193 miners need to ride the elevator to the bottom of the mine. The elevator can hold 7 miners at a time. How many miners will be in the elevator on its last trip?

Here's the solution ~

There are total of 193 mimers and each time the elevator can carry 7 miners, now divide the numbers and the remainder ww get gives the number of people left at last, that is also equal to the number of miners in the last trip since they are the last group among them.

On dividing, we get quotient = 27 and remainder as 4

therefore, 28th trip will be the last trip carrying 4 miners.

I hope that helps ~

answer;

the correct answer

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the answer will be d: 2