JAMB UTME - Mathematics - 1993

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

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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

If x is negative, what is the range of values of x within which $\frac{x+1}{3}$ > $\frac{1}{X+3}$

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$\frac{x+1}{3}$ > $\frac{1}{X+3}$ = $\frac{x+1}{3}$ > $\frac{x+3}{X+3}$
= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)
x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0
= x(x + 3)9x + 4) > 0
Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0
= x > -4 (solution only)
Case 2 (+, -, -) = x > 0, x + 4 < 0

= x > 0, x < -3, x < -4 = x < -3(solution only)

Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)

Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)

combining the solutions -4 < x < -3

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

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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

If two angles of a triangle are 30o each and the longest side is 10cm. Calculate the length of each of the other sides

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102 = 2x2 - 2x2 cos 120o (Cosine rule)
100 = 242 - 2x2 - 2x2 x -$\frac{1}{2}$
100 = 3x2 + x2
= 3x2
x = $\sqrt{\frac{100}{3}}$
= $\frac{10}{\sqrt{3}}$ x $\frac{3}{\sqrt{3}}$
x = 10$\frac{3}{3}$cm

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

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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 = $\frac{2}{3}$

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

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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${}^{\circ }$

Simplify $\frac{1}{p}$ - $\frac{1}{q}$ + $\frac{p}{q}$ - $\frac{q}{p}$

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$\frac{1}{p}$ - $\frac{1}{q}$ + $\frac{p}{q}$ - $\frac{q}{p}$ = $\frac{q?p}{pq}$ + $\frac{p^2?q^2}{pq}$
$\frac{q?p}{pq}$ x $\frac{pq}{p^2{q}^2}$ = $\frac{q?p}{p^2{q}^2}$
$\frac{?(p?q)}{(p+q)(p?q)}$
= $\frac{?1}{p+q}$

The shaded portion in the Venn diagram is

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

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

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

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

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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?

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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.

In the diagram, PQRs is a circle with 0 as centre and PQ/RT. If RTS = ${}^{\circ }$. Find PSQ

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< PSO = $\frac{1}{2}$ < SOQ = $\frac{1}{2}$(180) = 90${}^{\circ }$

< RTS = < PQS = 32${}^{\circ }$ (Alternative angle)

< PSQ = 90 - < PSQ = 90${}^{\circ }$ - 32${}^{\circ }$

= 58${}^{\circ }$

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

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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.

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?

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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 $\frac{1}{2}$
= $\frac{1}{12}$

Divide the expression x3 + 7x2 - x - 7 by -1 + x2

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

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

Solve the inequality y2 - 3y > 18

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y2 - 3y > 18 = 3y - 18 > 0
y2 - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0
= (y + 3) (y - 6) > 0
Case 1 (+, +) $\to$ (y + 3) > 0, (y - 6) > 0
= y > -3 y > 6
Case 2 (-, -) $\to$ (y + 3) < 0, (y - 6) < 0

= y < -3, y < 6

Combining solution in case 1 and Case 2
= x < -3y < 6

= -3 < y < 6

Solve without using tables log5(62.5) - log5($\frac{1}{2}$)

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log5(62.5) - log5($\frac{1}{2}$)
= log5$\frac{\left(62.5\right)}{\frac{1}{2}}$ - log5(2 x 62.5)
= log5(125)
= log553 - 3log55
= 3

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

Antwoorddetails

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

If sin x = cos 50o, then x equals

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We know that sin x = cos 50°. To solve for x, we need to find the angle whose sine is equal to cos 50°. We can use the trigonometric identity that sin (90° - θ) = cos θ. So, sin x = cos 50° can be rewritten as sin x = sin (90° - 50°). Using the identity, we have: sin x = sin 40° Now, we need to find the angle x whose sine is equal to sin 40°. Since sine is a periodic function, there are multiple angles whose sine is equal to a given value. One such angle is x = 40°. However, sine is also negative in the third and fourth quadrants. In the third quadrant, x = 180° - 40° = 140° and in the fourth quadrant, x = 360° - 40° = 320° also satisfy the equation sin x = sin 40°. However, since x has to be between 0° and 360°, we can eliminate the30°. Therefore, the possible values of x are 40°, 140°, and 320°. However, since we know that sin x = cos 50° and cos is positive in the first quadrant, x cannot be in the third or fourth quadrants. Therefore, the only possible value of x is x = 40°. Hence, the answer is x = 40°.

PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP

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

Find the median of the observation in the table given.

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A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime

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To solve this problem, we need to first determine the set of possible numbers that could be selected at random between 20 and 30. Since the problem states that both 20 and 30 are inclusive, the set of possible numbers is: {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} To find the probability that the selected number is prime, we need to determine the number of prime numbers in the set of possible numbers, and then divide that by the total number of possible numbers. Prime numbers are numbers that are only divisible by 1 and themselves. The prime numbers in the set of possible numbers are: { 23, 29 } Therefore, the probability that the selected number is prime is 2/11 or approximately 0.18.

The angle between latitudes 30oS and 13oN is

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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

If sin $\theta$ = cos $\theta$, find $\theta$ between 0o and 360o

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sin $\theta$ = cos $\theta$ 0 $\le$ $\theta$ $\le$ 360o
The acute angle where sin $\theta$ = cos $\theta$ = 45o
But at the fourth Quadrant Cos $\theta$ = +ve
at the 4th quadrant, value with respect to Q is
(360 - Q) where Q = acute angle
(360 - 45) = 315o
The two solution are 45o, 315o

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

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

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

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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.

