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Solve the following problems related to ratios and shares.

(a) Simplify the given ratios:

(i)
[tex]\[
4: 4 = 1 \frac{1}{2} \div 2 \frac{2}{3} = 1 \frac{5}{2}: 2 x^2 = \frac{3 \pi}{42}, \frac{12}{52} = \frac{15}{15}: \frac{24}{10} = 0.5: \frac{10}{8} \alpha + 3
\][/tex]

(ii)
[tex]\[
6:c = 0.105:0.50 = xa70.105:0.389 \times 1000 = 3105: 350 \quad 1 \div 35
\][/tex]

(b) Hence, find the ratio of [tex]\(a: b: c\)[/tex]:

[tex]\[
a: b: c = x a 5: 5: 3 = 15: 2 + \div 50
\][/tex]

(c) Alan, Bob, and Cathy share [tex]\(\$ 500\)[/tex] in the ratio [tex]\(a: b: c\)[/tex] found in (b). Find Alan's share, correct to the nearest cent.



Answer :

GinnyAnswer
Certainly! Let's solve the problem step by step.

We are given the ratios a : b : c as 15 : 2 : 50 from part (b).

(c) Alan, Bob, and Cathy share [tex]$500 in the ratio of a : b : c. Find Alan's share correct to the nearest cent. 1. Identify the given values: - Ratio of Alan (a) = 15 - Ratio of Bob (b) = 2 - Ratio of Cathy (c) = 50 - Total amount to be shared = $[/tex]500

2. Calculate the total ratio:
- Total ratio = a + b + c
- Total ratio = 15 + 2 + 50
- Total ratio = 67

3. Calculate each person's share:
- Alan's share = [tex]\(\left(\frac{a}{\text{Total ratio}}\right) \times \text{Total amount}\)[/tex]
- Bob's share = [tex]\(\left(\frac{b}{\text{Total ratio}}\right) \times \text{Total amount}\)[/tex]
- Cathy's share = [tex]\(\left(\frac{c}{\text{Total ratio}}\right) \times \text{Total amount}\)[/tex]

4. Substitute the values to find each share:
- Alan's share = [tex]\(\left(\frac{15}{67}\right) \times 500\)[/tex]
- Bob's share = [tex]\(\left(\frac{2}{67}\right) \times 500\)[/tex]
- Cathy's share = [tex]\(\left(\frac{50}{67}\right) \times 500\)[/tex]

5. Calculate the monetary amounts:
- Alan's share = [tex]\( \frac{15}{67} \times 500 \approx 111.94\)[/tex]
- Bob's share = [tex]\( \frac{2}{67} \times 500 \approx 14.93\)[/tex]
- Cathy's share = [tex]\( \frac{50}{67} \times 500 \approx 373.13\)[/tex]

Therefore, Alan's share, rounded to the nearest cent, is $111.94.

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