JAMB UTME - Mathematics - 1989

Akọwa Nkọwa

A rectangular has all sides and all angles equal. If each interior angle is 140o each exterior angle must be
180o - 140o = 40o
The number of sides must be $\frac{{360}^o}{{40}^o}$ = 9 sides
hence 2k + 1 = 9
2k = 9 - 1
8 = 2k
k = $\frac{8}{2}$
= 4

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The given equation is 4sin²x - 3 = 0. We need to find the value of x such that 0 ≤ x ≤ 90°. To solve this equation, we first need to isolate sin²x by adding 3 to both sides: 4sin²x = 3 Then, we divide both sides by 4: sin²x = 3/4 Now, we take the square root of both sides: sinx = ±√3/2 We have two possible values for sinx: √3/2 and -√3/2. To determine which of these values of sinx is valid for the given range of x, we need to examine the unit circle. The unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. It is used to visualize the values of sine, cosine, and tangent for all angles in standard position (angles whose vertex is at the origin and whose initial side lies along the positive x-axis). If we draw the unit circle and plot the points corresponding to sinx = √3/2 and sinx = -√3/2, we find that sinx = √3/2 corresponds to an angle of 60° (or π/3 radians) and sinx = -√3/2 corresponds to an angle of 300° (or 5π/3 radians). Since we are looking for a value of x such that 0 ≤ x ≤ 90°, the only valid solution is sinx = √3/2 and x = 60°. Therefore, the answer is x = 60°.

Akọwa Nkọwa

QMN = 60o

MRQ = 60o(angle in the alternate segment are equal)

MQN = 80o(angle sum of a ? $?$ = 180o)

60 = x = 80o(exterior angle = sum of opposite interior angles)

x = 80o - 60o = 20o

RMQ = 20o

Akọwa Nkọwa

$\frac{324-4x^2}{2x+18}$ = $\frac{{18}^2-(2x{)}^2}{2x+18}$
= $\frac{(18-2x)(18+2x)}{(2x+18)}$
18 - 2x = 2(9 - x)
or -2(x - 9)

Akọwa Nkọwa

$\frac{9}{10}\frac{4}{5}\frac{3}{4}\frac{17}{10}$ = $\frac{18,16,15,34}{20}$
$\frac{4}{5}\frac{9}{10}\frac{3}{4}\frac{17}{20}$ = $\frac{16,18,15,17}{20}$
$\frac{9}{10}\frac{17}{10}\frac{4}{5}\frac{3}{4}$ = $\frac{18,17,16,15}{20}$
∴ $\frac{4}{5}\frac{9}{10}\frac{17}{10}\frac{3}{4}$ is in descending order

Akọwa Nkọwa

Sum of nth term of a G.P = Sn = $\frac{a{r}^n-1}{r-1}$
sum of the first two terms = $\frac{a{r}^2-1}{r-1}$
x = a(r + 1)
sum of the last two terms = Sn - Sn - 2
= $\frac{a{r}^n-1}{r-1}$ - $\frac{\left(a{r}^{n-1}\right)}{r-1}$
= $\frac{a(r^n-1-r^{n-2}+1)}{r-1}$ (r2 - 1)
∴ $\frac{a{r}^{n-2}(r+1)(r-1)}{1}$= arn - 2(r + 1) = y
= a(r + 1)r^n - 2
y = xrn - 2
= yrn - 2
$\frac{y}{x}$ = r = ($\frac{y}{x}$)$\frac{1}{n-2}$

Akọwa Nkọwa

< PSQ = < SQR = < SRQ = 24${}^{\circ }$

< QSR = 180${}^{\circ }$ - 48${}^{\circ }$ = 132${}^{\circ }$

< PSQ + < QSR + y + 180 (angle on a straight lines)

24 + 132 + y = 180${}^{\circ }$ = 156${}^{\circ }$ + y = 180

y = 180${}^{\circ }$ - 156${}^{\circ }$

= 24${}^{\circ }$

Akọwa Nkọwa

The gradient of any curve is equal to zero at the turning points. i.e. maximum or minimum points. The points in the above curve are b, e, g, j, m

Akọwa Nkọwa

x = 10 cot35o - 10 cot55o
= 10(cot35o - cot55o)

Akọwa Nkọwa

Arrange the numbers in order, 1.6, 2.1, 2.1| 2.4, 2.6| 2.6, 2.6, 3.7
n = median = $\frac{2.4+2.6}{2}$
= 2.5
m = mode = 2.6
∴ (n, m) = (2.5, 2.6)

Akọwa Nkọwa

13 = 52
1 = $\frac{52}{13}$
= 4
5 + 12 + 13 = 30
Total perimeter = 30 x 4
= 120cm

