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. How many houses have only TV and washing machine but not AC?

• A. 65
• B. 119
• C. 184
• D. 185
• E. 213

Answer 1

Answer: (E)

Question goal: Find the number of houses that have only TV and washing machine but not AC.

Given data (total houses = 1500):

• Total with TV = 862
• Total with AC = 783
• Total with washing machine = 736
• Only TV = 95
• Only AC = 136
• Only washing machine = 88
• All three (TV & AC & W) = 398

Notation (for pairwise-but-not-all-three counts):

• a = # (TV & AC only) = (TV & AC) − (all three)
• b = # (AC & W only) = (AC & W) − (all three)
• x = # (TV & W only) = (TV & W) − (all three) — this is what we want

Write equations from the totals

1. From TV total:
   862 = Only_TV + a + x + All3
   ⇒ a + x = 862 − 95 − 398 = 369.
2. From AC total:
   783 = Only_AC + a + b + All3
   ⇒ a + b = 783 − 136 − 398 = 249.
3. From Washing total:
   736 = Only_W + x + b + All3
   ⇒ x + b = 736 − 88 − 398 = 250.

Solve the system

a + x = 369    (1)
a + b = 249    (2)
x + b = 250    (3)
  

Subtract (2) from (1): (a+x) − (a+b) = 369 − 249 ⇒ x − b = 120.

Add that to (3): (x − b) + (x + b) = 120 + 250 ⇒ 2x = 370 ⇒ x = 185.

Then b = 250 − x = 250 − 185 = 65 and a = 369 − x = 369 − 185 = 184.

Answer: The number of houses having only TV and washing machine (but not AC) is 185.

Note: The value 213 shown earlier is not supported by the correct algebra above; the consistent solution of the three simultaneous equations gives 185.