Assuming that each student in the school gets one piece of bread and one pack of jam, we can write the following inequality based on the supplied packets:
8x + 6y ≤ 550
This is because the total number of bread pieces (8x) and jam packets (6y) should be less than or equal to the maximum number of students (550).
Similarly, for the hospital, we can write:
8(3x) + 6z ≥ 900
Here, we assumed that the hospital has three times as many students as the school and assigned the number of jam packets supplied for the hospital as z. The total number of bread pieces and jam packets supplied to the hospital should be greater than or equal to the number of students in the hospital (900).
Simplifying the second inequality, we get:
24x + 6z ≥ 900
Dividing both sides by 6, we get:
4x + z ≥ 150
So, the agency needs to supply at least 150 packets of jam to the hospital to cater to its students.
Note that we have not used the information that the number of students in the school is between 500 and 550. This information can be used to further constrain the values of x and y, but it is not necessary to answer the question.
An agency supplies bread and jams to three places -a hospital, a bank and a school. Bread comes in a bunch of 8 pieces and Jam comes in a pack of 6 pieces. On a particular day, agency has supplied x packets of bread and y packets of jam to the school. On the same day, agency has supplied 3x packets of bread along with sufficient packets of jam to hospital. It is known that the number of students in the school are between 500 and 550.
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