Akọwa Nkọwa

4a2 - 12ab - C2 + 9b2
rearranges: (4a2 - 12ab + 9b2) - c2
(2a - 3b)(2a - 3b) - c2 = (2a - 3b)2 - c2
= (2a - 3b + c)(2a - 3b - c)

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Akọwa Nkọwa

2x-t - 3y-1 = 4, 4x-1 + y-1 = 1
Let x - 1 = a and y - 1= b
2a - 3b = 4 .......(i)
4a + b = 1 .........(ii)
(i) x 3 = 12a + 3b = 3........(iii)
2a - 3b = 4 ...........(i)
(i) + (iii) = 14a
= 7
∴ a = $\frac{7}{14}$ = $\frac{1}{2}$
From (i), 3b = 2a - 4
3b = 1 - 4
3b = -3
∴ x = -1
From substituting, 2-1 = x - 1
∴ x = 2
y-1 = -1, y = -1

Akọwa Nkọwa

This problem is asking us to find the number of years that Oke's ₦800.00 deposit was left in the bank at a certain interest rate, given that the total amount he received was one and a half times the original deposit amount. We can start by using the simple interest formula, which is: Simple Interest = Principal * Rate * Time Here, we know the principal is ₦800.00 and the rate is 12.12%. We don't know the time, which is what we're trying to find. Let's call it "t". We also know that the total amount Oke received after some time was one and a half times the original deposit amount, or: Total Amount = 1.5 * Principal Substituting the values we know, we get: Total Amount = 1.5 * ₦800.00 = ₦1200.00 We can use this equation to solve for "t" by first finding the simple interest: Simple Interest = Total Amount - Principal Simple Interest = ₦1200.00 - ₦800.00 = ₦400.00 Then, we can rearrange the simple interest formula to solve for "t": Time = Simple Interest / (Principal * Rate) Substituting the values we know, we get: Time = ₦400.00 / (₦800.00 * 0.1212) = 4 years (rounded to the nearest whole number) Therefore, Oke's money was left in the bank for 4 years.

Akọwa Nkọwa

When we round 0.007685 to three significant figures, we get 0.00768. The last digit (5) is rounded down because it is less than 5. Therefore, the difference between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal is: 0.007685 - 0.0077 = -0.000015 So the answer is -1.5 x 10^-5. Therefore, the correct option is: - 10-5

Akọwa Nkọwa

a : b = 5 : 8 = 2.5 : 40
x : y = 25 : 16
$\frac{a}{x}$ : $\frac{b}{y}$ = $\frac{25}{25}$ : $\frac{40}{16}$
= 1 : $\frac{40}{16}$
= 16 : 40
= 2 : 5

Akọwa Nkọwa

Since HK is parallel to QR, angle HPQ is equal to angle KPR, and angle HQP is equal to angle KRP. Therefore, triangles HPQ and KRP are similar triangles by the angle-angle (AA) criterion. The ratio of the lengths of the corresponding sides of similar triangles is equal. Therefore: KR/PR = HQ/PQ = 3/(3+4) = 3/7 So the ratio of KR:PR is 3:7. Therefore, the correct option is 3:7.

Akọwa Nkọwa

S = $\sqrt{\frac{2R+T}{2RT}}$
squaring both sides = S2 = $\frac{2R+T}{2RT}$
S2(2RT) = 2R + T
2S2RT - 2R = T
R(2S2T - 2) = T
R = R = $\frac{T}{2(T{S}^2-2)}$

Akọwa Nkọwa

Let's call the grandfather's age "G". We know that the family consists of 21 people, so if we exclude the grandfather, there are 20 people remaining. We're given that the average age of the family is 14 years, so we can write: Total age of the family = 21 x 14 We're also given that if we exclude the grandfather, the average age drops to 12 years, so we can write: Total age of the family (excluding grandfather) = 20 x 12 We can set up an equation with these two expressions and solve for G: 21 x 14 - G = 20 x 12 294 - G = 240 G = 54 Therefore, the age of the grandfather is 54 years. So the correct option is: 54 years.

Akọwa Nkọwa

$\frac{1}{2}$(3y - 4x)2 = (8x2 + kxy + Ly2)
$\frac{1}{2}$(9y2 - 24xy + 16x2) = 8x2 + kxy + Ly2
$\frac{9}{2}$y2 - 12xy) = kxy + Ly2
k = -12 ∴ L = $\frac{9}{2}$

Akọwa Nkọwa

Y $\to$ 30o
X $\to$ 120o
Z $\to$ 30o
$\frac{3}{\text{sin 30}}$ = $\frac{YZ}{\text{sin 120}}$
YZ = $\frac{3\text{sin 120}}{\text{sin 30}}$
3 x $\frac{\sqrt{3}}{2}$ x 2 = 3$\sqrt{3}$

