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001 JAMB CBT 2024 MATHEMATICS QUESTIONS AND ANSWERS

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INSTRUCTIONS: Answer All Questions In This Part
===========================

(1a)
Principal P= 15,000
Amount = (5/4)P = 5/4 x 15,000
=18,750
Simple interest = A – P
= 18,750 – 15,000
= ₦3750
I = PRT/100
T = 100I/PR
= (100×3750) /(15000 x25)
= 4 x 1/4
= 1year

(1b)
A = P(1 + R/100)^T
A = 4000(1 + 5/100)²
= 4000 x (1.05)²
= 4000 x 105/100 x 105/100
= ₦4,410
Compound interest = A – P
= 4,410 – 4000
= ₦410
===========================

(2a)
3/x = 9/(8x +20)
3(8x + 20) = 9x
24x – 9x = – 60
15x = – 60
x = – 60/15
x = – 4

(2b)

===========================

(3a)

Using SOHCAHTOA
Tanθ = Opposite/Adjacent
Tan 40°= H-1.8/3.5
H – 1.8 = 35 x Tan40°
H – 1.8 = 29.3685
H = 29.3685 + 1.8
H = 31.1685
Height of building = 31.2m

(3b)

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Using Sine Rule
QR/sin20° = 20/sin50°
QR = 20sin20°/sin50°
QR = 20×0.3420/0.7660
QR = 8.93km
===========================

(4ai)
∑f(x)= 5+6+9+7+3 =30
Prob(both are 18)
= 9/30 x 8/29
= 12/145

(4aii)
Prob(16 & 20)
= (5/30 x 3/29) + (3/30 x 5/29)
= 1/58 + 1/58
= 1/29

(4b)
Average age = (16×5) + (17×6) + (18×9) + (19×7) + (20×3)/30
= (80 + 102 + 162 +133 +60)/30
= 537/30
= 17.9years
===========================

(5a)
y = (x² + 1)³
u = x² + 1
y= u³
du/dx = 2x
dy/du = 3u²
using chain rule
dy/dx = (du/dx) x (dy/du)
= 2x x 3u²
= 6xu²
dy/dx = 6x(x² + 1)²

(5b)
y = 3x² + 2x + 1
∫³₂ (3x² + 2x + 1)dx
[3³ + 3² + 3 + c] – [2³ + 2² + 2 + c]
= 27 + 9 + 3 + c – 8 – 4 – 2 – c
= 27 + 12 – 8 – 6
= 25
===========================

PART II
INSTRUCTIONS: Answer Five(5) Questions Only In This Part
===========================

(6ai)
X = {x: – 5, -4, -2, 3}
Y = {x: -2, -1, 1, 2}
Z = {x: – 5, -2, -1, 3}

(6aii)
(i) XnZ = {-5, -2, 3}
(ii) XnY = {-5, -4, -2, -1, 1, 2, 3}
(iii) XnYnZ = {-2}

(6aiii)
(X∩Y) ∪ (X∩Z) = X∩(Y∪Z)
{-2}U{-5,-2,3} = {-5,-4,-2,3}∩{-5,-2,-1,1,2,3}
{-5, – 2, 3} = {-5, – 2, 3}

(6bi)
Actual Area = 50×40 = 2000cm²
Measured Area = 43×35 = 1505cm²
%error = |Measured – Actual|_/|Actual| x 100%
= |1505 – 2000|_/|2000| x 100%
= 24.75%

(6bii)
%error in length= |43-50|/|50| x 100%
= 7/50 x 100%
= 14%
===========================

(7ai)
Gradient, S = 3 – 5/(-3-2)
= – 2/-5
= 2/5
Angle = Tan-¹(2/5)
= 21.8°

(7aii)
Point of intersection
5x + 6 = 3x + 2
5x – 3x = 2 – 6
2x = – 4
x = – 4/2
x = – 2
y = 5x + 6 = 3x +2
y = 3(-2) + 2 = – 6 + 2 = – 4
Point of intersection (- 2, – 4)

(7aiii)
dy/dx = 10x + 3
y = ∫(10x + 3)dx
y = 10x²/2 +3x + c
y= 5x² + 3x + c
at (1, 0), f(1) = 5(1)² + 3(1) +c
c = – 8
Curve is y = 5x² + 3x – 8

(7b)
5x + y < 5 ……..(1)
4x – y < 1……… (2)
Adding both equations
9x < 6
x < 6/9 < 2/3
5x + y < 5
y < 5 – (5x 2/3)
y < 5 – 10/3 < 5/3
===========================

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(9a)

(9bi)
Let the marked price be ₦x
Cash discount = (₦x) x (10x/100)
Trade discount = (₦x) x (15/100)
x – (10x/100) – (15x/100) = 8100
100x – 10x – 15x = 810,000
75x = 810,000
x = 810,000/75
x = ₦10,800

(9bii)
Cash discount = ₦10,800 x 10/100
= ₦1,080

(9biii)
Trade discount = ₦10,800 x 15/100
= ₦1,620
===========================

(10a)
(6x+5y)/(x+5y) = 5
(6x/y +5)/(x/y +5) = 5
6(x/y) + 5 = 5(x/y) + 25
(x/y) = 25 – 5 = 20
x = 20y
. : (x² – y²) /(2xy) = (20y)² – y²/2(20y)y
= (400y² – y²)/40y²
= 399y²/40y²
= 9.975

(10b)
Sin(x+y) = (√3)/2
Cos(x-y) = (√3)/2
x + y = Sin-¹(√3/2) = 60 ….. (1)
x – y = Cos-¹(√3/2) = 30 …..(2)
Adding equ(1) and equ(2)
2x = 90
x =90/2
x = 45
y = 60 – x
y = 60 – 45
y = 15
===========================

(12)
TABULATE

Class boundary | 0.5-10.5 | 10.5-20.5 | 20.5-30.5 | 30.5-40.5 | 40.5-50.5

Frequency | 8 | 10 | 18 | 30 | 14

X | 5.5 | 15.5 | 25.5 | 35.5 | 45.5

fx | 44 | 155 | 459 | 1065 | 637

(x- xÌ„) | – 24 | -14 | – 4 | 6 | 16

(x-x̄) ² | 576 | 196 | 16 | 36 | 256

f(x-x̄)² | 4608 | 1960 | 288 | 1080 | 3584

∑f = 80
∑fx = 2360
∑f(x-x̄)² ⁼ 11,520

Mean(x̄) = ∑fx/∑f
= 2360/80
=29.5

(i) Variance = ∑f(x-x̄)²/∑f
= 11,520/80
= 144

(ii) Standard deviation = √variance
= √144
= 12

(iii) Modal Mark = 30.5 + (30-18/2(30)-18-14) x10
= 30.5 + 4.2757
= 34.8

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