IAS 2022 — CSAT Maths

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.