Diagnostic Test: CAT - 2 - CAT MCQ

Initially, the arrangement is as follows.

Now, in the first transition, 1 moves out, 6 replaces 1 and 11 replaces 6. 

Now in the second transition, 5 moves out, 10 replaces 5, and 12 replaces 10.

Now, in the third transition, 9 moves out, 4 replaces 9, and 13 replaces 4.

In the fourth transition, 3 moves out, 8 replaces 3, and 14 replaces 8.

In the fifth transition, 7 moves out, 2 replaces 7 and 15 replaces 2.

In the sixth transition, 6 moves out, 11 replaces 6 and 16 replaces 11.

In the sixth transition, 6 moves out, 11 replaces 6 and 16 replaces 11.
So if we observe now, in the sixth transition, a person again moves out of seat A. This is similar to that of 1st transition, so we can assume a set of 5 transitions because, after each set of 5 transitions, the pattern repeats itself.
Now we need to see who is seating where initially and after the 5th transition so that we can use that information to deduce who is sitting where after each 5n transition.

So, after 5 transitions, 6, who sits at F replaces 1, hence after 10 transitions, 11 who sits at F replaces 6.
After 5 transitions, 15, who is the 5th person to enter the table replaces 1, hence after 10 transitions, 20 who is the 5th person to enter in the second 5 rounds replaces 15.
After 5 transitions, 8, who sits at H replaces 1, hence after 10 transitions, 14 who sits at H replaces 8.
After 5 transitions, 13, who is the 3rd person to enter the table replaces 4, hence after 10 transitions, 18 who is the 3rd person to enter in the second 5 rounds replaces 13.
Similarly, we can calculate for the remaining people, as to who would be seating where after the 10 transitions, and we get this.

In a similar way, we can calculate, who would be sitting where after the 15th transition.

Hence, out of the given options, 17 remain.

Initially, the arrangement is as follows.

Now, in the first transition, 1 moves out, 6 replaces 1 and 11 replaces 6. 

Now in the second transition, 5 moves out, 10 replaces 5, and 12 replaces 10.

Now, in the third transition, 9 moves out, 4 replaces 9, and 13 replaces 4.

In the fourth transition, 3 moves out, 8 replaces 3, and 14 replaces 8.

In the fifth transition, 7 moves out, 2 replaces 7 and 15 replaces 2.

In the sixth transition, 6 moves out, 11 replaces 6 and 16 replaces 11.

So, after the 6th transition, 2, 4, 8, 10, 11, 12,13, 14, 15 and 16 remain seated.

It is clear that a batsman can score a minimum of 0 runs in an innings. When he remains not out, his average remains the same (with 0 runs) and does not decrease. Therefore, if a batsman is not out in an innings, his average after the innings will not drop in any case.
Using this, we can infer that the batsman must have been out in 2nd, 5th and 8th match. In the rest of the matches, he can either be out or not out, both.
It is given that before the start of the tournament, the batsman averaged 58 in 20 innings. So, his total runs= 58 x 20 = 1160

