This problem involves calculating the weight of an individual, B, given the average weights of different groups. We need to use the definition of average to find the total weights and then isolate B's weight.
The average of a set of numbers is calculated by dividing the sum of the numbers by the count of the numbers. The formula is:
Average $= \frac{\text{Sum of values}}{\text{Number of values}}$
This can be rearranged to find the sum of values:
Sum of values $= \text{Average} \times \text{Number of values}$
Let $W_A$, $W_B$, and $W_C$ represent the weights of individuals A, B, and C, respectively.
We are given that the average weight of A, B, and C is 45 kg.
Average weight of (A, B, C) $= \frac{W_A + W_B + W_C}{3} = 45$ kg
Using the rearranged formula, the total weight of A, B, and C is:
$W_A + W_B + W_C = 45 \times 3 = 135$ kg
We are given that the average weight of A and B is 37 kg.
Average weight of (A, B) $= \frac{W_A + W_B}{2} = 37$ kg
The total weight of A and B is:
$W_A + W_B = 37 \times 2 = 74$ kg
We are given that the average weight of B and C is 47 kg.
Average weight of (B, C) $= \frac{W_B + W_C}{2} = 47$ kg
The total weight of B and C is:
$W_B + W_C = 47 \times 2 = 94$ kg
We have the following sums:
We can find the weight of B using these equations. One method is to add Equation 2 and Equation 3:
$(W_A + W_B) + (W_B + W_C) = 74 + 94$
$W_A + 2W_B + W_C = 168$
Now, subtract Equation 1 from this new equation:
$(W_A + 2W_B + W_C) - (W_A + W_B + W_C) = 168 - 135$
This simplifies to:
$W_B = 33$ kg
Alternatively, we can find the weight of C first. Substitute Equation 2 into Equation 1:
$74 + W_C = 135$
$W_C = 135 - 74 = 61$ kg
Now substitute the value of $W_C$ into Equation 3:
$W_B + 61 = 94$
$W_B = 94 - 61 = 33$ kg
Both methods confirm that the weight of B is 33 kg.
Average of 40 numbers is 71, if the number 100 replaced by 140, then average is increased by
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