To solve this problem, we must determine how many jumps the frog needs to make to get out of the 30-meter deep well.
- The frog jumps up \(70 \text{ cm}\) on each jump and then slides back down \(30 \text{ cm}\).
- The effective height gained after each jump is: \(70 \text{ cm} - 30 \text{ cm} = 40 \text{ cm}\)
- First, convert the depth of the well to centimeters: \(30 \text{ m} = 3000 \text{ cm}\)
- Determine how many effective jumps of \(40 \text{ cm}\) each are needed to reach just below the rim of the well: \(\frac{3000 \text{ cm}}{40 \text{ cm per jump}} = 75 \text{ jumps}\)However, with 75 full jumps, let's consider the scenario:
- After 74 jumps, the frog would have climbed: \(74 \times 40 \text{ cm} = 2960 \text{ cm}\)
- This means at the start of the 75th jump, the frog is 2960 cm up, requiring one more \(70 \text{ cm}\) jump to finish the climb without slipping back again.
Therefore, the minimum number of jumps required for the frog to completely get out of the well is 75.
Summary: The frog effectively gains \(40 \text{ cm}\) per jump, and after 74 jumps, it is at 2960 cm. The 75th jump allows it to reach the top.
Important Questions from Quick Math
A is 120% of B and B is 65% of C. If the sum of A, B and C is 121.5, then the value of C - 2B + A is:
- A college hostel mess has provisions for 25 days for 350 boys. At the end of 10 days, when some boys were shifted to another hostel, it was found that now the provisions will last for 21 more days. How may boys were shifted to another hostel?
- Some students (only boys and girls) from different schools appeared for an Olympiad exam. 20% of the boys and 15% of the girls failed the exam. The number of boys who passed the exam was 70 more than that of the girls who passed the exam. A total of 90 students failed. Find the number of students that appeared for the exam.
The price of an item is reduced by 20%. As a result, customers can get 2 kg more of it for ₹360. Find the original price (in ₹) per kg of the item.
A tyre has 3 punctures. The first puncture alone would have made the tyre flat in 9 minutes, the second alone would have done it in 18 minutes, the third alone would have done it in 6 minutes. If the air leaks out at a constant rate, then how long (in minutes) does it take for all the punctures together to make it flat?
