Given: Average = 1 mill

[1] NS, only tells us one of three values so there could be a number of values that will avg to 1 mill given that one value is 1.3
[2] NS, as in [1], only tells us one of three values

[1]+[2] SUFF
Using residuals...average is 1, so residual of [1] is +0.3 mill and residual of [2] is -0.3, resulting in a net residual of 0. Therefore, the third account has to be 1 mill (the net residual of all terms has to remain 0). NO, the value is not less than 500k.

Alternate method is to solve the average equation. However, using residuals sometimes makes for simpler calculations (especially if there are larger values and more numbers to add).

Can you please provide the solution?

Thx!

martie_11 wrote:Given: Average = 1 mill

[1] NS, only tells us one of three values so there could be a number of values that will avg to 1 mill given that one value is 1.3
[2] NS, as in [1], only tells us one of three values

[1]+[2] SUFF
Using residuals...average is 1, so residual of [1] is +0.3 mill and residual of [2] is -0.3, resulting in a net residual of 0. Therefore, the third account has to be 1 mill (the net residual of all terms has to remain 0). NO, the value is not less than 500k.

Alternate method is to solve the average equation. However, using residuals sometimes makes for simpler calculations (especially if there are larger values and more numbers to add).

Can you please provide the solution?

Thx!

Recall that the sum of the 3 accounts is $3 million. If the largest is $1.3 million, then the sum of the other two accounts must be $1.7 million. Since the second-largest account could be $1.29 million, then the smallest account could be $0.41 million, which is less than $0.5 million. Hence, Statement 1 by itself is insufficient.

However, if we know the value of 2 of the variables (as we do if we read statements 1 and 2 together), and the average value of all three variables (as we do from reading the question), then we can determine the value of the third variable. If the largest account is worth $1.3 million and the smallest is worth $0.7 million, then the third variable must be $1.0 million. With the value of all 3 variables, we can quickly determine the value of the smallest variable. Statements 1 and 2 together give us the information needed to answer the question.

[spoiler]C

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