Total bill to be paid = Rs. 1840. Denominations – Rs. 10, Rs. 20 and Rs. 50.

Statement 1 –

•   If 25 notes are Rs. 50 are used, it will give us 25 × 50 = Rs. 1250.

•   Which means remaining 1840 – 1250 = Rs. 590 to be paid in Rs. 20 and Rs. 10 denominations.

•   But we have remaining number of notes = 50 – 25 = 25.

•   Now even if we take all these 25 notes of Rs. 20 we can pay maximum 25 × 20 = Rs. 500. So, Rs. 590 can’t be paid. Statement 1 is not correct.

Statement 2 –

•   If 25 notes are Rs. 50 are used, it will give us 35 × 20 = Rs. 700.

•   Which means remaining 1840 – 700 = Rs. 1140 to be paid in Rs. 50 and Rs. 10 denominations.

•   Whereas we have remaining number of notes = 50 – 35 = 15.

•   Now even if we take all these 15 notes of Rs. 50 we can pay maximum 15 × 50 = Rs. 750. So, Rs. 1140 can’t be paid. Statement 2 is not correct.

Statement 3 –

•   If 25 notes are Rs. 50 are used, it will give us 10 × 20 = Rs. 200.

•   Which means remaining 1840 – 200 = Rs. 1640 to be paid in Rs. 50 and Rs. 20 denominations.

•   Whereas we have remaining number of notes = 50 – 20 = 30.

•   Now even if we take all these 30 notes of Rs. 50 we can pay maximum 30 × 50 = Rs. 1500. So, Rs. 1640 can’t be paid. Statement 3 is not correct.

Thus All three statements are NOT correct.

•   First 150 persons will have 150/2 = 75 matches

•   75 persons will have (75 – 1)/2 = 37 matches

•   37 + 1 (left in the last step) = 38 person will have 38/2 = 19 matches

•   19 persons will have (19 – 1)/2 = 9 matches

•   9 + 1 (left in the last step) = 10 persons will have 10/2 = 5 matches

•   5 persons will have (5 – 1)/2 = 2 matches

•   2 + 1 (left in the last step) = 3 persons will have (3 - 1)/2 = 1 match

•   Then this 1 person will have a match with a person left in the last step = 1 match

•   Total matches = 75 + 37 + 19 + 9 + 5 + 2 + 1 + 1 = 149.

•   The statement meant that one of those two actions were bound to happen - either picnic or trekking.

•   Option (b) is irrelevant.

•   Options (a) and (c) are correct - do not falsify the claim.

•   Option (d) negates both.

•   Hence, best answer is option (d).

•   Question does not say anything about the gender composition of the group invited for party.

•   Further, we do not know which specific female faculty knew music.

•   So either all in party were males, or some males and some females who did not know music were present in the party.

•   All options have some problem except (d) which is definitely correct.

•   5 hobby groups - photography, yachting, chess electronics and gardening.

•   Gardening - Meets every second day hence = 90 times in 180 days.

•   Electronics group meets every third day = 60 times / 180

•   Chess group meets every 4th day = 45 times / 180

•   Yachting group meets every 5th day = 36 times / 180

•   Photography group meets every 6th day = 30 times / 180

•   We have to find highest common factor of these 5 numbers 90, 60, 45, 36 and 30. It is 3. Hence option (d) is correct answer.

A > = B, D > = C, B > C. This implies that A > C. Hence A is older than C. Answer is (d).

•   We have to list the total number of cases which are distinct from each other.

•   Such cases will be 2 x 2 x 2 x 2 x 2 = 2⁵

•   In all these cases, no two candidates have marked answers identically.

•   Hence, Maximum number of candidates is 25 = 32

Option D: 150 m
To solve this problem, we need to calculate the total distance the competitor runs to collect all 6 apples and drop them in the bucket. Let's break it down step by step:
  ● Distance from the bucket to the first apple: 5 meters  
  ● Distance between each subsequent apple: 3 meters  
 Step-by-step calculation:
 1. First Apple:
         ○ Distance to the first apple: 5 meters
         ○ Round trip distance (to the apple and back to the bucket): \(5 \times 2 = 10\) meters
 2. Second Apple:
         ○ Distance to the second apple: \(5 + 3 = 8\) meters
         ○ Round trip distance: \(8 \times 2 = 16\) meters
 3. Third Apple:
         ○ Distance to the third apple: \(5 + 3 \times 2 = 11\) meters
         ○ Round trip distance: \(11 \times 2 = 22\) meters
 4. Fourth Apple:
         ○ Distance to the fourth apple: \(5 + 3 \times 3 = 14\) meters
         ○ Round trip distance: \(14 \times 2 = 28\) meters
 5. Fifth Apple:
         ○ Distance to the fifth apple: \(5 + 3 \times 4 = 17\) meters
         ○ Round trip distance: \(17 \times 2 = 34\) meters
 6. Sixth Apple:
         ○ Distance to the sixth apple: \(5 + 3 \times 5 = 20\) meters
         ○ Round trip distance: \(20 \times 2 = 40\) meters
 Total Distance:
      ○ Sum of all round trip distances: \(10 + 16 + 22 + 28 + 34 + 40 = 150\) meters
 Thus, the total distance the competitor has to run is 150 meters.

