Explanation : 

A can X contains 399 liters of petrol

Let’s approach this step-by-step:

1) To find the possible bottle sizes, we need to find the common factors of 399 and 532.

2) First, let’s find the factors of 399:
399 = 3 × 7 × 19

3) Now, let’s find the factors of 532:
532 = 2² × 7 × 19

4) The common factors will be those that appear in both factorizations:
Common factors: 1, 7, 19, and 7 × 19 = 133

5) Therefore, the possible bottle sizes are:
1 liter
7 liters
19 liters
133 liters

6) Count the number of possible bottle sizes:
There are 4 different possible bottle sizes.

Therefore, the correct answer is b) 4.

Explanation :

421 and 427 

Let’s approach this step-by-step:

1) First, we need to understand what the question is asking. We’re looking for a number that, when dividing both 421 and 427, leaves a remainder of 1 in both cases.

2) Mathematically, we can express this as:
421 = k * d + 1
427 = m * d + 1
Where d is our divisor, and k and m are some integers.

3) Subtracting these equations:
427 – 421 = (m – k) * d
6 = (m – k) * d

4) This means that d must be a factor of 6.

5) The factors of 6 are: 1, 2, 3, and 6.

6) Let’s check each:

– If d = 1:
421 ÷ 1 = 421 remainder 0
427 ÷ 1 = 427 remainder 0
This doesn’t work.

– If d = 2:
421 ÷ 2 = 210 remainder 1
427 ÷ 2 = 213 remainder 1
This works!

– If d = 3:
421 ÷ 3 = 140 remainder 1
427 ÷ 3 = 142 remainder 1
This works!

– If d = 6:
421 ÷ 6 = 70 remainder 1
427 ÷ 6 = 71 remainder 1
This works!

7) Therefore, there are 3 numbers (2, 3, and 6) that can be used as the divisor to get the same remainder 1 for both 421 and 427.

The correct answer is c) 3.

Explanation : 

Let’s approach this step-by-step:

1) We can break down 30^30 into (3^30) * (10^30).

2) 10^30 will give us 30 zeros at the end of the number.

3) So, we need to find the last digit of 3^30.

4) Let’s look at the pattern of last digits when we raise 3 to different powers:
3^1 = 3
3^2 = 9
3^3 = 27 (last digit 7)
3^4 = 81 (last digit 1)
3^5 = 243 (last digit 3)

5) We see that the pattern of last digits repeats every 4 powers: 3, 9, 7, 1.

6) To find the last digit of 3^30, we can divide 30 by 4:
30 ÷ 4 = 7 remainder 2

7) This means that 3^30 will have the same last digit as 3^2, which is 9.

Therefore, the rightmost digit preceding the zeros in 30^30 is indeed 9.

The correct answer is d) 9.

Explanation : 

Percent of water must be mixed with honey

Let’s approach this step-by-step:

1) Let’s say we start with 100 units of honey.

2) We want to add water to this honey and sell the mixture at the same price as 100 units of pure honey, but make a 20% profit.

3) This means our mixture should be equivalent to 120 units of honey in terms of value (100 + 20% of 100).

4) But we’re selling it at the price of 100 units of honey. So our mixture must contain 100 units of honey.

5) If our mixture contains 100 units of honey and is equivalent to 120 units in value, the total mixture must be 120 units.

6) So we added 20 units of water to 100 units of honey.

7) To calculate the percentage:
(20 / 120) * 100 = 16.67% water in the final mixture

8) However, the question asks for the percent of water mixed with honey, not the percent in the final mixture.
So we calculate: (20 / 100) * 100 = 20%

Therefore, the correct answer is indeed a) 20%.

Explanation : 

222^333 + 333^222

1) Let’s start with the expression: 222^333 + 333^222

2) Factor out the common factors:
222 = 2 × 3 × 37
333 = 3 × 111

3) Now we can rewrite our expression:
(2 × 3 × 37)^333 + (3 × 3 × 37)^222

4) Let’s focus on the divisibility by each number:

For 2:
– The first term (2 × 3 × 37)^333 is clearly even (divisible by 2).
– The second term (3 × 111)^222 is odd (not divisible by 2).
– Even + Odd = Odd
– Therefore, the sum is not divisible by 2.

For 3:
– Both terms contain 3 as a factor.
– Therefore, the sum is divisible by 3.

For 37:
– Both terms contain 37 as a factor.
– Therefore, the sum is divisible by 37.