Akọwa Nkọwa

gy - 4 x 3y + 3 = 0
Put 3y = x2 - 4x + 3 = 0
factorize (x - 3)(x - 1) = 0
x = 3 or 1
? 3y = 3, y = 1, 3y = 1, y = 0

Akọwa Nkọwa

Volume of cylinder pip (V) = π $\pi$h(R2 ${}^2$ - r2 ${}^2$)

= 100π $\pi$(7.22 ${}^2$ - 2.82 ${}^2$)

= 100π $\pi$(51.84 - 7.84)

= 100π $\pi$ x 44

= 440π $\pi$cm3

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Akọwa Nkọwa

To express the given expression as a single algebraic fraction, we first need to find a common denominator for all the terms. The common denominator is (x+1)(2-x). Then, we can simplify each term by multiplying the numerator and denominator by the missing factor in the denominator to obtain: 1(x+1)(2-x) / (x+1) + 1(2-x) / (2-x)(x+1) - 1x(x+1) / (2-x)(x+1) Simplifying further by combining like terms, we get: [(x+1)(2-x) + 1(1-x)(x+1) - x(x+1)] / [(2-x)(x+1)] Simplifying the numerator by distributing, we get: [-x^2 + 3x - 1] / [(2-x)(x+1)] Therefore, the expression 1x+1 / (x+1) + 1 / (2-x) - 1x?2 / (x+1)(2-x) can be simplified to -[-x^2 + 3x - 1] / [(2-x)(x+1)]. So, the answer is (a) -3(x+1)(2-x) / [-x^2 + 3x - 1].

Akọwa Nkọwa

To find the angle of the sector which represents meat in the pie chart, we first need to find the total weight of the ingredients used: Total weight = 75g + 40g + 105g + 20g = 240g Now, we can find the percentage of meat used: Percentage of meat = (Weight of meat / Total weight) x 100% = (105g / 240g) x 100% = 43.75% To find the angle of the sector in the pie chart, we use the formula: Angle of sector = Percentage of sector / 100% x 360 degrees Substituting the percentage of meat, we get: Angle of sector = 43.75 / 100 x 360 degrees = 157.5 degrees Therefore, the answer is 157.5 degrees.

Akọwa Nkọwa

PR2 + QS2 = KPQ2
SQ2 = SR2 + RQ2
PR2 + SQ2 = PQ2 + SR2 + 2RQ2
= 2PQ2 + 2RQ2
= 4PQ2
∴ K = 4

Akọwa Nkọwa

log3P + 3log3q = 3
log3(Pq3) = 3
Pq3 = 33
P = ($\frac{3}{P}$)3

Akọwa Nkọwa

To solve this problem, we need to find the area of the rectangular floor and divide it by the area of one tile. The area of the rectangular floor is: 7.2m x 4.2m = 30.24 m² The area of one square tile is: 30cm x 30cm = 0.09 m² Now we can divide the area of the floor by the area of one tile to get the number of tiles needed to cover the floor: 30.24 m² / 0.09 m² = 336 Therefore, 336 square tiles will cover the rectangular floor. So, the answer is 336.

Akọwa Nkọwa

xz2 = 62 + 62
36 + 36 = 72
xz = $\sqrt{72}$
6$\sqrt{2}$ = xT
$\frac{6\sqrt{2}}{2}$ = $\frac{3}{\sqrt{2}}$
OT2 = 52 + (3$\sqrt{2}$)2 = 25 + 18
OT = 4$\sqrt{3}$

Akọwa Nkọwa

This histogram is transferred into this frequency table
$\begin{array}{cccccc}Marks& 20& 40& 60& 80& 100\\ students& 9& 7& 6& 6& 2\end{array}$

Students who scored more than 40 = 6 + 6 + 2 = 14

i.e. $\frac{14}{30}$ x 100% = 46$\frac{3}{4}$%

Akọwa Nkọwa

Surface Area of Sphere A = 4$\pi r^2$
∴ A = 4$\pi$$\frac{(D{)}^2}{2}$
= $\frac{(D{)}^2}{2}$
= $\pi$D2
When increased by 44% A = $\frac{144\pi D^2}{100}$
$\pi$$\frac{(12D{)}^2}{10}$ = $\pi$$\frac{(6D{)}^2}{5}$
Increase in diameter = $\frac{6D}{5}$ - D = $\frac{1}{5}$D
Percentage increase = $\frac{1}{5}$ x $\frac{1}{100}$%
= 20%

Akọwa Nkọwa

y3 -4xy + xy3 - 4y = y3(1 + x) - 4y(1 + x)
(y3 - 4y)(1 + y) = (y3(1 + x) - 4y(1 + x))
∴ = y(1 + x)(y + 2)(y - 2)

