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प्रश्न
| The number of bacteria in a certain culture doubles every hour. There were 30 bacteria present in the culture originally. |
1. The number of bacteria that would be present at the end of 4th hour is:
(a) 240
(b) 480
(c) 960
(d) 450
2. The number of bacteria that would be present at the end of 10th hour is:
(a) 30720
(b) 15360
(c) 61440
d) 27540
3. The number of bacteria that would be present at the end of nth hour is:
(a) `30 xx 2^((n - 1))`
(b) `30 xx 2^((n + 1))`
(c) `30 xx 2^n`
(d) none of these
4. What would be the sum of the first 5 terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
(a) 450
(b) 930
(c) 1890
(d) none of these
5. What would be the sum of the first n terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
(a) `30 xx 2^n - 1`
(b) `30 xx (2^(n - 1) - 1)`
(c) `30 xx (2^(n + 1) - 1)`
(d) `30 xx 2^n - 30`
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उत्तर
1. The correct option is (b)
Reasoning:
Given that the initial count of bacteria at \[t = 0\] is 30.
Since the bacteria count doubles every hour, the common ratio is \[r = 2\].
At the end of the 1st hour: \[30 \times 2 = 60\]
At the end of the 2nd hour: \[30 \times 2^2 = 120\]
At the end of the 4th hour: \[30 \times 2^4 = 30 \times 16 = 480\]
Therefore, while options (a), (c) and (d) are incorrect.
2. The correct option is (a)
Reasoning:
The number of bacteria present at the end of the \[n^{\text{th}}\] hour is given by: \[N_n = 30 \times 2^n\]
For the end of the 10th hour \([n = 10]\):
\[N_{10} = 30 \times 2^{10}\]
\[= 30 \times 1024\]
\[= 30720\]
Therefore, while options (b), (c) and (d) represent incorrect powers or calculations.
3. The correct option is (c)
Reasoning:
The initial population is 30.
After 1 hour, population = \[30 \times 2^1\]
After 2 hours, population = \[30 \times 2^2\]
After [n] hours, population = \[30 \times 2^n\]
Hence, option (a) represents the value at the beginning of the \[n^{\text{th}}\] hour, and option (b) represents the value at the end of the \[(n + 1)^{\text{th}}\] hour.
4. The correct option is (b)
Reasoning:
Taking the standard G.P. defined by the bacteria growth model with first term \[a = 30\] and common ratio \[r = 2\]: \[S_n = \frac{a(r^n - 1)}{r - 1}\]
For \[n = 5\]: \[S_5 = \frac{30(2^5 - 1)}{2 - 1}\]
\[= 30(32 - 1)\]
\[= 30 \times 31\]
\[= 930\]
Therefore, option (b) is the correct answer.
5. The correct option is (d)
Reasoning:
For a G.P. with first term \[a = 30\] and common ratio \[r = 2\]:
\[S_n = \frac{a(r^n - 1)}{r - 1}\]
\[S_n = \frac{30(2^n - 1)}{2 - 1}\]
\[= 30(2^n - 1)\]
\[= 30 \times 2^n - 30\]
Thus, options (a), (b) and (c) do not match the algebraic simplification of \[S_n\].