After match 1
CASE 1- He is out. Total times, batsman was out = 20+1= 21.
Total runs after this match= Average after this match x Total number of times he was out= 61.5 × 21= 1291.5, which is not possible (He cannot score runs in decimal).
CASE 2- He was not out in that innings. Total times, the batsman was out = 20
Total runs after this match= Average after this match ×Total number of times he was out=61.5 x 20 = 1230, which is possible. Therefore, Case 2 is true and he scored (1230-1160)= 70 not out in the first match.
Match 2- As discussed earlier, the batsman cannot remain not out when the average drops. He must have been out in this match. 
Total times, the batsman was out = 20+1= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 59×21= 1239.
So, he scored a total of 1239- 1230= 9 runs in this match.
Match 3- 
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 62 x 22 = 1364.
Runs scored if he was out= 1364- 1239= 125, which is not possible because his highest score in the tournament was that of 115.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out=62 x 21 = 1302.
Runs scored in this match= 1302- 1239= 63 not out. 
Match 4-
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×22= 1397.
Runs scored in this match= 1397-1302=95.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×21= 1333.5, which is not possible.
Hence, case 1 is true and he scored 95 in this match before getting out.
Match 5- It is clear that the batsman was out in this match as there is a drop of average after this match. So, total number of times the batsman was out after this match= 22+1= 23.
Total runs after this match= Average after this match ×Total number of times he was out= 62 ×23= 1426.
Runs scored in this match= 1426-1397=29.
Match 6-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×24= 1608.
Runs scored in this match= 1608-1426= 182, which is not possible because his highest score was 115.
CASE 2- The batsman was not out. Total number of times he was out= 23 
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×23= 1541.
Runs scored in this match= 1541-1426= 115, which is possible. So, case 2 is true and he scored 115 not out in this match.
Match 7-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×24= 1620.
Runs scored in this match= 1620-1541= 79, which is possible.
CASE 2- The batsman was not out. Total number of times he was out= 23
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×23= 1552.5, which is not possible as runs cannot be in decimals.
So, in match 7, the batsman scored 79 before getting out.
Match 8- As discussed earlier, he was out in this match as there is a drop in average. Total number of times he was out= 24+1= 25.
Total runs after this match= Average after this match ×Total number of times he was out= 65 ×25= 1625.
Runs scored in this match= 1625-1620= 5.
Match 9- 
CASE 1-  The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×26= 1742.
Runs scored in this match= 1742-1625= 117, which is not possible as his highest score is of 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×25= 1675.
Runs scored in this match= 1675-1625= 50.
So, he scored 50 not out in the 9th match of the tournament.
Match 10-
CASE 1- The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×26= 1781.
Runs scored in this match= 1781-1675= 106, which is possible as his highest score is 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×25= 1712.5, which is not possible as the runs cannot be in decimal.
.'. In the last match, the batsman scored a total of 106 runs before getting out.
The final scorecard of the batsman in the tournament will look like:

* Shows that the batsman was not out.

Initially, the arrangement is as follows.

Now, in the first transition, 1 moves out, 6 replaces 1 and 11 replaces 6. 

Now in the second transition, 5 moves out, 10 replaces 5, and 12 replaces 10.

Now, in the third transition, 9 moves out, 4 replaces 9, and 13 replaces 4.

In the fourth transition, 3 moves out, 8 replaces 3, and 14 replaces 8.

In the fifth transition, 7 moves out, 2 replaces 7 and 15 replaces 2.


But in this question, we have been given that 4 moves out in the first transition.
Now we can take 2 approaches. First, we can repeat what we did when 1 moved out in the first transition to the case when 4 moved out in the first transition, and follow similar steps to find out who sits where after the fifth transition. 
The second and better approach is that since we have already worked out who sits where after the fifth transition (when 1 moves out first), we can simply apply the final results, adjusting them in a way that 4 is the first one to move. This approach follows the logic that irrespective of the person, the one who is sitting on A moves out first, so we can adjust in a way that 4 sits on A in the initial arrangement.

Now, in the first table, we see that 6 who was sitting on F is replacing 1, who was sitting on A. SImilarly, in table 2, 9 who is sitting on F will replace 4 who is sitting on A.
In the first table, 15(the person joining in the fifth transition) is replacing the one sitting at B. Similarly in table 2, 15 will be replacing 5.
In the first table, we see that 8 who was sitting on H is replacing 3, who was sitting on C. SImilarly, in table 2, 1 who is sitting on H will replace 6 who is sitting on C.
In the first table, 13(the person joining in th 3rd transition) is replacing the one sitting on D. Similarly in table 2, 13 will be replacing 7.
We can continue in a similar fashion to find out the positions of all people after the 5th transition.

Hence the initial position of 2 and the final position of 7 are the same.

Initially, the arrangement is as follows.

Now, in the first transition, 1 moves out, 6 replaces 1 and 11 replaces 6. 

Now in the second transition, 5 moves out, 10 replaces 5, and 12 replaces 10.

Now, in the third transition, 9 moves out, 4 replaces 9, and 13 replaces 4.

In the fourth transition, 3 moves out, 8 replaces 3, and 14 replaces 8.

Hence, 3 moves out of the table in the 4th transition.

