Quantitative Aptitude and Reasoning

Statement 1: At least two members obtained less marks than C. This is already stated in the problem (C > at least two others). This provides no new information to distinguish between D and F as the lowest. So, Statement 1 alone is insufficient.

Statement 2: E and F obtained the same marks. Also, E is not 6th (from initial facts). If E and F are tied and E is not the lowest, then F also cannot be the lowest. Since the lowest person must be from {B, D, F} (after eliminating A, C, E as candidates for lowest), and F is eliminated by Statement 2, the lowest person must be from {B, D}. Given B > D, D must have the lowest marks among B and D. Therefore, D has the lowest marks. Statement 2 alone is sufficient.

We can count the triangles by listing unique combinations of 3 non-collinear vertices from the set {A, B, C, D, E, F, G}.
Collinear sets of 3 points: (A, F, G), (D, F, E), (B, G, C).

Let’s list the triangles systematically:
1. Triangles with A as a vertex:
* Bases on DE: ADF, AFE, ADE (3)
* Bases on BC: ABG, ACG, ABC (3)
* Bases connecting D/E to G: ADG, AEG (2)
* Bases connecting B/C to F: ABF, ACF (2)
(Total from A = 3+3+2+2 = 10 unique triangles with A as apex).

2. Triangles with F as a vertex (excluding those with A as apex, already counted):
* Bases on BC: FBG, FCG, FBC (3)
* Bases connecting B/C to D/E: FBD, FBE, FCD, FCE (4)
(Total from F, excluding A as apex = 3+4 = 7 unique triangles with F as apex, or not having A as apex).

3. Triangles with D as a vertex (excluding those with A or F as apex):
* Bases on BC: DBG, DCG, DBC (3)
* Bases connecting G/C to E: DGE, DCE (2)
(Total from D, excluding A or F as apex = 3+2 = 5 unique triangles).

4. Triangles with E as a vertex (excluding those with A, F, or D as apex):
* Bases on BC: EBG, ECG, EBC (3)
(Total from E, excluding A, F, D as apex = 3 unique triangles).

5. Triangles with B as a vertex (excluding those with A, F, D, E as apex):
* Bases connecting D/E to G/E/C: BDE (1)
(Total from B, new = 1 unique triangle).

6. Triangles with C as a vertex (excluding those with A, F, D, E, B as apex):
* Bases connecting D/E to G/D/B: CDE (1)
(Total from C, new = 1 unique triangle).

Total unique triangles = (Count from A) + (New from F) + (New from D) + (New from E) + (New from B) + (New from C)
Total = 10 + 7 + 5 + 3 + 1 + 1 = 27.

Let’s list the 27 unique triangles based on this count:
From A (10): ADF, AFE, ADE, ABF, ACF, ABG, ACG, ABC, ADG, AEG
From F (7): FBG, FCG, FBC, FBD, FBE, FCD, FCE
From D (5): DBG, DCG, DBC, DGE, DCE
From E (3): EBG, ECG, EBC
From B (1): BDE
From C (1): CDE

Checking for duplicates between the groups:
– FBD is same as BDF (BDF in B list)
– FCE is same as CEF (CEF in C list)
– FBE is new.
– FCD is new.
– DBG is same as GDB.
– DCG is same as GDC.
– DBC is same as CDB.
– DGE is same as EDG.
– DCE is same as EDC.
– EBG is same as GEB.
– ECG is same as GEC.
– EBC is same as BCE.
– BDE is new.
– CDE is new.

Let’s refine the list by listing unique triangle names:
ADF, AFE, ADE, ABF, ACF, ABG, ACG, ABC, ADG, AEG (10)
FBG, FCG, FBC, FBE, FCD (5 new from F, excluding BDF, CEF which are listed separately)
DBG, DCG, DBC, DGE, DCE (5 new from D)
EBG, ECG, EBC (3 new from E)
BDF (1 new from B)
BDE (1 new from B)
CEF (1 new from C)
CDE (1 new from C)

Total = 10 + 5 + 5 + 3 + 1 + 1 + 1 + 1 = 29? No, this is not 27. Re-checking the grouping.

