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Difficulty: HardPercentiles and Quartiles

In a corporate firm of NN employees, each employee earns a distinct annual salary. An employee earning a salary of $78,000\$78,000 is at the 60th60\text{th} percentile of all salaries in the firm. The firm subsequently hires 1010 new employees, each of whom earns an annual salary strictly less than $78,000\$78,000. If $78,000\$78,000 is now at the 70th70\text{th} percentile of all salaries in the expanded firm, how many employees were originally in the firm?

  1. 30Answer
  2. B
    40
  3. C
    50
  4. D
    70
  5. E
    100

Answer

30
The percentile rank of a score indicates the percentage of values in the set that are less than or equal to that score. Originally, 60%60\% of NN employees earned ≤$78,000\le \$78,000, giving 0.60N0.60N employees. Adding 1010 employees who all earn under $78,000\$78,000 brings the count of employees earning ≤$78,000\le \$78,000 to 0.60N+100.60N + 10, while the total workforce becomes N+10N + 10. Since $78,000\$78,000 is at the 70th percentile of the new distribution, 0.60N+10=0.70(N+10)0.60N + 10 = 0.70(N + 10). Expanding and solving yields 0.60N+10=0.70N+70.60N + 10 = 0.70N + 7, so 0.10N=30.10N = 3, which gives N=30N = 30.

Step-by-Step Solution

1
Define the initial number of employees earning at or below $78,000.
Initially, 0.60N0.60N employees earn ≤$78,000\le \$78,000.
By definition of percentile rank with distinct values, being at the 60th percentile of NN scores means 60%60\% of the total NN salaries are less than or equal to $78,000\$78,000.
2
Determine the new count of employees and the updated count of employees earning at or below $78,000.
New total employees =N+10= N + 10; new count earning ≤$78,000\le \$78,000 is 0.60N+100.60N + 10.
All 10 newly hired employees earn salaries strictly less than $78,000\$78,000, increasing the count of salaries ≤$78,000\le \$78,000 by 10.
3
Set up an equation using the new percentile rank.
0.60N+10=0.70(N+10)0.60N + 10 = 0.70(N + 10)
The salary $78,000\$78,000 is now at the 70th percentile of the updated total group size of (N+10)(N + 10).
4
Solve the algebraic equation for NN.
0.60N+10=0.70N+7  ⟹  3=0.10N  ⟹  N=300.60N + 10 = 0.70N + 7 \implies 3 = 0.10N \implies N = 30.
Subtracting 0.60N0.60N and 77 from both sides gives 0.10N=30.10N = 3, which yields N=30N = 30.

Key Concept

Percentile Rank in Expanding Data Sets
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Topics

Percentiles and Quartiles

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