GMAT Problem Solving Discussions

MBA
1985
A----------B---------C----------D


Is
CD > BC ?
(1) AD = 20
(2) AB = CD


A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D EACH Statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.

Answer is E as explained by most other pguys.
1986
Answer is E as explained by most other pguys.


yes ans is E


st 1 nothing say @ CD & BC distance
st 2 nothing say @ BC distance

both combine
AB=20
AB=CD

AB=BC=CD=6.33 possible
and in another case CD>BC,CD
1987
Hi,
A toughie here from GmatPrep.
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosena t random,what is the probablity that their product will be of the form x^2-(by)^2,where b is an integer.
A.1/2
B.1/3
C.1/4
D.1/5
E.1/6

Please discs your method of solving.WIll disclose OA later.Thanks


product can be taken in 6 ways.the said result comes,when,(x+y)(x-y) is done.

so,1/6 is the answer.
option(E)
1988
Hi,
A toughie here from GmatPrep.
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosena t random,what is the probablity that their product will be of the form x^2-(by)^2,where b is an integer.
A.1/2
B.1/3
C.1/4
D.1/5
E.1/6

Please discs your method of solving.WIll disclose OA later.Thanks



Here is another solution :
total products are :6
Possible results for x^2-(by)^2 can be :

x+y*x-y
x+5y*x-y
x+5y*5x-y = 5x^2 - 5y^2 = x^2 - (1y)^2 since b is an integer and 1 is integer as well.

Hence, why isn't the answer 1/2.
1989
Here is another solution :
total products are :6
Possible results for x^2-(by)^2 can be :

x+y*x-y
x+5y*x-y
x+5y*5x-y = 5x^2 - 5y^2 = x^2 - (1y)^2 since b is an integer and 1 is integer as well.

Hence, why isn't the answer 1/2.


(x+5y)*(5x-y)
5x^2+24xy-5y^2
it is not of the form:x^2-(by)^2
so,not considered.

hope that helps.
1990
shud be E....both are insufficient and in combination also they tell u nothing....
ab=cd=5 bc=10 cdbc yes

no specific answer.

The answer is E
Taking stem 1: AB=20 => nothing can be concluded about CD and BC
Stem 2: AB=CD => nothng is known about BC
Combining, BC = 20-2x, x = AB=CD
if x = 6, BC= 8, BC>CD
if x = 7, BC=6, Bc
1991
Hi,
Can someone help me understand this problem that I faced on Gmatprep?
A certain candy manufcaturer decreases the weight of candy M by 20% but leaves the price unchanged.What is the resulting increase in price per ounce of the candy M?
A.5%
B.10
C.15
D.20
D.25

OA is 25.Thanks in advance.

let wt of candy be x ounce and price be C
cost/ounce of candy = c/x

new wt of candy = 0.8x
price/ounce = C/0.8x
increase in price/ounce = (C/x(1/0.8-1))/C/x
25%
hence E

btw suruchi, pl do not post the OAs along with questions

thanks
1992
suruchee Says
This does not seem to be the OA!!

my pick is B

x^2 -(by)^2 = (x+by)(x-by)
now x+by can be picked in 2 ways while x-by can be picked in only 1 way
hence prob = 2*1/4C2 = 2/6 = 1/3

wats the OA?
thanks
1993
my pick is B

x^2 -(by)^2 = (x+by)(x-by)
now x+by can be picked in 2 ways while x-by can be picked in only 1 way
hence prob = 2*1/4C2 = 2/6 = 1/3

wats the OA?
thanks

sory for the above post!!
i checked again n there is only 1 way to choose to arrive at (x+by)(x-by)

thanks
1994
adiben Says
I think it should be C!

my pick is also C
but wats the OA?
1995
Hi TOny

The OA is correct, but the explanation should be as follows:-

If you read the question, among 40 students in the class, we are given the following info:-
1) 9/20 are boys, ie, 9/20*40= 18 of the students are boys
2) 4/5 are right handed, ie 4/5 * 40 = 32 students are right handed.

The important difference is in information snippet 2). What seems to be indicated is 4/5 of the students and not 4/5 of the boys are right handed.

So based on 1 and 2, we can find a best case and worst case as follows:-
Best case: Boys are a subset of the right handed students.
In other words, all 18 boys are among the 32 right handed students.
So in this scenario, we can have 18 right handed boys.

Worst case: Assume all the left handed students(40-32=8 ) are boys. So in this case,
the number of right handed boys = (Total boys)- (left handed boys)
= 18-8
=10 right handed boys

Hence we can see that the number of right handed boys can be anywhere between 10 to 18 (Answer choice (c)). The exact number cannot be determined based on the information given in the question, but the answer choices are adequate to enable a unique correct choice.

a tricky question indeed!!
1996
suruchee Says
Yes!!Thanks so much.


please don't post for saying only thanx

there's a button below the post to thanx
1997

Sorry duplicate post

1998

This is continuation from thread

1)How many natural numbers below 100 will remainder 2 when divided by 6?

a) 15 b)16 c)17

d)18 E)19

Answer given

1. 100 / 6 = 16 factors of 6 less than 100. Add number 2 to this count, as 2/6 will also give 2 as reminder. So answer should be 17.

Add 2 to the count???I want to know how 17 was arrived at

1999

Not sure of the above method. I also cannot understand the above method.
I have my own way of solving this question, may be it helps.
==============================================
According to the question, I can frame an inequality as follows:

6x + 2
Solving the above inequality leads us to
x
Since x has to be a non-negative integer, x can be any integer from 0 to 16.

So by substituting different value for x in 6x + 2 will give you different numbers less than 100.

x=0, 6x + 2 = 2
x=1, 6x + 2 = 8
x=2, 6x + 2 = 14
x=3, 6x + 2 = 20
x=4, 6x + 2 = 26
......
......
x=15, 6x + 2 = 92
x=16, 6x + 2 = 98

So in total there are 17 numbers.

2000
This is continuation from thread

1)How many natural numbers below 100 will remainder 2 when divided by 6?

a) 15 b)16 c)17

d)18 E)19

Answer given

1. 100 / 6 = 16 factors of 6 less than 100. Add number 2 to this count, as 2/6 will also give 2 as reminder. So answer should be 17.

Add 2 to the count???I want to know how 17 was arrived at



there are 16 multiples of 6 less than 100. now on adding2 to all we get 16 numbers which are leaving a remainder 2 when divided by 6 but there you have left 2 also which is less than 100 and leaves a remainder 2 when divided by 6.


so 17 is the answer.
2001
there are 16 multiples of 6 less than 100. now on adding2 to all we get 16 numbers which are leaving a remainder 2 when divided by 6 but there you have left 2 also which is less than 100 and leaves a remainder 2 when divided by 6.


so 17 is the answer.

but there you have left 2 also which is less than 100 and leaves a remainder 2 when divided by 6.

Explain this to me a with some more detail please I am confused.
2002
This is continuation from thread

1)How many natural numbers below 100 will remainder 2 when divided by 6?

a) 15 b)16 c)17

d)18 E)19

Answer given

1. 100 / 6 = 16 factors of 6 less than 100. Add number 2 to this count, as 2/6 will also give 2 as reminder. So answer should be 17.

Add 2 to the count???I want to know how 17 was arrived at

the numbers are from the set {2,8,14....6n+2} where,n is a whole number.

maximum possible value below 100 is 98.
so,n=16.
as we are considering whole numbers,0 is also included as n,hence.total numbers are 17.

option (C)

hope that helps.
2003
but there you have left 2 also which is less than 100 and leaves a remainder 2 when divided by 6.

Explain this to me a with some more detail please I am confused.



the smallest number which when divided by any other number N and leaves a remainder R is the number R itself.

suppose we have to find the smallest number which on dividing by 7 leaves a remainder 4. the answer would be 4 not 11.

hope this helps you.

if you didn't get the concept then tell me i will further explain it.
2004
the numbers are from the set {2,8,14....6n+2} where,n is a whole number.

maximum possible value below 100 is 98.
so,n=16.
as we are considering whole numbers,0 is also included as n,hence.total numbers are 17.

option (C)

hope that helps.

yes the OA is C....

the trap here was whether to include 2 or not...