Kaplan Problem

aimscore wrote:List L: 3, 7, 24, 26, x

If x < 5, which of the following describes all the values of x such that the average (arithmetic mean) of list L is less than the median of list L ?
a. x < −45
b. x < −25
c. −45 < x < −25
d. −25 < x < −10
e. x < −5

Ans: b

I chose the correct answer but later on changed it. The final equn boils down to [(60+x)/5 ] < 7. Hence from the equality I get that x < -25.
But if I substitute some value of x < - 45, say -55, I will still get [(60-55)/5] = 1 and 1 < 7.
Can some one pls explain why x <- 45 is wrong? Am I making any silly mistake here?

Thanks

When x < -45, x is also < -25.
If x =-30 Av = 6 < 7 (x !<-45)
Don't you think you are missing this one when x <-45 ?

aimscore wrote:List L: 3, 7, 24, 26, x

If x < 5, which of the following describes all the values of x such that the average (arithmetic mean) of list L is less than the median of list L ?
a. x < −45
b. x < −25
c. −45 < x < −25
d. −25 < x < −10
e. x < −5

Ans: b

I chose the correct answer but later on changed it. The final equn boils down to [(60+x)/5 ] < 7. Hence from the equality I get that x < -25.
But if I substitute some value of x < - 45, say -55, I will still get [(60-55)/5] = 1 and 1 < 7.
Can some one pls explain why x <- 45 is wrong? Am I making any silly mistake here?

Thanks

Hi aimscore,

One of the most important steps in Kaplan's methods for solving Problem Solving questions is to make sure you state the task. In this case, here is what the questions wants:
"...which of the following describes all the values of x..."
In other words, this question asks us for the definition that includes 100% of the numbers that cause the mean to be less than the median.
Now, you were absolutely correct in your testing--if you pick a number less that -45, which is also less than -25, you get a valid solution. However, you can pick a number greater than -45 and also get a valid solution! Try it out with -30, for instance.
So all values less than -45 satisfy the conditions of this problem. But not all numbers that satisfy those conditions are less than -45. Thus, choice (A) is the right answer to the wrong question, and choice (B) is the correct answer.

Hope this helps!

KapTeacherEli wrote:

aimscore wrote:List L: 3, 7, 24, 26, x

If x < 5, which of the following describes all the values of x such that the average (arithmetic mean) of list L is less than the median of list L ?
a. x < −45
b. x < −25
c. −45 < x < −25
d. −25 < x < −10
e. x < −5

Ans: b

I chose the correct answer but later on changed it. The final equn boils down to [(60+x)/5 ] < 7. Hence from the equality I get that x < -25.
But if I substitute some value of x < - 45, say -55, I will still get [(60-55)/5] = 1 and 1 < 7.
Can some one pls explain why x <- 45 is wrong? Am I making any silly mistake here?

Thanks

Hi aimscore,

One of the most important steps in Kaplan's methods for solving Problem Solving questions is to make sure you state the task. In this case, here is what the questions wants:
"...which of the following describes all the values of x..."
In other words, this question asks us for the definition that includes 100% of the numbers that cause the mean to be less than the median.
Now, you were absolutely correct in your testing--if you pick a number less that -45, which is also less than -25, you get a valid solution. However, you can pick a number greater than -45 and also get a valid solution! Try it out with -30, for instance.
So all values less than -45 satisfy the conditions of this problem. But not all numbers that satisfy those conditions are less than -45. Thus, choice (A) is the right answer to the wrong question, and choice (B) is the correct answer.

Hope this helps!

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