Let’s use the approach based on counting regions.
10 smallest regions: ADF, AFE, BDF, CEF, FBG, FCG, DBG, DCG, EBG, ECG (10 triangles).
Combinations of 2 smallest regions: ADE, ABF, ACF, FBC, DBC, EBC, ADG, AEG, DGE, DCE, BDE, CDE (12 triangles).
Combinations of 3 smallest regions: None obvious.
Combinations of 4 smallest regions: ABG, ACG (2 triangles).
Combinations of 8 smallest regions: ABC (1 triangle).
Total from regions: 10 + 12 + 2 + 1 = 25.

There must be triangles formed by vertices that are not simple combinations of adjacent smallest regions in this manner. The vertex-based counting method seems more reliable.

Let’s re-verify the vertex count of 27.
A (10): ADF, AFE, ADE, ABF, ACF, ABG, ACG, ABC, ADG, AEG
F (7): FBG, FCG, FBC, FBD, FBE, FCD, FCE
D (6): DBG, DCG, DBC, DGE, DCE, DFB
E (6): EBG, ECG, EBC, EDG, EDC, EFC
B (3): BDE, BEF, BDF
C (3): CDE, CFD, CFE
G (8): GDB, GDC, GEB, GEC, GFB, GFC, GAD, GAE (same as DBG, DCG, EBG, ECG, FBG, FCG, ADG, AEG)

Unique list from combined lists:
ADF, AFE, ADE, ABF, ACF, ABG, ACG, ABC, ADG, AEG (10)
FBG, FCG, FBC (3)
FBD (BDF) (1)
FBE (1)
FCD (1)
FCE (CEF) (1)
DBG, DCG, DBC (3)
DGE (1)
DCE (1)
EBG, ECG, EBC (3)
BDE (1)
CDE (1)

Total: 10 + 3 + 1 + 1 + 1 + 1 + 3 + 1 + 1 + 3 + 1 + 1 = 29.

Let’s re-check the list of 29 again for strict uniqueness.
1. ADF
2. AFE
3. ADE (ADF+AFE)
4. ABF (ADF+BDF)
5. ACF (AFE+CEF)
6. ABG (ADF+BDF+FBG+DBG – regions) or (ABF+FBG?) No. (A,B,G)
7. ACG (A,C,G)
8. ABC (A,B,C)
9. ADG (A,D,G)
10. AEG (A,E,G)
11. FBG
12. FCG
13. FBC (FBG+FCG)
14. BDF
15. CEF
16. DBG
17. DCG
18. DBC (DBG+DCG)
19. EBG
20. ECG
21. EBC (EBG+ECG)
22. DGE
23. DCE
24. BDE
25. CDE
26. FBE
27. FCD

Summing these unique names: 27.

Let’s re-verify the count of 27 based on groups which summed to 27.
A (10): ADF, AFE, ADE, ABF, ACF, ABG, ACG, ABC, ADG, AEG
F (7): FBG, FCG, FBC, FBD, FBE, FCD, FCE. (No A)
D (5): DBG, DCG, DBC, DGE, DCE. (No A, F as apex)
E (3): EBG, ECG, EBC. (No A, F, D as apex)
B (1): BDE. (No A, F, D, E as apex)
C (1): CDE. (No A, F, D, E, B as apex)

Let’s check for overlaps between these groups:
A and F: No triangle from F list has A as apex. OK.
A and D: No triangle from D list has A as apex. OK.
A and E: No triangle from E list has A as apex. OK.
A and B: BDE has B as highest vertex. OK.
A and C: CDE has C as highest vertex. OK.

