GMAT Problem Solving Discussions

Let the two machine rates are x and y and let it takes machine y T days to produce W.
framing equations-
Ty=(T+2)x=W=>w=2xy/(x-y)------1
from second condition-3(x+y)=5/4W......2
solving the above two equations-we get
6x^2-5xy-6y^2=0,solving will give 2x=3y

substitute the value of y in eq-2,
x=w/4

hence if it takes t days to make 2w,tx=2w=t*w/4=2w=>t=8 days


T=4 and X can produce w widgets for 6 days.So x can produce 2w widgets in 12 days.
A certain characteristic in a large population has a distribution that is symmetric about the mean
m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%

A certain characteristic in a large population has a distribution that is symmetric about the mean
m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?

A. 16%
B. 32%
C. 48%
D. 84%
E. 92%





For SD, on GMAT we can assume the 34-14-2 probability distribution percentages

So, straigtaway we can say 84 % data is less than m+d

Else, lets takeit this way :
symmetric implies 50 % is above mean and other 50 % is below mean and 68 % within one deviation mean half of the values are above and below mean i.e 34% on each side

So, 50 % below mean and 34 % between mean and m+d, hence 84% below m+d ..Ans D
For SD, on GMAT we can assume the 34-14-2 probability distribution percentages

So, straigtaway we can say 84 % data is less than m+d

Else, lets takeit this way :
symmetric implies 50 % is above mean and other 50 % is below mean and 68 % within one deviation mean half of the values are above and below mean i.e 34% on each side

So, 50 % below mean and 34 % between mean and m+d, hence 84% below m+d ..Ans D


thanks for the expln man.. I was somehow getting 50%.. now I knw why.. I forgot to consider m+d.. πŸ˜ƒ
For SD, on GMAT we can assume the 34-14-2 probability distribution percentages

So, straigtaway we can say 84 % data is less than m+d

Else, lets takeit this way :
symmetric implies 50 % is above mean and other 50 % is below mean and 68 % within one deviation mean half of the values are above and below mean i.e 34% on each side

So, 50 % below mean and 34 % between mean and m+d, hence 84% below m+d ..Ans D


one doubt-34-14-2 means the %ages after mean,
so it should not be-2*(34+14)=96% and since 92 is less then that, opt for 92.
I am still trying to get hand on the sigma and found urs previous post useful.
one doubt-34-14-2 means the %ages after mean,
so it should not be-2*(34+14)=96% and since 92 is less then that, opt for 92.
I am still trying to get hand on the sigma and found urs previous post useful.


34-14-2 means resp distribution within 1st, 2nd and 3rd deviation on either side of mean ...

Hence, 68 % is within one deviation ( 2*34)
Also, 96 % is within 2 deviations of mean
And 100 % within 3 deviations

Now, 96 % also includes data between m+d and m+2d, something that we dont want to be accounted ...

Ok..to make it a very simple trick ...draw a symmetric bell curve and divide it into 3 parts on each side of centre and allocate % distribution ...It would be easily to visualise that we need the entire left part about the symmetric axis (i.e mean ) and 1st portion of right !

Hence, 50 +34 = 84 !!

Q- K is a three digit number such that the ratio of the number to the sum of its digits is least.

i) What is the difference between the hundreds and the tens digit of K?
ii) What can be said about the difference between the tens and the units digit?
iii) For how many values of K will the ratio be highest?

Q- K is a three digit number such that the ratio of the number to the sum of its digits is least.

i) What is the difference between the hundreds and the tens digit of K?
ii) What can be said about the difference between the tens and the units digit?
iii) For how many values of K will the ratio be highest?


lets say xyz is the number
100x+10y+z/x+y+z is the least for the ratio-11
and the numbers corresponding to this is-
198
1.8
2.7
3.dont understand?asking for the highest value of K
Q- K is a three digit number such that the ratio of the number to the sum of its digits is least.

i) What is the difference between the hundreds and the tens digit of K?
ii) What can be said about the difference between the tens and the units digit?
iii) For how many values of K will the ratio be highest?


Well well ...very interesting question !!
Not really sure if it can be expected on GMAT ..

Let the digits be x,y,z resp .
Hence, no is 100x+10y+z
and sum of digits is x+y+z

ratio = (100x+10y+z)/(x+y+z)

Not really sure what is the quickest way mathematically to calculate the min ratio, but logically ratio is least for lower numerator and higher denominator ...

So ratio would be least when no is in 100's ...for 200 to 999 ratio keeps increasing coz sum of digits cannot increase prop.

SO no should be 199 ..