It is clear that a batsman can score a minimum of 0 runs in an innings. When he remains not out, his average remains the same (with 0 runs) and does not decrease. Therefore, if a batsman is not out in an innings, his average after the innings will not drop in any case.
Using this, we can infer that the batsman must have been out in 2nd, 5th and 8th match. In the rest of the matches, he can either be out or not out, both.
It is given that before the start of the tournament, the batsman averaged 58 in 20 innings. So, his total runs= 58 x 20 = 1160

After match 1
CASE 1- He is out. Total times, batsman was out = 20+1= 21.
Total runs after this match= Average after this match x Total number of times he was out= 61.5 × 21= 1291.5, which is not possible (He cannot score runs in decimal).
CASE 2- He was not out in that innings. Total times, the batsman was out = 20
Total runs after this match= Average after this match ×Total number of times he was out=61.5 x 20 = 1230, which is possible. Therefore, Case 2 is true and he scored (1230-1160)= 70 not out in the first match.
Match 2- As discussed earlier, the batsman cannot remain not out when the average drops. He must have been out in this match. 
Total times, the batsman was out = 20+1= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 59×21= 1239.
So, he scored a total of 1239- 1230= 9 runs in this match.
Match 3- 
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 62 x 22 = 1364.
Runs scored if he was out= 1364- 1239= 125, which is not possible because his highest score in the tournament was that of 115.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out=62 x 21 = 1302.
Runs scored in this match= 1302- 1239= 63 not out. 
Match 4-
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×22= 1397.
Runs scored in this match= 1397-1302=95.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×21= 1333.5, which is not possible.
Hence, case 1 is true and he scored 95 in this match before getting out.
Match 5- It is clear that the batsman was out in this match as there is a drop of average after this match. So, total number of times the batsman was out after this match= 22+1= 23.
Total runs after this match= Average after this match ×Total number of times he was out= 62 ×23= 1426.
Runs scored in this match= 1426-1397=29.
Match 6-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×24= 1608.
Runs scored in this match= 1608-1426= 182, which is not possible because his highest score was 115.
CASE 2- The batsman was not out. Total number of times he was out= 23 
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×23= 1541.
Runs scored in this match= 1541-1426= 115, which is possible. So, case 2 is true and he scored 115 not out in this match.
Match 7-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×24= 1620.
Runs scored in this match= 1620-1541= 79, which is possible.
CASE 2- The batsman was not out. Total number of times he was out= 23
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×23= 1552.5, which is not possible as runs cannot be in decimals.
So, in match 7, the batsman scored 79 before getting out.
Match 8- As discussed earlier, he was out in this match as there is a drop in average. Total number of times he was out= 24+1= 25.
Total runs after this match= Average after this match ×Total number of times he was out= 65 ×25= 1625.
Runs scored in this match= 1625-1620= 5.
Match 9- 
CASE 1-  The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×26= 1742.
Runs scored in this match= 1742-1625= 117, which is not possible as his highest score is of 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×25= 1675.
Runs scored in this match= 1675-1625= 50.
So, he scored 50 not out in the 9th match of the tournament.
Match 10-
CASE 1- The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×26= 1781.
Runs scored in this match= 1781-1675= 106, which is possible as his highest score is 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×25= 1712.5, which is not possible as the runs cannot be in decimal.
.'. In the last match, the batsman scored a total of 106 runs before getting out.
The final scorecard of the batsman in the tournament will look like:

* Shows that the batsman was not out.

It is clear that a batsman can score a minimum of 0 runs in an innings. When he remains not out, his average remains the same (with 0 runs) and does not decrease. Therefore, if a batsman is not out in an innings, his average after the innings will not drop in any case.
Using this, we can infer that the batsman must have been out in 2nd, 5th and 8th match. In the rest of the matches, he can either be out or not out, both.
It is given that before the start of the tournament, the batsman averaged 58 in 20 innings. So, his total runs= 58 x 20 = 1160

