The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?
Step-by-step solution
Idea: Each hour multiplies the number by 2, so after n hours the original 30 has been doubled n times.
- Start: 30. After 1 hour: 60; after 2 hours: 120; after 3 hours: 240; after 4 hours: 480.1 mark
- The numbers 30, 60, 120, 240, … form a GP with ratio 2, where the original 30 is ‘hour 0’.1 mark
- After n hours the 30 has doubled n times: 30 × 2n.1 mark
Check: n = 4: 30 × 2⁴ = 30 × 16 = 480 ✓.
Answer to write in the exam
Number at the start = 30; doubles every hour
End of 1st hour = 60, 2nd hour = 120, 3rd hour = 240, 4th hour = 480
End of nth hour = 30 × 2n
∴ 120, 480 and 30 × 2n
Common mistakes that cost marks
- Using 30 × 2n−1 (giving 60 after 2 hours). The 30 is the count at the start, before any doubling; after n hours there have been n doublings.
- Adding 30 each hour (60, 90, 120, …). Doubling means multiplying by 2.
- Writing 2 × 30n. It is 30 multiplied by 2n.
How this can come in the exam
A population of 50 insects triples every week. After 3 weeks there are
- 150
- 450
- 1350
- 4050
Show answer
(C) 1350
50 × 33 = 50 × 27 = 1350.
A video is shared by 5 people on Monday. Each day, the number of people who have shared it doubles.
(i) How many people have shared it on Wednesday? (ii) Write the number after n days (Monday is day 0). (iii) On which day does it first pass 300? (iv) Is this an AP or a GP? Give the common ratio or difference.
Show answer
(i) Tuesday 10, Wednesday 20 (1 mark). (ii) 5 × 2n (1 mark). (iii) 5 × 26 = 320 > 300 while 5 × 25 = 160: day 6, which is Sunday (1 mark). (iv) GP with common ratio 2 (1 mark).Try one yourself
A sum of ₹1000 doubles every 8 years. How much will it be after 32 years?
Show answer
32 years = 4 doublings: 1000 × 24 = ₹16,000.
More questions like this
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- The sum of the first three terms of a GP is 1312 and their product is −1. Find the common ratio and the terms.
- If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
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