@mampu the answer for Q) 6 is not coming 6 it is 4 for sure everyone has confirmed
@sparsh_temani in the first question firstly you should be writing down the given array with the index written on the top for ease and then u should write the i % 5 value on the top of each index i so that you know what you can choose. Then lastly select the maximum value from each unique i % 5 and then choose the maximum values.(remember even in the first part you can’t choose 6 elements in fact you can never choose more than 5 as there are only five unique possible values of i % 5 i.e, 0, 1, 2, 3, 4):
P.S if this solved your problem please give me some reputation points
Does someone remember Q.3. with values ?
Also, is the answer for Q.4. a) 49 OR 61 ?
I got the 6th answer as 5.
I know how one could get 61. When you calculate the number of partitions where the size of the sets are 2,2 you may forget to divide by 2. That is, instead of 4!/2 you write 4!. So you would get an answer 12 more than the actual one for this case. 61 is 12 more than 49.
I remember for the 2nd and 3rd parts of the first question, for remainder® = 3- max number was 23, for r = 2 - max number was 22, for r=1 - max number was 21 and for r=0 it was 20, these were the max four which gave the sum 86.
You have to take the maximum number corresponding to the index, not the index itself.
I got 2 for the 4th Question.
I separated the balls which count more than 10 and the balls which count less than 10.
Then i started filling the boxes.
First i filled the boxes with balls which count more than 10, keeping the extra ball which don’t fit in aside then I filled the remaining boxes with the ball with balls that count less than 10 and then i started filling the left over ball from the last box.
And there I got 2 as the answer for the 4th question.
I hope I was right!
Didn’t we had to take consecutive indices in first question else the question would be straight forward?
Don’t understand how the second question is 3, 2, 4. I got completely different answers.
Can anyone please explain how they got the first one, I’m getting different answers as compared to majority.
The answers are as follows
1.) 74 2.) 67 3.) 65 4.) 3 5.) 2 6.) 4 7.) 20 8.) 16 9.) 96 10.) 61 11.) 321 12.) 2671
The paper was pretty easy this time.
Easier than last year.
I just goofed up in one questions. Many of my friends are getting all correct.
I believe the cutoff would be
12:th 70-75
11th 65-70
10 th 60
9th and below :55
I believe that I would be correct regarding the cutoff as the paper was very very easy this year.
For 6th one I did like this:
Actually I saw that the numbers above 10 were 15, 21, 22 and 11. So we have 29 extras. As B was equal to 13. So there was an extra B set left. I filled it up with 10. So I have remaining 19. I fill 6 in the set containing 4. So I have 13 left. I fill 5 in the set of 5. Next I fill 4 in the set of 6. So I am left with 4. As I remember there was three sets having 7. I fill one set of 7 with 3 and another with 1. So I have got 5 impure sets. If it is wrong then please explain.
For question 10, 11, and 12, the answers are not 61, 321, and 2671. You get these answers by over counting.
1 Like
Expected cutoff after asking a number of people…
9th and below----30-35
10th grade----35-40
11th grade----45-50
12th grade----50-55
Please post your scores here so that estimated cutoff can be calculated.