JAMB UTME - Mathematics - 1993

In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles with OS = 5cm. QT = 2cm and OR = 14cm. Find ST

Answer Details

SQ2 + OS2 = OQ2 + 52 = 142

SQ2 = 142 - 52

196 - 25 = 171

ST2 + TQ2 = SQ2

ST2 + 22 = 171

ST2 = 171 - 4

= 167

ST = $\sqrt{167}$

= 12.92 = 12.9cm

Which of the following is a factor of 15 + 7x - 2x2

Answer Details

Factorize 15 + 7x - 2x2
(5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0
∴ (5 - x)(3 + 2x) = 0
x = 5 or x = -$\frac{3}{2}$
Since 5 is a root, then (x - 5) is a factor

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

Answer Details

Solve for y in the equation 101 x 5(2x - 2) x 4(y - 1) = 1

Answer Details

10y x 5(2y - 2) x 4(y - 1) = 1
but 10y - (5 x 2)y = 5y x 2y
= (Law of indices)
5y x 2y x 5(2y - 2) x 4(y - 1) = 1
but 4(y - 1) = 22(y - 1)
= 2y - 2 (Law of indices)
5y x 5(2y -2) x 2(- 2) = 1
5(3y -2) x 2y x 2(2y -2) = 1
= 5(3y -2) x 2(3y -2) = 1
But ao = 1
10(3y -2) = 10o
3y - 2 = 0
? y = $\frac{2}{3}$

Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of 22$\frac{1}{2}$o at the centre of the circle

Answer Details

Diameter = 8cm
∴ Radius = 4cm
Length of arc = $\frac{\theta }{360}$ x 2 $\pi$r but Q = 22$\frac{1}{2}$
∴ Length $\frac{22\frac{1}{2}}{360}$ x 2 x $\pi$ x 4
= $\frac{22\frac{1}{2}\times 8\pi }{360}$
= $\frac{180}{360}$
= $\frac{\pi }{2}$

Find the area of the sector of a circle with radius 3m, if the angle of the sector is 60o

Answer Details

To find the area of a sector, we use the formula: Area of sector = (θ/360) x πr² Where θ is the angle of the sector in degrees, r is the radius of the circle, and π (pi) is a mathematical constant approximately equal to 3.14. In this problem, the radius is given as 3m and the angle of the sector is 60°. Substituting these values into the formula, we get: Area of sector = (60/360) x π(3)² = (1/6) x π(9) = (1/6) x 28.27 = 4.71m² (rounded to two decimal places) Therefore, the area of the sector of the circle with radius 3m and angle 60° is approximately 4.71m². The answer is option C.

In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.

Answer Details

Using angle bisector theorem: line PS bisects angle QPR

QS/QP = SR/PR

3/4 = SR/g

4SR = 27

SR = $\frac{27}{4}$

= 6$\frac{3}{4}$cm

Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

Answer Details

In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)

The shaded portion in the Venn diagram is

Answer Details

The shaded part exists on x $\cap$ z but not in y

$\begin{array}{cccccc}Weight\left(s\right)& 0-10& 10-20& 20-30& 40-50& \\ \text{Number of coconuts}& 10& 27& 19& 6& 2\end{array}$

Estimate the mode of the frequency distribution above.

Answer Details

Mode = a + $\frac{(b-a)(F_m-F_b)}{2F_m-F_a-F_b}$
= $L_1+\frac{{\Delta }_1{x}^{\text{c}}}{{\Delta }_1+{\Delta }_2}$
= $10+\frac{(20-10)(27-10)}{2\left(27\right)-10-19}$
= 10 + $\frac{170}{25}$
= 10 + 6.8
= 16.8

The chances of three independent events X, Y, Z occurring are $\frac{1}{2}$, $\frac{2}{3}$, $\frac{1}{4}$ respectively. What are the chances of Y and Z only occurring?

Answer Details

Chance of x = $\frac{1}{2}$
Change of Y = $\frac{2}{3}$
Chance of Z = $\frac{1}{4}$
Chance of Y and Z only occurring
= Pr (Y ? Z ? Xc)
where Xc = 1 - Pr(X)
1 = $\frac{1}{2}$ = 1$\frac{1}{2}$
= Pr(Y) x Pr(Z) x Pr(Xc)
= $\frac{2}{3}$ x $\frac{1}{4}$ x $\frac{1}{2}$
= $\frac{1}{12}$

In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.

