number of boys who play both sports) = (7) + (10) - (4) = 13 Therefore, there are 13 boys who play at least one of the two sports. Hence, the correct answer is option (B) 21 is incorrect.
Tambaya 16 Rahoto
Let \(p\) be a probability function on set \(S\), where \(S = (a_1, a_2, a_3, a_4)\). Find \(P(a_1)\) if \(P(a_2) = \frac{1}{3}\), \(p(a_3) = \frac{1}{6}\) and \(p(a_4) = \frac{1}{5}\)
Bayanin Amsa
Tambaya 17 Rahoto
For what value of x is the tangent to the curve \(y = x^2 - 4x + 3\) parallel to the x-axis?
Bayanin Amsa
To find the value of x for which the tangent to the curve y = x^2 - 4x + 3 is parallel to the x-axis, we need to find the point on the curve where the slope of the tangent line is zero. The slope of the tangent to a curve y = f(x) at a point (a, f(a)) is given by f'(a), the derivative of f(x) evaluated at x = a. So, we need to find the derivative of y = x^2 - 4x + 3, which is y' = 2x - 4. Setting y' equal to zero, we get: 2x - 4 = 0 Solving for x, we get: x = 2 Therefore, the value of x for which the tangent to the curve y = x^2 - 4x + 3 is parallel to the x-axis is 2.
Tambaya 18 Rahoto
Finds a positive value of p if the expression 2x2 - px + p leaves a remainder 6 when divided by x - p and q
Bayanin Amsa
x - p,x = p
2p2 - p2 + p = 6
= p2 + p - 6
= 0
p = 3, 2
Tambaya 19 Rahoto
The mean and the range of the set of numbers 1.20, 1.00, 0.90, 1.40, 0.80, 1.20, and 1.10 are m and r respectively. Find m + r
Bayanin Amsa
To find the mean of a set of numbers, we add up all the numbers in the set and divide by the total number of numbers. To find the range, we subtract the smallest number from the largest number in the set. In this case, we have the set of numbers: 1.20, 1.00, 0.90, 1.40, 0.80, 1.20, and 1.10. To find the mean, we add up all the numbers and divide by 7 (the total number of numbers in the set): mean = (1.20 + 1.00 + 0.90 + 1.40 + 0.80 + 1.20 + 1.10) / 7 = 1.14 To find the range, we subtract the smallest number (0.80) from the largest number (1.40): range = 1.40 - 0.80 = 0.60 Therefore, m + r = 1.14 + 0.60 = 1.74. So the correct answer is 1.69.
Tambaya 20 Rahoto
A worker's present salary is ₦24,000 per annum. His annual increment is 10% of his basic salary at the beginning of the third year?
Bayanin Amsa
The worker's annual increment is 10% of his basic salary at the beginning of the third year. This means that at the beginning of the third year, his salary will increase by 10% of his present salary. The third year's salary will be: = 24,000 + 10% of 24,000 = 24,000 + 2,400 = 26,400 Therefore, at the beginning of the third year, the worker's basic salary will be ₦26,400 per annum. Now, we need to calculate the worker's salary after the annual increment at the beginning of the third year. Salary after increment = Basic salary at the beginning of the third year + 10% of basic salary at the beginning of the third year = ₦26,400 + 10% of ₦26,400 = ₦26,400 + ₦2,640 = ₦29,040 Therefore, the worker's salary after the annual increment at the beginning of the third year is ₦29,040 per annum. Hence, the answer is ₦29,040.
