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Shortcut Trick
Possible sums m without 5-coin: 1+2+10=13, 1+2+20=23, 1+10+20=31, 2+10+20=32.
Check Statement 2: Sum of digits for these is {4, 5, 4, 5}. None are > 5. All sums with digit sum > 5 must contain a 5-coin. (Sufficient)
Check Statement 1: m=32 is not prime and has no 5-coin, while m=8 is not prime and has a 5-coin. (Insufficient)
∴ The correct answer is statement 2 alone is sufficient.
Alternate Method
Given: Coins = {1, 2, 5, 10, 20}. Number of coins = 3. Total amount = m.

Step 1: Analyze Statement 1 (m is not a prime number)
⇒ Composite values of m: {8, 16, 26, 27, 32, 35}
⇒ If m = 32 (2 + 10 + 20), X does not have a 5-coin.
⇒ If m = 8 (1 + 2 + 5), X has a 5-coin.
⇒ Statement 1 alone is not sufficient.
Step 2: Analyze Statement 2 (Sum of digits of m > 5)
⇒ m values with digit sum > 5: {8, 16, 17, 26, 27, 35}
⇒ Case 1: m = 8 → {1, 2, 5}
⇒ Case 2: m = 16 → {1, 5, 10}
⇒ Case 3: m = 17 → {2, 5, 10}
⇒ Case 4: m = 26 → {1, 5, 20}
⇒ Case 5: m = 27 → {2, 5, 20}
⇒ Case 6: m = 35 → {5, 10, 20}
⇒ Every possible case contains a 5-denomination coin.
⇒ Statement 2 alone is sufficient.
∴ The correct answer is statement 2 alone is sufficient.
Additional Information
Data Sufficiency Logic
If one statement leads to a unique 'Yes' or 'No' answer, it is sufficient. If it leads to both possibilities, it is insufficient.
Prime and Composite Numbers
A prime number has exactly two factors (1 and itself). A composite number (not prime) has more than two factors.
Sum of Digits Property
The sum of digits is often used in divisibility rules (e.g., divisibility by 3 or 9) and can help uniquely identify sets of numbers in constrained problems.

Shortcut Trick

Statement 2 implies √x is real and less than 1, so 0 ≤ x < 1.

For any value in the range 0 < x < 1, the mathematical property x² < x < √x always holds.

Combining both statements results in the range x² < y < √x.

Since the variable x lies between the lower bound (x²) and upper bound (√x) of y, y can be either smaller or larger than x.

Because multiple possibilities exist (y < x or y > x), the question cannot be determined uniquely.

∴ The correct answer is cannot be answered even using both statements.

Alternate Method

Given: x and y are distinct real numbers.

From Statement 2: y < √x < 1

⇒ For √x to be a real number < 1, x must be in the range 0 ≤ x < 1.

From Statement 1: x² < y < 1

Combining both statements for the range 0 < x < 1:

⇒ x² < y < √x

Testing with values to verify indeterminacy:

⇒ Let x = 0.25 (then x² = 0.0625 and √x = 0.5).

⇒ The range for y is 0.0625 < y < 0.5.

⇒ Case 1: If y = 0.1, then y < x (0.1 < 0.25).

⇒ Case 2: If y = 0.4, then y > x (0.4 > 0.25).

⇒ Both cases satisfy all given conditions, meaning no unique comparison is possible.

∴ The correct answer is cannot be answered even using both statements.

Additional Information

Inequalities for Proper Fractions

For any real number x such that 0 < x < 1, the order of powers is xⁿ < x < x^1/n for all n > 1.

Data Sufficiency Logic

A question is considered 'answerable' only if the statements lead to a unique and consistent conclusion (always x > y or always y > x).

Domain of Square Root

The expression √x is defined for all x ≥ 0. If √x < 1, it directly restricts the base variable to 0 ≤ x < 1.

