GMAT Data Sufficiency Discussions

MBA GMAT
1221
siddharthaduggirala


Why do u say so buddy ?
from the question we have original SD of 10 ..

St 1: My point was (somebody correct if its wrong) 30 % reduction of each value ,reduces new mean by 30 % and hence also the SD is reduced by 30 % ...So new SD = 7 ...Sufficeint ...Ans A

P.S : i am loving the cross correction, it only helps to get better !! ..Pls correct me again dude ..

Cheers

P.S :This post also makes me hardcore Pagal !!
1222
Why do u say so buddy ?
from the question we have original SD of 10 ..

St 1: My point was (somebody correct if its wrong) 30 % reduction of each value ,reduces new mean by 30 % and hence also the SD is reduced by 30 % ...So new SD = 7 ...Sufficeint ...Ans A

P.S : i am loving the cross correction, it only helps to get better !! ..Pls correct me again dude ..

Cheers

P.S :This post also makes me hardcore Pagal !!


I also like this kind of discussion.

U got this right man:

If you multiply all terms by x then SD =x times old SD and mean = x times old mean
1223

this is something that tricks everyone I guess, including me... SD as a concept is very delicate, and change in single entity can change the whole value of SD.. and for the candidate to find the value he should have the individual water amounts, range etc..

1) For each tank, 30 percent of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.
- the 30% decrease might mean diff values for diff initial tank values. NOT SUFF

2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons.
- Since we dont know the inidividual values of tank waters, we cant solve using this either. NOT SUFF

Try both together.. still doesnt help.. since you dont get any value as such

I would have marked option E






PS: If the question would have been something like, DOES THE SD CHANGE AT ALL, then the options would have helped to conlude something substantial I guess.

During an experiment, some water was removed from each of 6 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 10 gallons, what was the standard deviation of the volumes of water in the tanks at the end of the experiment?
1) For each tank, 30 percent of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.
2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons.
I found this is one of the earlier threads but the soln wasn't there. Please provide the explaantory soln.
Thanks.
1224
this is something that tricks everyone I guess, including me... SD as a concept is very delicate, and change in single entity can change the whole value of SD.. and for the candidate to find the value he should have the individual water amounts, range etc..

1) For each tank, 30 percent of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.
- the 30% decrease might mean diff values for diff initial tank values. NOT SUFF

2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons.
- Since we dont know the inidividual values of tank waters, we cant solve using this either. NOT SUFF

Try both together.. still doesnt help.. since you dont get any value as such

I would have marked option E






PS: If the question would have been something like, DOES THE SD CHANGE AT ALL, then the options would have helped to conlude something substantial I guess.

HI Puys,
Thanks for the explanations. But I do not have the OA wid me, so can't tell which option A/E is correct. If some one knows the source of this problem, please post the OA here for our reference.
Thanks a lot!
1225

Tumhi ki bolchi dada.. no OA!!

HI Puys,
Thanks for the explanations. But I do not have the OA wid me, so can't tell which option A/E is correct. If some one knows the source of this problem, please post the OA here for our reference.
Thanks a lot!
1226
this is something that tricks everyone I guess, including me... SD as a concept is very delicate, and change in single entity can change the whole value of SD.. and for the candidate to find the value he should have the individual water amounts, range etc..

1) For each tank, 30 percent of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.
- the 30% decrease might mean diff values for diff initial tank values. NOT SUFF
2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons.
- Since we dont know the inidividual values of tank waters, we cant solve using this either. NOT SUFF

Try both together.. still doesnt help.. since you dont get any value as such

I would have marked option E

PS: If the question would have been something like, DOES THE SD CHANGE AT ALL, then the options would have helped to conlude something substantial I guess.


Hey Varun ,
On a GMAT SD is very rarely asked and is often a topic left out for discussion ..And we are not expected to calculate SD in GMAT, and it is tricky for sure ...

However ,
When every qty changes by same %, Irresepective of initial values, New SD also changes by same % ...THERE is no doubt regarding this concept ..