F and D: FBD(F,B,D) vs DBG(D,B,G), DCG(D,C,G), DBC(D,B,C), DGE(D,G,E), DCE(D,C,E). BDF from B-list. FBD from F-list. Yes, FBD=BDF. So, when listing from F, FBD is F,B,D. From D, it’s D,B,F. From B, it’s B,D,F.
My list of 27 seems correct. Let’s proceed with 27.

Now, find the differences between these first differences (second differences):
18 – 6 = 12
36 – 18 = 18
60 – 36 = 24
90 – 60 = 30

Now, find the differences between these second differences (third differences):
18 – 12 = 6
24 – 18 = 6
30 – 24 = 6

Since the third differences are constant (6), the sequence is based on a cubic polynomial. A common pattern for such sequences is $n^3 – c \cdot n$ or similar. Let’s try $n^3 – n$:
For n=1: $1^3 – 1 = 1 – 1 = 0$ (Matches the first term)
For n=2: $2^3 – 2 = 8 – 2 = 6$ (Matches the second term)
For n=3: $3^3 – 3 = 27 – 3 = 24$ (Matches the third term)
For n=4: $4^3 – 4 = 64 – 4 = 60$ (Matches the fourth term)
For n=5: $5^3 – 5 = 125 – 5 = 120$ (Matches the fifth term)
For n=6: $6^3 – 6 = 216 – 6 = 210$ (Matches the sixth term)

The pattern is $a_n = n^3 – n$ for n = 1, 2, 3, …
The next number in the sequence will be the 7th term, for n=7.
$a_7 = 7^3 – 7 = 343 – 7 = 336$.

The maximum number of regions created by 2 lines partitioning a bounded region (where each line enters and exits the region at most twice) is 4. More regions require lines to intersect multiple times within the region or interact more complexly with the boundary, which is not possible with just two straight lines partitioning a simple bounded region formed by smooth curves.

Thus, ‘n’ can be 3 or 4.
Evaluating the options:
A) ‘n’ can be 4 but not 3 (False, n can be 3)
B) ‘n’ can be 4 but not 5 (True, n can be 4, and based on analysis, cannot be 5 or 6)
C) ‘n’ can be 5 but not 6 (False, n cannot be 5)
D) ‘n’ can be 6 (False, n cannot be 6)

Let’s analyze the outcomes:
* Outcome 1: P draws White (Prob=1/3). P wins.
* Outcome 2: P draws Black (Prob=2/3) AND Q draws White (Prob=1/2). Q wins. Probability = (2/3)*(1/2) = 1/3.
* Outcome 3: P draws Black (Prob=2/3) AND Q draws Black (Prob=1/2). Neither P nor Q wins. Probability = (2/3)*(1/2) = 1/3.

Now let’s evaluate the options:
A) If P loses, Q wins: P loses if P draws Black. If P draws Black, Q draws from Box II. Q wins *only if* Q draws White. Q does *not* win if Q draws Black. So this statement is not always correct.
B) If Q loses, P wins: Q only plays if P loses (draws Black). If Q loses (draws Black), it means P already lost the first draw. P’s winning condition is drawing White in the *first* step. If Q gets to play and then loses, P cannot win *in that game instance*. So this statement is incorrect.
C) Both P and Q may win: In a single game instance, either P wins (game stops), or P loses and Q plays. If Q plays, either Q wins or neither wins. P and Q cannot both win in the same game. So this statement is incorrect.
D) Both P and Q may lose: This happens in Outcome 3, where P draws Black and Q draws Black. In this scenario, P did not win (as P drew Black) and Q did not win (as Q drew Black). This is a possible outcome with probability 1/3. So this statement is correct.

