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Three people A, B and Care standing in a queue. There are 5 people between A andbetween B and C. If there be 3 people ahead of C and 21 people behind A what could beumber of people in the queue?(a) 20(b) 21(c) 25(d) 28 |
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Answer» option 28 Three persons A, B, C can be arranged in a queue in six different ways, ie 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, ie CBA and CAB. We may consider the two cases as under: Case I:←3C↔8B↔5A→21 Clearly, number of persons in the queue = (3+1+8+1+5+1+21=) 40 Case II:←3CA↔5B Number of persons between A and C = (8 - 6) = 2 ∵[C↔8BA→21B] Clearly number of persons in the queue = (3+1+2+1+21) = 28 Now, 28 < 40. So, 28 is the minimum number of persons in the queue. |
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