For Fun and learning try it out on excel ...take 6 random values and calculate SD by stdev function on excel ..Change the initial values to 70 % and calculate the SD again ....You would notice that new SD is 70% of original ..IF there is still a niggling doubt, change original values again, yet New SD changes by same % of Original ...

Given an original SD and an equal change of each values, we can calculate the new SD irrespective of original values

If there is still an doubt lets understand the reason behind it :

P.S : we are not expected to know or calculate SD for exam .This is for concept building only.

Let X1, X2 ,....X6 be the 6 original values .
Mean = X bar = (X1+X2+...+X6)/6
SD = sq root ( (X1-Xbar)^2 + (X2-X bar)^2 +......+(X6-Xbar)^2)/6)

Now new values are 0.7X1, 0.7X2 .......0.7X6
New mean = new X bar = 0.7(X1+X2+...+X6)/6 = 0.7 original X bar

New SD = sq root ((0.7^2(X1-Xbar)^2 +0.7^2(X2-X bar)^2 +........+0.7^2(X6-Xbar)^2)/6)

=0.7 original SD

Hence, irrespective of original values, for same % change, relation between original and new SD remains the same ..

Hence Ans A

Hey Arijit, does not matter if u do not know OA, we can discuss it to length to make sure there is no error..
1227

HI Puys,
SD is a freq topic.And rule of thumb is

If you multiply all terms by x then SD =x times old SD and mean = x times old mean

This might help to get to the answer quickly.

Siddhartha

1228

HI puys,
I got an idea yesterday.Shall we start a thread where we can discuss all the DS OG questions at length? this is just a proposal we can do the same in this thread also.

Siddhartha

PS: I'm an expert pagal now.

1229

I tried the entire calculation thing.. and got the facts straightened.. thanks man!! ;)

Hey Varun ,
On a GMAT SD is very rarely asked and is often a topic left out for discussion ..And we are not expected to calculate SD in GMAT, and it is tricky for sure ...

However ,
When every qty changes by same %, Irresepective of initial values, New SD also changes by same % ...THERE is no doubt regarding this concept ..

For Fun and learning try it out on excel ...take 6 random values and calculate SD by stdev function on excel ..Change the initial values to 70 % and calculate the SD again ....You would notice that new SD is 70% of original ..IF there is still a niggling doubt, change original values again, yet New SD changes by same % of Original ...

Given an original SD and an equal change of each values, we can calculate the new SD irrespective of original values

If there is still an doubt lets understand the reason behind it :

P.S : we are not expected to know or calculate SD for exam .This is for concept building only.

Let X1, X2 ,....X6 be the 6 original values .
Mean = X bar = (X1+X2+...+X6)/6
SD = sq root ( (X1-Xbar)^2 + (X2-X bar)^2 +......+(X6-Xbar)^2)/6)

Now new values are 0.7X1, 0.7X2 .......0.7X6
New mean = new X bar = 0.7(X1+X2+...+X6)/6 = 0.7 original X bar

New SD = sq root ((0.7^2(X1-Xbar)^2 +0.7^2(X2-X bar)^2 +........+0.7^2(X6-Xbar)^2)/6)

=0.7 original SD

Hence, irrespective of original values, for same % change, relation between original and new SD remains the same ..

Hence Ans A

Hey Arijit, does not matter if u do not know OA, we can discuss it to length to make sure there is no error..
1230

well ... i havent seen many questions on SD being posted ...n by the flow of last question many concepts of SD are yet unexplored ..

So, here is 1 question from me ..its a good one ...give it a try ...will post the approach and answer after few replies .

A school surveyed 100 students to determine the average number of minutes studied per night. If 84 percent of the students studied for at least 114 minutes per night and the students responses had a normal distribution, what is the average number of minutes studied per night?

(1) The new average would be 10 percent greater, If 20 more students averaging 192 minutes per night were added to the sample,
(2) Only 2 students studied at least 132 minutes per night.

Cheers !!

1231

@Bhavin..
I have no clue what this question is about.. but if I have to guess in the exam, I would mark option C

Could you please explain the question.. and also share the means to solve this.. 😉

1232
@Bhavin..
I have no clue what this question is about.. but if I have to guess in the exam, I would mark option C

Could you please explain the question.. and also share the means to solve this.. ;)


No varun ..Answer is not C ..
Will post up a detailed approach and solution by evening ...in a bit rush now 😉
1233

So, here is 1 question from me ..its a good one ...give it a try ...will post the approach and answer after few replies .

