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A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.
If a1 = a2 = a3 = . . = a10 = 150 , and
a10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes is
If a1 = a2 = a3 = . . = a10 = 150 , and
a10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes is
- a)24 minutes
- b)34 minutes
- c)125 minutes
- d)135 minutes
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A person is to count 4500 notes. Let an denote the number of notes he ...
Let us assume that total minutes = x
For first nine minutes = 150 × 9
Let the remaining minutes = y = x − 9 Now,
(150 × 9)
[2 × 150 + (y − 1)(−2)] = 4500
(302 − 2y) = 3150 ⟹ y2 − 151y + 3150 = 0
⟹ (y − 126)(y − 25) = 0 ⟹ y = 25 or 126
(Rejected)
So, x = y + 9 = 25 + 9 = 34 minutes
For first nine minutes = 150 × 9
Let the remaining minutes = y = x − 9 Now,
(150 × 9)
[2 × 150 + (y − 1)(−2)] = 4500 (302 − 2y) = 3150 ⟹ y2 − 151y + 3150 = 0⟹ (y − 126)(y − 25) = 0 ⟹ y = 25 or 126
(Rejected)
So, x = y + 9 = 25 + 9 = 34 minutes
Most Upvoted Answer
A person is to count 4500 notes. Let an denote the number of notes he ...
Let's break down the problem step by step:
We are given that a1 = a2 = a3 = . . = a10 = 150. This means that for the first 10 minutes, the person counts 150 notes per minute.
Now, let's find the common difference (d) for the arithmetic progression (AP) starting from the 10th minute.
We know that a10 + d = a11, a11 + d = a12, and so on.
Since a10 = 150, we can substitute this value to get:
150 + d = a11
150 + 2d = a12
We are also given that a11, a12, and so on are in AP with a common difference of d. This means that the difference between any two consecutive terms in the sequence is always d.
Now, let's find the total number of notes counted after n minutes.
The person counts 150 notes per minute for the first 10 minutes, so the total number of notes counted in the first 10 minutes is:
a1 + a2 + a3 + ... + a10 = 150 + 150 + 150 + ... + 150 (10 terms)
Since there are 10 terms, the sum is:
10 * 150 = 1500
After the 10th minute, the person counts a11, a12, and so on, with a common difference of d. So, the total number of notes counted after n minutes is:
1500 + a11 + a12 + ... + an
Since a11, a12, and so on are in AP with a common difference of d, we can use the formula for the sum of an arithmetic series to find the sum of these terms:
Sum = (n - 10) / 2 * (2a + (n - 10 - 1) * d)
where n is the number of minutes.
So, the total number of notes counted after n minutes is:
1500 + (n - 10) / 2 * (2a + (n - 10 - 1) * d)
In this case, a = a10 = 150 and d is the common difference.
Therefore, the formula becomes:
1500 + (n - 10) / 2 * (2 * 150 + (n - 10 - 1) * d)
Since we are counting 4500 notes in total, we can set this formula equal to 4500 and solve for n:
1500 + (n - 10) / 2 * (2 * 150 + (n - 10 - 1) * d) = 4500
Simplifying this equation will give us the value of n, which is the number of minutes it takes to count 4500 notes.
We are given that a1 = a2 = a3 = . . = a10 = 150. This means that for the first 10 minutes, the person counts 150 notes per minute.
Now, let's find the common difference (d) for the arithmetic progression (AP) starting from the 10th minute.
We know that a10 + d = a11, a11 + d = a12, and so on.
Since a10 = 150, we can substitute this value to get:
150 + d = a11
150 + 2d = a12
We are also given that a11, a12, and so on are in AP with a common difference of d. This means that the difference between any two consecutive terms in the sequence is always d.
Now, let's find the total number of notes counted after n minutes.
The person counts 150 notes per minute for the first 10 minutes, so the total number of notes counted in the first 10 minutes is:
a1 + a2 + a3 + ... + a10 = 150 + 150 + 150 + ... + 150 (10 terms)
Since there are 10 terms, the sum is:
10 * 150 = 1500
After the 10th minute, the person counts a11, a12, and so on, with a common difference of d. So, the total number of notes counted after n minutes is:
1500 + a11 + a12 + ... + an
Since a11, a12, and so on are in AP with a common difference of d, we can use the formula for the sum of an arithmetic series to find the sum of these terms:
Sum = (n - 10) / 2 * (2a + (n - 10 - 1) * d)
where n is the number of minutes.
So, the total number of notes counted after n minutes is:
1500 + (n - 10) / 2 * (2a + (n - 10 - 1) * d)
In this case, a = a10 = 150 and d is the common difference.
Therefore, the formula becomes:
1500 + (n - 10) / 2 * (2 * 150 + (n - 10 - 1) * d)
Since we are counting 4500 notes in total, we can set this formula equal to 4500 and solve for n:
1500 + (n - 10) / 2 * (2 * 150 + (n - 10 - 1) * d) = 4500
Simplifying this equation will give us the value of n, which is the number of minutes it takes to count 4500 notes.
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A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.If a1 = a2 = a3 = . . = a10 = 150 , anda10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes isa)24 minutesb)34 minutesc)125 minutesd)135 minutesCorrect answer is option 'B'. Can you explain this answer?
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A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.If a1 = a2 = a3 = . . = a10 = 150 , anda10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes isa)24 minutesb)34 minutesc)125 minutesd)135 minutesCorrect answer is option 'B'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.If a1 = a2 = a3 = . . = a10 = 150 , anda10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes isa)24 minutesb)34 minutesc)125 minutesd)135 minutesCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.If a1 = a2 = a3 = . . = a10 = 150 , anda10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes isa)24 minutesb)34 minutesc)125 minutesd)135 minutesCorrect answer is option 'B'. Can you explain this answer?.
A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.If a1 = a2 = a3 = . . = a10 = 150 , anda10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes isa)24 minutesb)34 minutesc)125 minutesd)135 minutesCorrect answer is option 'B'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.If a1 = a2 = a3 = . . = a10 = 150 , anda10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes isa)24 minutesb)34 minutesc)125 minutesd)135 minutesCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A person is to count 4500 notes. Let an denote the number of notes he counts in the nth minute.If a1 = a2 = a3 = . . = a10 = 150 , anda10 , a11 , a12 . .. are in AP with the common difference −2 , then the time taken by him to count all the notes isa)24 minutesb)34 minutesc)125 minutesd)135 minutesCorrect answer is option 'B'. Can you explain this answer?.
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