Therefore, the correct answer remains b) 3 and 37 but not 2.

Explanation : 

Calculating final savings

Let’s approach this step-by-step:

1) First, let’s calculate the savings for each day:
Jan 1: 1
Jan 2: 1 + 2 = 3
Jan 3: 3 + 2 = 5
Jan 4: 5 + 2 = 7
Jan 5: 7 + 2 = 9
Jan 6: 9 + 2 = 11
Jan 7: 11 + 2 = 13
Jan 8: 13 + 2 = 15
Jan 9: 15 + 2 = 17

2) Now, let’s calculate the total savings for each day:
Jan 1: 1
Jan 2: 1 + 3 = 4
Jan 3: 4 + 5 = 9
Jan 4: 9 + 7 = 16
Jan 5: 16 + 9 = 25
Jan 6: 25 + 11 = 36
Jan 7: 36 + 13 = 49
Jan 8: 49 + 15 = 64
Jan 9: 64 + 17 = 81

3) We’re looking for a number that is both a perfect square and a perfect cube.

4) From our list, we can see that 64 (on Jan 8) is both a perfect square (8^2) and a perfect cube (4^3).

Therefore, the correct answer is b) 8 January, 2023.

Explanation : 

Minimum amount of time to complete work

Let’s approach this problem step by step:

1) First, let’s calculate how much work each person can do in one hour:
X: 1/6 of the work
Y: 1/8 of the work
Z: 1/8 of the work

2) Now, let’s consider the constraints:
– Only one person can work each hour
– Nobody can work for two consecutive hours

3) To minimize time, we should use the most efficient worker (X) as much as possible, alternating with Y and Z.

4) Let’s see how much work can be done in 6 hours:
Hour 1: X does 1/6
Hour 2: Y does 1/8
Hour 3: X does 1/6
Hour 4: Z does 1/8
Hour 5: X does 1/6
Hour 6: Y does 1/8

5) After 6 hours, the total work done is:
(3 * 1/6) + (2 * 1/8) + (1 * 1/8) = 3/6 + 3/8 = 1/2 + 3/8 = 7/8

6) This means 7/8 of the work is done after 6 hours. We need to find how long it takes to do the remaining 1/8.

7) X is the fastest worker and can do 1/6 of the work in an hour. To do 1/8 of the work, X would need:
(1/8) / (1/6) = 3/4 of an hour = 45 minutes

Therefore, the minimum time to complete the work is 6 hours 45 minutes.

The correct answer is c) 6 hours 45 minutes.

Explanation : 

Calculating Consecutive zeros

To determine the number of consecutive zeros at the end of the integer obtained from the product 1^2 × 2^4 × 3^6 × 4^8 × … × 25^50, we need to find the highest power of 10 that divides this product. Since 10 = 2 × 5, we need to count the number of factors of 2 and 5 in the product and take the minimum of these counts.

First, let’s express the product in a more general form:
Product from k=1 to 25 of k^(2k)

Counting the Factors of 5:
We need to count the number of factors of 5 in each term k^(2k).

For k = 1 to 25:
– k = 5: 5^10 contributes 10 factors of 5.
– k = 10: 10^20 = (2 × 5)^20 contributes 20 factors of 5.
– k = 15: 15^30 = (3 × 5)^30 contributes 30 factors of 5.
– k = 20: 20^40 = (2^2 × 5)^40 contributes 40 factors of 5.
– k = 25: 25^50 = (5^2)^50 contributes 100 factors of 5.

Summing these contributions:
10 + 20 + 30 + 40 + 100 = 200

Multiple of 2 is definitely more.
So, Answer is 200

Explanation : 

Let’s approach this step-by-step:

1) Let’s say the original number of men is x.

2) We can express the relationship between men, days, and work as:
(Number of men) × (Number of days) = (Amount of work)

3) For the original scenario:
x × 6k = Amount of work

4) For the new scenario, let y be the new number of men:
y × 5k = Amount of work

5) Since the amount of work is the same in both cases:
x × 6k = y × 5k

6) We can express y in terms of x:
y = (x × 6k) / 5k = (6/5)x

7) The increase in the number of men is:
y – x = (6/5)x – x = (1/5)x

8) To express this as a percentage:
((1/5)x / x) × 100 = (1/5) × 100 = 20%

Therefore, the number of men should be increased by 20%.

The correct answer is c) 20%.

Explanation : 

Smallest non-negative value of the expression

5 – 4 – 3 + 2 × 1 = 0