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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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16 Aug 2015, 02:21
we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ?



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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16 Aug 2015, 05:17
karthickhari wrote: we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ? Be careful to use information from the other statement when you are evaluating each statement along in a DS question. You are correct that statement 1 is sufficient but 2 in itself (and without using the information from statement 1!) is NOT sufficient. Statement 2 says that umber of .15$ tickets = number of .29 tickets. Now this 'equal' number can be anything from 1 to a million even. Thus we do not get 1 unique value with statement 2 alone. The first step in a DS question is to evaluate the information provided in the 2 statements individually and ONLY when you do not get a sufficient answer individually with either of the 2 statements, should you combine the statements. Hope this helps.



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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17 Sep 2015, 13:19
Bunuel wrote: (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Why is that so clear that only one integer combination fits this? Very difficult to spot...
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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17 Sep 2015, 13:35
reto wrote: Bunuel wrote: (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Why is that so clear that only one integer combination fits this? Very difficult to spot... This is quite representative of a GMAT like question and thus for such questions wherein you are asked number of tickets, number of people, number of toys etc wherein only integer values can work, make sure to try to find a few sets of values for both the variables that will satisfy the given equation which in this case is 15x+29y=440. Once you create the equation above, you can see that you could also write it as 15x=44029y which means that 44029y MUST be a multiple of 15 (as the other side is 15x). Thus once you start by recognizing this fact, you will see that only y=10 satisfies this. For all other values you will not get an integer value of x or get a value <0 (this is not acceptable as number of tickets can not be <0). Hope this helps.



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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28 Sep 2015, 11:12
Bunuel wrote: udaymathapati wrote: 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. Let \(x\) be the # of $0.15 stamps and \(y\) the # of $0.29 stamps. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\) (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Sufficient. (2) She bought an equal number of $0.15 stamps and $0.29 stamps > \(x=y\). Not sufficient. Answer: A. So when we have equation of a type \(ax+by=c\) and we know that \(x\) and \(y\) are nonnegative integers, there can be multiple solutions possible for \(x\) and \(y\) (eg \(5x+6y=60\)) OR just one combination (eg \(15x+29y=440\)). Hence in some cases \(ax+by=c\) is NOT sufficient and in some cases it is sufficient. For more on this type of questions check: eunicesoldseveralcakesifeachcakesoldforeither109602.htmlmarthaboughtseveralpencilsifeachpencilwaseithera100204.htmlarentalcaragencypurchasesfleetvehiclesintwosizesa105682.htmljoeboughtonlytwentycentstampsandthirtycentstamps106212.htmlacertainfruitstandsoldapplesfor070eachandbananas101966.htmljoannaboughtonly015stampsand029stampshowmany101743.htmlatanamusementparktomboughtanumberofredtokensand126814.htmlcollectionsconfusedneedahelp81062.htmlHope it helps. Bunuel, How did you arrive at the 1010 combination in statement 1? Trial and error?



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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28 Sep 2015, 11:15
manhattan187 wrote: Bunuel wrote: udaymathapati wrote: 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. Let \(x\) be the # of $0.15 stamps and \(y\) the # of $0.29 stamps. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\) (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Sufficient. (2) She bought an equal number of $0.15 stamps and $0.29 stamps > \(x=y\). Not sufficient. Answer: A. So when we have equation of a type \(ax+by=c\) and we know that \(x\) and \(y\) are nonnegative integers, there can be multiple solutions possible for \(x\) and \(y\) (eg \(5x+6y=60\)) OR just one combination (eg \(15x+29y=440\)). Hence in some cases \(ax+by=c\) is NOT sufficient and in some cases it is sufficient. For more on this type of questions check: eunicesoldseveralcakesifeachcakesoldforeither109602.htmlmarthaboughtseveralpencilsifeachpencilwaseithera100204.htmlarentalcaragencypurchasesfleetvehiclesintwosizesa105682.htmljoeboughtonlytwentycentstampsandthirtycentstamps106212.htmlacertainfruitstandsoldapplesfor070eachandbananas101966.htmljoannaboughtonly015stampsand029stampshowmany101743.htmlatanamusementparktomboughtanumberofredtokensand126814.htmlcollectionsconfusedneedahelp81062.htmlHope it helps. Bunuel, How did you arrive at the 1010 combination in statement 1? Trial and error? 1 trick to remember in such questions or similar questions to number of fruits, number of kids, number of pencils etc, the actual numbers can only be integers. You can not have fractional pencils or stamps. Thus, based on the above question, \(15x+29y=440\) only has 1 pair of integer solution, (10,10). This is by trial and error based on the fact that both the values need to be integers.



