Answer:
Therefore a bus could carry 23 number of students and a van could carry 9 number of students.
Explanation:
Assume a van and a bus could carry x and y number of students respectively.
Given that,
For a trip, one high school rented and fits 114 students into 5 van and 3 buses.
So,
A van could carry x number of students
5 van could carry 5x number of students.
A buses van could carry y number of students.
3 buses van could carry 3y number of students.
∴5x+3y=114 ....(1)
Another high school instead fit its 247 students into 7 vans and 8 buses.
A van could carry x number of students
7 van could carry 7x number of students.
A buses van could carry y number of students.
8 buses van could carry 8y number of students.
∴7x+8y=247 ....(2)
∴5x+3y=114 ....(1)
and
7x+8y=247 ....(2)
7 times of equation (1) subtracts from 5 times of equation (2)
35x + 40y = 1235
35x + 21y = 798
- - -
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40y-21y=1235-798
⇒19y=437
Putting the value y in equation (1)
5x+3.23=114
⇒5x=114-69
⇒5x=45
⇒ x=9
Therefore a bus could carry 23 number of students and a van could carry 9 number of students.