After match 1
CASE 1- He is out. Total times, batsman was out = 20+1= 21.
Total runs after this match= Average after this match x Total number of times he was out= 61.5 × 21= 1291.5, which is not possible (He cannot score runs in decimal).
CASE 2- He was not out in that innings. Total times, the batsman was out = 20
Total runs after this match= Average after this match ×Total number of times he was out=61.5 x 20 = 1230, which is possible. Therefore, Case 2 is true and he scored (1230-1160)= 70 not out in the first match.
Match 2- As discussed earlier, the batsman cannot remain not out when the average drops. He must have been out in this match. 
Total times, the batsman was out = 20+1= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 59×21= 1239.
So, he scored a total of 1239- 1230= 9 runs in this match.
Match 3- 
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 62 x 22 = 1364.
Runs scored if he was out= 1364- 1239= 125, which is not possible because his highest score in the tournament was that of 115.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out=62 x 21 = 1302.
Runs scored in this match= 1302- 1239= 63 not out. 
Match 4-
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×22= 1397.
Runs scored in this match= 1397-1302=95.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×21= 1333.5, which is not possible.
Hence, case 1 is true and he scored 95 in this match before getting out.
Match 5- It is clear that the batsman was out in this match as there is a drop of average after this match. So, total number of times the batsman was out after this match= 22+1= 23.
Total runs after this match= Average after this match ×Total number of times he was out= 62 ×23= 1426.
Runs scored in this match= 1426-1397=29.
Match 6-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×24= 1608.
Runs scored in this match= 1608-1426= 182, which is not possible because his highest score was 115.
CASE 2- The batsman was not out. Total number of times he was out= 23 
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×23= 1541.
Runs scored in this match= 1541-1426= 115, which is possible. So, case 2 is true and he scored 115 not out in this match.
Match 7-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×24= 1620.
Runs scored in this match= 1620-1541= 79, which is possible.
CASE 2- The batsman was not out. Total number of times he was out= 23
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×23= 1552.5, which is not possible as runs cannot be in decimals.
So, in match 7, the batsman scored 79 before getting out.
Match 8- As discussed earlier, he was out in this match as there is a drop in average. Total number of times he was out= 24+1= 25.
Total runs after this match= Average after this match ×Total number of times he was out= 65 ×25= 1625.
Runs scored in this match= 1625-1620= 5.
Match 9- 
CASE 1-  The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×26= 1742.
Runs scored in this match= 1742-1625= 117, which is not possible as his highest score is of 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×25= 1675.
Runs scored in this match= 1675-1625= 50.
So, he scored 50 not out in the 9th match of the tournament.
Match 10-
CASE 1- The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×26= 1781.
Runs scored in this match= 1781-1675= 106, which is possible as his highest score is 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×25= 1712.5, which is not possible as the runs cannot be in decimal.
.'. In the last match, the batsman scored a total of 106 runs before getting out.
The final scorecard of the batsman in the tournament will look like:

* Shows that the batsman was not out.

It is clear that a batsman can score a minimum of 0 runs in an innings. When he remains not out, his average remains the same (with 0 runs) and does not decrease. Therefore, if a batsman is not out in an innings, his average after the innings will not drop in any case.
Using this, we can infer that the batsman must have been out in 2nd, 5th and 8th match. In the rest of the matches, he can either be out or not out, both.
It is given that before the start of the tournament, the batsman averaged 58 in 20 innings. So, his total runs= 58 x 20 = 1160

