[bhavin422]

Originally Posted by nuttyvarun

@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

[bhavin422]

Originally Posted by bhavin422

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 !!

Originally Posted by nuttyvarun

@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

[nuttyvarun]

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

Originally Posted by bhavin422

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

[nuttyvarun]

What is the product of positive integers P and Q?

(1) 18P + Q = 367

(2) Q < 18

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..

[deepakraam]

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

[Mcnabb_ISB]

Originally Posted by nuttyvarun

What is the product of positive integers P and Q?

(1) 18P + Q = 367

(2) Q < 18

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

[bhavin422]

Originally Posted by nuttyvarun

What is the product of positive integers P and Q?

(1) 18P + Q = 367

(2) Q < 18

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<18 (for that matter any value lesser than 18 as well)

Ans C

If St 2 had been q<19 (for that matter, if q is lesser than any value greater than 18 ) ..

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

[nuttyvarun]

How to address such questions..??.. CG makes me go nuts..

=====
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.

=====

[guy with guts]

Originally Posted by nuttyvarun

How to address such questions..??.. CG makes me go nuts..

=====
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.

[siddharthaduggirala]

Originally Posted by nuttyvarun

How to address such questions..??.. CG makes me go nuts..

=====
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 ques the ans is D(IMO). From 1) u can calc slope. so A alone is suff. From 2) we can calc the slope from y=mx+C so slope is +ve.


For the 2nd ques

we want the slope of line l.

Take 1)we can only know that the 2 lines are perpendicular.

SO 1) alone is insuff

Take 2) we have no info on line l so insuff.

take 1) and 2) here we have 2 points for k so we can calc slope of K and so we can calc slope for l since k and l are perpendicular.

So C.

Whats the answer for both?