three people a,b & c are standing in a queue.there are 5 people between a and b and 8 people between b and c . if there be 3 people ahead of c and 21 people behind a ,what could be the minimum number of people in the queue
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as per the question there are 5 people between a and b so,
a + 5 + b where a is the 1 person and b is the 7
there are 8 people between b and c
b + 8 + c where c is the 16
there are 3 people ahead of c
c + 3
so total 19 people
behind a there are 21 peole
by solving (19+21) we get that there are 40 oeople in the queue
a + 5 + b where a is the 1 person and b is the 7
there are 8 people between b and c
b + 8 + c where c is the 16
there are 3 people ahead of c
c + 3
so total 19 people
behind a there are 21 peole
by solving (19+21) we get that there are 40 oeople in the queue
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