Option D: Cannot be predicted
       1. Understanding the Problem:
            ○ There are twelve members in the club.
            ○ Each month, one member is chosen to host a dinner.
            ○ The question asks how many dinners a particular member will host in one year.
     2. Analyzing the Situation:
            ○ There are 12 months in a year.
            ○ Each month, one of the 12 members is randomly chosen to host a dinner.
            ○ Over the course of a year, each member has an equal chance of being chosen each month.
     3. Probability and Distribution:
            ○ In an ideal scenario where each member is chosen exactly once, each member would host one dinner per year. However, the selection is random.
            ○ It is possible for some members to be chosen more than once, while others might not be chosen at all.
     4. Conclusion:
            ○ Since the selection is random and independent each month, it is not possible to predict exactly how many times a particular member will host a dinner in a year.
            ○ Therefore, the number of dinners a particular member has to host in one year cannot be predicted with certainty.
     ● Final Answer: The correct answer is Option D: Cannot be predicted.  

From Statement 2, the male members are not at home. It is given that the daughters are invited for the feast, but It does not explicitly state that they left home.

Hence, none among (a), (c) and (d) can be said to be true. Nothing is specified about the man's wife. Hence, the statement that the man's wife is likely to be left at home is true.

Option C: C has more than what A and D have together.
Let's calculate the final amount each person has after the transactions:
     ● Initial Amounts:  
           ○ A: ₹100
           ○ B: ₹100
           ○ C: ₹100
           ○ D: ₹100
     ● Transactions:  
       1. A pays ₹20 to B:
              ○ A: ₹100 - ₹20 = ₹80
              ○ B: ₹100 + ₹20 = ₹120
       2. B pays ₹10 to C:
              ○ B: ₹120 - ₹10 = ₹110
              ○ C: ₹100 + ₹10 = ₹110
       3. C gets ₹30 from D:
              ○ C: ₹110 + ₹30 = ₹140
              ○ D: ₹100 - ₹30 = ₹70
     ● Final Amounts:  
           ○ A: ₹80
           ○ B: ₹110
           ○ C: ₹140
           ○ D: ₹70
     ● Analysis of Options:  
       ● Option A: C is the richest.  
             ○ Correct, as C has ₹140, which is more than anyone else.
       ● Option B: D is the poorest.  
             ○ Correct, as D has ₹70, which is less than anyone else.
       ● Option C: C has more than what A and D have together.  
             ○ Incorrect, as A and D together have ₹80 + ₹70 = ₹150, which is more than C's ₹140.
       ● Option D: B is richer than D.  
             ○ Correct, as B has ₹110, which is more than D's ₹70.
     Therefore, the statement that is not correct is Option C: C has more than what A and D have together.

Option D: _cauliflower is tastier than cabbage_.
         ○ Let's analyze the given statements:
       1. Lady's finger is tastier than cabbage.
       2. Cauliflower is tastier than lady's finger.
       3. Cabbage is not tastier than peas.
         ○ From statement 1, we know: Lady's finger > Cabbage.
         ○ From statement 2, we know: Cauliflower > Lady's finger.
         ○ Combining statements 1 and 2, we get: Cauliflower > Lady's finger > Cabbage.
         ○ From statement 3, we know: Peas ≥ Cabbage.
         ○ Now, let's evaluate the options:
       ● Option A: Peas are as tasty as lady's finger.  
             ○ This cannot be concluded because we only know that peas are not less tasty than cabbage, and we have no direct comparison between peas and lady's finger.
       ● Option B: Peas are as tasty as cauliflower and lady's finger.  
             ○ This cannot be concluded because we have no information comparing peas directly with cauliflower or lady's finger.
       ● Option C: Cabbage is the least tasty of the four vegetables.  
             ○ This is possible but not definitively concluded because peas could be equally as tasty as cabbage.
       ● Option D: Cauliflower is tastier than cabbage.  
             ○ This is directly supported by combining statements 1 and 2, which show that cauliflower is tastier than both lady's finger and cabbage.
         ○ Therefore, the most logical conclusion based on the given statements is that cauliflower is tastier than cabbage.