After match 1
CASE 1- He is out. Total times, batsman was out = 20+1= 21.
Total runs after this match= Average after this match x Total number of times he was out= 61.5 × 21= 1291.5, which is not possible (He cannot score runs in decimal).
CASE 2- He was not out in that innings. Total times, the batsman was out = 20
Total runs after this match= Average after this match ×Total number of times he was out=61.5 x 20 = 1230, which is possible. Therefore, Case 2 is true and he scored (1230-1160)= 70 not out in the first match.
Match 2- As discussed earlier, the batsman cannot remain not out when the average drops. He must have been out in this match. 
Total times, the batsman was out = 20+1= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 59×21= 1239.
So, he scored a total of 1239- 1230= 9 runs in this match.
Match 3- 
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 62 x 22 = 1364.
Runs scored if he was out= 1364- 1239= 125, which is not possible because his highest score in the tournament was that of 115.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out=62 x 21 = 1302.
Runs scored in this match= 1302- 1239= 63 not out. 
Match 4-
CASE 1- The batsman was out. Total number of times he was out= 21+1= 22.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×22= 1397.
Runs scored in this match= 1397-1302=95.
CASE 2- The batsman remained not out. Total number of times he was out= 21.
Total runs after this match= Average after this match ×Total number of times he was out= 63.5 ×21= 1333.5, which is not possible.
Hence, case 1 is true and he scored 95 in this match before getting out.
Match 5- It is clear that the batsman was out in this match as there is a drop of average after this match. So, total number of times the batsman was out after this match= 22+1= 23.
Total runs after this match= Average after this match ×Total number of times he was out= 62 ×23= 1426.
Runs scored in this match= 1426-1397=29.
Match 6-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×24= 1608.
Runs scored in this match= 1608-1426= 182, which is not possible because his highest score was 115.
CASE 2- The batsman was not out. Total number of times he was out= 23 
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×23= 1541.
Runs scored in this match= 1541-1426= 115, which is possible. So, case 2 is true and he scored 115 not out in this match.
Match 7-
CASE 1- The batsman was out. Total number of times he was out= 23+1= 24.
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×24= 1620.
Runs scored in this match= 1620-1541= 79, which is possible.
CASE 2- The batsman was not out. Total number of times he was out= 23
Total runs after this match= Average after this match ×Total number of times he was out= 67.5 ×23= 1552.5, which is not possible as runs cannot be in decimals.
So, in match 7, the batsman scored 79 before getting out.
Match 8- As discussed earlier, he was out in this match as there is a drop in average. Total number of times he was out= 24+1= 25.
Total runs after this match= Average after this match ×Total number of times he was out= 65 ×25= 1625.
Runs scored in this match= 1625-1620= 5.
Match 9- 
CASE 1-  The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×26= 1742.
Runs scored in this match= 1742-1625= 117, which is not possible as his highest score is of 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 67 ×25= 1675.
Runs scored in this match= 1675-1625= 50.
So, he scored 50 not out in the 9th match of the tournament.
Match 10-
CASE 1- The batsman was out. Total number of times he was out= 25+1= 26.
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×26= 1781.
Runs scored in this match= 1781-1675= 106, which is possible as his highest score is 115.
CASE 2- The batsman was not out. Total number of times he was out= 25
Total runs after this match= Average after this match ×Total number of times he was out= 68.5 ×25= 1712.5, which is not possible as the runs cannot be in decimal.
.'. In the last match, the batsman scored a total of 106 runs before getting out.
The final scorecard of the batsman in the tournament will look like:

* Shows that the batsman was not out.

Since the total number of rolls = 12, the number of rounds in the game = 12/4 = 3
This implies that in all 3 rounds, the same player wins, in order to complete the game in round 3.
Let us suppose Virat starts the first round.
Now, in the first round, Virat's sum = 3 + 6 = 9, and Sourav's sum = 5 + 4 = 9.
In case the sum is the same, Sourav wins. Hence in Round 2 and Round 3 also, Sourav wins.
For Sourav to win in Round 2, he must roll a minimum sum of 7 so that he can match the sum of Virat and win the round. Hence, A can only be 6.
For Sourav to win the third round as well, Virat cannot exceed the sum of Sourav.
Sourav's sum = 1 + 2 = 3.
Hence, Virat can roll (1,1), (1,2) or (2,1). Hence B and C can together take 3 different ordered pairs as values.
Hence, the required count = 1 x 3 = 3
Let us suppose Sourav starts the first round.
Now, in the first round, Sourav's sum = 3 + 6 = 9, and Virat's sum = 5 + 4 = 9.
In case the sum is the same, Sourav wins. Hence in Round 2 and Round 3 also, Sourav wins.
Sourav rolls a sum of 2 + 5 = 7 in the second round. So Virat has to roll a sum less than or equal to 7. So, A can take 1 to 6. Hence, 6 values.
In round 3, Virat rolls a sum of 3. Sourav has to throw a minimum of a total of 3. So he can roll anything except (1,1). Hence number of possibilities = 36 - 1 = 35.
Hence, the required count = 35 x 6 = 210.
Hence, total count = 210 + 3 = 213.

