Using and Managing Data and Information

This document covers various tasks related to using and managing data and information. It includes calculations, equations, and problem-solving techniques. The tasks cover topics such as power calculations, root calculations, mathematical operations, regression lines, perimeter calculations, algebraic equations, and more.

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