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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01 Oct 2015, 10:56
Bunuel wrote: udaymathapati wrote: 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. Let \(x\) be the # of $0.15 stamps and \(y\) the # of $0.29 stamps. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\) (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Sufficient. (2) She bought an equal number of $0.15 stamps and $0.29 stamps > \(x=y\). Not sufficient. Answer: A. So when we have equation of a type \(ax+by=c\) and we know that \(x\) and \(y\) are nonnegative integers, there can be multiple solutions possible for \(x\) and \(y\) (eg \(5x+6y=60\)) OR just one combination (eg \(15x+29y=440\)). Hence in some cases \(ax+by=c\) is NOT sufficient and in some cases it is sufficient. For more on this type of questions check: eunicesoldseveralcakesifeachcakesoldforeither109602.htmlmarthaboughtseveralpencilsifeachpencilwaseithera100204.htmlarentalcaragencypurchasesfleetvehiclesintwosizesa105682.htmljoeboughtonlytwentycentstampsandthirtycentstamps106212.htmlacertainfruitstandsoldapplesfor070eachandbananas101966.htmljoannaboughtonly015stampsand029stampshowmany101743.htmlatanamusementparktomboughtanumberofredtokensand126814.htmlcollectionsconfusedneedahelp81062.htmlHope it helps. Hi Bunuel, Could you tell me how can we determine in which case ax+by=c is sufficient? We cannot test all the cases in 2 minutes In this case, we can easily see that x = y = 10 is a solution, but it seems that nothing can guarantee that there are no other solutions to this Diophantine equation. Many thanks



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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01 Oct 2015, 11:00
yenh wrote: Bunuel wrote: udaymathapati wrote: 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. Let \(x\) be the # of $0.15 stamps and \(y\) the # of $0.29 stamps. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\) (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Sufficient. (2) She bought an equal number of $0.15 stamps and $0.29 stamps > \(x=y\). Not sufficient. Answer: A. So when we have equation of a type \(ax+by=c\) and we know that \(x\) and \(y\) are nonnegative integers, there can be multiple solutions possible for \(x\) and \(y\) (eg \(5x+6y=60\)) OR just one combination (eg \(15x+29y=440\)). Hence in some cases \(ax+by=c\) is NOT sufficient and in some cases it is sufficient. For more on this type of questions check: eunicesoldseveralcakesifeachcakesoldforeither109602.htmlmarthaboughtseveralpencilsifeachpencilwaseithera100204.htmlarentalcaragencypurchasesfleetvehiclesintwosizesa105682.htmljoeboughtonlytwentycentstampsandthirtycentstamps106212.htmlacertainfruitstandsoldapplesfor070eachandbananas101966.htmljoannaboughtonly015stampsand029stampshowmany101743.htmlatanamusementparktomboughtanumberofredtokensand126814.htmlcollectionsconfusedneedahelp81062.htmlHope it helps. Hi Bunuel, Could you tell me how can we determine in which case ax+by=c is sufficient? We cannot test all the cases in 2 minutes In this case, we can easily see that x = y = 10 is a solution, but it seems that nothing can guarantee that there are no other solutions to this Diophantine equation. Many thanks Let me try to answer. There is no other way to answer your question but to test out some pairs of values keeping in mind the values of x and y MUST be POSITIVE INTEGERS. For DS questions, you only need to get 2 values to make a statement or a combination of statements insufficient, stop, mark E and move onto next question at this point. Initially, yes this process will look extremely time consuming but once you practice a few similar question you will see that you dont need to spend a lot of time on such questions. Hope this helps.



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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11 Nov 2015, 04:56
udaymathapati wrote: 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. Pretty @!)*, let's start: (1) 15a+29b=440 > \(b=\frac{44015a}{29}\) > \(\frac{5(883a)}{29}\) so if we want this one to be an integer 883a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ?
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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11 Nov 2015, 05:31
BrainLab wrote: udaymathapati wrote: 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. Pretty @!)*, let's start: (1) 15a+29b=440 > \(b=\frac{44015a}{29}\) > \(\frac{5(883a)}{29}\) so if we want this one to be an integer 883a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ? This is the most straightforward way for these questions wherein the applicable values can only be positive ingeters. Similarly, you can not have 0.5 apples or 0.33 bananas etc and hence if the question had asked about certain fruits or number of men etc , your method will still be applicable to such questions.



