Explanation : 

To determine who received the least amount among X, Y, and Z, let’s analyze the given statements:

Statement I: Given: X = 4/5(Y + Z)

Implication: The total amount distributed, T = X + Y + Z. Substituting from the statement:
T = 4/5(Y + Z) + Y + Z = 9/5(Y + Z)
Y + Z = 5/9T
X = 4/9T
Analysis: While we know X = 4/9T and Y + Z = 5/9T, we cannot determine the individual values of Y and Z. Without knowing how Y and Z are divided, we cannot conclusively identify who received the least.

Statement II: Given: Y = 2/7(X + Z)

Implication:
Let X + Z = a. Then, Y = 2/7a and T = a + 2/7a = 9/7a
Y = 2/9T
X + Z = 7/9T
Analysis: Even though we know Y = 2/9T, without the individual values of X and Z, it’s unclear whether Y is the least or if X or Z could be lesser based on their distribution.

Using Both Statements Together:

From Statement I: X = 4/9T
From Statement II: Y = 2/9T and X + Z = 7/9T
Substituting:
Y = 2/7(X + Z) = 2/7 × 7/9T = 2/9T
X = 4/9T
Z = 3/9T
Conclusion: The distribution ratios are X:Y:Z = 4:2:3. Clearly, Y receives the least amount.

Explanation : 

To determine the unique values of x and y (distinct natural numbers) that satisfy the given conditions, let’s analyze each statement individually and then consider them together.

Statement I: x/y is odd.

Interpretation: For x/y to be an odd integer, x must be an odd multiple of y. This implies that y must be a divisor of x, and the quotient x/y must be an odd number.

Limitations: Without additional information, there are multiple pairs (x, y) that can satisfy this condition. For example:
– x = 3, y = 1 (since 3/1 = 3 is odd)
– x = 9, y = 3 (since 9/3 = 3 is odd)

Conclusion: Statement I alone is insufficient to determine unique values of x and y.

Statement II: xy = 12.

Interpretation: The product of x and y is 12. Considering that x and y are distinct natural numbers, the possible pairs are:
– (1, 12)
– (2, 6)
– (3, 4)

Limitations: Multiple valid pairs satisfy this condition.

Conclusion: Statement II alone is insufficient to determine unique values of x and y.

Combining Statements I and II:

Now, let’s use both statements together.

1. From Statement II, the possible pairs are (1, 12), (2, 6), and (3, 4).

2. Applying Statement I (x/y is odd) to each pair:
– (1, 12): 1/12 is not an integer, so this pair is invalid.
– (2, 6): 2/6 = 1/3 is not an integer, so this pair is invalid.
– (3, 4): 3/4 is not an integer, so this pair appears invalid at first glance.

However, there’s a misunderstanding here. For x/y to be an odd integer, x must be a multiple of y, and the quotient must itself be odd.

Re-examining the possible pairs:
– (12, 1): 12/1 = 12 (Even)
– (6, 2): 6/2 = 3 (Odd)
– (4, 3): 4/3 is not an integer.

Valid Pair: (6, 2) since 6/2 = 3 is an odd integer.

Conclusion: Using both statements together, the unique values are x = 6 and y = 2.

Therefore, option c is the correct choice.

Explanation : 

Initial Movement:
Direction: West
Action: Walks 100 meters West.

First Turn:
Turn: Left
From West, turning left means facing South.
Action: Walks 100 meters South.

Second Turn:
Turn: 225° clockwise from South
Understanding the Turn:
– Starting Direction: South (which is 180° on a compass)
– Clockwise Turn: Adding 225° to the current direction.
– Calculation: 180° (South) + 225° = 405°
– Since 405° is equivalent to 45° (405° – 360°), the final direction is 45°, which corresponds to North-East on a compass.

Final Direction: North-East, Answer: d) North-East

Explanation : 

Let’s analyze the person’s movements step by step to determine the initial direction:

1. Initial Direction: Suppose the person starts by walking South.

2. First Movement:
– Walks 100 m South.
– Position: (0, -100).

3. First Turn (Left):
– From South, turning left means facing East.
– Walks 100 m East.
– Position: (100, -100).

4. Second Turn (Left):
– From East, turning left means facing North.
– Walks 300 m North.
– Position: (100, 200).

5. Third Turn (Right):
– From North, turning right means facing East again.
– Walks 100 m East.
– Final Position: (200, 200).

The final position (200, 200) places the office North-East of the house, which matches the given condition.

Answer: c) South

Explanation : 

Let p and q be positive integers

To determine the smallest value of k that does not uniquely determine the positive integers p and q (with p < q and p + q = k), let’s evaluate each option:

1. k = 3
– Possible pairs: (1, 2)
– Unique pair: Yes.

2. k = 4
– Possible pairs: (1, 3)
– Unique pair: Yes. (Note: (2, 2) is invalid since p < q.)

3. k = 5
– Possible pairs: (1, 4) and (2, 3)
– Unique pair: No. There are two distinct pairs.

4. k = 6
– Possible pairs: (1, 5) and (2, 4)
– Unique pair: No. There are two distinct pairs.

The smallest k where multiple pairs (p, q) satisfy the conditions is k = 5.

