The average weight of A, B C is 40 Kg, the average weight of B, D, E is 42 kg and the weight of F is equal to that of B. what is the average weight…

Correct Option (3)

The problem requires calculating the average weight of six individuals (A, B, C, D, E, F) based on given average weights of two groups and a specific weight equivalence.

• Given, the average weight of A, B, C is 40 Kg.
  This implies: $\frac{A + B + C}{3} = 40$3A+B+C​=40
  Therefore, $A + B + C = 120$A+B+C=120 (Equation 1)
• Given, the average weight of B, D, E is 42 Kg.
  This implies: $\frac{B + D + E}{3} = 42$3B+D+E​=42
  Therefore, $B + D + E = 126$B+D+E=126 (Equation 2)
• From Equation 1, we can express B as: $B = 120 - A - C$B=120−A−C
• Substitute this expression for B into Equation 2:
  $(120 - A - C) + D + E = 126$(120−A−C)+D+E=126
  $D + E = 126 - 120 + A + C$D+E=126−120+A+C
  $D + E = 6 + A + C$D+E=6+A+C (Equation 3)
• Given, the weight of F is equal to that of B, i.e., $F = B$F=B.
• The average weight of A, B, C, D, E, F is calculated as:
  $\frac{A + B + C + D + E + F}{6}$6A+B+C+D+E+F​
• Substitute $F = B$F=B into the expression:
  $\frac{A + B + C + D + E + B}{6}$6A+B+C+D+E+B​
• Now, substitute Equation 1 ($A + B + C = 120$A+B+C=120) and Equation 3 ($D + E = 6 + A + C$D+E=6+A+C) into the sum:
  The sum can be written as: $(A + B + C) + (D + E) + B$(A+B+C)+(D+E)+B
  Substitute the known values: $120 + (6 + A + C) + B$120+(6+A+C)+B
  Rearrange to group terms: $120 + 6 + (A + C + B)$120+6+(A+C+B)
  Since $A + C + B = 120$A+C+B=120 (from Equation 1), the total sum is:
  $120 + 6 + 120 = 246$120+6+120=246 Kg
• Finally, calculate the average weight:
  $\frac{246}{6} = 41$6246​=41 Kg

Incorrect Options:

Options 1 (40.5 kg), 2 (40.8 kg), and 4 (Cannot be determined as data inadequate) are incorrect. The provided data is sufficient to determine the average weight of all six individuals through systematic algebraic substitution, leading to a precise value of 41 kg. Therefore, stating that the data is inadequate or arriving at any other numerical value is erroneous.