GMAT Problem Solving Discussions

MBA
3641
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


______________________

Option B .
3642

Option D

try values 0.5 and 0.8 and both the options satisfy.

3643
IIM_Pune Says
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?




Let the models be 1, 2, 3, 4, 5, 6, & 7 in the order of their height
Suppose the arrangement is _, 4, 5, 6 & 7....... there are three candidates to take the first spot.
Suppose the arrangement is 1, _, 5, 6 & 7....... there are three candidates to take the second spot.
---
----
Suppose the arrangement is 1, 2, 3, 4 & _....... there are three candidates to take the fifth spot.
Hence total 15 possible arrangements.
Now the arrangements where 4 and 6 are together.
1, 2, 3, 4, 6
1, 2, 4, 6, 7
1, 3, 4, 6, 7
2, 3, 4, 6, 7
So the total possible arrangements = 15 - 4 = 11.
Please post the OA.
3644
Let the models be 1, 2, 3, 4, 5, 6, & 7 in the order of their height
Suppose the arrangement is _, 4, 5, 6 & 7....... there are three candidates to take the first spot.
Suppose the arrangement is 1, _, 5, 6 & 7....... there are three candidates to take the second spot.
---
----
Suppose the arrangement is 1, 2, 3, 4 & _....... there are three candidates to take the fifth spot.
Hence total 15 possible arrangements.
Now the arrangements where 4 and 6 are together.
1, 2, 3, 4, 6
1, 2, 4, 6, 7
1, 3, 4, 6, 7
2, 3, 4, 6, 7
So the total possible arrangements = 15 4 = 11.
Please post the OA.



My original approach is incorrect. After writing down the possible arrangements I am coming up with 20 possible arragements. Deducting 4 arrangements where 4 is adjacent to 6, it leaves 16 as the answer. There should be a straight forward approach to this problem
3645

Im getting 17 different ways.....??... is there any catch....:lookround:

3646

I second you Deepakaraam... IMO option 1.

3647

hi

If n N then n^(13)-n is divisible by

a.13
b.14
c.15
d.none

3648
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



my ans would be A) NONE.

AS if x=1.0 then X^2 = 1/X if x=1.1 then 1/X if x=0.5 then X^2
3649

If n N then n^(13)-n is divisible by

for (A^P-A) where a is any natural number and P is any prime number. the form (A^P-A) is always divisible by P

hence, answer is 13


3650
If n N then n^(13)-n is divisible by

for (A^P-A) where a is any natural number and P is any prime number. the form (A^P-A) is always divisible by P

hence, answer is 13



Hello

Agreed

But by the way if i forgot for a second (A^P-A) as per the conditions is divisible by P and by using hit and trial method and then by using induction n^(13)-n is also divisible by 15. CHECK THIS OUT YOURSELF.
Now which answer would mark????????????
3651
______________________

Option B .

I didnt login PG for past few days ..
OA for above Q is D - i.e I & II both could be true but not III
3652
apex1umt Says
Im getting 17 different ways.....??... is there any catch....:lookround:


OA for .... "Model Arrangement for photograph" is - 17
3653
Q. T is a set of y integers, where 0
(a) 0 (b) x (c) -x (d) y/3 (e) 2y/7

Q. a, b and c are integers and a


answer for 1ts one is E. give the OA if correct will post my explanation.

Can we have options for que 2.
3654
Q. T is a set of y integers, where 0
(a) 0 (b) x (c) -x (d) y/3 (e) 2y/7

Q. a, b and c are integers and a


My Approach -

Set with only one integer is possible , thus in a set that just has 0 or X as an element median will be 0 and x resp.

Consider (c) ie -x here avg is x say 3 and median -3 consider set has 3 elements Thus -3 is middle one & total of them should be 9

possible combination -4, -3, 16 Thus c is possible

for ( d ) take no of elements as either 3 or 6 let us take 6 thus median 2

-5,-4 ,1 ,3 ,4 , 5 possible

Since a b c d are possible e must be ans
3655
Q. T is a set of y integers, where 0
(a) 0 (b) x (c) -x (d) y/3 (e) 2y/7

Q. a, b and c are integers and a

SOLN TO 2nd problem -

Note here that entire set has all consecuative integers

Let us begin with c=16 , median of later set is 14

therefore these conditions represent later set as

12, 13,14, 15, 16 This automatically makes b = 12

thus median of former set 9

Thus former set

6, 7, 8, 9, 10, 11, 12

This makes set r = 6 7 8 9 10 11 12 13 14 15 16

Median 11

Ans 11/16

PLS CONFIRM ANS ...
3656
SOLN TO 2nd problem -
Note here that entire set has all consecuative integers
Let us begin with c=16 , median of later set is 14
therefore these conditions represent later set as
12, 13,14, 15, 16 This automatically makes b = 12
thus median of former set 9
Thus former set
6, 7, 8, 9, 10, 11, 12
This makes set r = 6 7 8 9 10 11 12 13 14 15 16
Median 11
Ans 11/16
PLS CONFIRM ANS ...



Hi IIM_Pune
can you plz explain this line?
Note here that entire set has all consecuative integers . how can we be sure abt this one?
3657

If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c
(A) yj (B) yj > 0
(C) yj = 0
(D) j = 0
(E) y = 0

3658
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c
(A) yj (B) yj > 0
(C) yj = 0
(D) j = 0
(E) y = 0

yj=0 should be the answer.
Reasoning : According to equality 1 y=0 or u-c =0
equality 2 : j=0 or u-k=0

since c and k are not equal so both u-k and u-c can't be zero.

So either x or y must be zero, but xy must be zero.
3659
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c
(A) yj (B) yj > 0
(C) yj = 0
(D) j = 0
(E) y = 0


Ans (C) yj = 0 AS

j [ y ( u - c ) ] = 0 -------(1)
y [ j ( u - k ) ] = 0 -------(2)
yj ( u - c - u + k ) = 0 by (1) - (2)
yj ( k - c ) = 0
now c
3660
Hi IIM_Pune
can you plz explain this line?
Note here that entire set has all consecuative integers . how can we be sure abt this one?


Hi Avanika,

Q tells us that -

S is the set of all integers from a to b, inclusive
Q is the set of all integers from b to c inclusive
R is the set of all integers from a to c inclusive

Give attention to word ALL ...

Many times I do this mistake of not reading question Q carefully.

And even though you can solve the q accurately.. u either cant reach answer, feeling that inadequate info is given or even if u get ans thats wrong...