GMAT Problem Solving Discussions

Few ques from my side:


1) On the number line, the distance between x and y is greater than the distance between x and z. does z lie between x and y on the number line?
a) Xyz
b) Xy

2) For the students of class a, range of the heights is r cms, the greatest height is g cms,
For student sof class b, range- s cms and greatest height- h cms
Is the least height of the students in class a greater than least height of b?
a) rh
(I hate these type of ques, always got them wrong is og)

3) is a,b,k and m are positive integers, is a^k a factor of b^m?
a) a is a factor of b
b) k

4) lines n and p intersect lie in the xy-plane. Is the slope of line n less than the slope of p?
a) lines intersect at the point (5,1)
b) the y-intercept of line n is greater than the y-intercept of line p
guys ple reply asap till tomo evening.

thanks
neha

Ps- sorry in hurry, posted DS ques on PS thread. i'm really sorry.

1. E
2. C
3. C
4. E

Pls let me the OA.

If x

I y > 0

II x/y > -1

III 1/x+ 1/y

Rather than the answer I am more interested in elaborated approach at arriving answer

Enequalities sometimes scare me signed inequalities are yet more scary

Pls help
If x

I y > 0

II x/y > -1

III 1/x+ 1/y

Rather than the answer I am more interested in elaborated approach at arriving answer

Enequalities sometimes scare me signed inequalities are yet more scary

Pls help

1. Y>0; 0y+1y should be -ve and as x is -ve so y should be +ve, so y>0 must be true.
2. 0y+1y we get -1yy>-1 must be true.
3. 1/x+1/y gives (x+y)/xy and as per inequality x/y+1 is +ve value which means (x+y)/y is +ve value and dividing this value with x, which -ve will result in -ve result.

Whats the oA?
1. Y>0; 0y+1y should be -ve and as x is -ve so y should be +ve, so y>0 must be true.
2. 0y+1y we get -1yy>-1 must be true.
3. 1/x+1/y gives (x+y)/xy and as per inequality x/y+1 is +ve value which means (x+y)/y is +ve value and dividing this value with x, which -ve will result in -ve result.

Whats the oA?

It should be I and II

OA is I , II III

This Q is from princeston .

In case of II X/Y Cannot have any positive value ( which is also GREATER than -1 ) , but it can have values from -0.99 to -0.01 . Therefore according to me II is partially true .. Any thoughts ?

OA is I , II III

This Q is from princeston .

In case of II X/Y Cannot have any positive value ( which is also GREATER than -1 ) , but it can have values from -0.99 to -0.01 . Therefore according to me II is partially true .. Any thoughts ?



My answer is I and II and here is my approach.

Stat 2: 0 subtract 1 from all three, the statement becomes
-1
Since x is less than zero, and for the statement to hold true y has to be positive. Hence I is true.
We can see above that x/y > -1, hence II is true.

Since -1 => -y => -1/x => 0 Assume x = -2 and y = 3. This statement holds true. Hence III is not true.

'If equation x/2 + y/2 = 5 encloses a region on the coordinate plane, what is the area of this region?

a) 20
b) 50
c) 100
d) 200
e) 400

'If equation x/2 + |y/2 = 5 encloses a region on the coordinate plane, what is the area of this region?

a) 20
b) 50
c) 100
d) 200
e) 400


My Answer :-

This can be considered as X|+Y = 10

This MEANS sum of abs values of x and y coordinates is 10.

Therefore area will be bound by 4 line-segments each one placed in each Quadrant.

End points will be (10,0), (-10,0), (0,10), (0,-10)

Thus this will be a Square with side 10*2^1/2

Therefore total area = 200

Pls provide OA

Friends !!

