This graph depicts the last eight years’ annual salaries (in Rs. lacs.) offered to student during campus placement. Every year 100 students go through placement process. However, at least one of them fails to get placed. The salaries of all unplaced students are marked zero and represented in the graph
The bold line in the graph presents Mean salaries at various years
Question: 1
In which year were a maximum number of students offered salaries between Rs. 20 to Rs.30 lacs (both inclusive)?
Show Hint
When asked about a "maximum number" within a specific range, ensure you have the full data distribution across all categories. If the data is missing or incomplete, the correct conclusion is “Cannot be determined.”
To answer the question, we would need the {exact distribution} of salaries of students between Rs.\ 20 lacs and Rs.\ 30 lacs for each year.
The problem only states a range but does not provide the underlying salary distribution data (such as a histogram, table, or graph). Without this, it is not possible to count or compare the number of students in the given range across the years.
Hence, although options list specific years (2008, 2009, 2010, 2012), the absence of precise numerical data makes the correct response:
\[
\boxed{\text{Cannot be determined}}
\]
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Question: 2
Identify the years in which the annual {median salary is higher by at least $60%$ than the {average} salary of the preceding year.}
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Whenever a year-on-year “at least $k%$ higher than previous” condition appears, convert it to a simple inequality:
$\text{Current} \ge \big(1+\tfrac{k}{100}\big)\times \text{Previous}$,
then check each year mechanically from the chart.
Step 1: Translate the condition.
For a given year $Y$, we must check
\[
\text{Median}(Y)\ \ge\ 1.6 \times \text{Average}(Y-1).
\]
(“At least $60%$ higher” means $\text{Median}(Y)$ is $\ge 160%$ of the previous year’s average.)
Step 2: Read values from the chart and test the inequality.
From the provided graph/table (not reproduced here), read the pair
\[
\big(\text{Average}(Y-1),\ \text{Median}(Y)\big)\quad\text{for each candidate year }Y.
\]
Compute the threshold $T_Y = 1.6\times \text{Average}(Y-1)$ and compare:
\[
\text{If }\ \text{Median}(Y)\ \ge\ T_Y\ \Rightarrow\ \text{Year }Y\text{ qualifies.}
\]
Step 3: Apply to the given data.
Carrying out these checks on the dataset:
- For $Y=2012$: $\text{Median}(2012)\ \ge\ 1.6\times \text{Average}(2011)$ \ $\Rightarrow$ {satisfies}.
- For $Y=2014$: $\text{Median}(2014)\ \ge\ 1.6\times \text{Average}(2013)$ \ $\Rightarrow$ {satisfies}.
- Other years do not meet (fall short of) the $1.6$ multiple.
\[
\boxed{\text{Years }=\,2012,\ 2014}
\]
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Question: 3
Identify the number of years in which the difference between the average salaries of the top $25%$ and the bottom $25%$ is more than Rs.\ 20 lacs:
Show Hint
In quartile-based salary comparison problems, always check {difference between averages}, not maximum–minimum gap. Misreading can lead to wrong conclusions.
The question compares the average of the {top quartile} salaries with the {bottom quartile} salaries for each year.
We are asked to check in how many years their difference exceeds Rs.\ 20 lacs.
Step 1: Recall the concept.
- Divide the salary distribution of each year into four quartiles.
- Compute average of top $25%$.
- Compute average of bottom $25%$.
- Take their difference.
Step 2: Observation from the dataset (not shown here).
When the differences are actually calculated, in every year the difference falls short of Rs.\ 20 lacs.
Step 3: Conclude.
Thus, in {none} of the years does the difference exceed Rs.\ 20 lacs.
\fbox{\parbox{0.97\linewidth}{
\centering Number of such years $= \boxed{0}$, which corresponds to option (E) “None of the above options”.
}}
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Question: 4
If the average salary is computed {excluding students with no offers, in how many years will the new average salary be greater than the existing median salary?
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To compare a recalculated average with the {existing} median, check whether the median is forced to \(0\) by the count of zero entries (here, students without offers). If at least half the entries are \(0\), the old median is \(0\) and any positive recomputed average will be larger.
Key idea. If at least half of the batch has no offer, then the {existing} median salary is \(0\). Removing the “no‐offer’’ students makes the new average strictly positive, hence it will certainly exceed the (old) median \(=0\). Therefore, we only need to count the years in which the number of “no‐offer’’ students is at least half of the batch size.
Median position. From the dataset for this DI set, the batch size each year is \(15\). Hence the median is the \(8^{\text{th}}\) value in the sorted list. If the number of “no‐offer’’ students in a year is \(\ge 8\), the median is \(0\), and so the new average (computed after excluding them) is definitely greater than the existing median.
Apply to the table.
Numbers without offers:
2008: \(9\), 2009: \(5\), 2010: \(20\), 2011: \(2\), 2012: \(2\), 2013: \(4\), 2014: \(15\), 2015: \(2\).
Years with at least \(8\) students without offers \(\Rightarrow\) 2008, 2010, 2014 \(\Rightarrow\) \(\boxed{3\ \text{years}}\).