Tambaya 21 Rahoto
Solve the inequality \( (x - 3)(x - 4) \le 0 \)
Bayanin Amsa
To solve this inequality, we need to find the values of x that make the expression (x - 3)(x - 4) less than or equal to 0. To do this, we can use a method called the "sign chart" or "interval notation". First, we can find the critical values of x by setting the expression equal to 0: (x - 3)(x - 4) = 0. This gives us x = 3 and x = 4. Next, we can create a sign chart by dividing the number line into intervals using the critical values of x. We then choose a test point from each interval and evaluate the expression (x - 3)(x - 4) using that test point. If the result is positive, the expression is positive in that interval. If the result is negative, the expression is negative in that interval. Intervals: (-∞, 3), (3, 4), (4, ∞) Test point: -1, 3.5, 5 Expression: (-)(-)=+ (+)(-)= - (+)(+)=+ From the sign chart, we can see that the expression is less than or equal to 0 when x is between 3 and 4, inclusive. Therefore, the solution is: 3 ≤ x ≤ 4 So the correct option is: - 3 ≤ x ≤ 4
Tambaya 22 Rahoto
Find T in terms of K, Q and S if S = 2r\( \pi \)QT + K)
Bayanin Amsa
s24r2
= QTπ
+ KT
s24r2
- kπ
= QTπ
T = s24Qπr2 - k
Tambaya 25 Rahoto
The 4th term of an A.P is 13 while the 10th term is 31. Find the 21st term
Bayanin Amsa
a + 3d = 13
a + 9d = 31
6d = 18
= d = 3
a = 13 - 9
= 4
a + 20d = 4 + (20 x 3)
= 64
Tambaya 28 Rahoto
The pie chart above shows the distribution of students in a secondary school class. If 30 students offered French, how many offered CRK?
Bayanin Amsa
Tambaya 29 Rahoto
Two variables x and y are such that \( \frac{dy}{dx} = 4x - 3 \) and y = 5 when x = 2. Find y in terms of x
Bayanin Amsa
∫dy = ∫(4x - 3)dx, y = 2x2 - 3x + C
when y = 5, x = 2, C = 3
y = 2x2 - 3x + 3
Tambaya 30 Rahoto
| Class Interval | Frequency | Class boundaries | Class Mid−point |
| \(1.5-1.9\) | \(2\) | \(1.45-1.95\) | \(1.7\) |
| \(2.0-2.4\) | \(21\) | \(1.95-2.45\) | \(2.2\) |
| \(2.5-2.9\) | \(4\) | \(2.45-2.95\) | \(2.7\) |
| \(3.0-2.9\) | \(15\) | \(2.95-3.45\) | \(3.2\) |
| \(3.5-3.9\) | \(10\) | \(3.45-3.95\) | \(3.7\) |
| \(4.0-4.4\) | \(5\) | \(3.95-4.45\) | \(4.2\) |
| \(4.5-4.9\) | \(3\) | \(4.45-4.95\) | \(4.7\) |
Find the mode of the distribution above to find the mode of the distribution.
Bayanin Amsa
Mode = a + (b - a)(fm - Fb)
2Fm - Fa - Fb
= 3.0 + (3.4?3)(15?4)2(15)?4?10
= 3 + (6.4)(11)30?14
= 3 + 4.416
= 3 + 0.275
= 3.275
= 3.3cm
Tambaya 31 Rahoto
| Class Interval | Frequency | Class boundaries | Class Mid-point |
| \(1.5 - 1.9\) | \(2\) | \(1.45 - 1.95\) | \(1.7\) |
| \(2.0 - 2.4\) | \(21\) | \(1.95 - 2.45\) | \(2.2\) |
| \(2.5 - 2.9\) | \(4\) | \(2.45 - 2.95\) | \(2.7\) |
| \(3.0 - 2.9\) | \(15\) | \(2.95 - 3.45\) | \(3.2\) |
| \(3.5 - 3.9\) | \(10\) | \(3.45 - 3.95\) | \(3.7\) |
| \(4.0 - 4.4\) | \(5\) | \(3.95 - 4.45\) | \(4.2\) |
| \(4.5 - 4.9\) | \(3\) | \(4.45 - 4.95\) | \(4.7\) |
The median of the distribution above is
Bayanin Amsa
Median = a+bfm
(12∑f−CFb
)
= 2.95 + 0.515
(2.-7)
= 2.95 + 0.515
x 13
= 2.95 + 0. 43
= 3.38
= 3.4
Tambaya 32 Rahoto
A die has four of it's faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive time, the top face will be white in both cases?
