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Difficulty: Very hardWeighted Average and Combined Sets

A corporation has three regional branches: Branch X, Branch Y, and Branch Z. The ratio of the number of employees in Branch X to Branch Y is 1:31 : 3. The average monthly salary of employees in Branch Y is 40%40\% higher than the average monthly salary of employees in Branch X. Branch Z has twice as many employees as Branch Y, and the average monthly salary of employees in Branch Z is x%x\% lower than the combined average monthly salary of employees in Branches X and Y. If the overall average monthly salary across all three branches combined is $910\$910, and the average monthly salary in Branch X is $1000\$1{}000, what is the value of xx?

  1. A
    35
  2. B
    40
  3. 50Answer
  4. D
    70
  5. E
    100

Answer

50
The combined average salary of Branches X and Y is calculated by weighting Branch X's salary ($1000\$1{}000) with 1 unit of weight and Branch Y's salary ($1400\$1{}400) with 3 units of weight, giving 1000+42004=$1300\frac{1{}000 + 4{}200}{4} = \$1{}300. For the entire company of 1010 units of employees with an average salary of $910\$910, the total payroll is 10×910=$910010 \times 910 = \$9{}100. Subtracting the combined payroll of Branches X and Y ($5200\$5{}200) leaves $3900\$3{}900 for Branch Z's 66 units of employees. Thus, Branch Z's average salary is 39006=$650\frac{3{}900}{6} = \$650. Comparing $650\$650 to $1300\$1{}300 gives a percentage decrease of 1300−6501300×100%=50%\frac{1{}300 - 650}{1{}300} \times 100\% = 50\%.

Step-by-Step Solution

1
Determine the average monthly salary of Branch Y.
Branch Y average salary = 1000×1.40=$14001{}000 \times 1.40 = \$1{}400.
Branch Y's average salary is given as 40%40\% higher than Branch X's average salary of $1000\$1{}000.
2
Express employee counts in terms of a single variable nn.
Branch X has nn employees, Branch Y has 3n3n employees, and Branch Z has 2×3n=6n2 \times 3n = 6n employees. Total employees = n+3n+6n=10nn + 3n + 6n = 10n.
The ratio of Branch X to Branch Y employees is 1:31:3, and Branch Z has twice as many employees as Branch Y.
3
Calculate the combined average salary of Branches X and Y.
Combined average salary SXY=n(1000)+3n(1400)n+3n=5200n4n=$1300S_{XY} = \frac{n(1{}000) + 3n(1{}400)}{n + 3n} = \frac{5{}200n}{4n} = \$1{}300.
The combined average of two sets is total combined earnings divided by total combined size.
4
Calculate total company payroll and determine Branch Z's average salary.
Total payroll = 10n×910=9100n10n \times 910 = 9{}100n. Branch Z payroll = 9100n−5200n=3900n9{}100n - 5{}200n = 3{}900n. Average salary for Branch Z SZ=3900n6n=$650S_Z = \frac{3{}900n}{6n} = \$650.
Total payroll is total employees times overall mean. Subtracting the combined payroll of X and Y yields Branch Z's payroll.
5
Compute the percentage by which Branch Z's average salary is lower than the combined average salary of Branches X and Y.
x = \frac{1{}300 - 650}{1{}300} \times 100\% = 50\%.
Percentage decrease is calculated as (Base Value - New Value) / Base Value.

Key Concept

Weighted Averages of Combined Sets
Estimated Time:2m 0s
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Topics

Weighted Average and Combined Sets

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