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Around the World in 80 Questions (Day 10): In a set of k consecutive

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Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:00
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In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)


 


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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:27
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Given: In a set of k consecutive integers, where k is two-digit number, the largest number is 21.
Asked: What is the average (arithmetic mean) of the set in terms of k ?

First term a = 21
common difference = -1
Last term l = 21 + (k-1)(-1) = 21 + 1-k = 22 -k

Arithmetic mean = Sum / k = k/2(a+l)/k = (21 + 22 - k)/2 = (43-k)/2

IMO E
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:27
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Bunuel wrote:
In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)


 


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The average of a set of consecutive integers is the middle number if the count of numbers (k) is odd, or the average of the two middle numbers if k is even.
In this case, the largest number is 21 and there are k consecutive numbers. So, the smallest number is 21 - k + 1.
The average of the set would be (smallest number + largest number) / 2 = (21 + 21 - k + 1) / 2 = (42 - k + 1) / 2 = (43 - k) / 2.
So, the correct answer is E. (43−k)/2.
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:23
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In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?
suppose k is 3 hence (19+20+21)/3 = 20
try to plug in
a = 21*3/2 =31.5
b= (21-3)/2 = 9
c=(41-3)/2 = 19
d = (42-3)/2=19.5
e=(43-3)/2 = 20
the answer is E
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:33
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Let \(k=10\)
Numbers are
\(12,13,14,15....21 \)(\(10 \)numbers)
\(S=\frac{(21*22)}{2}-\frac{(11*12)}{2}=165\)
Avg=\(\frac{165}{10}=16.5\)

Lets check the options:
A) 21*10/2 (wrong)
B) (21-10)/2=11/2 (wrong)
C) (41-10)/2=15.5 (wrong)
D) (42-10)/2=16 (wrong)
E) (43-10)/2=16.5 (right)

Hence E)
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:38
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In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

let k = 10
largest = 21
21, 20, 19, 18, 17, 16, 15, 14, 13, 12
first term = 12, last term = 21
average = \frac{(12+21)}{2} = 33/2

only answer that matches this fraction is E
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:39
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Bunuel wrote:
In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)



Largest term = \(21\)

Total k terms, so first term = \(21 - (k-1)\)

So average = \(\frac{{21-(k-1)+21}}{2}\)

= \(\frac{43 - k}{2}\)

Ans E
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:40
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The Answer is E
We can assume the value of k = 10
Since we know the value of the last term, we can find the sum of the 10 terms by n/2 (2a+(n-1)d which comes out to be 165
we can find the mean= 165/10=16.5
Plug the value of k in the answer options and see if it is equal to the mean. Only Option E =16.5 Hence correct.
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:40
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In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

constraints given :
k consecutive integers.
k is two-digit number.
the largest number is 21.

What's to find :
What is the average (arithmetic mean) of the set in terms of

So if the largest number is 21 and k is a 2 digit number.
So let k = 10.
and k consecutive numbers so
21 20 19 ....... 12
sum of 12 to 21 = 165
and average = 165/10 = 16.5

Now use the options :

43−k/2
k = 10
43-10/2
33/2 = 16.5

Hence IMO E.
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:45
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given that
In a set of k consecutive integers, where k is two-digit number, the largest number is 21

smallest value of k is 10 and largest is 21
total value of k is 12
AM of set will be 15+16 /2 ; 15.5
plug in values of k in options
\(\frac{43 - k}{2}\)

OPTION E is correct

Bunuel wrote:
In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)


 


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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 08:53
In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

_____________________________
If k is two-digit number, it means that there are at least 10 numbers.
21,20,19,18,17,16,15,14,13,12... or more.
SO, C is our winner.
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:01
The largest number is 21 -> The smallest number is 21-k
Sum of k consecutive integers = (21+21-k)*k/2=(42-k)*k/2
Average = Sum/k = (42-k)/2

Answer D
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:01
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Bunuel wrote:
In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)


 


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Assume k = 11

Smallest Value = 21 - 11 + 1 = 11

11 12 13 14 15 16 17 18 19 20 21

In consecutive numbers --> mean = median. In this example mean = median = 16.

A. \(\frac{21k}{2}\) = (21*11)/2 = Not 16 : Eliminate.

B. \(\frac{21 - k}{2}\)= (21-11)/2 = Not 16 : Eliminate.

C. \(\frac{41 - k}{2}\) = (41-11)/2 = Not 16 : Eliminate.

D. \(\frac{42 - k}{2}\) = (42-11)/2 = Not 16 : Eliminate.

E. \(\frac{43 - k}{2}\)= (43-11)/2 = 32/2 = 16 : Our Answer.

IMO E
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:06
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In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)

Answer:
K consecutive integers
Largest number = 21
so 1st number = 21- K +1= 22 - K
Sum of K consecutive numbers. = K/2(21+ 22-K) = K/2*(43-K)
So average of k consecutive numbers = (43 -K )/2

E is the answer
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:16
Bunuel wrote:
In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)


 


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[(21-1)+(21-2)+...+(21-k)]/k
= [21k-(1+2+...+k)]/k
={21k-[(1+k)*(k/2)]}/k
= 21-(1+k)/2
= 42/2 - (1+k)/2
= (42 - 1 - k )/2
= (41-k)/2

C is the answer.
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:18
We have given k consecutive terms that means
d =1 and 21 will be the last term
Sn =n/2(2a+(n-1)d)

Also if last value is known
Sn =n/2(a+last term)

Sk= k/2(a+21)

Also ak=a+(k-1)d

21=a+(k-1)*1

Gives a=22-k

Putting in sum and dividing by k
Gives (43-k)/2

Largest no is 21 hence it will be last

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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:21
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Bunuel wrote:
In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A. \(\frac{21k}{2}\)

B. \(\frac{21 - k}{2}\)

C. \(\frac{41 - k}{2}\)

D. \(\frac{42 - k}{2}\)

E. \(\frac{43 - k}{2}\)


 


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For A.P. we know that an = a1+ (n-1) * d

Here, d=1 (consecutive numbers), an = 21 (Largest number will be the last number) and n=k

So, 21 = a1 + (k-1) *1
=> a1 = 22 - k

Now Arithmetic mean = 1/2 * (a1 + an)
=> A.M. = 1/2 * (22 - k + 21) = \(\frac{43 - k}{2}\)

E is correct choice.
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:21
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Let's the smallest number in the set "x". Since the largest number in the set is 21, we know that x + (k - 1) = 21. Solving for x, we get x = 22 - k. The average of the set is the sum of the numbers in the set divided by the number of numbers in the set. The sum of the numbers in the set is x + (x + 1) + (x + 2) + ... + (x + k - 1), which simplifies to kx + (1 + 2 + ... + k - 1). We know that 1 + 2 + ... + k - 1 = k(k - 1)/2. Substituting x = 22 - k and simplifying, we get that the average of the set is (43 - k)/2.
Answer is E
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:21
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In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?
Largest No 21
If we put K=10, the no will be 12,13,14,15,16,17,18,19,20,21 - AM =(5th Digit +6th Digit)/2=16.5
If we put k-11, AM=16
If we put the values only E Option i.e (43−k)/2 is correct

Option E is correct
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink] New post   28 Jul 2023, 09:28
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For a set of consecutive integers, the average is the sum of the first and last digit divided by 2.

To find: The first integer
For an AP. The nth term = a + (n-1)d
In the above substitute: the nth term = 21, d = 1, n=k
We get a = 22-k

Thus the average is (21+22-k)/2 = (43-k)/2
Thus E
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Re: Around the World in 80 Questions (Day 10): In a set of k consecutive [#permalink]
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