Number, Ranking and Time Sequence Test Questions

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:  ←3C A↔5B

Number of persons between A and C                  

= (8 - 6) = 2   

∵[C↔8B A→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.                                                     

Initialy from left 7th = kumar
from right 9th = pawan
Interchange kumar = 11th from left, from right kumar = 9th;
11+8 = 19;
8 is 1,2,3,4,5,6,7(pawan),8,9,10,11(kumar),12,13,14,15,16,17,18,19;
simply count the 9th place from right we will get total as 19.

The possibe two-digit numbers are: 88, 85, 82, 81, 87, 86, 58, 28, 18, 78, 68

These are 11 in number

Here the series is:

987654321, 87654321, 7654321, 65...

Number of boys in the class = (18 + 1 +18) = 37

Only one such symbol is there in the given arrangement.

It is 4@5.

Dates on which Tuesday fall are 3,10,17, and 24

Hence, the 27th day of the month will be (Tuesday + 3 days=)Friday

Required day of the month = Friday - 4 = Monday

Number of boys in the row = (6 + 10 + 8) = 24