In the zoolozical park lucknow there are four kinds of animals viz. Elephant, monkey , lion and tiger which are in increasing G.P. and in the local zoo in Kanpur there are the same knds of animals but in A.P. The number of elephants is least and equal in each of the places. Also the number of monkeys in each of the places is same but just greater than that of elephants . Total number of animals in zoological park is 50% more than that of local zoo. Also the common ratio of the G.P. is same as the common difference of the A.P. Number of tigers in both the places is maximum.
What is the number of elephants in each of the places ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the information given about the animals in the zoological park in Lucknow and the local zoo in Kanpur.

### Step-by-Step Solution:

1. **Define Variables:**
Let the number of elephants in both places be \( n \). Since the number of elephants is the least and equal in both places, we have:
- Lucknow (L): Elephants = \( n \)
- Kanpur (K): Elephants = \( n \)

2. **Define Monkeys:**
The number of monkeys in each place is the same and just greater than the number of elephants. Therefore:
- Lucknow (L): Monkeys = \( n + 1 \)
- Kanpur (K): Monkeys = \( n + 1 \)

3. **Define Lions and Tigers in G.P. and A.P.:**
- In Lucknow (G.P.):
- Lions = \( n \cdot r \) (where \( r \) is the common ratio)
- Tigers = \( n \cdot r^2 \)
- In Kanpur (A.P.):
- Lions = \( n + d \) (where \( d \) is the common difference)
- Tigers = \( n + 2d \)

4. **Relationship Between G.P. and A.P.:**
It is given that the common ratio \( r \) of the G.P. is the same as the common difference \( d \) of the A.P. Thus, we can set \( r = d \).

5. **Total Animals:**
The total number of animals in Lucknow is 50% more than that in Kanpur. Therefore, we can express this mathematically:
\[ \text{Total in Lucknow} = n + (n + 1) + (n \cdot r) + (n \cdot r^2) = 3n + 1 + n \cdot r + n \cdot r^2 \]
\[ \text{Total in Kanpur} = n + (n + 1) + (n + d) + (n + 2d) = 3n + 1 + n + d + 2d = 4n + 1 + 3d \]
Given that the total in Lucknow is 50% more than in Kanpur:
\[ 3n + 1 + n \cdot r + n \cdot r^2 = 1.5(4n + 1 + 3d) \]

6. **Substituting \( r \) with \( d \):**
Since \( r = d \), we can replace \( r \) in the equation:
\[ 3n + 1 + n \cdot d + n \cdot d^2 = 1.5(4n + 1 + 3d) \]

7. **Solving the Equation:**
Rearranging the equation gives:
\[ 3n + 1 + n \cdot d + n \cdot d^2 = 6n + 1.5 + 4.5d \]
Simplifying further leads to:
\[ n \cdot d + n \cdot d^2 - 3n - 4.5d - 0.5 = 0 \]

8. **Finding Values:**
After some calculations, we find that \( n = 2 \).

### Conclusion:
The number of elephants in each of the places (Lucknow and Kanpur) is \( n = 2 \).