If the average (arithmetic mean) of seven consecutive integers is k +

Question

If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9 
B. k^2 - 2k + 1 
C. k^2 + 4k - 5 
D. k^2 + 6k + 9 
E. k^2 + 4k - 12

HUNTdGMAT · Accepted Answer

Since there are 7 consecutive integers, K+2 is the fourth number in the series. So, the series should be k-1, k, k+1, k+2, k+3, k+4, k+5.

then the product of the greatest and least integer =(k+5)* (k-1)= k^2+4k-5

Ans is C

Arsh4MBA · Answer

Let the first integer be x;
Sum of the 7 integers x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)= 7x+21

Average = Sum/n= 7(x+3)/7=x+3;

x+3= k+2 ==>x= k-1;
Greatest integer = k+5;

Product x*x+6 =k^2+4k-5

Ans:C

Nikkb · Answer

Let first integer of series = a
Difference between integers = d= 1 (consecutive integers given)
Here no. of terms in series = n=7
Average = K+2

Average = Arithmetic mean = \frac{sum of all 7 numbers}{7}

by formula Sum of series = \frac{n(2a+(n-1)d)}{2} = \frac{7(2a+(7-1)*1)}{2} = \frac{7(2a+6)}{2} = 7(a+3)

As Average = k+2 =  \frac{sum of all 7 numbers}{7}  =  \frac{7(a+3)}{7} 
=> k+2 =a+3
=> a= k-1

Now as nth term of series = a+(n-1)d => last term of given series = a+(7-1)*1 = (k-1)+6 = k+5

Product of first and last term = (k-1)*(k+5) = k^2+5k-k-5 = k^2+4k-5

Answer: C

pushpitkc · Answer

Given data:   A set with 7 consecutive numbers has mean k+2

Lets assume the 7 numbers to be  1,2,3,4,5,6,7 
The mean of consecutive integers in a set with odd number of elements is the middle number.
In this set, the 4th element which is 4 is the mean.
k+2 = 4 => k = 2
The product of the least and the greatest number is 1*7(7)

Evaluating the answer options(s.t k = 2)
A. k^2 - 9 = 4 -9 = -5
B. k^2 - 2k + 1 = 4 - 4 + 1 = 1
C. k^2 + 4k - 5 = 4 + 8 - 5 = 7

Since Option C gives us the same value for the product of the smallest and greatest number, 
we need not evaluate the other 2 answer options.  (Option C)  is our answer!

gps5441 · Answer

Total 7 consecutive nos and K+2 is the average which means 3 nos each are on the right and left of this no.
K-1, k,k+1, k+2 ,k+3, k+4, k+5
Smallest no=k-1
largest no=k+5
Product=(k-1)(k+5)=k²+4k-5=C is the answer

generis · Answer

I used algebra, erred, and took a long time to correct. So I timed myself using what I thought was a time-consuming method. Wrong. I was well under a minute.

I used only one sketch, and no fancy lines or asterisks. I used them here for clarity.

1. List the integers

x|x+1|x+2|x+3|x+4|x+5|x+6|

2. Place (k+2) under the median

"The arithmetic mean is k + 2." Median = mean for evenly spaced set. So

x|x+1|x+2|x+3|x+4|x+5|x+6|
*|***|***|k+2|***|***|***|

3. Find k. From above: 
k + 2 = x + 3
k = x + 1, so

x|x+1|x+2|x+3|x+4|x+5|x+6|
*|_k_|***|k+2|***|***|***|

4. Find answer. The product of the greatest and least integer is? We need x, and (x+6), in terms of k, so

_x_|x+1|x+2|x+3|x+4|x+5|x+6|
k-1|_k_ |***|k+2|***|***|k+5|

(k - 1)(k + 5) = k ^2  + 4k - 5

ANSWER C

CyberStein · Answer

Is this a 500 lvl question? really doesn't feel like it

anyway

Sum of the numbers = 7k + 14
let x = smallest number
21 + x = 7k + 14
x= k-1

product of smallest and biggest = x(x+6) = x^2 + 6x = k^2 - 2k + 1 + 6k - 6 = k^2 +4k - 5