Logic followed here is based on the parity rules (even or odd properties) for the addition and multiplication of integers.

To determine if x is an even integer:

For Statement 1: x² y² is even →

If the product of two squared integers is even, then at least one of the base integers must be even. This leads to the following cases:

Since x can be either even or odd while satisfying the condition, Statement 1 alone is not sufficient.

For Statement 2: 1 + x² + y² is odd →

If 1 + x² + y² is odd, then the term (x² + y²) must be even (because odd − 1 = even). For the sum of two squares to be even, both x and y must share the same parity:

Since x can still be either even or odd, Statement 2 alone is not sufficient.

Combining both Statement 1 and Statement 2:

From Statement 2, we know that x and y must be either both even or both odd.

Now we apply Statement 1, which states that x² y² is even. We check the two possibilities from Statement 2:

• If x and y are both odd: x² × y² = Odd × Odd = Odd. This contradicts Statement

1. • If x and y are both even: x² × y² = Even × Even = Even. This satisfies Statement

2. The only case that satisfies both statements is when both x and y are even. Therefore, x is even.

Therefore, the question can be answered using both statements together, but cannot be answered using either statement alone.

Shortcut Trick

Statement 1: Sum of any two numbers is even ⇒ Numbers must be either all odd or all even.

Statement 2: (p − 1)q is even ⇒ If any q is odd, then (p − 1) must be even, so p is odd (making T all odd); or all q are even (making T all even).

Both statements confirm that T is a homogeneous set (contains only one type), but neither statement specifies whether that type is X (odd) or Y (even).

Since T could be {1, 3} (from X) or {2, 4} (from Y), we cannot determine if every number is from Y.

∴ The correct answer is the question cannot be answered even using any of the statements.

Alternate Method

Given: X = {certain odd numbers}, Y = {certain even numbers}, T ⊆ X ∪ Y.

Case Analysis:

⇒ Let Statement 1 be true: p + q = Even. This happens if (Odd + Odd) or (Even + Even). Thus, T ⊆ X OR T ⊆ Y. (Insufficient)

⇒ Let Statement 2 be true: (p − 1)q = Even.

⇒ If T ⊆ X (Odd), then (p − 1) is (Odd − 1) = Even. (Even × Odd = Even). Statement holds.

⇒ If T ⊆ Y (Even), then q is Even. (p − 1 × Even = Even). Statement holds.

⇒ If T contains both 2 (even) and 3 (odd): Let p = 2, q = 3. (2 − 1) × 3 = 1 × 3 = 3 (Odd). This violates the statement. So T must be homogeneous.

⇒ Combining both: T is still either all Odd or all Even. We cannot uniquely identify if it is Y.

∴ The correct answer is the question cannot be answered even using any of the statements.

Additional Information

Sum and Difference Properties

The sum or difference of two odd numbers is always even, and the sum or difference of two even numbers is always even.

Product Properties

The product of any number and an even number is even. The product of two odd numbers is always odd.

Data Sufficiency Logic

If two different sets (one subset of X and one subset of Y) both satisfy the conditions but lead to different answers to the question, the statements are insufficient.

Shortcut Trick

Factorize 116: 116 = 2² × 29. Since y > 1, y² must be 4, so x = 29 (Prime). Statement 1 is Sufficient.

Factorize 261: 261 = 3² × 29 = 9 × 29 or 3 × 87. Since x can be 9 (composite) or 3 (prime), Statement 2 is Insufficient.

∴ The correct answer is Statement 1 alone is sufficient.

Alternate Method

Given: x, y, z are integers > 1.

Analysis of Statement 1:

⇒ xy² = 116

⇒ Prime factorization: 116 = 2 × 2 × 29 = 2² × 29

⇒ Since y > 1, y² must be a perfect square factor of 116. The only such factor > 1 is 4 (where y = 2).

⇒ Therefore, x = 116 ÷ 4 = 29.