Akọwa Nkọwa

4 - $\frac{1}{2-\sqrt{3}}$ = 4 - $\frac{2+\sqrt{2}}{22-\left(\sqrt{3}\right)2}$
= 4 - $\frac{(2+3)}{4-3}$
= 4 - 2- $\sqrt{3}$
= 2 - $\sqrt{3}$

Akọwa Nkọwa

To solve this problem, we need to use the arithmetic progression formula: a_n = a_1 + (n-1)d where a_n is the nth term of the arithmetic progression, a_1 is the first term, n is the number of terms, and d is the common difference. We are given that -8, m, n, and 19 are in arithmetic progression. So we can set up the following equations: m = -8 + d n = -8 + 2d 19 = -8 + 3d We can solve for d by subtracting the first equation from the second equation: n - m = 2d - d n - m = d We can substitute this expression for d into the third equation: 19 = -8 + 3(n - m) Simplifying this equation gives: 27 = 3(n - m) 9 = n - m We can substitute this expression for n - m into the equation we derived earlier: n - m = d So we have: d = 9 Substituting this value of d into any of the earlier equations will allow us to solve for m and n. For example, using the equation: m = -8 + d gives: m = -8 + 9 m = 1 And using the equation: n = -8 + 2d gives: n = -8 + 2(9) n = 10 Therefore, the solution is (m, n) = (1, 10)

Akọwa Nkọwa

Since QS = QR

then, angle SQR = angle SRQ

2 SQR = 180 - 56, SQR = 1242 $\frac{124}{2}$ = 62o

QTP = 62o

QTP = 62o, corresponding angle

3y + 56 + 62 = 180 = 3y = 180 - 118

3y = 62 = 180

3y = 180 - 118

3y = 62

y = 623 $\frac{62}{3}$

= 2032

Akọwa Nkọwa

To find the prime factors of 2520, we need to divide it by its prime factors until we are left with prime numbers only. We start by dividing 2520 by 2, which gives 1260. 1260 is still an even number, so we divide it by 2 again to get 630. 630 is divisible by 2 and 3, so we divide it by 2 to get 315, and then by 3 to get 105. 105 is divisible by 3 and 5, so we divide it by 3 to get 35, and then by 5 to get 7. 7 is a prime number, so we can't divide it any further. Therefore, the prime factors of 2520 are 2, 2, 2, 3, 3, 5, and 7. So the answer is 2, 3, 5, 7.

Akọwa Nkọwa

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

Akọwa Nkọwa

$\frac{r}{a}$ + $\frac{r}{b}$ + $\frac{r}{c}$ > 1 = $\frac{bcr+acr+abr}{abc}$ > 1
r(bc + ac + ba > abc) = r > $\frac{abc}{bc+ac+ab}$

Akọwa Nkọwa

2(x3 - x2 - 2x) = 1
x3 - 2 - 2x = 0
x(x2 - x - 2) = 0
x2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = 2

Akọwa Nkọwa

To find f(x + 1), we need to replace x in the given function f(x) with (x + 1): f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3 Expanding the brackets and simplifying: f(x + 1) = 2(x^2 + 2x + 1) - 5x - 5 + 3 f(x + 1) = 2x^2 + 4x + 2 - 5x - 2 f(x + 1) = 2x^2 - x Therefore, the answer is option A: 2x^2 - x.

Akọwa Nkọwa

4x2 + 5x + Q
To make a complete square, the coefficient of x2 must be 1
= x2 + $\frac{5x}{4}$ + $\frac{Q}{4}$
Then (half the coefficient of x2) should be added
i.e. x2 + $\frac{5x}{4}$ + $\frac{25}{64}$
? $\frac{Q}{4}$ = $\frac{25}{64}$
Q = $\frac{4\times 25}{64}$
= $\frac{25}{16}$

Akọwa Nkọwa

We know that the cost of dinner for the group of students is partly constant and partly varies directly as the number of students. Let the constant cost be c and the cost per student be k. Therefore, the cost of dinner for 20 students is: c + 20k = ₦74.00 ...(1) Similarly, the cost of dinner for 30 students is: c + 30k = ₦96.00 ...(2) Now, we need to find the cost of dinner when there are 15 students. We can solve for the values of c and k using simultaneous equations: Subtracting equation (1) from (2) gives: 10k = ₦22.00 Therefore, k = ₦2.20 Substituting the value of k in equation (1) gives: c + 20(2.20) = ₦74.00 Therefore, c = ₦30.00 So, the cost of dinner when there are 15 students is: c + 15k = ₦30.00 + 15(2.20) = ₦63.00 Therefore, the answer is ₦63.00.

Akọwa Nkọwa

$\sqrt[]{(64r^6{)}^{\frac{1}{2}}}$
$=(64r^6{)}^{\frac{1}{2}\cdot \frac{1}{3}}$
$=64r^{6\cdot \frac{1}{6}}$
$=64r$