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A group of 630 students are to be arranged in a row such that each subsequent row contains 3 students more than the previous row. What number of rows is not possible?
- a)3
- b)5
- c)7
- d)6
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A group of 630 students are to be arranged in a row such that each sub...
Let n be the number of children in the least populated row. The subsequent rows will have n+3, n+6, n+9 ... etc children.
Start by POEing the options.
Option A, the distribution will be n, n+3, n+6 ---> n+n+3+n+6 = 630 --> n = Integer. Possible
Option B, the distribution will be n, n+3, n+6, n+9 ---> n+n+3+n+6+n+9 = 630 --> n = Integer. Possible
Option C, the distribution will be n, n+3, n+6, n+9, n+12 ---> n+n+3+n+6+n+9+ n+12 = 630 --> n = Integer. Possible
Option D, the distribution will be n, n+3, n+6, n+9, n+12, n+15 ---> n+n+3+n+6+n+9+ n+12+n+15 = 630 --> n ≠≠ Integer. NOT POSSIBLE.
Option E, the distribution will be n, n+3, n+6, n+9, n+12, n+15, n+18 ---> n+n+3+n+6+n+9+ n+12+n+15+n+18 = 630 --> n = Integer. possible.
D is the correct answer.
Most Upvoted Answer
A group of 630 students are to be arranged in a row such that each sub...
The most simple solution (can solve in few seconds ;D) :
Let's suppose there are 'x' number of students in 1st row.
Then the next row will contain 'x+3' students.
Further the next row will contain 'x+6' students and so on.
The sequence will become,
x, x+3, x+6, x+9,.........
You can observe as this is an Arithmetic Progression.
Summation of n number of terms in Arithmetic Progression is :
2*x + (n-1) * 3
Where x is the first term, 3 is the common difference and n is the number of terms.
The sum must be equal to 630.
Equating both we get,
2*x + (n-1) * 3 = 630
Where 'n' and 'x' both must be an integer,
Therefore, for x to be an integer (n-1) * 3 have to be even. That means (n-1) have to be even
Which ultimately means 'n' have to be odd for that.
But the ques asked is which is not possible, so the only possible answer is the value of n which is even.
If you think fast, you will get the answer in first 10-15 seconds.
Let's suppose there are 'x' number of students in 1st row.
Then the next row will contain 'x+3' students.
Further the next row will contain 'x+6' students and so on.
The sequence will become,
x, x+3, x+6, x+9,.........
You can observe as this is an Arithmetic Progression.
Summation of n number of terms in Arithmetic Progression is :
2*x + (n-1) * 3
Where x is the first term, 3 is the common difference and n is the number of terms.
The sum must be equal to 630.
Equating both we get,
2*x + (n-1) * 3 = 630
Where 'n' and 'x' both must be an integer,
Therefore, for x to be an integer (n-1) * 3 have to be even. That means (n-1) have to be even
Which ultimately means 'n' have to be odd for that.
But the ques asked is which is not possible, so the only possible answer is the value of n which is even.
If you think fast, you will get the answer in first 10-15 seconds.
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Community Answer
A group of 630 students are to be arranged in a row such that each sub...
Problem: A group of 630 students are to be arranged in a row such that each subsequent row contains 3 students more than the previous row. What number of rows is not possible?
Solution:
To find the number of rows that is not possible, let's first analyze the given information:
- We have a total of 630 students.
- Each subsequent row contains 3 students more than the previous row.
Now, let's assume the number of students in the first row as 'x'.
The second row will have 'x + 3' students, the third row will have 'x + 6' students, and so on.
We can form an arithmetic progression with the first term 'x' and the common difference '3'.
Since the sum of an arithmetic progression can be found using the formula:
Sum = (Number of terms / 2) * (First term + Last term),
we can set up the equation to find the total number of students:
630 = (Number of terms / 2) * (x + Last term)
Since we know the common difference is 3, the last term can be expressed as:
Last term = x + (Number of terms - 1) * 3
Substituting the value of the last term in the equation:
630 = (Number of terms / 2) * (x + x + (Number of terms - 1) * 3)
Simplifying the equation further:
630 = (Number of terms / 2) * (2x + 3(Number of terms - 1))
Now, we need to find the number of rows that is not possible. This means that the total number of students in a row should not exceed 630.
Let's consider the given options:
a) 3 rows:
For 3 rows, the total number of students will be (3/2) * (2x + 3(3-1)) = 9x + 18.
Since 9x + 18 > 630, this option is not possible.
b) 5 rows:
For 5 rows, the total number of students will be (5/2) * (2x + 3(5-1)) = 20x + 60.
Since 20x + 60 > 630, this option is not possible.
c) 7 rows:
For 7 rows, the total number of students will be (7/2) * (2x + 3(7-1)) = 35x + 105.
Since 35x + 105 > 630, this option is not possible.
d) 6 rows:
For 6 rows, the total number of students will be (6/2) * (2x + 3(6-1)) = 18x + 54.
Since 18x + 54 <= 630,="" this="" option="" is="">
Therefore, the correct answer is option 'C' - 7 rows is not possible.
Solution:
To find the number of rows that is not possible, let's first analyze the given information:
- We have a total of 630 students.
- Each subsequent row contains 3 students more than the previous row.
Now, let's assume the number of students in the first row as 'x'.
The second row will have 'x + 3' students, the third row will have 'x + 6' students, and so on.
We can form an arithmetic progression with the first term 'x' and the common difference '3'.
Since the sum of an arithmetic progression can be found using the formula:
Sum = (Number of terms / 2) * (First term + Last term),
we can set up the equation to find the total number of students:
630 = (Number of terms / 2) * (x + Last term)
Since we know the common difference is 3, the last term can be expressed as:
Last term = x + (Number of terms - 1) * 3
Substituting the value of the last term in the equation:
630 = (Number of terms / 2) * (x + x + (Number of terms - 1) * 3)
Simplifying the equation further:
630 = (Number of terms / 2) * (2x + 3(Number of terms - 1))
Now, we need to find the number of rows that is not possible. This means that the total number of students in a row should not exceed 630.
Let's consider the given options:
a) 3 rows:
For 3 rows, the total number of students will be (3/2) * (2x + 3(3-1)) = 9x + 18.
Since 9x + 18 > 630, this option is not possible.
b) 5 rows:
For 5 rows, the total number of students will be (5/2) * (2x + 3(5-1)) = 20x + 60.
Since 20x + 60 > 630, this option is not possible.
c) 7 rows:
For 7 rows, the total number of students will be (7/2) * (2x + 3(7-1)) = 35x + 105.
Since 35x + 105 > 630, this option is not possible.
d) 6 rows:
For 6 rows, the total number of students will be (6/2) * (2x + 3(6-1)) = 18x + 54.
Since 18x + 54 <= 630,="" this="" option="" is="">
Therefore, the correct answer is option 'C' - 7 rows is not possible.
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