Using and Managing Data and Information: Math Assignment Solutions

This document presents solutions to a mathematics assignment focused on using and managing data and information. The assignment covers a range of mathematical concepts, including calculations with exponents and roots, simplification of expressions, and solving various algebraic equations. It also involves problems related to geometry, such as calculating the perimeter of a shape. Furthermore, the assignment delves into problem-solving scenarios involving real-world applications like calculating student combinations, and financial problems. The solutions demonstrate the application of formulas and the step-by-step processes required to arrive at the correct answers, providing a comprehensive guide for students. The tasks include calculations, algebraic manipulations, and word problems designed to test the student's understanding of mathematical principles and their ability to apply these principles to solve practical problems. This assignment provides a solid foundation in mathematical concepts and problem-solving skills.

Using and Managing
Data and Information

TASKS
Task 1
(a) 1.64
To get the correct answer it is needed to enter 1.6 and in calculator with use of power
button for entering 4 power to 1.6.
6.5536
(b) 5√786 = 140.1784
It can be managed by enabling of using root tab throughout the calculator.
(c) 1 / (1+0.07) 3
Now first add 1 to 0.07 that will be 1.07. Later this amount control 3 of the 1.07 by
persistent power key within calculator in addition to resulted value will be 1.22. Then,
divide 1 from resulted figure of 1.22 to determine the suitable answer.
= 1/1.22
= 0.82

Task 2
Task 3
i) 53 × 55 ÷ 52 = (5 × 5 × 5) × (5 × 5 × 5 × 5 × 5) ÷ (5 × 5) = 125 * 3125 ÷ 25 = 125 *
125 = 15625
ii) 10-4 × 10-6 × 103 = 10 (-4 – 6 + 3) = 10 -7
iii) (1/2)2 × (1/2)5 = (1/2)7 = 0.0078125

iv) 2-3 ÷ 2-5 = 2 [-3-(-5)] = 22 = 4
Task 4
In the question, 5 students can we make from a class of 12 students= 12! /5! (12-5)!
= 12! /5!7!
= (12 * 11 * 10 * 8 * 7!)/ (5 * 4 * 3 * 2 * 1 * 7!)
= (12 * 11 *10 * 8) / (5 * 4 * 3 * 2 * 1)
= 792
Task 5
y= 2x + 1
As per the aforementioned equation 2 means the m which is actually the slop of regression
line and 1 will be the b that actually intercept of Y.
A) Description of diagram
i) Gradient of line:
ii) Y= 3m+ 1
iii) Value of y if x is 5
3*5 +1 = 16
Task 6
Size of width = (2x + 3)
Let the length = 2 (2x + 3) = (4x + 6)
Perimeter: 2 (l + w)
= 2 (4x + 6) (2x + 3)
= 2 (4x (2x + 3) + 6 (2x + 3)
= 2 (8x2 + 12x + 12x + 18)
= 2 (8x2 + 24x + 18)
16x2+ 48x + 36

Task 7
a) 5x + 22 = 32
5x= 32-22
X = 10/5 = 2
b) X / 5 = -4/1
X = 5*-4
X= -20
c) 5(x + 1) = 40
5x + 5 = 40
5x = 40-5
5x = 35
X = 35/5 = 7
d) 5 (x+2) + 6 (x + 3) = 39
5x + 10 + 6x + 18 = 39
11x + 28 = 39
11x = 39-28
11x = 11
X = 11/11 = 1
e) x+7 / 2 =-5
(x+7) = -10
X = -10-7
X =-17

f) x + 5 = 13 –x
x + x = 13 – 5
2x = 8
X = 8 / 2 = 4
g) 2(x +3) = 4x - 4
2x +6 = 4x - 4
2x-4x = -4 -6
-2x = -14
X = -14 / 2 = -7
h) 3 (2x +3) + (2x -1) = 7 (x-4)
6x + 9 + 2x – 1 = 7x - 28
8x + 10 = 7x – 28
8x-7x = -28-10
X = -38
i) 1/5 x – 3 = x
x/5- 3 = 7.5
j) 5-2x/5 = 1 + 4/5 x
x = 20
k) 2x + 7 / 6 = 5 – 5x /5
5 (2x + 7) = 6 (5 -5x)
10x + 35 = 30 – 30x

10 x + 30 x = 30 – 35
40x = -5
X = -5 / 40
= -1 / 8
l) 5x + 2 / 3 (x +3) = 1 / 4 (x – 1)
5x +2x + 6 /3 = x – 1 / 4
60x + 8x + 24 = 3x – 3
65 x = -27
X = -27 /65
Task 8
Let no. of table = x
And, no. of chair = y
5x + 4y = £220 equation (i)
5x + 1y = £130 equation (ii)
by solving these equation (i)-(ii):
3y = £220 - £130
3y = £90
y = 90/3 = £30
5x = £130
x = £130 / 5 = £ 26

Task 9
Let, the total amount in bank for each individual as follows:
Ali has = y
Banko = x
Cheng = z
From the question the sum of Ali is £1,500 more as compared with Banko
y = 1500 + x
As well as the amount of cheng is triple the sum of Banko
z = 3x
Thus the sun of money will be = 1500 + x + x + 3x
= 1500 + 4x
Task 10