Sure, let's tackle these ratio and proportion problems step-by-step:
### Question 8
#### Part a)
Given:
1. [tex]\( p: q = 1: 2 \)[/tex]
2. [tex]\( q: r = 4: 3 \)[/tex]
We need to find [tex]\( r \)[/tex].
First, express both ratios with the same [tex]\( q \)[/tex].
1. From [tex]\( p: q = 1: 2 \)[/tex], we have:
[tex]\[ p = 1 \text{ unit}, \quad q = 2 \text{ units} \][/tex]
2. From [tex]\( q: r = 4: 3 \)[/tex], we have:
[tex]\[ q = 4 \text{ units}, \quad r = 3 \text{ units} \][/tex]
To combine these ratios, find a common value for [tex]\( q \)[/tex]. The least common multiple (LCM) of 2 and 4 is 4.
Scale the first ratio [tex]\( p: q \)[/tex]:
[tex]\[ p = 1 \text{ unit} \quad \text{scaled by factor of 2:} \quad p = 2 \text{ units} \][/tex]
[tex]\[ q = 2 \text{ units} \quad \text{scaled by factor of 2:} \quad q = 4 \text{ units} \][/tex]
So, [tex]\( p: q: r = 2: 4: 3 \)[/tex].
Thus, [tex]\( r \)[/tex] in terms of [tex]\( p \)[/tex] is [tex]\( r = 3 \)[/tex].
### Part b)
Given:
1. [tex]\( a: b = 2: 3 \)[/tex]
2. [tex]\( b: c = 6: 5 \)[/tex]
We need to find [tex]\( c \)[/tex].
First, express both ratios with the same [tex]\( b \)[/tex].
1. From [tex]\( a: b = 2: 3 \)[/tex], we have:
[tex]\[ a = 2 \text{ units}, \quad b = 3 \text{ units} \][/tex]
2. From [tex]\( b: c = 6: 5 \)[/tex], we have:
[tex]\[ b = 6 \text{ units}, \quad c = 5 \text{ units} \][/tex]
To combine these ratios, find a common value for [tex]\( b \)[/tex]. The least common multiple (LCM) of 3 and 6 is 6.
Scale the first ratio [tex]\( a: b \)[/tex]:
[tex]\[ a = 2 \text{ units} \quad \text{scaled by factor of 2:} \quad a = 4 \text{ units} \][/tex]
[tex]\[ b = 3 \text{ units} \quad \text{scaled by factor of 2:} \quad b = 6 \text{ units} \][/tex]
So, [tex]\( a: b: c = 4: 6: 5 \)[/tex].
Thus, [tex]\( c \)[/tex] in terms of [tex]\( a \)[/tex] is [tex]\( c = 5 \)[/tex].
### Question 9
#### Part a)
Given the proportion:
[tex]\[ x: 3 = 4: 6 \][/tex]
Cross-multiply to solve for [tex]\( x \)[/tex]:
[tex]\[ x \cdot 6 = 4 \cdot 3 \][/tex]
[tex]\[ 6x = 12 \][/tex]
[tex]\[ x = 2 \][/tex]
#### Part b)
Given the proportion:
[tex]\[ 2: x = 6: 9 \][/tex]
Cross-multiply to solve for [tex]\( x \)[/tex]:
[tex]\[ 2 \cdot 9 = 6 \cdot x \][/tex]
[tex]\[ 18 = 6x \][/tex]
[tex]\[ x = 3 \][/tex]
#### Part c)
Given the proportion:
[tex]\[ 5: 4 = x: 12 \][/tex]
Cross-multiply to solve for [tex]\( x \)[/tex]:
[tex]\[ 5 \cdot 12 = 4 \cdot x \][/tex]
[tex]\[ 60 = 4x \][/tex]
[tex]\[ x = 15 \][/tex]
### Question 10
#### Part a)
The problem statement is incomplete, so it's not possible to provide a solution.
#### Part b)
Given:
- The ratio of length to breadth is [tex]\( 4: 3 \)[/tex].
- The breadth is [tex]\( 75 \)[/tex] meters.
We need to find the length.
From the ratio:
[tex]\[ \frac{\text{length}}{\text{breadth}} = \frac{4}{3} \][/tex]
Given the breadth:
[tex]\[ \text{length} = \frac{4}{3} \times 75 \][/tex]
[tex]\[ \text{length} = 100 \][/tex]
Thus, the length of the ground is [tex]\( 100 \)[/tex] meters.