From the figure, calculate TH in centimeters

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TH = tan 45${}^{\circ }$, TH = QH

$\frac{TH}{5+QH}$ = tan 30${}^{\circ }$

TH = (b + QH) tan 30${}^{\circ }$

QH = 56 (5 + QH) $\frac{1}{\sqrt{3}}$

QH(1 - $\frac{1}{\sqrt{3}}$) = $\frac{5}{\sqrt{3}}$

QH = $\frac{5\sqrt{3}}{\sqrt{3}}-\frac{1}{\sqrt{3}}$

= $\frac{5}{\sqrt{3}-1}$

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

Estimate the mode of the frequency distribution above.

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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

Quantities in the proportions 1, 4, 6, 7 are to be represented in a pie chart. Calculate the angle of the sector with proportion 7

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To calculate the angle of the sector with proportion 7, we first need to find the total of all the quantities in the proportion. Total quantity = 1 + 4 + 6 + 7 = 18 Next, we need to find the proportion of quantity 7 in relation to the total quantity. Proportion of quantity 7 = 7/18 To find the angle of the sector that represents this proportion, we need to use the formula: Angle of sector = Proportion of quantity * 360 degrees Angle of sector = (7/18) * 360 degrees Angle of sector = 140 degrees Therefore, the answer is option D, which is 140 degrees.

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

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The mean of the ages of ten secondary school pupils is 16 but when the age of their teacher is added to it the men becomes 19. Find the age of the teacher

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Let the age of the teacher be T. According to the problem, the mean age of the 10 pupils is 16. Therefore, the sum of their ages is 10 x 16 = 160. When the teacher's age is added to it, the sum becomes 160 + T. Also, it is given that the new mean age after adding the teacher's age is 19. So, we have: (160 + T) / 11 = 19 Multiplying both sides by 11, we get: 160 + T = 209 Subtracting 160 from both sides, we get: T = 49 Therefore, the age of the teacher is 49.

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

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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

The bar chart shows the distribution of marks in a class test. How many students took the test?

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Number of students that took the test = $\sum f$. Where $f$ is the frequencies
= 2 + 5 + 0 + 3 + 4 + 0 + 0 + 0 + 1 + 0 + 2 = 17

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

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OQ = OR = radii

< ROQ = 180 - 86 = 84${}^{?}$

$?$OQR = Isosceles

R = Q

R + Q + 84 = 180(angle in a $?$)

2R = 96 since R = Q

R = 48${}^{?}$

ORQ = 48${}^{?}$

simplify $\frac{1}{\surd 3-2}$ - $\frac{1}{\surd 3+2}$

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$\frac{1}{\surd 3-2}$ - $\frac{1}{\surd 3+2}$
L.C.M = (3- 2) (3 + 2)
∴ $\frac{1}{\sqrt{3-2}}$ - $\frac{1}{\sqrt{3-2}}$ = $\frac{\sqrt{3+2}-\sqrt{3-2}}{\sqrt{3-2}+\sqrt{3-2}}$

$\frac{\surd 3+2-\surd 3+2}{3-2\surd 3+2\surd 3-4}$ = $\frac{4}{3-2}$
= $\frac{4}{-1}$
= -4

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

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

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

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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.

Solve the following equation (3x - 2)(5x - 4) = (3x - 2)2

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(3x - 2)(5x - 4) = (3x - 2)2 = 5x2 - 22x + 6
= 9x2 = 12x + 4
6x2 - 10x + 4 = 0
6x2 - 6x - 4x + 4 = 0
6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0
x = 1 or $\frac{2}{3}$

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

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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

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

Antwoorddetails

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

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

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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
= $\frac{22\frac{1}{2}\times 8\pi }{360}$
= $\frac{180}{360}$
= $\frac{\pi }{2}$

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

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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

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

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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

The following marks were obtained by twenty students in an examination: 53, 30, 70, 84, 59, 43, 90, 20, 78, 48, 44, 60, 81, 73, 50, 37, 67, 68, 64, 52. Find the numbers of students who scored at least 50 marks

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To find the number of students who scored at least 50 marks, we need to count the number of marks that are greater than or equal to 50. Looking at the marks obtained by the students, we can see that there are 14 marks that are greater than or equal to 50. Therefore, the number of students who scored at least 50 marks is 14. We can count these marks by going through the list one by one, or by using a table or chart to organize the data. It's important to pay close attention to the wording of the question and to make sure we're answering the question being asked. In this case, the question is asking for the number of students, not the number of marks.

If 2log3 y + log3 x2 = 4, then y is

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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".

In the figure, the area of the square of the square PQRS is 100cm2. If the ratio of the area of the square TUYS to the area of the area of the square XQVU is 1 : 16, Find YR

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Since area of square PQRS = 100cm2

each lenght = 10cm

Also TUYS : XQVU = 1 : 16

lengths are in ratio 1 : 4, hence TU : UV = 1: 4

Let TU = x

UV = 1: 4

hence TV = x + 4x = 5x = 10cm

x = 2cm

TU = 2cm

UV = 8cm

But TU = SY and UV = YR

YR = 8cm

Evaluate (x + $\frac{1}{x}$ + 1)2 - (x + $\frac{1}{x}$ + 1)2

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(x + $\frac{1}{x}$ + 1)2 - (x + $\frac{1}{x}$ + 1)2
= (x + $\frac{1}{x}$ + 1 + x + $\frac{-1}{x}$ - 1)(x - $\frac{1}{x}$ + 1 - x + $\frac{1}{x}$ + 1)
= (2x) (2 + $\frac{2}{x}$) = 2x x 2(1 + $\frac{1}{x}$)
4x (1 + $\frac{1}{x}$) = 4x + 4
= 4(1 + x)

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

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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.

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

Antwoorddetails

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