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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12 Dec 2015, 14:18
Bunuel How did you systematically see that there was only one solution for the equation in statement 1? I got this answer correct but guessed. I knew it was either A or C and figured that because X was too obvious of a solution, it was A. But that leaves a lot to chance. Wondering how to best approach these diophantine eqns. Bunuel wrote: udaymathapati wrote: 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. Let \(x\) be the # of $0.15 stamps and \(y\) the # of $0.29 stamps. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\) (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Sufficient. (2) She bought an equal number of $0.15 stamps and $0.29 stamps > \(x=y\). Not sufficient. Answer: A. So when we have equation of a type \(ax+by=c\) and we know that \(x\) and \(y\) are nonnegative integers, there can be multiple solutions possible for \(x\) and \(y\) (eg \(5x+6y=60\)) OR just one combination (eg \(15x+29y=440\)). Hence in some cases \(ax+by=c\) is NOT sufficient and in some cases it is sufficient. For more on this type of questions check: eunicesoldseveralcakesifeachcakesoldforeither109602.htmlmarthaboughtseveralpencilsifeachpencilwaseithera100204.htmlarentalcaragencypurchasesfleetvehiclesintwosizesa105682.htmljoeboughtonlytwentycentstampsandthirtycentstamps106212.htmlacertainfruitstandsoldapplesfor070eachandbananas101966.htmljoannaboughtonly015stampsand029stampshowmany101743.htmlatanamusementparktomboughtanumberofredtokensand126814.htmlcollectionsconfusedneedahelp81062.htmlHope it helps.



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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12 Dec 2015, 16:56
dubyap wrote: Bunuel How did you systematically see that there was only one solution for the equation in statement 1? I got this answer correct but guessed. I knew it was either A or C and figured that because X was too obvious of a solution, it was A. But that leaves a lot to chance. Wondering how to best approach these diophantine eqns. Let me try to answer the question. For all such questions wherein you are asked to find the number fo fruits or stamps or pencils etc, remember: 1. The number must be an integer. 2. As the number must be an integer, the minimum value of such a number is 0.Coming back to the question at hand, from statement 1, you can generate the following equation: 0.15x+0.29y=4.40 and we need to find whether we get 1 unique value for x. We can rewrite the given equation as x=(4.400.29y)/0.15 or x=(44029y)/15 with x and y only taking integer values such that x,y \(\geq\) 0Now, based on that information, we can see that iteratively that 44029y can only be a factor of 15 when 44029*y gives you a number ending with 0 or 5. Also, a quick knowledge of what unit digits you get with a number such as A9 = 0/1/2/3/4/5/6/7/8/9 and you need to get a unit digit of 0 or 5 from the operation 440one of the digits of obtained from 29*y, the only possible values of y can be either 0,5,10,15, 20(not needed as this will make x<0). Out of these only y=10 will give you an integer value for 'x'. Hence the only possible value of (x,y) is (10,10). Hope this helps.



