The weight of three heaps of gold are in the ratio 5 : 6 : 7. By what fractions of themselves must the first two be increased so that the ratio of the weights may be changed to 7 : 6 : 5 ?

Options

A. `(24)/(25),(2)/(5)`
B. `(48)/(50),(4)/(5)`
C. `(48)/(50),(3)/(5)`
D. `(24)/(25),(3)/(7)`

Answer

To solve the problem, we need to find out by what fractions the first two heaps of gold must be increased to change the ratio from 5:6:7 to 7:6:5.

### Step 1: Define the weights of the heaps
Let the weights of the three heaps be represented as:
- Weight of the first heap (W1) = 5x
- Weight of the second heap (W2) = 6x
- Weight of the third heap (W3) = 7x

### Step 2: Set up the new ratio
We want to change the ratio to 7:6:5. Let’s denote the increases in the first and second heaps as fractions of their original weights:
- Increase in W1 = \( \frac{a}{5x} \times 5x = a \)
- Increase in W2 = \( \frac{b}{6x} \times 6x = b \)

The new weights will then be:
- New W1 = \( 5x + a \)
- New W2 = \( 6x + b \)
- W3 remains the same = \( 7x \)

### Step 3: Set up the equation based on the new ratio
According to the new ratio, we have:
\[ \frac{5x + a}{6x + b} = \frac{7}{6} \]
Cross-multiplying gives:
\[ 6(5x + a) = 7(6x + b) \]
Expanding both sides:
\[ 30x + 6a = 42x + 7b \]

### Step 4: Rearranging the equation
Rearranging the equation to isolate terms gives:
\[ 6a - 7b = 42x - 30x \]
\[ 6a - 7b = 12x \]

### Step 5: Set up the second equation based on W3
Since W3 remains the same, we also need to ensure that the ratio of W3 to the new weights holds:
\[ \frac{7x}{5x + a} = \frac{5}{6} \]
Cross-multiplying gives:
\[ 6(7x) = 5(5x + a) \]
Expanding both sides:
\[ 42x = 25x + 5a \]

### Step 6: Rearranging the second equation
Rearranging gives:
\[ 5a = 42x - 25x \]
\[ 5a = 17x \]
\[ a = \frac{17x}{5} \]

### Step 7: Substitute back to find b
Substituting \( a \) back into the first equation:
\[ 6\left(\frac{17x}{5}\right) - 7b = 12x \]
\[ \frac{102x}{5} - 7b = 12x \]
Multiplying through by 5 to eliminate the fraction:
\[ 102x - 35b = 60x \]
Rearranging gives:
\[ 35b = 102x - 60x \]
\[ 35b = 42x \]
\[ b = \frac{42x}{35} = \frac{6x}{5} \]

### Step 8: Find the fractions of increase
Now we can find the fractions of increase for W1 and W2:
- Fraction increase for W1 = \( \frac{a}{5x} = \frac{\frac{17x}{5}}{5x} = \frac{17}{25} \)
- Fraction increase for W2 = \( \frac{b}{6x} = \frac{\frac{6x}{5}}{6x} = \frac{1}{5} \)

### Final Answer
The fractions by which the first two heaps must be increased are:
- For W1: \( \frac{17}{25} \)
- For W2: \( \frac{1}{5} \)