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

Explanation : 

To find the sum of the first 28 terms of the sequence:

1, 1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, …

Step 1: Understand the Pattern

First, let’s observe the pattern in the sequence:
The sequence is built in groups, where each group starts with 1 and counts up to a certain number, then counts down to 2.
Here’s how the groups look:

Group 1: 1
Group 2: 1, 2
Group 3: 1, 3, 2
Group 4: 1, 4, 3, 2
Group 5: 1, 5, 4, 3, 2
Group 6: 1, 6, 5, 4, 3, 2
Group 7: 1, 7, 6, 5, 4, 3, 2

Each group n has n terms.

Step 2: Sum the Terms in Each Group

Now, let’s sum the terms in each group:
Sum = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

Step 3: Calculate the Total Sum

Add up the sums from each group:

Total Sum = 1 + 3 + 6 + 10 + 15 + 21 + 28 = 84

Explanation : 

To determine how many times the digit 5 appears in the page numbers of a 260-page book, we can analyze each digit place (units, tens, and hundreds) separately.

1. Units Place:

Every 10 numbers, the units digit cycles from 0 to 9.
Therefore, the digit 5 appears once in every set of 10 numbers.
Total sets of 10 in 260 pages: 260/10 = 26 sets.
Occurrences in units place: 26 × 1 = 26

2. Tens Place:

The digit 5 appears in the tens place for every number from 50-59, 150-159, and 250-259.
Each of these ranges contains 10 numbers where the tens digit is 5.
Total ranges with 5 in tens place: 3 (i.e., 50-59, 150-159, 250-259)
Occurrences in tens place: 3 × 10 = 30

3. Hundreds Place:

Since the book has only 260 pages, the hundreds digit can be 0, 1, or 2.
The digit 5 does not appear in the hundreds place within this range.
Occurrences in hundreds place: 0

Total Number of Fives:
Units place: 26
Tens place: 30
Hundreds place: 0
Total: 26 + 30 + 0 = 56

Therefore, the correct answer is b) 56.

Explanation : 

1. First, let’s understand how the clock hands move:

Hour hand makes a complete 360° rotation in 12 hours = 360°/12 = 30° per hour
Hour hand also moves 30°/60 = 0.5° per minute
Minute hand makes a complete 360° rotation in 60 minutes = 360°/60 = 6° per minute

At 4:25:

Hour hand:
For 4 hours: 4 × 30° = 120°
For 25 minutes: 25 × 0.5° = 12.5°
Total = 120° + 12.5° = 132.5°

Minute hand:
For 25 minutes: 25 × 6° = 150°

The angle between the hands:
|150° – 132.5°| = 17.5°

Therefore, the correct answer is c) 17.5°

Explanation : 

Let’s analyze each statement given the conditions:

Statement 1: (p – r)(qs) is even.

– (p – r): Since p is odd and r is even, p – r is odd.
– qs: q is odd and s is even, so qs is even.
– Therefore, (p – r)(qs) = odd × even = even.
– Conclusion: Statement 1 is correct.

Statement 2: (q – s) q s is even.

– (q – s): q is odd and s is even, so q – s is odd.
– q × s: q is odd and s is even, making qs even.
– Thus, (q – s)qs = odd × even = even.
– Conclusion: Statement 2 is correct.

Statement 3: (q + r)^2 (p + s) is odd.

– (q + r): q is odd and r is even, so q + r is odd. Squaring it, (q + r)^2, remains odd.
– (p + s): p is odd and s is even, so p + s is odd.
– Therefore, (q + r)^2 (p + s) = odd × odd = odd.
– Conclusion: Statement 3 is correct.

All three statements are correct.

Answer: d) 1, 2 and 3

Explanation : 

The calendar for the year 2025 is same as 2031.

Steps:
1. For calendars to be same, the number of days between the years must be divisible by 7

2. From 2025 to:
2029 = 4 years
2030 = 5 years
2031 = 6 years
2033 = 8 years

3. Leap years in between:
2025 to 2029: 1 leap year (2028)
2025 to 2030: 1 leap year (2028)
2025 to 2031: 1 leap year (2028)
2025 to 2033: 2 leap years (2028, 2032)

4. Total days:
2029: (4 × 365) + 1 = 1461 days
2030: (5 × 365) + 1 = 1826 days
2031: (6 × 365) + 1 = 2191 days
2033: (8 × 365) + 2 = 2922 days

5. Dividing by 7:
2029: 1461 ÷ 7 = 208.714…
2030: 1826 ÷ 7 = 260.857…
2031: 2191 ÷ 7 = 313 (exactly divisible)
2033: 2922 ÷ 7 = 417.428…

Answer: Option (c) 2031 is correct as 2191 days is exactly divisible by 7.

Explanation : 

Hour Hand & Minute Hand

1) First, let’s understand how the hands move:
Hour hand makes a complete 360° rotation in 12 hours, so it moves at 360°/12 = 30° per hour or 0.5° per minute
Minute hand makes a complete 360° rotation in 1 hour, so it moves at 360°/60 = 6° per minute

2) For hands to coincide, they must point at the same angle. Let’s use the formula:
Let ‘t’ be minutes after start time
Minute hand angle = 6t degrees
Hour hand angle = (initial hour angle + 0.5t) degrees
When they coincide: 6t = 30h + 0.5t
Where h is hours passed from start of clock (0 to 12)

3) From 10:00 AM to 2:00 PM is 4 hours
At 10:00, hour hand is at 300°
Coincidences happen when: 6t = 300 + 0.5t
5.5t = 300
t = 54.545… minutes after 10:00

4) Similarly, we can find all coincidences:
First: Around 10:54 AM
Second: Around 12:00 noon
Third: Around 1:05 PM

Therefore, in the given 4-hour period from 10:00 AM to 2:00 PM, the hour and minute hands coincide 3 times.

The answer is a) 3 times.