Bayanin Amsa
The probability of getting a white face on a single roll of the die is 4/6 or 2/3, since four of the six faces are white. Similarly, the probability of getting a white face on a second roll of the die is also 2/3. To find the probability of both events occurring (getting a white face on both rolls), we multiply the probabilities of each event together. So, the probability of getting a white face on the first roll AND getting a white face on the second roll is (2/3) x (2/3) = 4/9. Therefore, the correct answer is not provided in the options given.
Tambaya 33 Rahoto
Which of the following binary operations is cumulative in the set of integers?
Bayanin Amsa
Tambaya 34 Rahoto
The graph of f(x) = x2 - 5x + 6 crosses the x-axis at the points
Bayanin Amsa
To find where the graph of the function crosses the x-axis, we need to find the values of x when f(x) = 0. So we can set the function equal to 0 and solve for x: x^2 - 5x + 6 = 0 We can factor the quadratic equation as: (x - 2)(x - 3) = 0 So the solutions are x = 2 and x = 3. Therefore, the graph of the function crosses the x-axis at the points (2, 0) and (3, 0). Therefore, the correct option is: - (2, 0), (3, 0)
Tambaya 36 Rahoto
If \(x - 1\) and \(x + 1\) are both factors of the equation \(x^3 + px^2 + qx + 6 = 0\), evaluate \(p\) and \(q\)
Bayanin Amsa
Tambaya 37 Rahoto
In the diagram, the base diameter is 14cm while the height is 12cm. Calculate the total surface area if the cylinder has both a base and a top.[\( \pi \frac{22}{7} \)]
Bayanin Amsa
To calculate the total surface area of the cylinder with both a base and a top, we need to find the area of the two bases and the lateral surface area. The area of one circular base is given by the formula πr², where r is the radius of the base. The diameter of the base is given as 14cm, so the radius is half of that, which is 7cm. Therefore, the area of one base is π(7cm)² = 49πcm². Since the cylinder has both a base and a top, the total area of the two bases is 2 × 49πcm² = 98πcm². The lateral surface area of the cylinder is given by the formula 2πrh, where r is the radius of the base and h is the height of the cylinder. Using the values given, we have r = 7cm and h = 12cm. Therefore, the lateral surface area is 2π(7cm)(12cm) = 168πcm². To get the total surface area, we add the area of the two bases to the lateral surface area. Thus, the total surface area of the cylinder is 98πcm² + 168πcm² = 266πcm². Now we need to simplify this expression, using the approximate value of π given to us as 22/7. So, 266πcm² ≈ 266 × (22/7) cm² ≈ 836cm². Therefore, the total surface area of the cylinder is approximately 836cm². Answer is the correct choice.
Tambaya 38 Rahoto
A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank.
Bayanin Amsa
To solve this problem, we can use trigonometry. Let h be the height of the water tank, and d be the horizontal distance between the foot of the water tank tower and the position of the boy. Then, we have: tan(60°) = h / d We know that d = 30m, and tan(60°) = √3, so we can solve for h: h = d * tan(60°) h = 30m * √3 h = 30√3 m Therefore, the height of the water tank is 30√3 meters.
Tambaya 39 Rahoto
If \(a \ast b = +\sqrt{ab}\), evaluate \(2 \ast (12 \ast 27)\)
Bayanin Amsa
We can start by evaluating 12 ∗ ∗ 27 using the given formula: 12 ∗ ∗ 27 = + √(12 x 27) = + √(324) = 18 Then we can substitute this value in 2 ∗ ∗ (12 ∗ ∗ 27) as follows: 2 ∗ ∗ (12 ∗ ∗ 27) = 2 ∗ ∗ (18) = + √(2 x 18) = + √36 = +6 or -6 Therefore, the answer is either 6 or -6, depending on whether the sign is positive or negative, which is not specified in the question.
