What is the average (arithmetic mean) of a list of 6 consecutive two-d

Question: 6 consecutive integers -> \(x, x+1, x+2, x+3, x +4, x+5 = 6x+15\). \(\frac{6x+15}{6}=?\)
(1) 4th = 5q+3: could be 13 (is the first one n the list, as we are told there are only 2 digit integers in the list), 18, 23 etc. We would get different amounts while calculating the average. Not sufficient
(2) Largest (see above) \(= x+5\), smallest \(= x\)-->\(\frac{x+5}{x}=\frac{5}{4}\)
x=20 and the list looks like this 20, 21, 22, 23, 24, 25. Sufficient

Answer B

Hello,

I was pondering about this problem as well. Here is my bit:
I understand your reasoning but are wondering whether this is sufficient. To my knowledge, an integer can either be positive OR negative. Would not, in this sense, -20 - -25 and thus the average -22,5 also be a possibility? I see no reason why not (although PR comes to the same conclusion as you do...)...

(2) says that "the ratio of the largest integer to the smallest integer is 5:4." The set cannot be -25, -25, -23, -22, -21, -20 because in this case the ratio of the largest to smallest is -20/-25 = 4/5, not 5/4 as given in the statement.

Hope it's clear.

Crystal clear. Thanks a lot for the fast response!

Great Question.
This is what i did in this one =>
Six consecutive integers.
Let they be ->

N
N+1
N+2
N+3
N+4
N+5

We are told that they are all two digits.

we need the average => Average of evenly spaced set = Median = Average of its first and the last terms => \(\frac{2N+5}{2}\)
So we need N


Statement 1->
If a number leaves a remainder 3 when divided by 5 =>Its units digit can be 3 or 8
This statement tells us that the units digit of 4th term can be 3 or 8
Numerous cases re possible
25
26
27
28 -> Units digts= 8
29

OR
30
31
32
33=> Units digit =3
34

Hence not sufficient


Statement 2=>

\(\frac{N+5}{N}=\frac{5}{4}\)
Hence N=20

Thus => Average = 22.5
Hence Sufficient

Hence Sufficient

Hence B

Let L = the largest integer and S = the smallest integer.
If we know the values of L and S, we can determine the average of the 6 consecutive integers.
Question stem, rephrased:
What are L and S?

Statement 1: The remainder when the fourth integer is divided by 5 is 3
In other words, the fourth integer is 3 more than a multiple of 5:
5a + 3, where a is a nonnegative integer.
Two-digit options for the fourth integer such that all 6 integers will have two digits:
18, 23, 28...
Since there are multiple options for the fourth integer, the values of L and S cannot be determined.
INSUFFICIENT.

Statement 2: L:S = 5:4
Since there are 6 consecutive integers, L-S = 5.
The difference for the parts of the given ratio:
5-4 = 1.
Since required difference is 5 times as great, the parts of the given ratio must each be multiplied by a factor of 5:
L:S = (5*5) : (4*5) = 25:20, with the result that L=25, S=20 and L-S = 25-20 = 5.
SUFFICIENT.

Stem: Avg = 6n+15/6 = ?

1) n+3 / 5 = Q+R3
Take n= 20, n+3 = 23, 23/5 = 4+R3
Take n= 30, n+3 = 33, 33/5 = 6+R3
Insufficient.

2) n+5/n = 5/4
4n+20 = 5n
n = 20
Sufficient.

What is the average (arithmetic mean) of a list of 6 consecutive two-digit integers?

(1) The remainder when the fourth integer is divided by 5 is 3.
(2) The ratio of the largest integer to the smallest integer is 5:4

Option 1 says, X4 = 5a+3 => 13,16...

The consecutive nos. are -

-------------4th---------------
10 11 12 13 14 15
13 14 15 16 17 18 and so on... for each case the average should be different. INSUFFICIENT

Option 2 says, Largest : Smallest = 5 : 4 => 5x : 4x

This deduces to -

Smallest--------------------------------------------Largest
(1)-----(2)------(3)--------(4)-------(5)----------(6)

12-------------------------------------------------15
16-------------------------------------------------20
20-----21------22-------- 23--------24----------25 ........ this suffices. Now we can find mean. SUFFICIENT

Correct Answer B

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