wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Three persons A, B and C are standing in queue. There are five persons between A and B and eight persons between B and C. If there ne three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue ?

A
41
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
40
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
28
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
27
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 28

Three persons A, B, C can be arranged in a queue in six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements i.e., CBA and CAB. We may consider the two cases as under

3 8 5 21

Case I : .
Now, 28 < 40. So, 28 is the minimum number of perosn in the queue.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Multiples
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon