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Joanna bought only $0.15 stamps and $0.29 stamps. How many

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Joanna bought only $0.15 stamps and $0.29 stamps. How many

by mitzwillrockgmat » Sun May 23, 2010 10:22 am
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.

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 not C, as one would quickly jump to. it is a in fact. can someone please explain how??

.15x + .29y =4.4 => 15x + 29y = 440 i find whole no.s easier to work with

pls explain y i can only place 10 as x or y and how i can reach this conclusion in less than a minute?? thanks!
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by clock60 » Sun May 23, 2010 10:41 am
i don`t know what is optional way. at least how i think
15x+29y=440
15x=440-29y
x=(440-29y)/15
as 440-29y must be divisible by 15 it must be divisible by 5
so it makes sence to select the values of y so as multiplying with 9 they got 5 or 0 as last digit

for example if y=3 then 9*3=27. and 440-..7=..3, not divisible by 5 so by 15
y=5,10,15 only these value makes sence
but here i don`t see ways other than cheking
29*15=435, 440-435=5 not divisible by 15
29*10=290, 440-290=150 divisible by 15
29*5=145, 440-145=295 not divisible
so the only value is 10
hope it makes sence
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by Patrick_GMATFix » Sun May 23, 2010 12:30 pm
To simplify the math, let's deal in cents rather than dollars. The total she spent on stamps is 15x + 29y.

The first statement tells us that this total is 440, so we can write 15x + 29y = 440.

Typically, an equation like this cannot be solved. However, you will have sufficient data to solve under these conditions:
(1) unknowns are positive integers
(2) total (440) is less than the LCM of the coefficients (15 & 29) + the smaller coefficient.

That's it. Because 29 is prime, the LCM of 29 and 15 is 29*15. Since the total (440) is less than 29*15+15, there is only one possible solution. Statement (1) is sufficient.

If you own The Official Guide Companion, have a look at the comprehensive Take-Aways list found in the Reference chapter; it will highlight this as well as other little known intricacies of GMAT math. In addition you have access to the Solutions Engine, search for questions of topic="Algebraic Translations & Manipulations" and difficulty="700+" and you will see a few of these types of problems that most people have a hard time with.

Hope this helps.

-Patrick
Last edited by Patrick_GMATFix on Sat Jul 17, 2010 10:11 am, edited 1 time in total.
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by mitzwillrockgmat » Sun May 23, 2010 11:27 pm
Patrick_GMATFix wrote:To simplify the math, let's deal in cents rather than dollars. The total she spent on stamps is 15x + 29y.

The first statement tells us that this total is 440, so we can write 15x + 29y = 440.

Typically, an equation like this cannot be solved. However, you will have sufficient data to solve under these conditions:
(1) unknowns are integers
(2) total (440) is less than the LCM of the coefficients (15 & 29) + the smaller coefficient.

That's it. Because 29 is prime, the LCM of 29 and 15 is 29*15. Since the total (440) is less than 29*15+15, there is only one possible solution. Statment (1) is sufficient.

If you own The Official Guide Companion, have a look at the comprehensive Take-Aways list found in the Reference chapter; it will highlight this as well as other little known intricacies of GMAT math. In addition you have access to the Solutions Engine, search for questions of topic="Algebraic Translations & Manipulations" and difficulty="700+" and you will see a few of these types of problems that most people have a hard time with.

Hope this helps.

-Patrick
Because 29 is prime, the LCM of 29 and 15 is 29*15. Since the total (440) is less than 29*15+15, there is only one possible solution. Statment (1) is sufficient.

is the above a general statement that holds true in most cases ? can i apply this rule whenever a prime appears in a linear equation???? can you give me some examples pls? thanks! :)
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by Patrick_GMATFix » Sun May 23, 2010 11:40 pm
The rule isn't about the coefficient being prime, but the 2 conditions that I listed higher in my response. I also indicated how you could locate similar questions, but consider this example:

In a workout rooms there are bars that weigh 15 pounds each and bars that weigh 33 pounds each. How many bars are there in total?

(1) The total weight of all bars is 159 pounds

This statement would yield to the equation 15x + 33y = 159. The equation meets both of the conditions I set at the top: #1. x & y are integers (# of bars) and #2. 159 is less than (the LCM of 15 and 33) + 15 --> 159 is less than 165+15.