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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23 Feb 2017, 05:45
PROMPT ANALYSIS Let the number of $0.15 stamp be x and number of $0.29 stamp be y. Total value of stamp = 0.15x +0.29y. Superset X can be a nonnegative integer Translation In order to find the value of x we need: 1# exact value of x and y 2# two equations with x and y as variable to solve for x and y Statement analysis St 1: total value of stamp = 4.40 =0.15x +0.29y. One equation two variable. Since we know that x and y are positive integers. We can say 15x +29y =440: x =(44029y)/15. The only integer that will satisfy will be x = 10 , y = 10, ANSWER. Hence option b, c, e eliminated. St 2: x =y. Therefore the total will be 0.44x. We cannot calculate the value of x from this information. INSUFFICIENT. Option A.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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06 Feb 2019, 10:30
udaymathapati wrote: 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. Given: Joanna bought only $0.15 stamps and $0.29 stamps. Let C = number of $0.15 stamps purchased Let E = number of $0.29 stamps purchased Target question: What is the value of C? Statement 1: She bought $4.40 worth of stamps We can write the equation 0.15C + 0.29E = 4.40IMPORTANT: In high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable. However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables. To determine whether this is the case here, let's examine all possible values of E. If E = 0, then the entire $4.40 was spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 4.40, it cannot be the case that E = 0If E = 1, then $0.29 was spent on $0.29 stamps, leaving the remaining $4.11 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 4.11, it cannot be the case that E = 1If E = 2, then $0.58 was spent on $0.29 stamps, leaving the remaining $3.82 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 3.82, it cannot be the case that E = 2If E = 3, then $0.87 was spent on $0.29 stamps, leaving the remaining $3.53 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 3.53, it cannot be the case that E = 3IMPORTANT: At this point, we might speed up our solution by recognizing that, in order for 0.15 to divide evenly into a number, that number must end with 5 or 0. Also recognize that, in order for the resulting value to end with a 5 or 0, E must be divisible by 5 So, from this point on, we'll just check values of E that are divisible by 5. If E = 5, then $1.45 was spent on $0.29 stamps, leaving the remaining $2.95 to be spent on $0.15 stamps. NICE! $2.95 ends with a 5. So this MIGHT work. Unfortunately, 0.15 does NOT divide evenly into 2.95. So, it cannot be the case that E = 5Keep going! If E = 10, then $2.90 was spent on $0.29 stamps, leaving the remaining $1.50 to be spent on $0.15 stamps. 1.50/0.15 = 10 = C. So, one possible solution is E = 10 and C = 10 If E = 15, then $4.35 was spent on $0.29 stamps, leaving the remaining $0.05 to be spent on $0.15 stamps. Doesn't work. If E = 20, then $5.80 was spent on $0.29 stamps. Hmmm. Looks like we can stop here! So, there is only one possible scenario that meets the given conditions. So, it MUST be the case that E = 10 and C = 10Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: She bought an equal number of $0.15 stamps and $0.29 stamps.We have no idea how much money Joanna spent on stamps. As such, there are infinitely many scenarios that satisfy statement 2. Here are two: Case a: She bought 3 $0.29 stamps and 3 $0.15 stamps. In this case, the answer to the target question is C = 3Case b: She bought 8 $0.29 stamps and 8 $0.15 stamps. In this case, the answer to the target question is C = 8Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Answer: A Cheers, Brent
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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27 Apr 2019, 05:43
Bunuel wrote: udaymathapati wrote: 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. Let \(x\) be the # of $0.15 stamps and \(y\) the # of $0.29 stamps. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\) (1) She bought $4.40 worth of stamps > \(15x+29y=440\). Only one integer combination of \(x\) and \(y\) is possible to satisfy \(15x+29y=440\): \(x=10\) and \(y=10\). Sufficient. (2) She bought an equal number of $0.15 stamps and $0.29 stamps > \(x=y\). Not sufficient. Answer: A. So when we have equation of a type \(ax+by=c\) and we know that \(x\) and \(y\) are nonnegative integers, there can be multiple solutions possible for \(x\) and \(y\) (eg \(5x+6y=60\)) OR just one combination (eg \(15x+29y=440\)). Hence in some cases \(ax+by=c\) is NOT sufficient and in some cases it is sufficient. For more on this type of questions check: http://gmatclub.com/forum/eunicesolds ... 09602.htmlhttp://gmatclub.com/forum/marthabought ... 00204.htmlhttp://gmatclub.com/forum/arentalcar ... 05682.htmlhttp://gmatclub.com/forum/joeboughton ... 06212.htmlhttp://gmatclub.com/forum/acertainfru ... 01966.htmlhttp://gmatclub.com/forum/joannabought ... 01743.htmlhttp://gmatclub.com/forum/atanamuseme ... 26814.htmlhttp://gmatclub.com/forum/collectionsc ... 81062.htmlHope it helps. Bunuel, How are you sure that only one combination possible in equation 1.....Is it Trial error or any method to find....I always struggle in optimizing values... Please help!
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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30 Jun 2019, 15:14
Bought $0.15 and $0.29. How many $0.15?
Can be 0.15, 0.30, 0.45, 0.60, 0.75, 0.90, 1.05, 1.20, 1.35, 1.50, 1.65..
1. Bought $4.40 > 0.15A + 0.29B = 4.40 The amount of B need to be resulted in cost of x.x0, x.x5 B=10 > 2.9 > (4.42.9)/0.15 =10 B=20 > 5.8 cant (coz >4.4)
Since the only possibility is: 10*0.29=2.9 10*0.15=1.5 (2.9+1.5=4.4) Therefore 1 is Sufficient.
2. Sane amount of 0.15 & 0.29. Can be anything > Therefore 2 is inSufficient.
Cheers, Eyal




Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp
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