Let the number of cars parked be C and total number of two wheelers parked be T.

Then, according to the question

4C + 2 T = 1 00 + 2 (C + T)

⸫ C = 50

So, the number of cars parked is 50.

According to the given information, 

Usha > Kamala  ...(i)

Swati > Priti  ...(ii)

Kamala > Swati  ...(iii)

On combining Eqs. (i), (ii) and (iii), we get

Usha > Kamala > Swati > Priti

So, it is clear from the above arrangement that Priti is the slowest runner.

Let the number of males and females in the party be x and y, respectively.

Then. xy = 24  …(i)

⸫ Number of hugs among the males = x (x - 1) / 2

and number of hugs among the females = y (y - 1) / 2

⸫ Total number of hugs at the party = Numbers of hugs among the males + Number of hugs among the females

= x (x - 1) / 2 + y (y - 1) / 2  …(ii)

From Eq. (i), the value of xy can be (1, 24), (3, 8), (2, 12), (4, 6), (6, 4). (8, 3), (12, 2) and (24, 1).

By taking one of these values and putting in Eq. (ii), we get

Total number of hugs at the party = 6(6-1) / 2 + 4(4-1) / 2 = 15 + 6 = 21 

Given, total number of white marbles in the box = 10

⸫ Number of red marbles  = 10 + 3 = 13

and number of green marbles = 10 – 5 = 5

⸫ Total number of marbles = 10 + 13 + 5 = 28

Alternate Method

Given,W = 10

According to the given information,

R – W = 3  …(i)

And   W – G = 5  ...(ii)

Let the total number of marbles be x

then,  R + W + G = x

From Eqs. (i) and (ii),

we getR = 13 , G = 5

⸫ Total number of marbles = R + W + G = x

] x = 13 + 10 + 5 = 28

Hence, the total number of marbles in the box is 28.

Option B: 12
To solve this problem, we need to determine how many distinct points are marked on the line segment, considering the marking pattern from both ends and accounting for any overlap.
     Step-by-step Analysis:
     ● Marking from the Left End:  
           ○ The first point is at 1 cm from the left end.
           ○ The second point is at 1 + 2 = 3 cm from the left end.
           ○ The third point is at 3 + 3 = 6 cm from the left end.
           ○ The fourth point is at 6 + 4 = 10 cm from the left end.
           ○ The fifth point is at 10 + 5 = 15 cm from the left end.
           ○ The sixth point is at 15 + 6 = 21 cm from the left end.
           ○ The seventh point is at 21 + 7 = 28 cm from the left end.
           ○ The eighth point is at 28 + 8 = 36 cm from the left end.
     ● Marking from the Right End:  
           ○ The first point is at 1 cm from the right end, which is at 36 - 1 = 35 cm from the left end.
           ○ The second point is at 35 - 2 = 33 cm from the left end.
           ○ The third point is at 33 - 3 = 30 cm from the left end.
           ○ The fourth point is at 30 - 4 = 26 cm from the left end.
           ○ The fifth point is at 26 - 5 = 21 cm from the left end.
           ○ The sixth point is at 21 - 6 = 15 cm from the left end.
           ○ The seventh point is at 15 - 7 = 8 cm from the left end.
           ○ The eighth point is at 8 - 8 = 0 cm from the left end (which is the starting point and not counted).
     Combining Points:
         ○ From the left end, the points are at: 1, 3, 6, 10, 15, 21, 28, 36 cm.
         ○ From the right end, the points are at: 35, 33, 30, 26, 21, 15, 8 cm.
     Distinct Points:
         ○ Combining both sets and removing duplicates, we have the distinct points: 1, 3, 6, 8, 10, 15, 21, 26, 28, 30, 33, 35 cm.
     ● Total Number of Distinct Points: 12  
     Therefore, the number of distinct points marked on the line segment is 12.

Gardening group meets once in 2 days, electronics group meets once in 3 days, chess group meets once in 4 days, yachting group meets once in 5 days and the photography group meets once in 6 days. If they meet on the same day one time, the next time they will meet on the same day again will be the LCM of 2. 3, 4, 5 and 6 which equals 60. Hence, within 180 days all the five groups will have to meet on the same day 100/60 so = 3 times.