Question

Use the same InfiniteJobs.com table as above.

Year	Category	Number of Registrations	Number of Candidates who posted their CVs	Number of Candidates short-listed by Employers	Number of offered jobs
2004	Technical	61,205	59,981	684	181
2004	Managerial	19,236	15,389	138	48
2005	Technical	63,298	60,133	637	115
2005	Managerial	45,292	40,763	399	84

Statement X: The success rate of candidates getting short-listed based on their CVs is higher for the Managerial category than for the Technical category in 2005.
Statement Y: The success rate of candidates getting short-listed based on their CVs is better for the Managerial category in 2005 than in 2004.

• A. Only [X] is True
• B. Only [Y] is True
• C. Both [X] and [Y] are True
• D. Neither [X] nor [Y] is True

Hint:

Success rate = (candidates short-listed) divided by (candidates who posted CVs); compute it separately for each year/category before comparing.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept.
The "success rate of getting short-listed based on CVs" means the number short-listed divided by the number of CVs posted, expressed as a percentage.

Step 2: Test Statement X, comparing categories in 2005.
Managerial 2005: \(\frac{399}{40{,}763}\times 100 \approx 0.979\%\). Technical 2005: \(\frac{637}{60{,}133}\times 100 \approx 1.059\%\). Since \(0.979\%\) is less than \(1.059\%\), Managerial's success rate is actually lower than Technical's in 2005, not higher, so Statement X is False.

Step 3: Test Statement Y, comparing years within Managerial.
Managerial 2004: \(\frac{138}{15{,}389}\times 100 \approx 0.897\%\). Managerial 2005 (from Step 2): \(\approx 0.979\%\). Since \(0.979\%\) is greater than \(0.897\%\), the Managerial success rate did improve from 2004 to 2005, so Statement Y is True.

Step 4: Final Answer.
X is false and Y is true, so only Y is correct. \[ \boxed{\text{Only [Y] is True}} \]