Mr. Abhishek has a bag that contains red, black, and yellow marbles in the ratio 5:1:4. He gave some red and yellow marbles to Mr. Navneet. As a result, the ratio of the number of red, black, and yellow marbles left in the bag becomes 5:3:7. If minimum possible number of marbles with him could be 'n' and maximum number of marbles given by him to Navneet could be 'm' (where m is a 2 digit number), find the value of m/n.

Let's solve the problem step by step:

### Step 1: Define the initial conditions
Let the number of red, black, and yellow marbles initially be \(5x\), \(1x\), and \(4x\) respectively, where \(x\) is the common multiplier.

### Step 2: Define the final conditions after giving marbles to Mr. Navneet
After giving some red and yellow marbles to Mr. Navneet, the number of marbles left becomes such that the ratio of red, black, and yellow marbles is \(5:3:7\). Let the new numbers of red, black, and yellow marbles be \(5y\), \(3y\), and \(7y\) respectively, where \(y\) is the new common multiplier.

### Step 3: Set up equations based on the given information
- **Red marbles**: Initially, there are \(5x\) red marbles. After giving away some, \(5y\) red marbles remain.
  \[
  \text{So, number of red marbles given away} = 5x - 5y = 5(x - y)
  \]

- **Yellow marbles**: Initially, there are \(4x\) yellow marbles. After giving away some, \(7y\) yellow marbles remain.
  \[
  \text{So, number of yellow marbles given away} = 4x - 7y
  \]

- **Black marbles**: The number of black marbles remains the same, so \(x = 3y\).

### Step 4: Simplify and solve for \(x\) and \(y\)
From the black marbles, we have:
\[
x = 3y
\]
Substitute this into the expressions for red and yellow marbles:

- For red marbles:
  \[
  5(3y) - 5y = 10y
  \]

- For yellow marbles:
  \[
  4(3y) - 7y = 12y - 7y = 5y
  \]

So, the number of red marbles given away is \(10y\) and the number of yellow marbles given away is \(5y\).

### Step 5: Calculate the total number of marbles
- **Initial total number of marbles**:
  \[
  \text{Total initial} = 5x + 1x + 4x = 10x = 10(3y) = 30y
  \]

- **Final total number of marbles left**:
  \[
  \text{Total final} = 5y + 3y + 7y = 15y
  \]

So, \(n\) (the minimum number of marbles left) is \(15y\).

### Step 6: Calculate \(m\) (total number of marbles given away)
\[
m = 30y - 15y = 15y
\]

Given that \(m\) is a two-digit number, we have:
\[
15y = m \implies 10 \leq y \leq 6
\]

\(y\) should be a number such that \(15y\) gives a two-digit number, so \(y = 1, 2, 3, \dots, 6\). Therefore, the maximum \(m\) can be \(90\) (when \(y = 6\)) and minimum \(n\) can be \(15\) (when \(y = 1\)).

### Step 7: Compute the ratio \( \frac{m}{n} \)
\[
\frac{m}{n} = \frac{90}{15} = 6
\]

Thus, the value of \(\frac{m}{n}\) is \(6\).

Would you like more details, or do you have any other questions?

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**Related Questions:**
1. What if Mr. Navneet had taken away some black marbles as well?
2. How would the solution change if the initial ratio were different?
3. Can you derive a general formula for \(m/n\) given any initial and final ratios?
4. What if the ratio after removing marbles was different, such as 3:2:5?
5. How does the problem change if the total number of marbles is specified?
6. What would happen if \(y\) was not an integer?
7. Can this problem be solved using algebraic equations directly without ratios?
8. What would the result be if \(m\) wasn't restricted to a two-digit number?

**Tip:** When solving problems with ratios, expressing all quantities in terms of a common multiplier helps simplify the problem.

Explore how to solve a problem involving marbles in different ratios. Understand how to adjust the number of marbles and calculate the ratios effectively. Suitable for students in Grades 7-9.

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