Understanding the Problem: Dividing a Number into Specific Parts

The question asks us to divide the number 3740 into three distinct parts. Let's call these parts Part 1, Part 2, and Part 3. The condition given is that half of the first part, one-third of the second part, and one-sixth of the third part are all equal to each other.

We need to find the specific values of these three parts that satisfy both the sum condition (they add up to 3740) and the ratio condition (half of the first = one-third of the second = one-sixth of the third).

Setting Up the Equations

Let the three parts be represented by variables, say $A$, $B$, and $C$.

The first condition is that the sum of the three parts is 3740:

$\hspace{1cm} A + B + C = 3740$

The second condition relates the parts through a ratio:

$\hspace{1cm} \frac{A}{2} = \frac{B}{3} = \frac{C}{6}$

Solving for the Parts using Ratios

Since all three fractions are equal, we can set them equal to a common constant. Let this constant be $k$.

$\hspace{1cm} \frac{A}{2} = k \implies A = 2k$

$\hspace{1cm} \frac{B}{3} = k \implies B = 3k$

$\hspace{1cm} \frac{C}{6} = k \implies C = 6k$

Now we can substitute these expressions for $A$, $B$, and $C$ into the sum equation:

$\hspace{1cm} (2k) + (3k) + (6k) = 3740$

Combine the terms with $k$:

$\hspace{1cm} (2 + 3 + 6)k = 3740$

$\hspace{1cm} 11k = 3740$

Now, solve for $k$ by dividing 3740 by 11:

$\hspace{1cm} k = \frac{3740}{11}$

Let's perform the division:

So, the value of the constant $k$ is 340.

Calculating the Values of the Three Parts

Now that we have the value of $k$, we can find the values of $A$, $B$, and $C$:

Part 1 ($A$) = $2k = 2 \times 340 = 680$

Part 2 ($B$) = $3k = 3 \times 340 = 1020$

Part 3 ($C$) = $6k = 6 \times 340 = 2040$

The three parts are 680, 1020, and 2040.

Verification of the Solution

Let's check if these parts satisfy the given conditions:

1. Sum Condition: Do the parts add up to 3740?

$\hspace{1cm} 680 + 1020 + 2040 = 1700 + 2040 = 3740$

The sum is correct.

2. Ratio Condition: Is half of the first part equal to one-third of the second, and equal to one-sixth of the third?

• Half of the first part: $\frac{680}{2} = 340$
• One-third of the second part: $\frac{1020}{3} = 340$
• One-sixth of the third part: $\frac{2040}{6} = 340$

All three values are equal to 340. The ratio condition is also satisfied.

Conclusion

The three parts are 680, 1020, and 2040. This matches option C.

Revision Table: Key Concepts for Dividing Numbers

Additional Information: Understanding Proportional Parts

When a quantity is divided into parts based on a ratio, say $x:y:z$, the parts are $\frac{x}{x+y+z} \times \text{Total}$, $\frac{y}{x+y+z} \times \text{Total}$, and $\frac{z}{x+y+z} \times \text{Total}$.

In our problem, the condition $\frac{A}{2} = \frac{B}{3} = \frac{C}{6}$ directly implies that the parts are in the ratio $2:3:6$. This is because if $\frac{A}{2} = \frac{B}{3} = \frac{C}{6} = k$, then $A=2k, B=3k, C=6k$. The ratio $A:B:C$ is $2k:3k:6k$, which simplifies to $2:3:6$.

So, dividing 3740 in the ratio $2:3:6$ means the total number of "ratio units" is $2+3+6 = 11$.

Each ratio unit corresponds to $\frac{3740}{11} = 340$.

• Part 1: 2 units $\times 340 = 680$
• Part 2: 3 units $\times 340 = 1020$
• Part 3: 6 units $\times 340 = 2040$

This confirms the result obtained using the constant $k$. Both methods are essentially solving the same ratio problem.