Let the rate at which the whole seller sells the rice be Rs x per kg.
But the cunning merchant uses the faulty scale to trick the wholesaler. When the wholeseller puts in 1kg of rice, the machine will show 0.8kg of rice. Thus merchant will get 1 kg of rice for the price of 0.8x0.8x and thus Actual selling price for the wholeseller = Actual cost-price for the merchant = Rs0.8x = 4x/5 per kg.
The rate at which he sells the rice = Rs 1.2x per kg.
When the merchant puts 1 kg of rice on the weighing machine, the machine will show 1.1 kg.
As per the rate the customer has to pay 1.2x x 1.1 = 1.32x  for 1 kg of rice. 
Actual selling price for the merchant= Rs 1.32x per kg
Let Sherlock by y kgs of rice.
Cost price to the merchant = 0.8xy
Actual selling price for the merchant = 1.32 xy
Discount = d
In order to be no-profit no loss

Let the cost price be x
When markup percent is m% and discount given is d% then (1+m%)(1−d%)x=1.08x
Upon expanding we get 

When markup percent is d% and discount given is m% then
(1+d%)(1−m%)x = 0.88x
Upon expanding we get

Adding (I) and (II) we get,

Substracting  (I) and (II) we get,

or m-d =10
Squaring both sides, m² + d² - 2md = 100
m² + d² - 400 = 100
m² + d² = 500

We have to find values of N such that N and N³ both have the same remainder when divided by 7
Thus (N mod 7) =(N³ mod 7)
It can be written as 
(N mod 7) = (N x N x N) mod7 = (Nmod7) x (Nmod7) x (Nmod7).
Thus it can be implied that
(N mod 7) = (N mod 7)³
When N = 1 then (Nmod7) = (Nmod7)³ = 1 so N =1 is a solution
When N = 2 then (Nmod7) = 2 and (Nmod7)³ = 1 so N ≠ 2

When N = 7 then the remainder is 0. Thus N has to be of form 7k, 7k+1 and 7k+6 to meet the required criteria
N< 1500 of from 7k = 7,14, .... 1498 which is 214 terms
N<1500 of from 7k+1 = 1, 8,15, .... 1499 which is again 215 terms
N<1500 of from 7k+6 = 6,15, .... 1497 which is again 214 terms
Total terms = 214+215+214 = 643.

There are a total of 11 alphabets in VENKATESHAN which when arranged in ascending order are A, A, E, E, H K, N, N, S, T and V.
Total possible permutation are =  
Now we need to find the cases where all the "A"s are before "E"s. There are total of 2 A and 2 E. If we pick places for A and E, they can be placed in one way. 4 places can be picked in ¹¹C₄ ways.
Remaining 7 alphabets can be arranged in 7!/2! ways.
The number of ways possible  = 
Probability = 
Probability = 

The divisibility rule of 3 states that the sum of digits has to be divisible by 3 in order to be divisible by 3.
Such numbers are 1002,1005,1008.....9999 : 3000 numbers.
The divisibility rules of 8 imply that the last 3 digits have to be divisible by 8: Thus "dcba" has to be divisible by 8. Such number are : 1000,1008,..9992 : 1125.
We have included the number which are divisible by 24 twice. Thus we have to remove those extra numbers.
Such number are : 1008,1032,...9984: 375 numbers.
The numbers of 4-digit number  which are divisible by 3 or 8: 3000+1125-375 = 3750.

Let the profit be P and cost price be C.P
Then 10 P = C.P ....(I)
Selling Price = Cost Price + Profit = 10 P +P = 11P
Discount given  = 3P.
Marked price = Selling Price + Discount = 11P+3P = 14P.
Let the discount given to ensure profit of 20% be d%.