A school surveyed 100 students to determine the average number of minutes studied per night. If 84 percent of the students studied for at least 114 minutes per night and the students responses had a normal distribution, what is the average number of minutes studied per night?

(1) The new average would be 10 percent greater, If 20 more students averaging 192 minutes per night were added to the sample,
(2) Only 2 students studied at least 132 minutes per night.

Cheers !!

@Bhavin..
I have no clue what this question is about.. but if I have to guess in the exam, I would mark option C

Could you please explain the question.. and also share the means to solve this.. ;)


Ans is D

Maybe some concepts would be handy before solving:

Whenever data is said to be normal distributed about the mean this is the deal :

Normal distribution has 2 parameters viz mean and standard deviation.

Valid approximations for GMAT:

34 % of the data falls within 1 standard deviation above the mean and 34 % data falls within 1 standard deviation below the mean...OR 68 % of the data is within 1 std deviation from mean..

14 % of the data is between 1st and 2nd deviation above mean and 14 % of the data is between 1st and 2nd deviation below mean... OR 96 % of the data is within 2 deviations from mean..

2 % of the data is between 2nd and 3rd deviation above mean and 2% of the data is between 2nd and 3rd deviation below mean... OR 100% of the data is within 3 deviations from mean..

this is very much a valid assumption for GMAT..actually 99.xx % data is within 3 std deviations from mean...on GMAT u can approximate it to 100% and use 34 % , 14 % and 2% rule..

( these values are valid close approximation obtained from normal distribution table or z table, basically for a given mean and SD, every value is associated with a corresponding % distribution)

Getting back to the sum :
We know, strength = 100 and from above learning 84% corresponds to people above 1st deviation below mean
So, M-SD = 114

St 1 :
(Here no concept of SD is used, just averages)
Let original average be x
So, (100X + 192*20)/(100+20) = 1.1x
We can actually calculate x =120 OR realise its a linear eqn in x, hence a definite answer and no need to solve ..Sufficient

St 2 :
Back to SD
2 student study more than 132 minutes per night ...
Now, 2 in a batch of 100 corresponds to 2%, which we now know corresponds to 2nd deviation above the mean ..

So, M+2SD = 132
And we already have M-SD = 114
Solve, to get SD = 6 and M=120 OR realise 2 eqn 2 unknowns, definite value , no need to solve ...Sufficient ..

Hence, Ans D.
Had to type evrtng, so looks diff, else its just 1 concept.

Long post, hope that helps !!

Cheers 😉
1234

Had to read it twice to understand

amazing fundas though.. u lived upto ur words.. this is truly good example hat covers lots of basics.. thanks man.. appreciate it ;)

Ans is D

Maybe some concepts would be handy before solving:

Whenever data is said to be normal distributed about the mean this is the deal :

Normal distribution has 2 parameters viz mean and standard deviation.

Valid approximations for GMAT:

34 % of the data falls within 1 standard deviation above the mean and 34 % data falls within 1 standard deviation below the mean...OR 68 % of the data is within 1 std deviation from mean..

14 % of the data is between 1st and 2nd deviation above mean and 14 % of the data is between 1st and 2nd deviation below mean... OR 96 % of the data is within 2 deviations from mean..

2 % of the data is between 2nd and 3rd deviation above mean and 2% of the data is between 2nd and 3rd deviation below mean... OR 100% of the data is within 3 deviations from mean..

this is very much a valid assumption for GMAT..actually 99.xx % data is within 3 std deviations from mean...on GMAT u can approximate it to 100% and use 34 % , 14 % and 2% rule..