Take a look at below thread ...you may post Quick math techniques gained thru experience..

http://www.pagalguy.com/discussions/speed-raising-techniques-on-quant-section-25050113/1883625

2 COUPLES AND ONE single person are seated at random in a row of 5 chairs. what is the probability that neither of the couple sits together in adjacent chairs

1/5, 1/4, 1/2, 3/8, 2/5

As part of a game, four people each must secretly chooose an integer between 1 to 4 inclusive. what is the approx. likelihood that all 4 people will choose diffeent no.

9% 12% 16% 20% 25%



As part of a game, four people each must secretly chooose an integer between 1 to 4 inclusive. what is the approx. likelihood that all 4 people will choose diffeent no.

9% 12% 16% 20% 25%


Is the answer 9%?

Reasoning is : For 1st member probability of choosing distinct number is 1
For 2nd member probability of choosing distinct number is 3/4
For 3rd member probability of choosing distinct number is 2/4
For 4th member probability of choosing distinct number is 1/4

Probability of the event is 1*3/4*1/2*1/4=9%(approx)
Is the answer 9%?

Reasoning is : For 1st member probability of choosing distinct number is 1
For 2nd member probability of choosing distinct number is 3/4
For 3rd member probability of choosing distinct number is 2/4
For 4th member probability of choosing distinct number is 1/4

Probability of the event is 1*3/4*1/2*1/4=9%(approx)


Yes Correct !! U solved in most simple way
My Answer :-

This can be considered as X+Y = 10

This MEANS sum of abs values of x and y coordinates is 10.

Therefore area will be bound by 4 line-segments each one placed in each Quadrant.

End points will be (10,0), (-10,0), (0,10), (0,-10)

Thus this will be a Square with side 10*2^1/2

Therefore total area = 200

Pls provide OA


U nailed it abosolutely, didnt u?... r u not confident about that? come on, its the right answer... i made a mistake in not reading the mode sign...
2 COUPLES AND ONE single person are seated at random in a row of 5 chairs. what is the probability that neither of the couple sits together in adjacent chairs

1/5, 1/4, 1/2, 3/8, 2/5



Total number of arrangements possible : 5!=120

Atleast 1st couple sit together arrangement 4!*2=48
Atleast 2nd couple sit together arrangement 4!*2=48
Both couple sit together 3!*2*2=24
P(A,B)=P(A) + P(B) - P(A B)
Total arrangements when atleast 1 couple sit together 2*48-24=72

Probability of atleast 1 couple sit together=72/120.
Probability of no couple sit together 1-72/120=2/5

Whats the OA? btw gr8 question

yes, I support this. I also derived I and II

Total number of arrangements possible : 5!=120

Atleast 1st couple sit together arrangement 4!*2=48
Atleast 2nd couple sit together arrangement 4!*2=48
Both couple sit together 3!*2*2=24
P(A,B)=P(A) + P(B) - P(A B)
Total arrangements when atleast 1 couple sit together 2*48-24=72

Probability of atleast 1 couple sit together=72/120.
Probability of no couple sit together 1-72/120=2/5

Whats the OA? btw gr8 question


Yes Nitin 2/5 is right ans.

Out of 7 models, all of diff height, 5 will be chosen to pose for a photograph. if the 5 models are to stand in a row from shortest to tallest, and the fourth-tallest and sixth-tallest models can't be adjacent, how many diff arrangements of 5 models are possible?

I got following question on GMATPrep :- Pls give attention to word COULD

If x is +ive which of the following could be the could be correct ordering of 1/x, 2x, and x^2

I X^2
II X^2
III 2x
A) NONE B) I C) III D) I & II E) I II & III

I got following question on GMATPrep :- Pls give attention to word COULD

If x is +ive which of the following could be the could be correct ordering of 1/x, 2x, and x^2

I X^2
II X^2
III 2x
A) NONE B) I C) III D) I & II E) I II & III



considering x as +ve,I will explain 2 cases here.

Case 1 :x > 1 .For e.g x=2,3,4 etc
Case 2 : 0
For case 1,none of the option matches.
For case 2, I'm finding option 1 is matching

So, option B will be my take for this Q.

Pls post the OA