A set consists of 19 elements with an average of a. If addition

Question

A set consists of 19 elements with an average of  a . If addition of a new element increases the average by  k\% , what is the value of the new element?

A.  a(1 + \frac{k}{5})

B.  a*\frac{k}{100} - 20a

C.  20a(1 + \frac{k}{100})

D.  20(1 + \frac{k}{100}) - 19a

E.  a*\frac{k}{5} - 19a

Bunuel · Answer

A set consists of 19 elements with an average of  a . If addition of a new element increases the average by  k\% , what is the value of the new element?

A.  a(1 + \frac{k}{5}) 
B.  a*\frac{k}{100} - 20a 
C.  20a(1 + \frac{k}{100}) 
D.  20(1 + \frac{k}{100}) - 19a 
E.  a*\frac{k}{5} - 19a

Let  x  represent the value of the new element. 
 
The new average is given by  \frac{	ext{new sum} }{20} = \frac{	ext{old sum} + x}{20} = \frac{19a + x}{20} . 
 
Given that the average increased by  k\% , we'd get  \frac{19a + x}{20} = (1 + \frac{k}{100})*a . 
 
 19a + x = a*(20 + \frac{k}{5}) .
 
 x = a*(20 + \frac{k}{5})-19a .
 
 x = a * (1 + \frac{k}{5}) .

Answer: A

unraveled · Answer

A set consists of 19 elements with an average of  a . If addition of a new element increases the average by  k\% , what is the value of the new element?

Testing values can also help. A little longer way to get to the answer, however, it needs to be verified by checking conditions also.

Let's say a = 5 and k = 5
Initial sum = 19a = 85
New sum = 85 + 5(1+5%)

A.  a(1 + \frac{k}{5}) 
 5(1 + \frac{5}{5}) = 10

Can be the answer. Keep it

B.  a*\frac{k}{100} - 20a 
 5*\frac{5}{100} - 20*5 = -99.75 
Negative value, eliminate

C.  20a(1 + \frac{k}{100}) 
 20*5(1 + \frac{5}{100}) = 105 
Can keep it but this needs to be verified since value seems to be too high considering the initial value of sum is only 85.

Two ways we can verify:
1. New sum we already know from above so this is not the right answer.

2. Also, if 105 is the value of new element then 
total sum becomes 190 which gives us average of 9.5.
9.5 is more than 5% the jump we are looking for. The jump now is 90%.
So, this can't be answer.

D.  20(1 + \frac{k}{100}) - 19a 
 20(1 + \frac{5}{100}) - 19*5 = -64  
Negative eliminate.

E.  a*\frac{k}{5} - 19a 
 5*\frac{5}{5} - 19*5 = -80 
Negative, eliminate.

Answer A.

gmatophobia · Answer

When the question presents us with generic information, we can also assume values and solve the question.

Assume that the set has 19 integers, each of which has a value of 10.

S = {10, 10 , 10 , ....... 10 , 10 }

The average of the set S = a = 10.

After, a new element is added the average increase by (let's assume) 10%. So k = 10

New average = 11

Value of the element added = b = (20*11) - (10*19) = 220 - 190 = 30.

Now let's use the options to find, which option yields the value 30 -

A.  a(1 + \frac{k}{5})

10(1 + \frac{10}{5}) = 30

Option A

Deeya007 · Answer

Why do we not take 0 as a positive integer? In a very similar question, we counted 0 as well.

Bunuel · Answer

Not sure how this is related to the question above, but 0 is most definitely NOT a positive number. Zero is neither positive nor negative, it is the only number with that property.

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.

3. Zero is  neither positive nor negative  (the only one of this kind)

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{0}{x} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x)

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

Hope it helps.

Vivek1707 · Answer

One faster method of doing this quesiton would be to find the increase in value and add that increase to the previous average.

For example say the original average was p the number of terms was n. When one person got added the average increased by 2. So the total extra amount brought by the new element would be 2 x (n+1)
Now this extra amount if you add it to the previous average ---> will give you the amount brought by the new element

Hence in this question
Since the average increased by k% 
The value of the increase would be k/100 *a

Hence the total amount extra bought by the new element would be k/100*a *20

Since this is the extra amount ---> the total amount bought by the new element would be
a + k/100*a*20 ---> Which is i a(k/5+1)

Hence answer A

A set consists of 19 elements with an average of a. If addition

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A set consists of 19 elements with an average of a. If addition

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards