Question: 1. A trader sold Grilled 1500g of 4 Avocado, fourgo for fried gatoram in the Jen days. On Grilled 32,500 and had a Profit of 20% a) Calculate the cost of each article ii) If she wanted 45% the profit made on the cost price, then much would he have sold each of the Grushes. 2. In a construction Company 50 workers had a Choice of beans, plantain, Rice. Of these workers 21 choose beans, 24 plantain and 15 rice. Also 3 choose beans only, 7 plantain only, 2 rice only and 5 choose all the 3 kinds of food a) Illustrate the information on Venn Diagram b) use the venn diagram to find the i) workers who choose plantain and beans only ii) Fourty two kind of the food iii) None of the 3 kinds of of food.
1. **Problem 1: Calculate cost price and selling price with given profit percentages**
2. **Given:**
- Selling price (SP) with 20% profit = 32,500
- Profit = 20% means SP = Cost Price (CP) + 20% of CP = 1.20 \times CP
3. **Calculate Cost Price (CP):**
$$
SP = 1.20 \times CP \implies CP = \frac{SP}{1.20} = \frac{32500}{1.20} = 27083.33
$$
4. **Calculate new selling price for 45% profit:**
$$
SP_{new} = CP \times 1.45 = 27083.33 \times 1.45 = 39270.83
$$
---
5. **Problem 2: Venn Diagram for workers choosing beans, plantain, rice**
6. **Given:**
- Total workers = 50
- Beans chosen = 21
- Plantain chosen = 24
- Rice chosen = 15
- Beans only = 3
- Plantain only = 7
- Rice only = 2
- All three kinds = 5
7. **Find intersections:**
- Let $x$ = workers choosing beans and plantain only
- Let $y$ = workers choosing plantain and rice only
- Let $z$ = workers choosing beans and rice only
8. **Use inclusion-exclusion principle:**
$$
21 = 3 + x + z + 5 \implies x + z = 21 - 3 - 5 = 13
$$
$$
24 = 7 + x + y + 5 \implies x + y = 24 - 7 - 5 = 12
$$
$$
15 = 2 + y + z + 5 \implies y + z = 15 - 2 - 5 = 8
$$
9. **Solve system:**
$$
x + z = 13
$$
$$
x + y = 12
$$
$$
y + z = 8
$$
10. Subtract second from first:
$$
(x + z) - (x + y) = 13 - 12 \implies z - y = 1
$$
11. From third equation:
$$
y + z = 8
$$
12. Add equations:
$$
(z - y) + (y + z) = 1 + 8 \implies 2z = 9 \implies z = 4.5
$$
13. Then:
$$
y = 8 - z = 8 - 4.5 = 3.5
$$
$$
x = 12 - y = 12 - 3.5 = 8.5
$$
14. **Calculate number of workers choosing exactly two kinds:**
- Beans and plantain only = $x = 8.5$
- Plantain and rice only = $y = 3.5$
- Beans and rice only = $z = 4.5$
15. **Calculate number of workers choosing none of the three:**
$$
\text{Total chosen} = \text{only beans} + \text{only plantain} + \text{only rice} + x + y + z + \text{all three}
$$
$$
= 3 + 7 + 2 + 8.5 + 3.5 + 4.5 + 5 = 33.5
$$
$$
\text{None} = 50 - 33.5 = 16.5
$$
16. **Answer to sub-questions:**
- (i) Workers who choose plantain and beans only = $8.5$
- (ii) "Fourty two kind of the food" is unclear, possibly a typo or misinterpretation, no direct answer.
- (iii) Workers who choose none of the 3 kinds = $16.5$
---
**Final answers:**
1. Cost price = 27083.33, New selling price for 45% profit = 39270.83
2. (i) Plantain and beans only = 8.5 workers
(ii) Not clear
(iii) None of the 3 kinds = 16.5 workers