Data Sufficiency

The average of three distinct positive integers x,y and z is 64.If a new number d is added,will the percent increase exceed 13%?
1. d is twice the value of one of the original numbers
2.d is half the value of the largest of the original numbers

[spoiler=]Ans:d[/spoiler]

Detailed explanation please!!

Why are the spoilers used by use "spoilt"
Never seem to work

Now i think there is some typo. Does it mean 13% increase in avg or total

NOt too sure..I posted the complete question!i am awaiting for one of the genius minds to come and help us!

I think it is asking for the increase in total and in that case the answer should be B. Can you please check the answer again?

(1) 1 64 127 avg 64
1 2 64 127 avg < 64
(2) 63 64 65 avg 64
32.5 63 64 64 avg < 64

IMO E

The question is extremely ambiguous...

tohellandback wrote:I think it is asking for the increase in total and in that case the answer should be B. Can you please check the answer again?

I find if the qn talks about avg % increase, then B will be the answer.

real2008 wrote:

tohellandback wrote:I think it is asking for the increase in total and in that case the answer should be B. Can you please check the answer again?

I find if the qn talks about avg % increase, then B will be the answer.

yes I guess.

The three numbers are positive.Hence they have to be >= 1. Also sum of 3 numbers = 64*3 = 192

Let the three numbers be 1 ,1 and 190.

if(d =2 , double of 1) then new avg= 194/5 .. The percentage increase is definetly < 13%

However if(d=380 double of 190) then new avg will be 13% more than the original avg.

Hence A not sufficient

Statement B

If the threee numbers are 64 , 64 and 64 then d= 32

New avg is 224/5 = 44.8 There is a percentage decrease in avg

However if the three numbers are 1 , 1 and 190 and d= 95 then new avg > 13%

Hence B is not sufficient


Combining both statements


We will get three of the numbers are 2d , d and d/2.. Let the fourth number be c

We need to find out if we can prove that d always takes values such that % increase > 13

I am not able to proceed after this.. In the exam I would guess C



Unable to understand how ans is D

raghavsarathy wrote:The three numbers are positive.Hence they have to be >= 1. Also sum of 3 numbers = 64*3 = 192

Let the three numbers be 1 ,1 and 190.

if(d =2 , double of 1) then new avg= 194/5 .. The percentage increase is definetly < 13%

However if(d=380 double of 190) then new avg will be 13% more than the original avg.

Hence A not sufficient

Statement B

If the threee numbers are 64 , 64 and 64 then d= 32

New avg is 224/5 = 44.8 There is a percentage decrease in avg

However if the three numbers are 1 , 1 and 190 and d= 95 then new avg > 13%

Hence B is not sufficient


Combining both statements


We will get three of the numbers are 2d , d and d/2.. Let the fourth number be c

We need to find out if we can prove that d always takes values such that % increase > 13

I am not able to proceed after this.. In the exam I would guess C



Unable to understand how ans is D

QN TALKS ABOUT THREE DISTINCT +VE INTEGERS

real2008 wrote:

raghavsarathy wrote:The three numbers are positive.Hence they have to be >= 1. Also sum of 3 numbers = 64*3 = 192

Let the three numbers be 1 ,1 and 190.

if(d =2 , double of 1) then new avg= 194/5 .. The percentage increase is definetly < 13%

However if(d=380 double of 190) then new avg will be 13% more than the original avg.

Hence A not sufficient

Statement B

If the threee numbers are 64 , 64 and 64 then d= 32

New avg is 224/5 = 44.8 There is a percentage decrease in avg

However if the three numbers are 1 , 1 and 190 and d= 95 then new avg > 13%

Hence B is not sufficient


Combining both statements


We will get three of the numbers are 2d , d and d/2.. Let the fourth number be c

We need to find out if we can prove that d always takes values such that % increase > 13

I am not able to proceed after this.. In the exam I would guess C



Unable to understand how ans is D

QN TALKS ABOUT THREE DISTINCT +VE INTEGERS

Oops .. made a mistake.. Thanks for that

Anywyas it does not change the answer.

WE can take the numbers are 1 , 3, and 188

Other set of numbers are 62 , 64 and 66.. The results do not change.

raghavsarathy wrote:

real2008 wrote:

raghavsarathy wrote:The three numbers are positive.Hence they have to be >= 1. Also sum of 3 numbers = 64*3 = 192

Let the three numbers be 1 ,1 and 190.

if(d =2 , double of 1) then new avg= 194/5 .. The percentage increase is definetly < 13%

However if(d=380 double of 190) then new avg will be 13% more than the original avg.

Hence A not sufficient

Statement B

If the threee numbers are 64 , 64 and 64 then d= 32

New avg is 224/5 = 44.8 There is a percentage decrease in avg

However if the three numbers are 1 , 1 and 190 and d= 95 then new avg > 13%

Hence B is not sufficient


Combining both statements


We will get three of the numbers are 2d , d and d/2.. Let the fourth number be c

We need to find out if we can prove that d always takes values such that % increase > 13

I am not able to proceed after this.. In the exam I would guess C



Unable to understand how ans is D

QN TALKS ABOUT THREE DISTINCT +VE INTEGERS

Oops .. made a mistake.. Thanks for that

Anywyas it does not change the answer.

