Question:

An electronic company conducts a survey of 1500 houses for their products. The survey suggested that 862 houses own TV, 783 houses has AC and 736 houses has washing machine. There were 95 houses having only TV, 136 houses having only AC and 88 houses having only washing machine. There were 398 houses having all the three equipments. What is the percentage of houses having all the three equipments to the houses having only two equipments? (Approx.)

Updated On: Dec 16, 2025
  • 65%
  • 76%
  • 79%
  • 84%
  • 92%
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is

Solution and Explanation

To solve this problem, we need to determine the percentage of houses having all three devices (TV, AC, Washing Machine) relative to those having only two devices. We'll break down the problem in steps:

  1. Identify the total number of houses with only TV, only AC, and only washing machine:
    • Houses with only TV = 95
    • Houses with only AC = 136
    • Houses with only Washing Machine = 88
  2. Total houses owning all three devices = 398.
  3. We need to calculate the total number of houses having exactly two devices (TV & AC, AC & Washing Machine, TV & Washing Machine).
  4. Use the principle of inclusion-exclusion for sets:
    • Number of houses owning at least one device = Total houses - Houses owning all three = 1500 - 398 = 1102.
  5. Using the inclusion-exclusion principle:
    • Let T = houses having a TV, A = houses having an AC, W = houses having a washing machine.
    • According to the inclusion-exclusion principle for three sets:
      • \( |T \cup A \cup W| = |T| + |A| + |W| - |T \cap A| - |A \cap W| - |T \cap W| + |T \cap A \cap W| \)
      • Substituting values: \[ 1102 = 862 + 783 + 736 - (T \cap A) - (A \cap W) - (T \cap W) + 398 \]
  6. Simplify to find the combined intersections of the sets having exactly two devices:
    • \( 862 + 783 + 736 = 2381 \)
    • \( 2381 - 1102 - 398 = (T \cap A) + (A \cap W) + (T \cap W) \)
    • \( 881 = (T \cap A) + (A \cap W) + (T \cap W) \)
  7. Percentage of houses having all three equipments to those having only two equipments:
    • \( \text{Percentage} = \left( \frac{398}{881} \right) \times 100\% \approx 45.17\% \)

Given options suggest checking calculations or possible setting errors, as none perfectly match. Verify all steps and survey conditions for misinterpretations to resolve disparities.

Was this answer helpful?
0
0