Since the radius of the circle is 10 and the chords are 14 units apart, both of them have to be on the opposite side of the centre, let CD = 12. It can be shown as

E and F are the midpoint of respective chords.
From the properties of the circle, we know that OF is the perpendicular bisector of DC. Thus DF = 6 and OD =10(radius). It gives OF = 8 (from Pythagoras theorem)
Since it is given that EF = 14, it implies that EO = 14-OF =14-8 = 6
EOB is also a right-triangle such that EO = 6 and OB = 10. This gives EB = 8 or AB = 16.
Since they are symmetrical the trapezium will look like this

From pythogoras we find that AD² = 14² + 2² = 196 + 4 = 200
AD = 10√2

Let us draw a rectangle on the coordinate axis. With (0,0) be one of the edge A. B can be (6,0), C(6,12) and D(0,12).  E and F divide the AC in 3 equal part. E has to be (2,4) and F (4,8)
G is  centroid of DEF, Coordinates of G will be 

Upon finding G the required diagram looks like this

Area DGBC = Area GBC + Area DCG
Area DGBC = 
Area DGBC = 24+12 = 36 sq units

Since ABC is an equilateral triangle and DEF is the midpoint. Thus they will divide the ABC in 4 equilateral triangle of side half of that of ABC
Let side of ABC = 2a units. Then the side of the smaller triangle is a unit.
Shaded region = Area of 3 small circles + Area DEF - Area of 1 small circle
Shaded region = Area of 2 small circles + Area DEF 
Circles are an incircle of equilateral triangle of side "a"
Radius of this incircle 
Area of 1 small circle = 
Area of 2 small circle 
Area of triangle DEF = 
Shaded area = 
Unshaded area = Total area - Shaded area
Unshaded area =  - Shaded area
Unshaded area = 
Unshaded area = 
Ratio = 

• Let us start by writing the N in empirical form which is 2⁵ x 3⁸ x 2⁴ x 5³ x 2 x 3 x 7²
• Which equals 2¹⁰ x 3⁹ x 5³ x 7²
• Let A = number of factors not divisible by 9
• B = Number of Factors which are odd
• We have to find 
• (A∩B) = Number of factors not divisible by 9 and are odd. It implies it does not have 2 as a factor and the highest power of 3 can be  1. Such type of factors are = (1+1)(3+1)(2+1) = 24
• A = Number of factors that are not divisible by 9. Thus the maximum power of 3 they can have is 1. Such number are = (10+1)(1+1)(3+1)(2+1) = 264
• Probability = 24/264 = 1/11
• m+n = 12.

Through the passage, the author argues against nuclear power generation as an alternative to fossil fuels especially in the context of climate change. The author then gives these two examples to highlight how much damage they did to people and the planet and continue to do so. These two examples are extreme examples of how nuclear power generation can go terribly wrong. Thus, the primary reason to introduce these examples is to highlight the impact of nuclear disasters on the planet and people. Hence, option D.

Though the author does say that with climate change, the nuclear fallout of these disasters is likely to get worse, these examples are not introduced for that purpose. By highlighting the negative impact of these disasters, the author is trying to make the larger point against nuclear power. Hence, we can eliminate option B.

Though option A is the purpose behind the whole passage, it is not the reason why these two examples are introduced. Hence, we can eliminate option A. For the same reason we can eliminate option C.

Through the passage, the author argues against nuclear power. Hence, options B and C which are not negative can be eliminated. Both options A and D are close. But option A misses the primary motivation behind the author's recommendation. The author feels that with climate change, the number of nuclear disasters is likely to increase and thus nuclear power should be avoided. Hence, option D.

Through the last paragraph, the author is trying to counter the argument in favour of nuclear energy. The author says that though they have the advantage of reliability and capability, as the plants age they become inefficient and risky. Hence, option A is the right answer. 
Options B, C and D miss out the point of how aging of plants makes them a worse bet. Hence, option A.

The MIT research showed conclusively that the frequent forest fires, surface runoffs and flooding spread radiation. The research concluded that we are still paying for the nuclear by-products that were made in the 20th century. Only option C captures this point and hence is the right answer.
Option D, though true, is not the main conclusion of the study.
Option A contains a distortion. The study does not mention that only areas with radiation will experience acid rain. Hence, we can eliminate option A.
Option B is an exaggeration and hence can be eliminated.
Thus option C is the correct answer. 