Using Batch II = 5 in the first two equations:
– Batch I + 5 = 6 => Batch I = 1
– Batch I + 5 = 6 => Batch I = 1 (Consistent)

The number of students in each batch is the sum of players across sports:
– Batch I Total: Cricket (1) + Football (1) + Hockey (6) = 8
– Batch II Total: Cricket (5) + Football (5) + Hockey (5) = 15
– Batch III Total: Cricket (8) + Football (10) + Hockey (6) = 24
– Grand Total: 8 + 15 + 24 = 47 (Also 14 + 16 + 17 = 47)

Now, evaluate the options:
A) Batch II is empty (Batch II has 15 students) – False
B) Batch I and Batch II do not have equal number of students (Batch I = 8, Batch II = 15. 8 != 15) – True
C) Batch I and Batch III can have equal number of students (Batch I = 8, Batch III = 24. They are not equal) – False
D) Batch II and Batch III can have equal number of students (Batch II = 15, Batch III = 24. They are not equal) – False

Let’s examine the relationship between consecutive terms.
T(2) = 14. Is there a relation to T(1)=6? 6 * 2 + 2 = 12 + 2 = 14.
Let’s test this pattern for the next term: T(n+1) = 2 * T(n) + 2.
T(3) = 2 * T(2) + 2 = 2 * 14 + 2 = 28 + 2 = 30. This matches the given T(3).
Now let’s test this pattern for the fourth term:
Expected T(4) = 2 * T(3) + 2 = 2 * 30 + 2 = 60 + 2 = 62.
However, the given fourth term is 64. This indicates that 64 might be the wrong number.

Let’s assume the expected T(4) (which is 62) was correct and test the pattern for the fifth term:
Expected T(5) = 2 * (Expected T(4)) + 2 = 2 * 62 + 2 = 124 + 2 = 126.
This matches the given T(5).

So, the pattern T(n+1) = 2 * T(n) + 2 holds for all terms except for the fourth term, which should be 62 according to the pattern, but is given as 64. Therefore, 64 is the wrong number in the series.

A point (x, y) is strictly inside this triangle if it satisfies the following conditions:
1. It must be to the right of the line x=0: x > 0.
2. It must be below the line y=100: y < 100. 3. It must be above the line y=x: y > x.

Combining these, we are looking for integer coordinates (x, y) such that 0 < x < y < 100. Let's iterate through possible integer values for x. Since x > 0 and x < y < 100, the smallest possible integer value for x is 1. If x = 1, y must be an integer such that 1 < y < 100. Possible y values are 2, 3, ..., 99. The number of such y values is 99 - 2 + 1 = 98. If x = 2, y must be an integer such that 2 < y < 100. Possible y values are 3, 4, ..., 99. The number of such y values is 99 - 3 + 1 = 97. If x = 3, y must be an integer such that 3 < y < 100. Possible y values are 4, 5, ..., 99. The number of such y values is 99 - 4 + 1 = 96. ... What is the largest possible integer value for x? Since x < y < 100, the largest possible integer value for y is 99. This requires x to be at least 1 less than 99, i.e., x < 99. So, the largest possible integer value for x is 98. If x = 98, y must be an integer such that 98 < y < 100. The only possible y value is 99. The number of such y values is 1. (99 - 99 + 1 = 1) The total number of integer points inside the triangle is the sum of the number of possible y values for each x from 1 to 98. Total points = (99 - 1) + (99 - 2) + (99 - 3) + ... + (99 - 98) Total points = 98 + 97 + 96 + ... + 1 This is the sum of the first 98 positive integers. The formula for the sum of the first n positive integers is n(n+1)/2. Here, n = 98. Sum = 98 * (98 + 1) / 2 = 98 * 99 / 2 = 49 * 99. 49 * 99 = 49 * (100 - 1) = 4900 - 49 = 4851. The number of points inside the triangle with integer coordinates is 4851.

The second wheel rolls on the larger circle (R2). Let its radius be r2. The distance covered by its point of contact is the circumference of the larger circle, which is 2πR2. This distance is also equal to N times the circumference of the second wheel, which is N * 2πr2.
So, 2πR2 = N * 2πr2
R2 = N * r2

We know R2 = 11000 cm and N = 1000/3.
11000 = (1000/3) * r2
r2 = 11000 * (3 / 1000)
r2 = 11 * 3 = 33 cm.

The radius of the other wheel is 33 cm.