Which of the following procedures is always equivalent to adding 5

Question

Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5 ?

I. Multiplying the 5 numbers and then finding the 5th root of the product.
II. Adding the 5 numbers, doubling the sum, and then moving the decimal point one place to the left.
III. Ordering the 5 numbers numerically and then selecting the middle number.

(A) None
 (B) I only
 (C) II only
 (D) III only
 (E) I and III

Skywalker18 · Accepted Answer

Let's choose the 5 numbers to be {1, 2, 3, 4, 5}

adding 5 given numbers and then dividing the sum by 5 , that is we find the average .
Average = (1+2+3+4+5)/5 = 3

1.Multiplying the 5 numbers and then finding the 5th root of the product.
Product = 1*2*3*4*5 = 120
Fifth root of 120 is not 3 . (3^5 = 243)

II. Adding the 5 numbers, doubling the sum, and then moving the decimal point one place to the left.
Sum = 15
2*Sum = 30 
then moving the decimal point one place to the left , that is answer is 3

Doubling the sum and moving the decimal point one place to the left is essentially same as by dividing number of terms(5) because we double the sum and then divide by 10 here (5*2)
Always true

III. Ordering the 5 numbers numerically and then selecting the middle number.
Our middle number is 3 , which is same as average . But this will only be true incase the numbers are equally spaced , i.e
have the same common difference.
If the numbers are 1,2,3,4,10 ; then average = 4 , but our middle term = 3 .

Answer C

suresh8055 · Answer

AS Given ,
Adding 5 number and dividing the sum by 5 which can be stated as = Sum/5 = Sum *0.2
From The Given Statements 
II- Adding the 5 numbers, doubling the sum, and then moving the decimal point one place to the left.
=Sum*0.2

Hence C.

Ngocanhvy · Answer

assume all the given numbers are x, or -x to simplize calculation 
(I): we know that we we add 5 numbers: x,-x.... they gonna cancel out each other, then the sum can be: 5,4,3, etc... then devides 5, we still have a wide range of result
When multiple 5 number then  finding the 5th root of the product, the result is either x, or -x 
 So 2 procedures dont generate equivalent result

(II) The procedure in the question: SUM/5 
 The procedure in (II): (2*SUM)/10 = SUM/5  
 always equivalent

(III) like median, and average problem, they are not always equivalent

bishal128 · Answer

1 case, it is valid for only same number, but not the different numbers 
In the second case,

Let the sum of five numbers be (5n+k) where n shows the initial number and K represents the constant of the sum, hence for the initial argument, the resulting value is (5n+k)/5

in second case getting it into the one decimal lef is after multiplying with 2 is similar to 
2* (5n+k)/10 = (5n+K)/5

In third case, the average value would be the middle value only if the sequence of the number has even gaps

thus only result consistent is II option

Which of the following procedures is always equivalent to adding 5

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Which of the following procedures is always equivalent to adding 5

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