1) Hence, diff between hundreds and tens is 8
2) diff between tens and units is 0
3) ratio would be highest when sum of digits is least ..so let y=z=0
Then ratio becomes 100x/x= 100
And x can be any no from 1 to 9 ...
Hence for 9 nos ratio is max (100, 200 ,.......900)

Hence , Ans are :
1) 8
2) 0
3) 9

Interesting question. And I would say it could appear in GMAT, if you are scoring in 750 range. People appearing in GMAT recently have reported that the GMAT is throwing difficult PS/DS questions now a days.

Anybody has any collection of tough PS/DS questions to go through ? Would really appreciate.

Q- K is a three digit number such that the ratio of the number to the sum of its digits is least.

i) What is the difference between the hundreds and the tens digit of K?
ii) What can be said about the difference between the tens and the units digit?
iii) For how many values of K will the ratio be highest?


Numerator to be minimum & denominator to be maximum..

Denominator to be maximum -> Sum of all the 3 digits shall be maximum

-> Maximum sum could be : 9
-> Numerator could be: 108

Ans 1: 1.
Ans 2: 8.
Ans 3:need to know all possible combination where the sum of the denominator is : 1

I guess only 1: 991 / 9 + 9 + 1 = 991

For any integer n greater than 0, n! denotes the product of all the integers from 1 to n, inclusive. How many multiples of 3 are there between 6!-6 and 6!+6, inclusive ?

A) One
B) Two
C) Three
D) Four
E) Five

For any integer n greater than 0, n! denotes the product of all the integers from 1 to n, inclusive. How many multiples of 3 are there between 6!-6 and 6!+6, inclusive ?

A) One
B) Two
C) Three
D) Four
E) Five


There are five multiples E.

The five are (6!-6)/3, (6!-3)/3,6!/3, (6!+3)/3,(6!+6)/3.

Whats the OA?
For any integer n greater than 0, n! denotes the product of all the integers from 1 to n, inclusive. How many multiples of 3 are there between 6!-6 and 6!+6, inclusive ?

A) One
B) Two
C) Three
D) Four
E) Five

Ans E ..

There are 5 multiples of 3 :
6!-6
6!-3
6!
6!+3
6!+6

@Deepak: Please disclose the OA of previous sum ...
For any integer n greater than 0, n! denotes the product of all the integers from 1 to n, inclusive. How many multiples of 3 are there between 6!-6 and 6!+6, inclusive ?

A) One
B) Two
C) Three
D) Four
E) Five


IMO it must be option E

range is from (X + 6) to (X - 6).. that means 4 factors..
plus
X = 6! = 720.. is a mulitple of 3.. hence 1 count..

thus total count = 4 + 1 = 5

whats the OA πŸ˜‰

OA for "For any integer 'n' greater than 0, n! d..........." is option E)Five.

OA for the 3-digit sum is

1) 8
2) 0
3) 9

Well well ...very interesting question !!
Not really sure if it can be expected on GMAT ..

Let the digits be x,y,z resp .
Hence, no is 100x+10y+z
and sum of digits is x+y+z

ratio = (100x+10y+z)/(x+y+z)

Not really sure what is the quickest way mathematically to calculate the min ratio, but logically ratio is least for lower numerator and higher denominator ...

So ratio would be least when no is in 100's ...for 200 to 999 ratio keeps increasing coz sum of digits cannot increase prop.

SO no should be 199 ..

1) Hence, diff between hundreds and tens is 8
2) diff between tens and units is 0
3) ratio would be highest when sum of digits is least ..so let y=z=0
Then ratio becomes 100x/x= 100
And x can be any no from 1 to 9 ...
Hence for 9 nos ratio is max (100, 200 ,.......900)

Hence , Ans are :
1) 8
2) 0
3) 9



Bhavin: Just one question, when they say SUM : 1 + 9 + 9 = 10 + 9 = 19 = 1+9 = 1?

hence 199 / 1 = 199

IS this not correct, or i summed up too much?
Bhavin: Just one question, when they say SUM : 1 + 9 + 9 = 10 + 9 = 19 = 1+9 = 1?

hence 199 / 1 = 199

IS this not correct, or i summed up too much?


hey amsey ...yes u have summed up too much ...
they need min ratio of no to sum of digits

So ratio is 199/ (1+9+9) = 199 /19 = 10.xx which is the least for any 3 digit no ...

Major task is to arrive at the no 199 ...
Now, ideally max sum of digits is 9+9+9 = 27 ...

However when no increases from 1XX to 9XX, no increases by 500 to 900 % ...sum of digits cannot increase by same prop ...

So. no had to be 1XX ...

And now, for any ratio a/b greater than 1 , (a+x)/(b+x) Hence, u max for 1XX series, so no had to be 199

Can someone post some int prblms?