In a nationwide Smart city Business Idia hunt the jury selected total 100 proposals from six cities Bengaluru, Mumbai, Delhi, Hydrabad, Chennai and Pune. Bengaluru outsmarted all other cities by getting accepted 50 proposals alone from it. Pune stood last as only 5 proposals were selected from this city.
The second highest number of proposals was selected from Mumbai. What is the minimum number of proposals that would have been selected from Mumbai if distinct number of proposals were selected from each of the six cities?

Options

A. 9
B. 10
C. 12
D. 13

Answer

To solve the problem, we need to determine the minimum number of proposals selected from Mumbai, given the constraints of the problem. Let's break it down step by step.

### Step 1: Understand the given information
- Total proposals selected: 100
- Proposals from Bengaluru: 50
- Proposals from Pune: 5
- Proposals from Mumbai: Second highest number
- Proposals from the other cities (Delhi, Hyderabad, Chennai): Must be distinct and different from each other.

### Step 2: Calculate remaining proposals
After allocating 50 proposals to Bengaluru and 5 to Pune, we have:
\[ 100 - 50 - 5 = 45 \]
So, there are 45 proposals left to be distributed among Mumbai, Delhi, Hyderabad, and Chennai.

### Step 3: Assign distinct values
Since Mumbai has the second highest number of proposals, we need to assign distinct values to the remaining cities (Mumbai, Delhi, Hyderabad, and Chennai) such that:
- Mumbai > Delhi, Hyderabad, Chennai
- All values are distinct and greater than 5 (since Pune has only 5).

### Step 4: Determine the minimum for Mumbai
To find the minimum for Mumbai, we can start by assigning the highest possible values to the other cities while ensuring Mumbai remains the second highest.

Let’s denote the number of proposals from:
- Mumbai = M
- Delhi = D
- Hyderabad = H
- Chennai = C

We know:
- M > D, H, C
- D, H, C must be distinct and greater than 5.

### Step 5: Assign values to D, H, and C
Let’s assign the highest possible distinct values to D, H, and C:
- Assign D = 12
- Assign H = 11
- Assign C = 10

This gives us:
\[ D + H + C = 12 + 11 + 10 = 33 \]

### Step 6: Calculate M
Now we can calculate M:
\[ M + 33 = 45 \]
Thus,
\[ M = 45 - 33 = 12 \]

However, since M must be greater than D (which is 12), we need to adjust our values.

### Step 7: Adjust values
Let’s try assigning:
- D = 11
- H = 10
- C = 9

Now we have:
\[ D + H + C = 11 + 10 + 9 = 30 \]

Calculating M:
\[ M + 30 = 45 \]
Thus,
\[ M = 45 - 30 = 15 \]

Now, we check if we can lower M while keeping it distinct:
If we try:
- D = 10
- H = 9
- C = 8

This gives:
\[ D + H + C = 10 + 9 + 8 = 27 \]
Calculating M:
\[ M + 27 = 45 \]
Thus,
\[ M = 45 - 27 = 18 \]

### Final Step: Check for minimum
We can keep adjusting downwards, but we find that the minimum distinct values that satisfy all conditions lead us to:
- D = 12
- H = 11
- C = 10
- M = 13 (as the second highest)

Thus, the minimum number of proposals that would have been selected from Mumbai is:
**M = 13.**