Explanation : 

32^5 + 2^27

To determine which of the given options divides 32^5 + 2^27, let’s simplify and analyze the expression step-by-step.

Simplify the Expression First, express everything in terms of powers of 2: 32 = 2^5 implies 32^5 = (2^5)^5 = 2^25

2^27 remains as is.

So, the expression becomes: 32^5 + 2^27 = 2^25 + 2^27 = 2^25(1 + 2^2) = 2^25 × 5

This shows that the expression is divisible by 2^25 and 5, and consequently by 2 × 5 = 10.

Explanation : 

Let’s analyze the problem step by step.

Let:
o = cost of one orange
m = cost of one mango
a = cost of one apple

Given:
The total cost of 4 oranges, 6 mangoes, and 8 apples is equal to twice the total cost of 1 orange, 2 mangoes, and 5 apples.

This can be written as:
4o + 6m + 8a = 2(1o + 2m + 5a)
4o + 6m + 8a = 2o + 4m + 10a

Simplifying:
Subtract 2o + 4m + 10a from both sides:
2o + 2m – 2a = 0
Divide by 2:
o + m = a
This confirms Statement 2: The total cost of one orange and one mango is equal to the cost of one apple.

Now, let’s evaluate Statement 1:
“The total cost of 3 oranges, 5 mangoes, and 9 apples is equal to the total cost of 4 oranges, 6 mangoes, and 8 apples.”

Expressed mathematically:
3o + 5m + 9a = 4o + 6m + 8a
Rearrange:
3o + 5m + 9a – 4o – 6m – 8a = 0
-o – m + a = 0
o + m = a

This is the same as what we derived earlier, hence Statement 1 is also correct.

Conclusion: Both statements 1 and 2 are correct.

Explanation : 

Three numbers x, y, z…

To find the number of triplets (x, y, z) from the first seven natural numbers (1 to 7) satisfying the conditions x > 2y > 3z, we’ll systematically examine possible values of z, y, and x.

Consider z = 1:

– 3z = 3
– 2y > 3z => 2y > 3 => y > 1.5 => y >= 2
– Possible y values: 2, 3, 4, 5, 6, 7
– Compute 2y for each y and find x such that x > 2y within the set {1, 2, …, 7}:
– y = 2 => 2y = 4 => x > 4 => x = 5, 6, 7
– y = 3 => 2y = 6 => x > 6 => x = 7
– For y >= 4, 2y >= 8, but x cannot be greater than 7, so no valid x exists.
– Valid triplets for z = 1:
– (5, 2, 1)
– (6, 2, 1)
– (7, 2, 1)
– (7, 3, 1)

Consider z >= 2:

– For z = 2:
– 3z = 6
– 2y > 6 => y > 3
– Possible y values: 4, 5, 6, 7
– 2y >= 8, but x > 2y would require x > 8, which is beyond 7.
– No valid triplets.
– Similarly, for z = 3 to 7, the required y and x values exceed 7. No valid triplets exist.

Conclusion:
Only for z = 1 do we find valid triplets, and there are four such triplets.

Answer: Four triplets

Explanation : 

We are given:
– Two two-digit numbers: AB and CD
– Their sum is a three-digit number: 1CE
– All letters represent distinct digits.

Step 1: Set Up the Addition

   A B
+ C D
———-
1 C E

Step 2: Analyze the Units Place (Rightmost Digit)

Adding the units digits:
– B + D = E (with possible carryover to the tens place)

Let k be the carryover from the units place addition (k can be 0 or 1 because the sum of two digits can’t produce a carryover greater than 1 in decimal addition).

So:
1. B + D = E + 10k
2. Equation (1): B + D = E + 10k

Step 3: Analyze the Tens Place

Adding the tens digits, plus any carryover from the units place:
– A + C + k = C + 10
(since the tens digit in the result is C, and there’s a carryover to make it the same C)

Simplify:
1. A + k = 10
2. Equation (2): A + k = 10

Step 4: Solve for A and k

From Equation (2):
– Since A is a single digit (0-9), and k is 0 or 1:
– If k = 0: A = 10 (invalid, as A must be a single digit)
– If k = 1: A = 9 (valid, as A is a single-digit number)

Therefore, A = 9 and k = 1.

Step 5: Conclusion

Using the carryover method simplifies the problem and directly leads us to the value of A:
– A = 9

Explanation : 

THE COST OF P

Given:
– R costs 20% more than Q
– P costs 25% more than R
– Total cost (P + Q + R) = ₹3,330

STEP 1: EXPRESS ALL COSTS IN TERMS OF Q
Cost of Q = Q
Cost of R = Q + 20% of Q = 1.20Q
Cost of P = R + 25% of R = 1.25 × 1.20Q = 1.50Q

STEP 2: FORM TOTAL COST EQUATION
P + Q + R = 3,330
1.50Q + Q + 1.20Q = 3,330
3.70Q = 3,330

STEP 3: SOLVE FOR Q
Q = 3,330 ÷ 3.70
Q = 900

STEP 4: CALCULATE P
P = 1.50 × Q
P = 1.50 × 900
P = 1,350

ANSWER: P costs ₹1,350