WE can take the numbers are 1 , 3, and 188

Other set of numbers are 62 , 64 and 66.. The results do not change.

max percentage increase will happen if the first two nos are least possible,i.e. 1 & 2 and the other one 189. (since the fourth one need not be integer as per statement 2)

so fourth no. is 189/2= 94.5 now avg of four nos. is 71.625
so% increase is (71.625-64)/64=11.91%

hence B

[@raghavsarathy: even if you take 1,3,188 the % increase will not cross 11.91%]

real2008 wrote:

raghavsarathy wrote:

real2008 wrote:

raghavsarathy wrote:The three numbers are positive.Hence they have to be >= 1. Also sum of 3 numbers = 64*3 = 192

Let the three numbers be 1 ,1 and 190.

if(d =2 , double of 1) then new avg= 194/5 .. The percentage increase is definetly < 13%

However if(d=380 double of 190) then new avg will be 13% more than the original avg.

Hence A not sufficient

Statement B

If the threee numbers are 64 , 64 and 64 then d= 32

New avg is 224/5 = 44.8 There is a percentage decrease in avg

However if the three numbers are 1 , 1 and 190 and d= 95 then new avg > 13%

Hence B is not sufficient


Combining both statements


We will get three of the numbers are 2d , d and d/2.. Let the fourth number be c

We need to find out if we can prove that d always takes values such that % increase > 13

I am not able to proceed after this.. In the exam I would guess C



Unable to understand how ans is D

QN TALKS ABOUT THREE DISTINCT +VE INTEGERS

Oops .. made a mistake.. Thanks for that

Anywyas it does not change the answer.

WE can take the numbers are 1 , 3, and 188

Other set of numbers are 62 , 64 and 66.. The results do not change.

max percentage increase will happen if the first two nos are least possible,i.e. 1 & 2 and the other one 189. (since the fourth one need not be integer as per statement 2)

so fourth no. is 189/2= 94.5 now avg of four nos. is 71.625
so% increase is (71.625-64)/64=11.91%

hence B

[@raghavsarathy: even if you take 1,3,188 the % increase will not cross 11.91%]

Yes.. Missed that.. Answer has to be B. Thanks.

uptowngirl92 wrote:The average of three distinct positive integers x,y and z is 64.If a new number d is added,will the percent increase exceed 13%?
1. d is twice the value of one of the original numbers
2.d is half the value of the largest of the original numbers

[spoiler]Ans:d[/spoiler]

What's the source of this question? As written, it's unanswerable. "Percent increase" without reference is ambiguous; you have to have a percent increase of something specific.

In any case, no matter how you slice it, D certainly isn't the correct answer.

Let's use our secret bad question decoder rings and postulate that the question should read:

The average of three distinct positive integers x,y and z is 64. If a new number d is added to the set, will the average increase by more than 13%?

(1) d is twice the value of one of the original numbers

Picking numbers quickly shows us that (1) is insufficient. Let's choose extreme values (often a good strategy).

We know that x, y and z sum to 192 (sum = avg*# of terms), so let's pick 1, 2 and 189.

If d = 2 (nothing says that d must also be distinct), then our average actually goes down.

If d = 378, then our average more than doubles.

Therefore, the answer to the original question could be "yes" or "no"... insufficient.

(2) d is half the value of the largest of the original numbers

Again, let's look at the most extreme case.

If x, y and z are 1, 2 and 189, then d = 94.5 (nothing says d must be an integer). Our new sum is 192 + 94.5 = 286.5 and our new average is 286.5/4 = 71.625.

Is 71.625/64 > 1.13? What an annoying question! However, long division (we don't need to finish the problem, as soon as we see that the answer starts with 1.11 we know that it's going to be less than 1.13) shows us that the answer is "no".

So, we made d as big as we possibly could and we got a "no" answer. Smaller values of d will lead to a smaller increase. Therefore, the increase will NEVER be greater than 13%... sufficient.

(2) is sufficient, (1) is not: choose B.

Yes the ans is B.
One of my friend's gave me this solution could somebody pls explain the bolded ortion?

It means that x+y+z=192 (64x3)
Question asks this is x+y+z+d > 290 (Note that 64x(1+%13)x4=289,3) (average increases by over %13 percent) It means that is d greater than or equal to 99.
If a, b and c are distinct positive integers. They may be between 1 and 189 (say the smallest 1 and middle 2; 192-1-2)
Thus
1)Insuff. It is twice of one of original numbers. But we do not know which one. If it is the twice of smallest. It may be 2; If it twice of the greatest. It may be 378 (189x2). So we can not know.

2)Suff. If it is half of the greatest. It could not be greater that 94 (188/2=94; 188 is the greatest possible even number that satisfies the problem). So it is less that 99 in every case.