⇒ 29 is a prime number. Statement 1 alone is sufficient.

Analysis of Statement 2:

⇒ xz = 261

⇒ Factor pairs of 261: (3, 87), (9, 29), (29, 9), (87, 3).

⇒ If x = 3, it is prime. If x = 9, it is not prime (composite). If x = 87, it is composite.

⇒ Multiple possibilities exist for x. Statement 2 alone is not sufficient.

∴ The correct answer is Statement 1 alone is sufficient.

Additional Information

Prime Number Definition

A prime number is a natural number greater than 1 that has exactly two factors: 1 and itself (e.g., 2, 3, 5, 7, 11, 29, etc.).

Perfect Square Factors

When an equation involves y² as a factor of a constant, identifying the perfect square factors (1, 4, 9, 16...) of that constant helps isolate possible values for y and x.

Data Sufficiency Logic

A statement is sufficient only if it leads to a unique 'Yes' or 'No' answer. If a statement allows for both 'Yes' and 'No' cases, it is insufficient.

The correct answer is Option 1.

 

Key Points

• Statement 1 is valid: The passage explicitly states that Kalagram unfolded as a mosaic of India's diverse regions, with each represented by seven meticulously crafted 'Sanskriti Angans'.
• Statement 2 is valid: The text mentions that these thematic zones were inspired by iconic temples like the Dakshineshwar Kali Temple and the Brahma Mandir, and they served as treasure troves of regional artistry.
• Statement 3 is invalid: While the passage mentions 230 master artisans shaping India's ancient history, it emphasizes that they used age-old techniques. The statement's use of the term contemporary craftsmanship contradicts the passage's focus on traditional methods.
• Statement 4 is invalid: The passage describes the village as a mosaic of diverse regions, but it does not claim that art forms from all regions of India were showcased. Using the word all makes the conclusion too absolute.

Therefore, the correct answer is 1 and 2 only.

 

Additional Information

• Kalagram was a cultural village set up at the Maha Kumbh Mela to showcase India's heritage.
• Specific art forms mentioned include Bengal's Pattachitra paintings, Assam's bamboo crafts, Tamil Nadu's Thanjavur paintings, and Madhya Pradesh's tribal sculpture.
• The Sanskriti Angans were described as living galleries, indicating an immersive and active display of traditional arts.
• The craftsmanship involved 230 master artisans who acted as the bridge between ancient history and visual creations to behold.

The correct answer is Option 1.

 

Key Points

• The passage mentions four specific states and their respective art forms: Bengal (Pattachitra paintings), Assam (bamboo crafts), Tamil Nadu (Thanjavur paintings), and Madhya Pradesh (tribal sculpture).
• Upon closer inspection, only two of these states are associated with paintings: Bengal and Tamil Nadu.
• The other two states are associated with bamboo crafts (Assam) and tribal sculpture (Madhya Pradesh), which are not categorized as paintings.
• Therefore, the statement that paintings from four states have been mentioned is incorrect.

Therefore, the correct answer is Option 1.

 

Additional Information

• Option 2: Mentioning that Kalagram is likened to stepping into another world is a common contextual description for such immersive cultural exhibitions (though it may appear in the full passage text).
• Option 3: The passage explicitly states that these art forms were showcased in living galleries, referring to the Angans where 230 master artisans worked.
• Option 4: The text mentions the exhibition was inspired by iconic temples like the Dakshineshwar Kali Temple and the Brahma Mandir, confirming that the zones/Angans were themed after well-known Indian temples.
• The passage highlights the use of age-old techniques by 230 master artisans to shape India's ancient history into contemporary creations.

Given:

A + B + C + D = 17

A + C = 6

P + Q + S + D = 15

P + Q + R + B = 17

P = R and Q = S

From:

A + B + C + D = 17

A + C = 6

So,

B + D = 17 − 6 = 11

Now use:

P + Q + S + D = 15

Since Q = S,

P + 2Q + D = 15 ..........