Answer Details

Join PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o

In the diagram, QP//ST:PQR = 34${}^{\circ }$ QRS = 73${}^{\circ }$ and RS = RT. Find SRT

Answer Details

Construction joins R to P such that SRP = straight line

R = 180${}^{\circ }$ - 107${}^{\circ }$

< p = 180${}^{\circ }$ - (107${}^{\circ }$ - 34${}^{\circ }$)

108 - 141${}^{\circ }$ = 39${}^{\circ }$

Angle < S = 39${}^{\circ }$ (corr. Ang.) But in $△$SRT

< S = < T = 39${}^{\circ }$

SRT = 180 - (39${}^{\circ }$ + 39${}^{\circ }$)

= 180${}^{\circ }$ - 78${}^{\circ }$

= 102${}^{\circ }$

If the function f is defined by f(x + 2) = 2x2 + 7x = 5, find f(-1)

Answer Details

The three sides of an isosceles triangle are length of lengths (x + 3), (2x + 3), (2x - 3) respectively. Calculate x.

Answer Details

2x + 3 $\ne$ 2x - 3 for any value of x
∴ for the $△$ to be isosceles, either
2x - 3 = x + 3 or 2x + 3 = x + 3
solve the two equations we arrive at
x = 6 or x = 0
When x = 6, the sides are 9, 15, 9
When x = 0, the sides are 3, 4, -3 since lengths of a $△$can never be negative then the value of x = 6

Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR

Answer Details

QPS - QRK = 31o
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o

Make x the subject of the relation $\frac{1+ax}{1-ax}$ = $\frac{p}{q}$

Answer Details

To make x the subject of the given relation, we need to isolate x on one side of the equation. First, we can simplify the expression by finding the common denominator of the two fractions in the numerator: (1 + ax)/(1 - ax) = (1 - ax + ax + a^2x)/(1 - ax) = (1 + a^2x)/(1 - ax) Now we can cross-multiply to get rid of the denominator: (1 + a^2x) = pq - axpq Next, we can move the term with x to the left-hand side of the equation: a^2x + axpq = pq - 1 Finally, we can factor out x from the two terms on the left-hand side: x(a^2 + apq) = pq - 1 Therefore, we can isolate x by dividing both sides by (a^2 + apq): x = (pq - 1)/(a^2 + apq) So the correct option is: p-q / a(p+q) which is not among the given options.

Calculate the standard deviation of the following data: 7, 8, 9, 10, 11, 12, 13.

Answer Details

To calculate the standard deviation, we need to first find the mean (average) of the data. Mean = (7 + 8 + 9 + 10 + 11 + 12 + 13) / 7 = 10 Next, we need to find the difference between each data point and the mean, and square the differences. (7-10)^2 = 9 (8-10)^2 = 4 (9-10)^2 = 1 (10-10)^2 = 0 (11-10)^2 = 1 (12-10)^2 = 4 (13-10)^2 = 9 Then, we add up all of these squared differences and divide by the total number of data points. (9 + 4 + 1 + 0 + 1 + 4 + 9) / 7 = 28 / 7 = 4 Finally, we take the square root of this result to find the standard deviation. Standard deviation = √4 = 2 Therefore, the standard deviation of the given data is 2.

A man's initial salary is ₦540.00 a month and increases after each period of six months by ₦36.oo a month. Find his salary in the eighth month of the third year

Answer Details

Initial salary = ₦540
increment = ₦36 (every 6 months)
Period of increment = 2 yrs and 6 months
amount(increment) = ₦36 x 5 = ₦180
The man's new salary = ₦540 = ₦180
= ₦720.00

Solve the following simultaneous equation for x. x2 + y - 5 = 0, y - 7x + 3 = 0

Answer Details

x2 + y - 5 = 0.....(i)
y - 7x + 3 = 0.........(ii)
y = 7x - 3, substituting the value of y in equation (i)
x2 + (7x - 3) - 5 = 0
x2 + 7x + 3 = 0
(x + 8)(x - 1) = 0
x = -8 or 1

If ₦225.00 yields ₦27.00 in x years simple interest at the rate of 4% per annum, find x

Answer Details

To solve this problem, we use the formula for simple interest: I = P*r*t Where: I = Interest earned P = Principal amount r = Rate of interest per year (as a decimal) t = Time period in years In this case, we are given: P = ₦225.00 (the principal amount) I = ₦27.00 (the interest earned) r = 4% = 0.04 (the rate of interest per year) Substituting these values into the formula, we get: 27 = 225 * 0.04 * t Simplifying the equation, we have: 27 = 9t Dividing both sides by 9, we get: t = 3 Therefore, the answer is 3 years. Explanation: The interest earned on the principal amount of ₦225.00 is ₦27.00. This means that for every year, the interest earned is 27/225 = 0.12 or 12% of the principal amount. We know that the rate of interest per year is 4%, which is much lower than the interest earned per year of 12%. This means that it will take more than 1 year for the interest earned to reach the rate of interest. In fact, it takes 3 years for the interest earned to reach the rate of interest of 4%, which is why the answer is 3.

