Certainly! Let's work through these questions step by step.
### Question 1
#### Part (i): Finding the Longest Possible Length for Each Piece
To determine the longest possible length that can evenly cut both pieces of wood (one 120 cm and the other 64 cm), we need to find the greatest common divisor (GCD) of the two lengths.
- Thus, the greatest common divisor (GCD) of 120 and 64 is 8 cm.
This means the longest length of each piece that can equally divide both pieces of wood without any leftovers is 8 cm.
#### Part (ii): Determining the Number of Pieces
Now, let's calculate how many pieces will be obtained from each piece of wood when they are cut into lengths of 8 cm:
- For the 120 cm piece:
[tex]\[
\text{Number of pieces from 120 cm} = \frac{120}{8} = 15
\][/tex]
- For the 64 cm piece:
[tex]\[
\text{Number of pieces from 64 cm} = \frac{64}{8} = 8
\][/tex]
Therefore, you will get 15 pieces from the 120 cm wood and 8 pieces from the 64 cm wood.
### Question 2: Finding the Third Number
We know:
- The lowest common multiple (LCM) of three numbers is 432.
- The greatest common factor (GCF) of the three numbers is 12.
- Two of the numbers are 36 and 48.
We need to find the third number.
Using the relationship between the LCM and GCF of multiple numbers:
[tex]\[
\text{LCM}(a, b, c) \times \text{GCF}(a, b, c) = a \times b \times c
\][/tex]
Let's substitute the given values and solve for the third number [tex]\(c\)[/tex]:
[tex]\[
432 \times 12 = 36 \times 48 \times c
\][/tex]
First, calculate the left-hand side:
[tex]\[
432 \times 12 = 5184
\][/tex]
Next, calculate [tex]\(36 \times 48\)[/tex]:
[tex]\[
36 \times 48 = 1728
\][/tex]
Now we solve for [tex]\(c\)[/tex]:
[tex]\[
5184 = 1728 \times c \\
c = \frac{5184}{1728} \\
c = 3
\][/tex]
So, the third number is 3.
### Summary of Answers
1. a) i. The longest possible length of each piece is 8 cm.
ii. Number of pieces obtained will be [tex]\(\text{15 pieces}\)[/tex] from the 120 cm wood and [tex]\(\text{8 pieces}\)[/tex] from the 64 cm wood.
2. The third number is [tex]\(\textbf{3}\)[/tex].