( these values are valid close approximation obtained from normal distribution table or z table, basically for a given mean and SD, every value is associated with a corresponding % distribution)

Getting back to the sum :
We know, strength = 100 and from above learning 84% corresponds to people above 1st deviation below mean
So, M-SD = 114

St 1 :
(Here no concept of SD is used, just averages)
Let original average be x
So, (100X + 192*20)/(100+20) = 1.1x
We can actually calculate x =120 OR realise its a linear eqn in x, hence a definite answer and no need to solve ..Sufficient

St 2 :
Back to SD
2 student study more than 132 minutes per night ...
Now, 2 in a batch of 100 corresponds to 2%, which we now know corresponds to 2nd deviation above the mean ..

So, M+2SD = 132
And we already have M-SD = 114
Solve, to get SD = 6 and M=120 OR realise 2 eqn 2 unknowns, definite value , no need to solve ...Sufficient ..

Hence, Ans D.
Had to type evrtng, so looks diff, else its just 1 concept.

Long post, hope that helps !!

Cheers ;)
1235
What is the product of positive integers P and Q?
(1) 18P + Q = 367
(2) Q

I am little confused with the OA.. it says option C is correct.. whereas I dont think the 2 eqns provided above help us find any unique solution.. so I marked option E.. plz share ur thots..
1236

Back of the envelope calc gives the below result...Considering statement 2 we can get only P=20 and Q=7.So this will give us unique solution.


P=21, Q will be -ve
P=20, Q=7
P=19,Q=25

1237
What is the product of positive integers P and Q?

(1) 18P + Q = 367
(2) Q
I am little confused with the OA.. it says option C is correct.. whereas I dont think the 2 eqns provided above help us find any unique solution.. so I marked option E.. plz share ur thots..


first statement says 18p+q = 367...insufficient information (p can be zero and q can be 367..they can take n no of values)
second statement says q is less than 18, which means q can be anywhere b/w 0 and 17..insufficient

combining both, 18p is always even..367 is always odd..hence q should be odd to satisfy the equation..

only P=20 and Q=7 satisfy that equation

IMO C
1238
What is the product of positive integers P and Q?
(1) 18P + Q = 367
(2) Q
I am little confused with the OA.. it says option C is correct.. whereas I dont think the 2 eqns provided above help us find any unique solution.. so I marked option E.. plz share ur thots..


Yes ..Ans sould be C ..
Individual statements are insufficient ..

When both statements are combined ,it is suff..
Here, nos are small ...so it wont take much time to conclude the only poss values of p and q ..

Even if nos had been very huge, no need to actually calculate the multiples of 18 closer to the no ..

The fact that q is lesser than 18 (the divisor) implies that p has to be the highest multiple of 18 closest to the no and q has to be the diff .Only one poss set ..Suff ..

eg if the statements had been
St 1 : 18p + q = 672324238943242432 (for that matter any big no)
St 2 : q
Ans C

If St 2 had been q
Then Ans would have been E, as there are 2 poss sets (since q is greater than divisor )

So, really no need to work with nos, thought to share this, maybe it could help to save time
1239
How to address such questions..??.. CG makes me go nuts..:banghead:

=====
In the xy-plane, is the slope of line k positive?
(1) Line k passes through the points (-1, -7) and (2, 5).
(2) Line k has equation y = 4x 3.

=====

In the xy-plane, line l and line k intersect at the point (16/5, 12/5). What is the slope of
line l?
(1) The product of the slopes of line l and line k is 1.
(2) Line k passes through the origin.

=====
1240
How to address such questions..??.. CG makes me go nuts..:banghead:


=====
In the xy-plane, is the slope of line k positive?
(1) Line k passes through the points (-1, -7) and (2, 5).
(2) Line k has equation y = 4x 3.
=====
In the
xy-plane, line l and line k intersect at the point (16/5, 12/5). What is the slope of
line

l?
(1) The product of the slopes of line

l and line k is 1.
(2) Line k passes through the origin.
=====


For the first one,compare the lines with the standard equation of y=mx+c and if m positive slope positive.
for first statement-
m=(5+7)/(2+1)=4
second statement,m=4
so either is suffeciet-D

2.answer is C.
from the point of intersection, the two equations will be-
k=>12=16m'+5c'
l=>12=16m"+5c"

from first statement m'*m"=-1
and from second statement 5C'=0,m'=3/4
so m"=-4/3.