JAMB UTME - Mathematics - 1995

$\left|\begin{array}{ccc}-2& 1& 1\\ 2& 1& k\\ 1& 3& -1\end{array}\right|$ = 23

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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.

The length of a notebook 15cm, was measured as 16.8cm. Calculate the percentage error to 2 significant figures.

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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?

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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.

$\begin{array}{cccccc}\text{Age in years}& 13& 14& 15& 16& 17\\ \text{No. of students}& 3& 10& 30& 42& 15\end{array}$

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

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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.

PT is a tangent to the circle TYZX. YT = YX and < PTX = 50o. Calculate < TZY

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For what value of x is the tangent to the curve y = x${}^2$ - 4x + 3 parallel to the x-axis?

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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.

Find the distance between two towns p(45o N, 30oW) and Q(15oS, 30oW) if the radius of the earth is 7000km. [$\pi =\frac{22}{7}$]

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Express 5x−12(x−2)(x−3) $\frac{5x-12}{(x-2)(x-3)}$ in partial fractions

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5x−12(x−2)(x−3)=Ax−2+Bx−3 $\frac{5x-12}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$

= A(x−3)+B(x−2)(x−2)(x−3) $\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}$

⟹5x−12=Ax−3A+Bx−2B $⟹5x-12=Ax-3A+Bx-2B$

A+B=5...(i) $A+B=5...(i)$

−(3A+2B)=−12⟹3A+2B=12...(ii) $-(3A+2B)=-12⟹3A+2B=12...(ii)$

From (i), A=5−B $A=5-B$

3(5−B)+2B=12 $3(5-B)+2B=12$

15−3B+2B=12⟹B=3 $15-3B+2B=12⟹B=3$

A+3=5⟹A=2 $A+3=5⟹A=2$

5x−12(x−2)(x−3)=2x−2+3x−3

The pie chart above shows the distribution of students in a secondary school class. If 30 students offered French, how many offered CRK?

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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

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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?

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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.

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m = 1.05, r = 0.6

m + r = 1.05 + 0.5

= 1.65

The graph of f(x) = x2 - 5x + 6 crosses the x-axis at the points

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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)

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.

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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.

Express the product of 0.0014 and 0.011 in standard form

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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

Finds a positive value of p if the expression 2x2 - px + p leaves a remainder 6 when divided by x - p and q

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x - p,x = p
2p2 - p2 + p = 6
= p2 + p - 6
= 0
p = 3, 2

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

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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.

Which of the following binary operations is cumulative in the set of integers?

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In the diagram, find PQ if the area of triangle PQR is 35cm2

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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".

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< PYU = < YPQ = 40°(Alternative) = PQ∥ $\parallel$UV

< XPQ = < YPQ = 40°, Since QX = QY

therefore < XPY = 80°

If $x–1$ and $x+1$ are both factors of the equation $x^3+p{x}^2+qx+6=0$, evaluate $p$ and $q$

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If a $\ast$ b = + $\sqrt{ab}$, evaluate 2 $\ast$(12 $\ast$ 27)

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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.

Class Interval1.5?1.92.0?2.42.5?2.93.0?2.93.5?3.94.0?4.44.5?4.9Frequency2214151053Class boundaries1.45?1.951.95?2.452.45?2.952.95?3.453.45?3.953.95?4.454.45?4.95ClassMid?point1.72.22.73.23.74.24.7
Find the mode of the distribution above to find the mode of the distribution.

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Mode = a + (b - a)(fm - Fb)

2Fm - Fa - Fb

= 3.0 + (3.4?3)(15?4)2(15)?4?10 $\frac{(3.4?3)(15?4)}{2(15)?4?10}$

= 3 + (6.4)(11)30?14 $\frac{(6.4)(11)}{30?14}$

= 3 + 4.416 $\frac{4.4}{16}$

= 3 + 0.275

= 3.275

= 3.3cm

In a triangle XYZ, < YXZ = 44° and < XYZ = 112°. Calculate the acute angle between the internal bisectors of < XYZ and < XZY.

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The variance of the scores 1, 2, 3, 4, 5 is

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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.

Evaluate $\frac{(18{)}^{\frac{3}{4}}-(27{)}^3}{3\times 2^3}$

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Calculate 33105 - 14425

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Convert 3.1415926 to 5 decimal places

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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.

$\begin{array}{cccc}class& 1-3& 4-6& 7-9\\ Frequency& 5& 8& 5\end{array}$

Find the standard deviation of the data using the table above

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$\begin{array}{ccccccc}\text{class intervals}& Fre\left(F\right)& \text{class-marks(x)}& Fx& (x-x)& (x-x{)}^2& F(x-x{)}^2\\ 1-3& 5& 10& 10& -3& 9& 90\\ 4-6& 8& 40& 40& 0& 0& 0\\ 7-9& 5& 40& 40& 3& 9& 360\\ & 18& & 90& & & 450\end{array}$
x = $\frac{\sum fx}{\sum f}$
= $\frac{90}{18}$
= 5
S.D = $\frac{\sum f}{}$