The author begins the passage by explaining how country music has been termed “hill-billy”. He then goes on to explain what it means. The second paragraph speaks of the negative perceptions that country music has and the author states that this image is not true. The author explains how country music has been more progressive right from the beginning. The remaining paragraphs cite examples of issues that country music has dealt with. It is clear from the passage that the main argument of the author is to show that country music is progressive and not conservative. 
Option A is incorrect since it contributes to the author’s argument that country music is progressive.
Option B is a possible answer. Cultural ambassadors have not been discussed in the passage, although them exhibiting a varied history of America would still be beneficial for country music.
Option C is correct. Furthering "Christian" music would bolster Country Music's conservative credentials and not its progressive ones. 
Option D is incorrect since it tackles social issues which have been discussed in the passage.

The claim in statement 1 is made in the first paragraph. “Hill billy” has been stereotypically linked to a supposedly degenerate and backwards culture. We can further infer this from Abel Green’s description.
The claim in statement 2 is the main argument of the passage. We can infer it from the second paragraph.
Statement 3 can be inferred from the lines “These kinds of negative projections of the people who have made country music, and have listened to it, linger even unto today. The stereotype is that they all harbour conservative political and social beliefs, setting them as sexist, racist, jingoistic and fundamentalist Christian by nature. But this image is a lie. For, right from the start, country music spoke up with a progressive voice.”
So, Option D is correct.

Through the passage, the author argues that country music has always had a progressive voice. To support this assertion, the author first gives the example of Blind Alfred Reed who spoke of racial equality and equitable society. Then he gives the example of Loretta Lynn's progressive stand on women's reproductive rights. Thus, through this paragraph, the author is trying to highlight the progressive voices in country music's history to show that country music always had a progressive voice. Thus, option A is closest to the answer.

Option B is incorrect because the author has not claimed that country music facilitated the human rights movement.

Option C is incorrect. Although the author does say that “The Pill” stands out that is not the main point. We must see why the author believes it stands out. Read the lines “In the country-music market, this song stands out as an unabashed and rather radical call for sexual liberation and biological control, challenging the man’s past sexual prerogative and presenting a situation where the woman may also enjoy a variety of sexual liaisons without the social/economic restrictions that come with pregnancy, childbirth and childcare.” Through this example, the author is trying to make a larger point.

Option D is true but only captures a part of what Loretta Lynn’s song indicates. Option A is more accurate.

The author says that “These kinds of negative projections of the people who have made country music, and have listened to it, linger even unto today. The stereotype is that they all harbour conservative political and social beliefs, setting them as sexist, racist, jingoistic and fundamentalist Christian by nature. But this image is a lie. For, right from the start, country music spoke up with a progressive voice.” However, from these lines we cannot infer that the entire audience was literate and liberal. Hence, option A is incorrect.

Similarly “every song” need not have been progressive. Option B is gross generalization of country music.

Option C is incorrect because it is a generalization as well. 

Option D is correct, we can infer this from the lines “These kinds of negative projections of the people who have made country music, and have listened to it, linger even unto today. The stereotype is that they all harbour conservative political and social beliefs, setting them as sexist, racist, jingoistic and fundamentalist Christian by nature.”

After reading all the sentences, we know that the paragraph is about the author's dislike for singing and his new found love for spoken poetry. Statement 2 is the opening sentence as it mentions the author's dislike for singing. The author also mentions how discovering spoken poetry changed his understanding of why he disliked singing. Statements 4,3 and 1 together describe the author's experience of the poem recited with the help of a stringed instrument.
Hence, the order is 2-4-3-1.
Hence, 2431 is the correct answer.

The correct order is 1523. 4 is the odd one out. The passage begins by saying that a building could possibly be built and styled by two different people as long as they worked in harmony. Sentence 5 goes on to explain why this is not so since the average engineer does not value the average designer. Sentence 2 explains what the priorities of the engineer are, while sentence 5 states what a designer feels. Option 4 is the odd one out since it speaks of contrasts in domestic architecture which is unrelated to the remaining lines.

All the five sentences talk about the relation between psychology and teaching. But the structure of sentence 4 indicates that the topic of "ingenuity" and "tact" has already been discussed in the preceding lines. Since, there is no mention of "ingenuity" and "tact" in any other sentences, sentence 4 cannot be a continuation of any of the sentences given. The other 4 sentences can be arranged in the sequence 3-1-5-2.