As a result, we have enough info to solve. In DS, don't spend the time to solve. In this case no matter how long you look, the only solution you will find is x=4, y=3

As I mentioned before: "If you own The Official Guide Companion, have a look at the comprehensive Take-Aways list found in the Reference chapter; it will highlight this as well as other little known intricacies of GMAT math. In addition you have access to the Solutions Engine, search for questions of topic="Algebraic Translations & Manipulations" and difficulty="700+" and you will see a few of these types of problems that most people have a hard time with. "

Best of luck
-Patrick
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by mitzwillrockgmat » Mon May 24, 2010 12:12 am
Patrick_GMATFix wrote:The rule isn't about the coefficient being prime, but the 2 conditions that I listed higher in my response. I also indicated how you could locate similar questions, but consider this example:

In a workout rooms there are bars that weigh 15 pounds each and bars that weigh 33 pounds each. How many bars are there in total?

(1) The total weight of all bars is 159 pounds

This statement would yield to the equation 15x + 33y = 159. The equation meets both of the conditions I set at the top: #1. x & y are integers (# of bars) and #2. 159 is less than (the LCM of 15 and 33) + 15 --> 159 is less than 165+15.

As a result, we have enough info to solve. In DS, don't spend the time to solve. In this case no matter how long you look, the only solution you will find is x=4, y=3

As I mentioned before: "If you own The Official Guide Companion, have a look at the comprehensive Take-Aways list found in the Reference chapter; it will highlight this as well as other little known intricacies of GMAT math. In addition you have access to the Solutions Engine, search for questions of topic="Algebraic Translations & Manipulations" and difficulty="700+" and you will see a few of these types of problems that most people have a hard time with. "

Best of luck
-Patrick
excellent got it! this is a very useful tip! thanks sooo much!!
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by Patrick_GMATFix » Mon May 24, 2010 12:17 am
No prob :-)
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by cmugeria » Mon May 31, 2010 9:21 am
Hi Patrick_GMATFix
thanks for the explanation - where in the gmat companion is this shown? Is it page 255 and onward
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by taneja.niks » Thu Jul 15, 2010 12:46 am
Patrick what ur conditions state are good for me to find the answer as I dont own a OG companion but I have a question Normal mathematics rules say tht when u have 2 variables (x,y) u need to equations to solve them whereas here u have only one???

What abt tht ?????
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by Patrick_GMATFix » Thu Jul 15, 2010 4:42 am
Hi taneja.niks,

What you called the "normal mathematics rules" assume that we don't know anything about the values of our x and y. Whenever that is the case, we do need 2 equations. That has not changed.

In the situations I described above we know that x and y are positive integers; this severely limits possibilities.

Consider this very simple example. Can we find x and y from 3x + 5y = 8? Well if we have no idea about the values of x and y, then we cannot because we need a 2nd equation. However if we know that x and y represent a number of people (they are positive integers), then there is only 1 solution. (x,y)=(1,1). You cannot plugin any other positive integers that satisfy that equation. IN fact the following conditions are met:

(1) unknowns are positive integers
(2) total (8) is less than the LCM of the coefficients (3 & 5) + the smaller coefficient.


Thus you can be certain that we have sufficiency.
-Patrick
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by taneja.niks » Sat Jul 17, 2010 9:47 am
ok... patrick lets see this question please


3x+4y = 11, Then what is the value of y???

1. X and Y are non zero Integers
2. y=-2x

What would u say say abt the first statement here????


Please help with the same

Thanks.....
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by Patrick_GMATFix » Sat Jul 17, 2010 10:10 am
As I mentioned in the last post, the variables should be positive integers. In the question you posted, all we knwo is that X & Y are non-zero. They could be negative.

The first statement is not sufficient we could have (x,y)=(1,2) but we could also have (x,y)=(5,-1). Thus in Statement 1 we don't have sufficient data to find the value of y

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by gmat_chanakya » Fri Sep 05, 2014 6:30 am
would the above method work for ax-by= c type of equation ?
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by Brent@GMATPrepNow » Fri Sep 05, 2014 6:34 am
gmat_chanakya wrote:would the above method work for ax-by= c type of equation ?
We can write 15x + 29y = 440, but this would typically have an infinite number of solutions. HOWEVER, since x and y must be POSITIVE INTEGERS, there is only one possible solution.

Cheers,
Brent
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by Brent@GMATPrepNow » Fri Sep 05, 2014 6:37 am
NOTE: The whole 1 equation 2 variables thing is a common GMAT trap (along with other common traps). For more information about this and other traps you can watch the following free videos:
https://www.gmatprepnow.com/module/gmat- ... cy?id=1105
https://www.gmatprepnow.com/module/gmat- ... cy?id=1106
https://www.gmatprepnow.com/module/gmat- ... cy?id=1107

Cheers,
Brent
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