GMAT Problem Solving Discussions

MBA
542
Wow! Let me explain.

How about x^2 = 4
=> x = +2 or -2.

Let x = (y)^1/2

So y = +2 or -2.

square root of 4 can be +2 or -2; of course, because (2)^2 = (-2)^2 = 4.


Cheers,
Sumit

This is a mis concept many has.
X^2= 4 implies X^2-4=0 (x+2)*(x-2)=0 implies x=+2 or -2

if x^2 =Y x= +sqrt Y or -sqrt Y . but sqrt Y is always positive
543

Hi all,
I got this doubt from GMAT prep test.
Should a circle touching only one vertex of a triangle be considered in intersecting the circle ????
Main question was : which one is the list containing the number of points at which a circle can intersect a triangle ???
2, 4
2, 4, 6
1, 2, 3
1, 2, 3, 4
1, 2, 3, 4, 5, 6

544
Hi all,
I got this doubt from GMAT prep test.
Should a circle touching only one vertex of a triangle be considered in intersecting the circle ????
Main question was : which one is the list containing the number of points at which a circle can intersect a triangle ???
2, 4
2, 4, 6
1, 2, 3
1, 2, 3, 4
1, 2, 3, 4, 5, 6

the circle and touch the triangle only at one of the vertices...so it touches 1 point in the entire triangle...

the circle can intersect any two sides of a triangle..so it can also touch 2 points

the incircle/circumcircle of a triangle touches 3 points
the circle of radius between the incircle and circumcircle touches the triangle at 6 points...

the only option satisfying the above values is option E..i havent evaluated the other options cuz there's no need to do so..
545
the circle and touch the triangle only at one of the vertices...so it touches 1 point in the entire triangle...

the circle can intersect any two sides of a triangle..so it can also touch 2 points

the incircle/circumcircle of a triangle touches 3 points
the circle of radius between the incircle and circumcircle touches the triangle at 6 points...

the only option satisfying the above values is option E..i havent evaluated the other options cuz there's no need to do so..



Yes Answer should be E
546

try this one..



1) X + Y > 4
2) X*Y > 1
3) X/Y + Y/X 4) (X - Y)^2 > 0
5) none of the above

can somebody please confirm whether '0' is not considered a positive integer?

547


1) (-1, 1)
2) (-1, 2)
3) (0, 1)
4)(0, 2)
5) (1, 2)
548


1) (-1, 1)
2) (-1, 2)
3) (0, 1)
4)(0, 2)
5) (1, 2)

Ans is (0,2)

mod(1-x) 1-x0

-(1-x) 0
549
try this one..



1) X + Y > 4
2) X*Y > 1
3) X/Y + Y/X 4) (X - Y)^2 > 0
5) none of the above

can somebody please confirm whether '0' is not considered a positive integer?


0 is neither positive nor negative. It is neutral. Right side to it on number system are difined as positive and laft as negative.
550
try this one..



1) X + Y > 4
2) X*Y > 1
3) X/Y + Y/X 4) (X - Y)^2 > 0
5) none of the above

can somebody please confirm whether '0' is not considered a positive integer?




choice 2 must be true, because atleast one of x,y has to be greater than 1 inorder that the given condition holds, therefore x*y has a minimum value of 2.
when x=y, say 3 then (X - Y)^2 equals to zero.

PS- zero is non negative and non positive
551
try this one..



1) X + Y > 4
2) X*Y > 1
3) X/Y + Y/X 4) (X - Y)^2 > 0
5) none of the above

can somebody please confirm whether '0' is not considered a positive integer?

answer is B....cuz if u take the numbers as 1 and 2, then only options 2 and 4 get satisfied..but when both x=y=2, then option 4 gets ruled out...
552
try this one..



1) X + Y > 4
2) X*Y > 1
3) X/Y + Y/X 4) (X - Y)^2 > 0
5) none of the above

can somebody please confirm whether '0' is not considered a positive integer?

for sure 0 is not considered +ive or -ive integer...
for the answer i think its 2
explanation:
range for x,y : 1,2,3,4......
from the condition 1/x + 1/y 1
to eliminate the remaining choices
1. take x,y 2,2
3. it simpilifies to x^2 + y^2 4. take x,y 2,2
553

Guys am really flat on the following attached questions..can u guyz help me out...
i really apreciate ur help...
thx a bunch
n btw da answers selected are all wrong... :|


*God helps those who help themselves*

554
Guys am really flat on the following attached questions..can u guyz help me out...
i really apreciate ur help...
thx a bunch
n btw da answers selected are all wrong... :|


*God helps those who help themselves*

1. e firstly it has to be linear, secondly the constant cannot be anything otherthan zero. Easy to visualize through co-ordinate geometry.
2. c either of te statement fail to ascertain where the projection of point s may fall on the other arm of the right triangle, both included angle(rsq) & ang(tsu) can be written in terms of angle r and t, sum of which is 90..........u'll get the answer if you make the equation.........
3. a ???????????????????????????????????????????????
5 sigma = 40, 58 + 2 sigma = 74
4. b ???????????????????????????!!!!!!!!!!!!!!!!!!!!!!
come on man( err....or is that a woman 😃 ), a triangle in semi circle is right angled. get the third side....find the area
555
Guys am really flat on the following attached questions..can u guyz help me out...
i really apreciate ur help...
thx a bunch
n btw da answers selected are all wrong... :|


