Students in a school are made to stand in a row and column formation. On an average day, the number
of rows and the number of columns is fixed. On one particular day, the row size was increased by 4.
which lead to the number of rows decreasing by 2. On another day, the row size was decreased by 4,
which led to the number of rows increasing by 4. How many students are there in the school.
Answers
Answer:
Step-by-step explanation:
Let the (normal) number of rows be x and the number of students per row be y:
⇒ the total number of students will be = xy
[Case 1] - The number of students per row was increased by 4
the new number of students per row became y+4, and the number of rows became x−2.
(terms of these new numbers of rows and columns) the total number of students is (x−2)(y+4)
total number of students did not change. so we have:
(x−2)(y+4)=xy
⇒ xy−2y+4x−8=xy
⇒ 2x−y−4=0 ...(1)
[Case 2] - When the number of students per row was decreased by 4
then the new number of students per row became y−4, and the number of rows became x+4.
(terms of these new numbers of rows and columns) the total number of students is (x+4)(y−4)
total number of students did not change. so we have:
(x+4)(y−4)=xy
⇒ xy+4y−4x−16=xy
⇒ x−y+4=0 ...(2)
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So we now have:
2x−y−4=0 ...(1)
x−y+4=0 ...(2)
Subtracting equation (2) from (1), we get:
x−8=0
⇒ x=8
On substituting x in equation (2), we get:
8−y+4=0
⇒ 12−y=0
⇒ y=12
So, the total number of students in the school is xy=8×12=96.
xy = 96
Answer:
Therefore, s = 2.5. Since s must be a positive integer, we can assume that each row originally had 3 students on an average day.
Therefore, the total number of students on an average day is:
Total number of students = r x c x s = 8 x 4 x 3 = 96
Therefore, there are 96 students in the school.
Step-by-step explanation:
Let's assume that on an average day, there are r rows and c columns of students, and each row has s students. Therefore, the total number of students on an average day would be:
Total number of students = r x c x s
On the first day, when the row size was increased by 4 and the number of rows decreased by 2, we can set up the following system of equations:
(r - 2) x c x (s + 4) = r x c x s .....(1)
Simplifying equation (1), we get:
4rc - 8c = 2rs .....(2)
On the second day, when the row size was decreased by 4 and the number of rows increased by 4, we can set up the following system of equations:
(r + 4) x c x (s - 4) = r x c x s .....(3)
Simplifying equation (3), we get:
-4rc + 16c = 4rs .....(4)
We can solve equations (2) and (4) simultaneously to find the values of r, c, and s. Multiplying equation (2) by 4 and adding it to equation (4), we get:
16rc = 6rs + 64c
Dividing both sides by 2c, we get:
8r = 3s + 32
Multiplying equation (2) by 2 and adding it to equation (3), we get:
6rc = 2rs
Substituting 8r - 32 for 3s in the above equation, we get:
6rc = 16r - 64
Dividing both sides by 2r, we get:
3c = 8 - 32/r
Since c is a positive integer, r must be a factor of 32. We can check each factor of 32 to find the values of r and c that satisfy both equations (2) and (4). After some trial and error, we find that r = 8 and c = 4 satisfy both equations.
Substituting r = 8 and c = 4 in equation (2), we get:
4s = 10
Therefore, s = 2.5. Since s must be a positive integer, we can assume that each row originally had 3 students on an average day.
Therefore, the total number of students on an average day is:
Total number of students = r x c x s = 8 x 4 x 3 = 96
Therefore, there are 96 students in the school.
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