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 : 

Letters A B C D and E are arranged

The problem asks for the number of arrangements of the letters A, B, C, D and E such that there are exactly two letters between A and E. There are four possible arrangements for A and E: A _ _ E _, _ A _ _ E, E _ _ A _, and _ E _ _ A.

For each of these arrangements, there are 3! = 6 ways to arrange the remaining letters B, C and D. So the total number of arrangements is 4 * 6 = 24.

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.

Explanation : 

To determine the average weight of the boys, let’s break down the given information:

1. Equality of Weights:
– Weight of 6 boys = Weight of 7 girls = Weight of 3 men = Weight of 4 women = W

2. Average Weight of Women:
– Average weight of one woman = 63 kg
– Total weight of 4 women = 4 × 63 = 252 kg
– Therefore, W = 252 kg

3. Calculating Average Weight of Boys:
– Total weight of 6 boys = W = 252 kg
– Average weight of one boy = 252/6 = 42 kg

Answer: b) 42 kg

Explanation : 

Two-Digit Number Sum by Interchanging Digits of X

Let’s analyze the problem step by step.

Given:
– Let X = 10a + b be a two-digit number where a is the tens digit (1-9) and b is the units digit (0-9).
– Let Y = 10b + a be the number formed by interchanging the digits of X.

Sum of X and Y:
X + Y = (10a + b) + (10b + a) = 11(a + b)

Objective:
– To maximize X + Y such that it remains a two-digit number.

Maximum Two-Digit Number:
– The greatest two-digit number is 99.
– So, 11(a + b) ≤ 99 which implies a + b ≤ 9.

Maximizing the Sum:
– To achieve the largest possible two-digit sum, set a + b = 9.
– Thus, X + Y = 99.

Possible Combinations:
We need to find pairs (a, b) where a + b = 9 and a ranges from 1 to 9 while b ranges from 0 to 9.

Here are the possible pairs:
(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)

Excluding Invalid Cases:
– When a = 9, b = 0 leads to Y = 09, which is not a valid two-digit number.
– Therefore, exclude (9,0).

Valid Pairs:
(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1)

Number of Possible Values of X:
– There are 8 valid pairs.

Answer: d) 8