Question

Ten students of the physics department decide to go on an educational trip. They hired a minibus for the trip, but the bus can only carry eight students at a time and each student goes at least once. Find the minimum number of trips the bus has to make so that each students can go for an equal number of trips.
 $
  (A)\,\,5 \\
  (B)\,\,4 \\
  (C)\,\,6 \\
  (D)\,\,8 \\
  $

Answer 1

Hint: First we find numbers of students left in one trip and then for each student can go for an equal number of trips, we divide total number of students to numbers of students left in one trip to get the required answer.

Complete step-by-step answer:
Number of students decided to go on educational trip = $ 10 $
Number of students can go at a time in bus = $ 8 $
Number of students left in one trip = $ 2 $
Hence, to carry each students at least once bus has to go one more time

In the second trip two seats are occupied by those students who left out in the first trip but other six are by any of eight students who already took the first tip and two students from them must be left out in the second case.

But, as it is given in the statement each student can go for an equal number of trips.
Hence, in the second trip some students are there who have taken a second trip and there are some students who have taken only one trip which is a contradiction of the statement given.

Hence, the bus has to make more tips so that each student can go for an equal number of trips.
This can be calculated by dividing total numbers of students by numbers of students left out in each trip.
Therefore numbers of trips bus has to go: $ \dfrac{{10}}{2} = 5 $

Hence, from above we see that the minimum numbers of trips the bus has to make are $ 10 $ so that each student can go for an equal number of trips.

Note: As, in every trip two students are left out so the bus number of trips made by bus will be equal to the ratio of total numbers to students decided to go on trip to the number of students left out in each trip.