*God helps those who help themselves*

1. E...just substitute a = b = x and evaluate the options..only last option validates the result..

2. C...either st. used alone do nt provide adequate data..however, when u both the st. together, then consider triangles RQS and SUT...measure of all angles of two triangles is 360...but angles R+T = 90..hence, 2 RSQ + 2 UST = 270...hence, RSQ + UST = 135...now RSQ + UST + x =180..u can get value of x

3. A...u can calculate like this:
s.d. above the mean (-ve if it is below) = deviation/s.d.
since s.d. is going to remain the same in both the cases, the only value of mean tht satisfies the above eq. is A
-2 = (58-74)/s.d.
s.d. = 8

3 = (98-74)/s.d.
s.d. = 8

4. B...triangle is right angled since it is inscribed in a semicircle..u can find area easily..
556
1. E...just substitute a = b = x and evaluate the options..only last option validates the result..

2. C...either st. used alone do nt provide adequate data..however, when u both the st. together, then consider triangles RQS and SUT...measure of all angles of two triangles is 360...but angles R+T = 90..hence, 2 RSQ + 2 UST = 270...hence, RSQ + UST = 135...now RSQ + UST + x =180..u can get value of x

3. A...u can calculate like this:
s.d. above the mean (-ve if it is below) = deviation/s.d.
since s.d. is going to remain the same in both the cases, the only value of mean tht satisfies the above eq. is A
-2 = (58-74)/s.d.
s.d. = 8

3 = (98-74)/s.d.
s.d. = 8

4. B...triangle is right angled since it is inscribed in a semicircle..u can find area easily..


Hi Ajay
You said u have 150 essays. Can u pla send me those ([email protected]) [email protected] or upload here.
Thanks
557
Hi Ajay
You said u have 150 essays. Can u pla send me those ([email protected]) [email protected] or upload here.
Thanks

mailed it on ur yahoo account
558

dude ...thaaaaaaaaaaaaaaanx a million man..it helped me a lot
i feel a lil embarassed here cuz ..i mean whn i did da questions maself bein a lil more vigiliant..i got dem...but i still am confused about dat functions one..cuz i mean..
if x=a=b ..how come f(x)= -3(x) justifies f(a+b)=f(a)+f(b)
i have dis princeton review gmat manual but it dosnt have ne topic relating to dese sorta functions n am blank at it...

n yeah man my gmats on 13th sept...my mock gmat scores(p.review/gmat-prep/p.prep/kaplan) r around 640-660 ...ne word of advice?

n am like realy pissed @ verbal..sumtymes i skore like 40-45 while sumtymes its like 34-35...am pretty confident on da SC..but da problem iz ..i get stressed out by lookin at da monitor continuosly for 4 hrs n in da end am like..f*** it...

iv been practicing da tests n everything since god know when..like last sptember...
am no genius unlike u guys but stil workin at it..

ill appreciate advices,if any
n yeah i have sum more questions..am sure deyll b lame n easy for u experts...but again il realy apreciate da help...don kno wht i wud hv done widout u guys...!! :)
god bless ya!

559

n hey guys i don know if its like breach of privacy/copyrite laws/jack shyt...
after am done on 13th..i have dis princetonreview online center workin for like 2 more munths n same is ma petersons...so u guys can go on n reset ma tests n give em...

am stil not sure if more den 1 person can log on simultan. but ...u guyz can chek it out
dis iz da least i can do 2 return da favor..thx

560

1)hmm can u xplain the sy plain question...
i have an idea dat dis question would be solved thru the equation
y=mx +b
but how..
what wud be the figure if v draw it...???
i mean da line"K" wud b da hypotenuse or what?
btw da ans iz "A"


2)the second one iz pretty stupid i admit...
well...if da triange is 2wice da area ..den is it spose to have 2ice da base..given a constant height.....
am ending up wid "E" but its "C"

n hey guys tell me...dere r only two kinda triangles ...30:60:90 n da 45:45:90..right?
iz dere like a 30:30:120 tri?
Killer thankz!!!:

561
1)hmm can u xplain the sy plain question...
i have an idea dat dis question would be solved thru the equation
y=mx +b
but how..
what wud be the figure if v draw it...???
i mean da line"K" wud b da hypotenuse or what?
btw da ans iz "A"


2)the second one iz pretty stupid i admit...
well...if da triange is 2wice da area ..den is it spose to have 2ice da base..given a constant height.....
am ending up wid "E" but its "C"

n hey guys tell me...dere r only two kinda triangles ...30:60:90 n da 45:45:90..right?
iz dere like a 30:30:120 tri?
Killer thankz!!!:



Answer1:
Dude! For equation y = mx + c; m is the slope which is positive if it goes from north-east to south-west i.e. draw a line on indian map from bihar to gujarat that is a line with +ve m.
now it gives you the intercept on x axis, and gives you the area, so the area would be 1/2 x 4 x (intercept on Y axis) = 12;
solve it and draw the line you will get (-6).

Answer2:
Try to put values. And BTW there are different kind of tri-angles not only 30-60-90 or 45-45-90; angles can be any value; just the sum has to be 180 deg.
Try with 45-45-90 values and solve the equations; you will get the answer.

PS: do not post the answers mate; let the people solve the problems before you post your answers.


Cheers,
Sumit
:robot: