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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
09 Feb 2013, 19:33
for statement 1, we get 15x + 29y = 440 as we need the value of x,so 15x = 440 -29y x = (440 - 29y)/15 It's clear from the equation that the value of y should be 5 or 10 to be divided by 15 since x is an integer. if y = 5,you cannot divide it with 15. Now try y =10, so x = 10. Statement 1 is sufficient.
Also you can see from the 2nd statement that x=y .Statement 2 itself is not sufficient as you don't know the total price.
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
22 Mar 2013, 06:37
I'd also like to add the following -
(1) She bought $4.40 worth of stamps. - We notice that the ending digit is 0, and one type of stamps costs 0.29. The only way we can have a "0" at the end with $0.29 is to multiply 0.29 with 10,20,30, etc. So if we have 10x0.29, its OK, meaning we have 2.9 worth of 0.29 stamps, and the rest worth $0.15. If we are to have 20, then we would surpass the total value of 4.40$ ( 20*0.29 = 5.8 ). So clearly she bought 10 stamps worth $0.29, and subsequently we can calculate the rest. That's why A is sufficient.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps. - Clearly insufficient.
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
22 Mar 2013, 23:47
Expert's post
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.
Just a quick tool for Diophantine equations.
Lets assume the equation 4x+5y = 36. Rearrange it as x =\(\frac{(36-5y)}{4}\) or y = \(\frac{(36-4x)}{5}\). Now, lets take the first re-arrangement. We can see that for y=0, we get an integral value for x as x=9 .Now all we have to do is keep adding 4 to the initial value of y=0. Thus, the next value of y which will give an integral solution for x is y=4,8,12 etc. One could also subtract 4 and get subsequent values for y = -4,-8,12 etc.
Taking the second rearrangement, we can see that x=-1, we have y=8. Thus, following the same logic, the next value of x, which will lead to an integral value for y is x= 4,9,14 etc.
Now back to the given problem:
We know that 15x+29y = 440.
Now, 29y = 440-15x
or y =\(\frac{440-15x}{29}\) .Now as because the statements on GMAT don't contradict each other, we know that one of the value of x and y for the above equation is x=y=10. Thus, for the above equation, keep adding 29 to x=10 for all the successive values for getting an integral solution for y. Thus, in accordance with the given sum, we can neglect negative integral values and state that x=y=10 is the only possible solution for both x,y>0. _________________
Diophantine Equations related Data Sufficiency. [#permalink]
16 Jun 2013, 16:32
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Here is the method to never fail to answer correctly Diophantine-equations-related Data Sufficiency problems.
1. First, be sure that the 2 variables must be non-negative integers or positive integers and that each statement provides a linear equation relating the 2 variables. Furthermore, be sure that the 2 equations are not equivalent (2x+3y=20 and 6x+9y=60 are equivalent) and are reduced to the form: ax + by = c whith integral coefficients and constant term such that GCF (a,b)=1.
2. Find an initial Solution: "Take advantage" of the fact that statements never contradict each other and thus system of equations constructed with both statements have always at least one solution. So resolve the system of equations.
3. Unicity: Once you arrive to a solution, say (x0, y0), go back to the first statement alone, for example, and check the unicity of the solution using only that statement by applying the test below. In case the solution is unique, statement 2 is superfluous and statement 1 is sufficient. The answer is A or D. In case the solution is not unique the answer is B, C or E.
Apply the test on statement (2). And update your answer.
If there is more than one solution using each statement alone then the answer is C.
------------------------------------------------------------------------------------------------------------------------------------------------- Now here is the rule that indicates whether or not a non-negative integer solution is unique to an equation: Suppose the equation be: ax+by=c (reduced with a, b, c positive integers. i.e. GCF(a,b)=1) If (x0-b)<0 AND (y0-a)<0 then there is no other non-negative integer solution than (x0, y0) and the corresponding statement is sufficent. If (x0-b)>=0 OR (y0-a)>=0 then other non-negative integers solutions exist and the statement is not sufficient.
If the variables must be positive the test is: If (x0 - b)<=0 AND (y0 - a)<=0 then there is no other positive integer solution than (x0, y0) and the corresponding statement is sufficent. If (x0 - b)>0 OR (y0 - a)>0 then other positive integers solutions exist and the statement is not sufficient.
Note: The test is to subtract each coefficient from the solution found for the opposite variable. -------------------------------------------------------------------------------------------------------------------------------------------------- Let's apply this to a real GMAT problem: A man buys some juice boxes. The boxes are from two different brands, A and B. How many boxes of brand A did the man buy if he bought $5.29 worth of boxes? (1) The price of brand A box is $0.81 and the price of brand B box is $0.31 (2) The total amount of boxes is 9
Variables here must be positive integers -number of juice boxes- since it is suggested that some juice boxes are from brand A and the rest from brand B. 1. The equations provided are: (1) 0.81A + 0.31B = 5.29. (2) A + B = 9 Which are reduced to: (1) 81A + 31B = 529 (2) A + B = 9, which is a system of reduced, linear, non-equivalent equations.