A rectangular polygon has 150o as the size of each interior angle. How many sides has the polygon?

Answer Details

A rectangular polygon has each interior angle to be 150o
let the polygon has n-sides
therefore, Total interior angle 150 x n = 150n
hence 150n = (2n - 4)90
150n = 180n - 360
360 = (180 - 150)n
30n = 360
n = 12

If $\sqrt{x^2+9}$ = x + 1, solve for x

Answer Details

$\sqrt{x^2+9}$ = x + 1
x2 + 9 = (x + 1)2 + 1
0 = x2 + 2x + 1 - x2 - 9
= 2x - 8 = 0
2(x - 4) = 0
x = 4

The figure represents the graphs of y = x(2 - x) and y = (x - 1)(x - 3). What are the x-coordinates of P, Q and F respectively?

Answer Details

To find the x-coordinates of P, Q, and F, we need to solve the system of equations formed by setting the two equations equal to each other: x(2 - x) = (x - 1)(x - 3) Expanding the right-hand side and simplifying, we get: 2x - x^2 = x^2 - 4x + 3 Rearranging and simplifying, we get a quadratic equation in standard form: x^2 - 6x + 3 = 0 Using the quadratic formula, we get: x = (6 ± sqrt(6^2 - 4*1*3)) / (2*1) = 3 ± sqrt(3) Therefore, the x-coordinates of P, Q, and F are 1, 3, and 3 - sqrt(3) or approximately 1.268, respectively. To see why, we can plot the two functions and visually determine the x-coordinates of the points where they intersect. Point P is where the blue line intersects the x-axis, which occurs at x = 1. Point Q is where the red line intersects the x-axis, which occurs at x = 3. Point F is where the two lines intersect above the x-axis, which occurs at x = 3 - sqrt(3) or approximately 1.268. Therefore, the answer is 1, 2, 3.

The angle between latitudes 30oS and 13oN is

Answer Details

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes.
∴ Total angle difference = (30 + 13) = 43o

$\begin{array}{cc}Class& Frequency\\ 1-5& 2\\ 6-10& 4\\ 11-15& 5\\ 16-20& 2\\ 21-25& 3\\ 26-30& 2\\ 31-35& 1\\ 36-40& 1\end{array}$

Find the median of the observation in the table given.

Answer Details

If 2log3 y + log3 x2 = 4, then y is

Answer Details

The given equation is 2log3 y + log3 x2 = 4. We can simplify this equation using the laws of logarithms. First, we can use the power rule of logarithms to rewrite x2 as (x)2. Then, we can use the product rule of logarithms to combine the two logarithms on the left-hand side of the equation: 2log3 y + log3 (x)2 = log3 y2 + log3 (x)2 Now, we can use the sum rule of logarithms to combine the two logarithms on the right-hand side of the equation: log3 y2(x)2 = log3 (x2y2) We can now rewrite the original equation as: log3 (x2y2) = 4 Using the definition of logarithms, we know that log3 (x2y2) = 4 is equivalent to 3^4 = x2y2. Therefore, we have: x2y2 = 81 Taking the square root of both sides, we get: xy = ±9 Since y is a positive real number, we can discard the negative solution. Therefore, we have: xy = 9 Finally, we can solve for y: y = 9/x So, the answer is "±9/x".

find the radius of a sphere whose surface area is 154cm2 ($\pi =\frac{22}{7}$)

Answer Details

Surface area = 154cm2 (area of sphere)
4$\pi$r2 = 154
r$\sqrt{\frac{154}{4\pi }}$
= 3.50cm

A binary operation $\ast$ is defined on a set of real numbers by x $\ast$ y = xy for all real values of x and y. If x $\ast$ 2 = x. Find the possible values of x

Answer Details

x $\ast$ y = xy
x $\ast$ 2 = x2
x $\ast$ 2 = x
∴ x2 - x = 0
x(x - 1) = 0
x = 0 or 1

In the diagram, O is the centre of the circle and POQ a diameter. If POR = 96${}^{?}$, find the value of ORQ.

Answer Details

OQ = OR = radii