Find the appropriate relation for quantity 1 and quantity 2 in the following question :
Quantity 1 : In an examination, Ankita scored 35 marks less than Puneeta. Puneeta scored 65 more marks than Meenakshi. Rakhi scored 115 marks, which is 20 marks more than Meenkashi's . Simpy scored 108 marks less than the maximum marks of the test. What approximate percentage of marks did Simpy score in the examination, if she got 67 marks more than Ankita ?
Quantity 2 : The length of a rectangle is increased by `60%`. By what percent would the width have to be decreased to maintain the same area ?

Options

A. A)Quantity 1 `lt` Quantity 2
B. B)Quantity 1 `le` Quantity 2
C. C)Quantity 1 `ge` Quantity 2
D. D)Quantity 1 `gt` Quantity 2

Answer

To solve the problem, we will break it down into two parts: Quantity 1 and Quantity 2.

### Quantity 1:

1. **Let Ankita's score be K.**
- Therefore, Ankita's score = K.

2. **Puneeta's score:**
- Ankita scored 35 marks less than Puneeta, so:
\[ \text{Puneeta's score} = K + 35 \]

3. **Meenakshi's score:**
- Puneeta scored 65 more than Meenakshi, so:
\[ \text{Meenakshi's score} = (K + 35) - 65 = K - 30 \]

4. **Rakhi's score:**
- Rakhi scored 115 marks, which is 20 marks more than Meenakshi:
\[ \text{Rakhi's score} = \text{Meenakshi's score} + 20 = (K - 30) + 20 = K - 10 \]
- Setting Rakhi's score equal to 115 gives:
\[ K - 10 = 115 \implies K = 125 \]

5. **Simpy's score:**
- Simpy scored 67 marks more than Ankita:
\[ \text{Simpy's score} = K + 67 = 125 + 67 = 192 \]

6. **Total marks in the examination:**
- Simpy scored 108 marks less than the maximum marks:
\[ \text{Maximum marks} = \text{Simpy's score} + 108 = 192 + 108 = 300 \]

7. **Calculating the percentage of marks Simpy scored:**
\[ \text{Percentage of marks} = \left(\frac{\text{Simpy's score}}{\text{Total marks}}\right) \times 100 = \left(\frac{192}{300}\right) \times 100 \]
\[ = 64\% \]

### Quantity 2:

1. **Let the original length of the rectangle be L and the original width be W.**
- The length is increased by 60%, so:
\[ \text{New length} = L + 0.6L = 1.6L \]

2. **Area before the increase:**
\[ \text{Original Area} = L \times W \]

3. **Area after the increase (to remain constant):**
\[ \text{New Area} = \text{New length} \times \text{New width} = 1.6L \times W' \]
- Setting the areas equal gives:
\[ L \times W = 1.6L \times W' \]
- Dividing both sides by L (assuming L ≠ 0):
\[ W = 1.6W' \]
- Rearranging gives:
\[ W' = \frac{W}{1.6} = \frac{W}{\frac{8}{5}} = \frac{5W}{8} \]

4. **Calculating the percentage decrease in width:**
- The original width is W and the new width is \( \frac{5W}{8} \).
- The decrease in width is:
\[ \text{Decrease} = W - \frac{5W}{8} = \frac{3W}{8} \]
- The percentage decrease is:
\[ \text{Percentage decrease} = \left(\frac{\text{Decrease}}{\text{Original Width}}\right) \times 100 = \left(\frac{\frac{3W}{8}}{W}\right) \times 100 = \frac{3}{8} \times 100 = 37.5\% \]

### Conclusion:

- **Quantity 1:** 64%
- **Quantity 2:** 37.5%

Since \( 64\% > 37.5\% \), we conclude that **Quantity 1 is greater than Quantity 2**.