2. Find an initial Solution: (1) 81A + 31B = 529. GCF(31, 81)=1. (2) A + B = 9
Mutliplying (2) by 31 and subtracting it from (1) we get: 50A=250 so A=5 and B=4. An initial solution is (5, 4)
3. Unicity: Unicity for statement (1): 81(5) + 31(4) = 529 Since (5 - 31) <=0 AND (4 - 81)<=0 then there no other positive solution than (5, 4) so statement (1) is sufficient.
Unicity for statement (2): It is obvious that statement (2) alone is not sufficient but the test is still applicable. 1(5) + 1(4) = 9 Since (5 - 1) >0 OR (4 - 1) > 0 then there are other positive solutions than (5, 4) so statement (2) is not sufficient. Answer A.
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
18 Jun 2013, 17:45
I'd like to provide a demonstration for the rule above.
Let's ax+by=c...................................(1) an equation with a, b, c positive integers and GCD(a, b)=1. If (x0, y0) satisfies (1) then a(x0) + b(y0) = c Any other solution can be arrived at by varying in opposite directions x0 and y0 so the ax + by keep adding to c. a(x0 + k) + b(y0 - m) = c a(x0) + ak + b(y0) - bm = c a(x0) + b(y0) + ak - bm = c c + ak - bm = c ak = bm it follows that ak is multiple of b (and of course of a) so it is multiple of LCM(a, b)=ab since GCD(a, b) = 1. In other words ak can be written as ab*z for some integer z. The same applies to bm. So, ak = bm =ab*z So k = b*z and m = a*z. Returning back to our equation and reemplazing k and m by their expressions in terms of b and a we get: a(x0 + b*z) + b(y0 - a*z) = c Which means that the generating formula for all the solutions for (1) is: [(x0 + b*z), (y0 - a*z)] z is an integer. the first solutions generated are: etc z=-2-->(x0-2*b, y0+2*a) z=-1-->(x0-b, y0+a) z=0-->(x0, y0) z=1-->(x0+b, y0-a) z=2-->(x0+2*b, y0-2*a) etc So the minimum variation from (x0, y0) is: (x0+b, y0-a) and (x0-b, y0+a) In the case that solutions must be non-negative, then if both of the above 2 solutions is not non-negative then (x0, y0) is the only non-nergative one.
In the case that solutions must be positive, then if both of the above 2 solutions is not positive then (x0, y0) is the only positive one.
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
10 Jan 2014, 08:46
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.
We are given: 0.15*x + 0.29*y = TC
1) Gives us TC..... And we assume this is insufficient, but that's not true. Look closely: the least common multiple between these two numbers has to be a multiple of 5, so we basically have a pretty heavy restriction for the possible combinations of the two given TC = 4.40... Also, if we take one of each, we are given: 0.15 + 0.29 = 0.44... Multiply this by 10 (which is a multiple of 5), and thus we have 4.40.. If we multiply by 5, we only get 2.20, if we multiply by 15, we get 6.60, by 20 we get 8.80.. Etc, etc... So the ONLY combination of x and y that works is x = y = 10. We have 10 of each.
2) Clearly insufficient, this only tells us that x = y, but without TC that could be any multiple of 5.
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 non-negative 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.
Hope it helps.
Hi Bunel,
How to find the equation has more number of solutions or has single solution?
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
13 Aug 2014, 19:15
Hi, It looks like we can use the property that a (multiple of x) + (multiple of x) = multiple of x. And from that, we can see that since 4.40 and .15 are multiples of 5, .29 has to be a multiplied by a multiple of 5 as well.
Can someone comment as to whether this property holds true for subtraction/division/multiplication?
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
16 Aug 2015, 01:21
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 ?
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
16 Aug 2015, 04:17
karthickhari wrote:
we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ?
Be careful to use information from the other statement when you are evaluating each statement along in a DS question. You are correct that statement 1 is sufficient but 2 in itself (and without using the information from statement 1!) is NOT sufficient.
Statement 2 says that umber of .15$ 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.
Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
17 Sep 2015, 12:19
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 [#permalink]
17 Sep 2015, 12:35
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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=440-29y which means that 440-29y 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).
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
28 Sep 2015, 10:12
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 non-negative 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.
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
28 Sep 2015, 10:15
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 non-negative 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.
How did you arrive at the 10-10 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. _________________
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
01 Oct 2015, 09:56
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 non-negative 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.
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.
Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
01 Oct 2015, 10:00
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 non-negative 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.
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.
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
11 Nov 2015, 03:56
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 -> \(b=\frac{440-15a}{29}\) --> \(\frac{5(88-3a)}{29}\) so if we want this one to be an integer 88-3a 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 [#permalink]
11 Nov 2015, 04:31
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 -> \(b=\frac{440-15a}{29}\) --> \(\frac{5(88-3a)}{29}\) so if we want this one to be an integer 88-3a 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. _________________
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
12 Dec 2015, 13:18
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 non-negative 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.
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
12 Dec 2015, 15:56
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=(4.40-0.29y)/0.15 or x=(440-29y)/15 with x and y only taking integer values such that x,y \(\geq\) 0
Now, based on that information, we can see that iteratively that 440-29y can only be a factor of 15 when 440-29*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 440-one 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).
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