Class 12th
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New answer posted
5 months agoContributor-Level 10
From the above 2 cases, it follows case (I) and number of persons sitting between P2 and P4 is either 1 or 3.
New answer posted
5 months agoContributor-Level 10
From the above 2 cases.
In case (I) if P12 is sitting opposite to P8 then P6 is sitting opposite to P7. In case (II) if P12 is sitting opposite to P8 then P10 is sitting opposite to P7.
New answer posted
5 months agoContributor-Level 9
Input are :
(0, 0) ; (0, 1); (1, 0); (1, 1).
Thus, the output y is : (1, 0) s
A | B | P | Q | Y |
0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 |
New answer posted
5 months agoContributor-Level 10
From the given information we can conclude that (P3) and (P4) are at seat numbers 7 and 6, respectively. And (P11) is the only person between (P3) and (P10) while (P1) is opposite to (P11). Hence, (P1), (P11) and (P10) must be at seat numbers 5, 8 and 9 respectively.
Then we have following two cases:
New answer posted
5 months agoContributor-Level 10
According to the information provided following arrangement of the schedule will be formed.
Wed | Thu | Fri | Sat | Sun | Mon | Tue |
Z6, Z2 | Z8, Z5 | Z9, Z4 | Z3 | * | Z1, Z7 | Z10 |
New answer posted
5 months agoContributor-Level 10
Since, Kathak has one more dancer than Bhangra, there are 4 dancer in Kathak S4, S5, S6 and S9. Remaining dancers are in Bharatnatyam. So, option (a) is correct
New answer posted
5 months agoContributor-Level 10
If the Bharatnatyam team has just one member S10, then the Kathak team will surely have S4, S5 and S6 as dancers.
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