Maneela, Raghu and Aravind have some jems with each of them. Five times the number of jems with Raghu equals seven times the number of jems with Maneela while five times the number of jems with Maneela equals seven times the number of jems with Aravind. What is the minimum number of jems that can be there with all three of them put together ?

Options

A. A) 108
B. B) 107
C. C) 109
D. D) 110

Answer

To solve the problem, we need to establish the relationships between the number of gems each person has based on the information given.

Let:
- The number of gems with Raghu = R
- The number of gems with Maneela = M
- The number of gems with Aravind = A

From the problem, we have the following relationships:

1. Five times the number of gems with Raghu equals seven times the number of gems with Maneela:
\[ 5R = 7M \]
Rearranging gives us:
\[ \frac{R}{M} = \frac{7}{5} \]

2. Five times the number of gems with Maneela equals seven times the number of gems with Aravind:
\[ 5M = 7A \]
Rearranging gives us:
\[ \frac{M}{A} = \frac{7}{5} \]

Now, we can express R, M, and A in terms of a common variable. Let's set:
- \( M = 5k \) (where k is a common multiplier)

Using the first relationship:
\[ R = \frac{7}{5}M = \frac{7}{5}(5k) = 7k \]

Using the second relationship:
\[ A = \frac{5}{7}M = \frac{5}{7}(5k) = \frac{25}{7}k \]

Now we have:
- \( R = 7k \)
- \( M = 5k \)
- \( A = \frac{25}{7}k \)

To find a common multiple for R, M, and A, we need to express them with a common denominator. The least common multiple of the denominators (1 and 7) is 7.

Thus, we can express:
- \( R = 7k \) remains as is.
- \( M = 5k \) can be expressed as \( \frac{35k}{7} \).
- \( A = \frac{25}{7}k \) remains as is.

Now we can express all in terms of k:
- \( R = 7k \)
- \( M = \frac{35k}{7} \)
- \( A = \frac{25k}{7} \)

Now, to find the total number of gems:
\[ \text{Total gems} = R + M + A = 7k + 5k + \frac{25}{7}k \]
To combine these, we need a common denominator:
\[ = \frac{49k}{7} + \frac{35k}{7} + \frac{25k}{7} = \frac{49k + 35k + 25k}{7} = \frac{109k}{7} \]

To find the minimum number of gems, we set \( k = 7 \) (the smallest integer that makes A an integer):
\[ \text{Total gems} = \frac{109 \times 7}{7} = 109 \]

Thus, the minimum number of gems that can be there with all three of them put together is **109**.