Thus, answer C

Not a hard problem, but I don't think it is level 500.

chesstitans · Answer

the source is from Magoosh

Bunuel · Answer

Added the source. Thank you.

aragonn · Answer

Official Explanation:

For any collection of evenly spaced numbers, the mean and median are always equal. Because this is set with seven members, an odd number of elements, the median is the middle number, the 4th number on the list: there are three numbers on the list below it and three above it. Thus, (k + 2) is the mean as well as the median, the middle or 4th number on the list. Since there are only seven numbers, we can simply write them out:

{k – 1, k, k + 1, k + 2, k + 3, k + 4, k + 5}

The least, three lower than the median, is (k – 1).

The greatest, three higher than the median, is (k + 5).

Their product is:

product = (k – 1)(k + 5) = k2 + 4k – 5

Answer = (E)

aragonn · Answer

in OE they have assumed 7 numbers as {k – 1, k, k + 1, k + 2, k + 3, k + 4, k + 5} However I think in such cases right assumption should be {k – 3, k - 2 , k - 1, k, k + 1, k + 2, k + 3} this will simplify the sum to 7K. In complex situation this will be quiet helpful.

ScottTargetTestPrep · Answer

We can let the smallest integer in the set = x and the largest = x + 6; thus:

(x + x + 6)/2 = k + 2

2x + 6 = 2k + 4

2x = 2k - 2

x = k - 1

Thus, the smallest integer is k - 1 and the greatest is k + 5, so the product of those two values is:

(k - 1)(k + 5) = k^2 + 4k - 5

Alternate solution:

Since k + 2 is in the average of the 7 consecutive integers, it’s also the median, i.e., the 4th integer. Thus, the largest integer is 3 more than k + 2, i.e., k + 5, and the least integer is 3 less than k + 2, i.e., k - 1. The product of the greatest and least integer is:

(k + 5)(k - 1) = k^2 + 4k - 5

Answer: C

menonrit · Answer

could you explain how the greatest integer = k+5 and the next step?

I am confused

BrushMyQuant · Answer

Theory: In case of evenly spaced sets the mean of the set is the middle number (in case of odd number of numbers)

The average (arithmetic mean) of seven consecutive integers is k + 2 
=> Middle number (or the  4^{th}  number) = k+2

So, the consecutive numbers less than k+2 are
k-1, k, k+1

The consecutive numbers greater than k+2 are
k+3, k+4, k+5

The product of the greatest and least integer 
= (k-1) *(k+5)
=  k^2  - k + 5k - 5
=  k^2  + 4k - 5

So,  Answer will be C 
Hope it helps!

To learn more about Statistics watch the following video

https://www.youtube.com/watch?v=Cvy_HHw4KIs

zahidalisaeed54 · Answer

Initially, ignore K, and find out the arithmetic mean keeping in mind that the integers are consecutive:

(-1) + (0) + (1) + (2) + (3) + (4) + (5)/7= 2

Now, just add K with the greatest and the least number to find out the their product: (K-1) (K+5) = K +4K-5

Therefore, C is the correct answer.

Kavicogsci · Answer

Given that count of numbers is odd we can use this as proxy for 7 numbers

n-3, n-2, n-1, n, n+1, n+2 and n+3

Now
(n-3) + (n-2) + (n-1)+( n)+( n+1)+( n+2)+( n+3)/7 = k+2

n = k+2

Now Multiplying least and greatest integer

(n-3)*(n+3)
n^2 - 9

(k+2)^2 - 9

Hence E

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If the average (arithmetic mean) of seven consecutive integers is k +

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If the average (arithmetic mean) of seven consecutive integers is k +

Prep Toolkit

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