More Questions from Ratio and Proportion

Directions: These questions are based on the following information: Given that $a : b = 5 : 3$ and $b : c = 2 : 5$. $(c - a)$ will be equal to I. $10(a + c)$ II. $10a + 25b$

Aptitude Ratio and Proportion Difficulty: Medium
Choose an option
  • A
    Only I is true
  • B
    Only II is true
  • C
    Both I and II are true
  • D
    Both I and II are false

Answer

Correct Answer: Both I and II are false

Explanation

### Concept & Algebraic Evaluation of Ratios To determine if statements regarding variables in a ratio are true, express all variables in terms of a single proportionality constant ($k$) and substitute them into the given algebraic expressions to test for equivalence. ### Step-by-Step Solution 1. **Given:** The combined continuous ratio is $a : b : c = 10 : 6 : 15$. 2. Let the values be represented as $a = 10k$, $b = 6k$, and $c = 15k$. 3. First, evaluate the target expression $(c - a)$: $$(c - a) = 15k - 10k = 5k$$ 4. Now, evaluate Statement I: $10(a + c)$ $$10(10k + 15k) = 10(25k) = 250k$$ Since $5k \neq 250k$, Statement I is not equal to $(c - a)$. It is false. 5. Evaluate Statement II: $10a + 25b$ $$10(10k) + 25(6k) = 100k + 150k = 250k$$ Since $5k \neq 250k$, Statement II is also not equal to $(c - a)$. It is false. 6. Because neither algebraic statement evaluates to $5k$, both are false. ### Exam Strategy & Shortcut Test logic using the simplest integers. Let $a=10, b=6, c=15$. The target $(c-a)$ equals $5$. Statement I is $10(10+15) = 250$. Statement II is $10(10) + 25(6) = 100 + 150 = 250$. Since $5 \neq 250$, both are demonstrably false. ### Common Pitfall A student might notice that Statement I ($250k$) and Statement II ($250k$) are mathematically equal to each other and mistakenly select "Both I and II are true", forgetting that the question specifically asks if they are equal to the initial expression $(c - a)$. ### Final Answer Therefore, the correct answer is **Both I and II are false**.
Discussion & Comments
No comments yet. Be the first to comment!
Join Discussion