Directions: These questions are based on the following information: Given that $a : b = 5 : 3$ and $b : c = 2 : 5$. If $c = 50$, the value of $a + b + c$ will be
Aptitude
Ratio and Proportion
Difficulty: Medium
Choose an option
-
Amore than 50 but less than 100
-
Bmore than 100 but less than 103
-
Cmore than 103 but less than 105
-
Dmore than 105
Answer
Correct Answer: more than 103 but less than 105
Explanation
### Concept & Proportional Distribution
Once a unified ratio is established, finding the absolute value of one component allows you to determine the scale factor (constant $k$). This scale factor can then be applied to the sum of the ratio parts to find the total sum efficiently.
### Step-by-Step Solution
1. **Given:** From previous calculations, the unified ratio is $a : b : c = 10 : 6 : 15$.
2. Let the actual values be $a = 10k$, $b = 6k$, and $c = 15k$.
3. We are given that $c = 50$. Set up the equation:
$$15k = 50$$
$$k = \frac{50}{15} = \frac{10}{3}$$
4. The requested expression is the sum $a + b + c$:
$$a + b + c = 10k + 6k + 15k = 31k$$
5. Substitute the exact value of $k$ into the sum expression:
$$31 \cdot \left(\frac{10}{3}\right) = \frac{310}{3}$$
6. Convert the fraction to a decimal to check the ranges:
$$\frac{310}{3} = 103.333...$$
7. Analyzing the options, $103.33$ is strictly greater than $103$ and less than $105$.
### Exam Strategy & Shortcut
Do not waste time calculating the individual absolute values of $a$ and $b$ before adding. Sum the ratio proportions ($10 + 6 + 15 = 31$) and multiply by the multiplier ($\frac{50}{15}$) in one single step: $31 \cdot \frac{50}{15} = 103.33$.
### Common Pitfall
Converting the scale factor $k = \frac{10}{3}$ into a rounded decimal like $3.33$ too early. Doing so yields $31 \cdot 3.33 = 103.23$, which could cause hesitation when distinguishing between narrow bounds, especially if rounding errors push the value into an incorrect bucket. Always keep fractions until the final step.
### Final Answer
Therefore, the correct answer is **more than 103 but less than 105**.