Tambaya 40 Rahoto
| Age in years | 13 | 14 | 15 | 16 | 17 |
| No. of students | 3 | 10 | 30 | 42 | 15 |
The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is
Bayanin Amsa
To find the angle corresponding to the 15-year-olds in the pie chart, we need to first calculate the total number of students in the school. From the frequency distribution, we can see that there are: 3 + 10 + 30 + 42 + 15 = 100 students in total. Next, we need to find the fraction of students that are 15 years old. From the frequency distribution, we can see that there are 30 students who are 15 years old. Therefore, the fraction of students that are 15 years old is: 30/100 = 0.3 To convert this fraction to an angle, we need to multiply it by 360 degrees (the total angle in a pie chart). Therefore, the angle corresponding to the 15-year-olds in the pie chart is: 0.3 x 360 = 108 degrees Therefore, the answer is (D) 108o.
Tambaya 42 Rahoto
\[ \begin{vmatrix} -2 & 1 & 1 \\ 2 & 1 & k \\ 1 & 3 & -1 \end{vmatrix} = 23 \]
Bayanin Amsa
To solve the absolute value equation ∣∣ ∣∣−21121k13−1∣∣ ∣∣ = 23, we need to consider two cases: when the expression inside the absolute value bars is positive and when it is negative. Case 1: When the expression inside the absolute value bars is positive: −21121k13−1 = 23 Simplifying the equation, we get: −21121k13 = 24 Multiplying both sides by -1, we have: 21121k13 = -24 Multiplying both sides by 13, we have: 21121k = -312 Therefore, k = -312/21121 Case 2: When the expression inside the absolute value bars is negative: −(−21121k13−1) = 23 Simplifying the equation, we get: 21121k13−1 = 23 Adding 1 to both sides, we have: 21121k13 = 24 This is the same equation we obtained in Case 1. Therefore, the value of k is the same in both cases: k = -312/21121 Therefore, the answer is 2.
Tambaya 43 Rahoto
Convert 3.1415926 to 5 decimal places
Bayanin Amsa
To convert a number to a specific number of decimal places, we need to look at the digit in the next decimal place and round the original number accordingly. In this case, the digit in the sixth decimal place is 2, which is less than 5. Therefore, we do not need to round up the fifth decimal place, which is 9. Thus, the correct answer is 3.14159, which is the same as the original number but rounded to 5 decimal places.
Tambaya 44 Rahoto
Use the graph of the curve \(y = f(x)\)to solve the inequality \(f(x) \le 0\)
Bayanin Amsa
-1 ≥ x ≥ 1, 1 < x ≥ 2
Combining solutions
= x ≤ 1; 1 ≥ x ≥ 2
Tambaya 45 Rahoto
If \(x = \begin{pmatrix}1 & 2 \\ 0 & 3\end{pmatrix}\) and \(y = \begin{pmatrix}2 & 1 \\ 4 & 3\end{pmatrix}\). Find \(xy\).
Bayanin Amsa
To find xy, we need to perform matrix multiplication of x and y. To do this, we take the dot product of each row of x with each column of y, as follows: First row of x: (1 2 0 3) First column of y: ( 2 1 4 3 ) Dot product of the first row of x with the first column of y: 1*2 + 2*1 + 0*4 + 3*3 = 2 + 2 + 9 = 13 This gives us the first element of the product matrix. We repeat this process for the remaining rows and columns to get: xy = ( 1*2 + 2*1 + 0*4 + 3*3 1*1 + 2*4 + 0*3 + 3*2 1*4 + 2*3 + 0*2 + 3*1 1*3 + 2*2 + 0*1 + 3*4 0*2 + 3*1 + 1*4 + 2*3 0*1 + 3*4 + 1*3 + 2*2 0*4 + 3*3 + 1*2 + 2*1 0*3 + 3*2 + 1*1 + 2*4 ) = ( 13 11 10 17 11 19 13 14 ) Hence, xy = (107129). Therefore, the correct option is (A).
Tambaya 46 Rahoto
PT is a tangent to the circle TYZX. YT = YX and < PTX = 50o. Calculate < TZY
Tambaya 47 Rahoto
Bayanin Amsa
< PYU = < YPQ = 40°(Alternative) = PQ∥
UV
< XPQ = < YPQ = 40°, Since QX = QY
therefore < XPY = 80°