(1) Also,

P + Q + R + B = 17

Since P = R,

2P + Q + B = 17 .......... (2) Adding (1) and (2):

3P + 3Q + B + D = 32

We know B + D = 11

So,

3P + 3Q + 11 = 32

3(P + Q) = 21

P + Q = 7

And,

A + C = 6

Therefore,

P + Q = 7 > 6 = A + C

Hence, P and Q together weigh more than the total weight of A and C.

Hence, the correct answer is "Option 2".

Shortcut Trick

Bind Q and U together as a single block (QU) because Q must always be followed by U.

Remaining letters from QUEUE are E, U, E.

Total units to arrange = 4: (QU), E, U, E.

Since E is repeated 2 times, Total arrangements = 4! ÷ 2!.

Calculation: 24 ÷ 2 = 12.

∴ The correct answer is 12 words.

Alternate Method

Given word: QUEUE (Letters: Q, U, E, U, E)

Constraint: Q is always followed by U.

Method: String Method (Grouping)

⇒ Total letters in QUEUE = 5 (Q × 1, U × 2, E × 2).

⇒ Treat (QU) as 1 entity. The remaining letters are U, E, E.

⇒ Total entities (n) = 4.

⇒ Repetitions: E repeats 2 times (p = 2).

⇒ Total ways = n! ÷ p!

⇒ Total ways = 4! ÷ 2! = (4 × 3 × 2 × 1) ÷ (2 × 1)

⇒ Total ways = 24 ÷ 2 = 12.

∴ The correct answer is 12 words.

Additional Information

Permutation of Multiset

When 'n' objects have 'p' identical items of one kind, 'q' of another, etc., the total arrangements are given by n! ÷ (p! × q!).

The String Method (Grouping)

Used when specific items must always stay together. Treat the group as a single unit and multiply by internal arrangements if the order within the group is not fixed.

Factorial Notation

n! represents the product of all positive integers up to n. For example, 4! = 4 × 3 × 2 × 1 = 24.

Shortcut Trick

The common mark interval is the LCM of jump lengths 4, 6, and 5.

LCM(4, 6, 5) = 60 feet.

Common marks follow the form: 199 + 60k (where k is an integer).

Range condition: 195 ≤ 199 + 60k ≤ 1000 ⇒ −4 ≤ 60k ≤ 801.

Possible values for k are 0, 1, 2, ..., 13 (Total 14 values).

∴ The correct answer is 14 times.

Alternate Method

Given:

Jump length of X = 4'

Jump length of Y = 6'

Jump length of Z = 5'

Initial common mark = 199' at 8 AM

Target range = [195', 1000']

Calculations:

⇒ Common landing occurs every LCM(4, 6, 5) feet.

⇒ 4 = 2², 6 = 2 × 3, 5 = 5

⇒ LCM = 2² × 3 × 5 = 60'

⇒ Since they land on 199' at 8 AM, other marks are 199 ± 60k.

⇒ Checking marks below 199': 199 − 60 = 139' (Outside range [195, 1000]).

⇒ Checking marks above 199': 199 + 60k ≤ 1000

⇒ 60k ≤ 1000 − 199 ⇒ 60k ≤ 801

⇒ k ≤ 801 ÷ 60 ⇒ k ≤ 13.35

⇒ Integer values of k are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.

⇒ Total count = 14.

∴ The correct answer is 14 times.

Additional Information

LCM (Least Common Multiple)

The smallest number that is a multiple of two or more numbers. Used to find synchronized events or common points in periodic cycles.

Arithmetic Progression (AP)

The common landing marks form an AP: a, a+d, a+2d... where a = 199 and d = 60. The number of terms n is found using T_n = a + (n − 1)d.

Relative Distance and Time

The time mentioned (8 AM and 9 AM) ensures the physical possibility of the event but does not affect the spatial distribution of the marks.