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Intern
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Re: Eunice sold several cakes. If each cake sold for either
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18 Dec 2016, 02:42
gmat1220 wrote: How about this? Write the cases and apply the unit digit test? 5 sets of (a,b) values (2,6)  Unit digit 4 + 4 = 8 (3,5)  Unit digit 1 + 5 = 6 (4,4) OUT too high (5,3)  Unit digit 5 + 7 = 12 i.e. 2 (6,2)  Unit digit 2 + 8 = 10 i.e. 0> Voila! subhashghosh wrote: Answer is B.
I was thinking perhaps it can be like, say a multiple of 19 and a multiple of 17 that add up to 0 in unit's place, so by only checking the unit's place i got 38 (19*2) and 102 (17*6). Opinions ? I would use the same "Unit Digit" approach. Though, without taking A into consideration, how can you assume the total number of cakes are 8 though? Or are we essentially taking info from A and just to try if there is only 1 combination for condition B?



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Re: Eunice sold several cakes. If each cake sold for either
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23 Mar 2017, 22:59
Hello Friends, My 2 cents...
In such scenarios i cheat: i use S1 info in S2 and try to solve  if i am able to solve viola we have to answer otherwise NOT
For example, here  we know that a + b = 8 and we need to find b therefore a = 8 b in other equation: a = 17 cents b = 19 cents 17a+19b = 140 17*(8b) + 19b = 140 136  17b + 19b = 140 2b = 4 b = 2  we have our answer!!!
PLEASE NOTE: i ONLY do it in such scenarios as usually we SHOULD NOT carry data from S1 to S2.



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Re: Eunice sold several cakes. If each cake sold for either
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23 Mar 2017, 23:00
fluke wrote: Eunice sold x $17 cakes Eunice sold y $19 cakes
A. Eunice sold 8 cakes. Those cakes can be all 17$ cakes or all $19 cakes. Not sufficient.
B. 17x+19y = 140
Whenever we get such type of questions and we know that x and y can only be integers, it is a must we try all integral values for x and y that may satisy the equation.
Write multiples of 19 that are less than 140; 0,19,38,57,76,95,114,133
Start with x=0; x=0; 19y = 1400 = 140; not a multiple of 19. x=1; 19y = 14017=123; not a multiple of 19. x=2; 19y = 14034=106; not a multiple of 19. x=3; 19y = 14051=89; not a multiple of 19. x=4; 19y = 14068=72; not a multiple of 19 x=5; 19y = 14095=55; not a multiple of 19 x=6; 19y = 140112=38; is a multiple of 19 x=7; 19y = 140129=11; not a multiple of 19 x=8; 19y = 140146=6; not a multiple of 19 and < 0. We can stop here.
We see that the above equation satisfies only for x=6; y=38/19=2; Sufficient.
Ans: "B"
I wonder if there is an easier way to find this one!!! Hello Friends, My 2 cents... In such scenarios i cheat: i use S1 info in S2 and try to solve  if i am able to solve viola we have to answer otherwise NOT For example, here  we know that a + b = 8 and we need to find b therefore a = 8 b in other equation: a = 17 cents b = 19 cents 17a+19b = 140 17*(8b) + 19b = 140 136  17b + 19b = 140 2b = 4 b = 2  we have our answer!!! PLEASE NOTE: i ONLY do it in such scenarios as usually we SHOULD NOT carry data from S1 to S2.



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Re: Eunice sold several cakes. If each cake sold for either
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29 Mar 2017, 18:23
The way I looked at it 
1) Not enough but true 2) we need 17x + 19y = 140.
I was going to check for (x*7)+(x*9) = 0, only one combination worked. Tried it out and it worked. So hence B.



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Re: Eunice sold several cakes. If each cake sold for either
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10 Nov 2017, 08:37
this is never an official problem in og, we meet similar but different problem, in which oa is b but it is a very simple math to find that there is one couple x,y which meet choice B. we never have to do a lot of computation like in this problem.



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Re: Eunice sold several cakes. If each cake sold for either
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15 Dec 2017, 11:16
gmatpapa wrote: fluke wrote: Eunice sold x $17 cakes Eunice sold y $19 cakes
A. Eunice sold 8 cakes. Those cakes can be all 17$ cakes or all $19 cakes. Not sufficient.
B. 17x+19y = 140
Whenever we get such type of questions and we know that x and y can only be integers, it is a must we try all integral values for x and y that may satisy the equation.
Write multiples of 19 that are less than 140; 0,19,38,57,76,95,114,133
Start with x=0; x=0; 19y = 1400 = 140; not a multiple of 19. x=1; 19y = 14017=123; not a multiple of 19. x=2; 19y = 14034=106; not a multiple of 19. x=3; 19y = 14051=89; not a multiple of 19. x=4; 19y = 14068=72; not a multiple of 19 x=5; 19y = 14095=55; not a multiple of 19 x=6; 19y = 140112=38; is a multiple of 19 x=7; 19y = 140129=11; not a multiple of 19 x=8; 19y = 140146=6; not a multiple of 19 and < 0. We can stop here.
We see that the above equation satisfies only for x=6; y=38/19=2; Sufficient.
Ans: "B"
I wonder if there is an easier way to find this one!!! Not in my knowledge. Even I use trial and error method to solve such problems. However, now whenever i see such problems I instantly pick B because all such questions I have seen have the same structure and very predictable answer choice!! :D But I'm sure if I see one on GDay, I'd rather solve it.. Not true always, if a question offers several sets of possibilites then answer wont always be B. it would be C or E instead. here is an example of such question. https://gmatclub.com/forum/joeboughto ... 06212.html



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Re: Eunice sold several cakes. If each cake sold for either
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15 Dec 2017, 11:53
fluke wrote: Eunice sold x $17 cakes Eunice sold y $19 cakes
A. Eunice sold 8 cakes. Those cakes can be all 17$ cakes or all $19 cakes. Not sufficient.
B. 17x+19y = 140
Whenever we get such type of questions and we know that x and y can only be integers, it is a must we try all integral values for x and y that may satisy the equation.
Write multiples of 19 that are less than 140; 0,19,38,57,76,95,114,133
Start with x=0; x=0; 19y = 1400 = 140; not a multiple of 19. x=1; 19y = 14017=123; not a multiple of 19. x=2; 19y = 14034=106; not a multiple of 19. x=3; 19y = 14051=89; not a multiple of 19. x=4; 19y = 14068=72; not a multiple of 19 x=5; 19y = 14095=55; not a multiple of 19 x=6; 19y = 140112=38; is a multiple of 19 x=7; 19y = 140129=11; not a multiple of 19 x=8; 19y = 140146=6; not a multiple of 19 and < 0. We can stop here.
We see that the above equation satisfies only for x=6; y=38/19=2; Sufficient.
Ans: "B"
I wonder if there is an easier way to find this one!!! this would take so long to do!



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Re: Eunice sold several cakes. If each cake sold for either
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07 Mar 2018, 12:55
I used following approach:
140 is divisible by 5 and 10 and is even. 19 is obviously a prime. How can we transfer the unit digit into a 0? E.g. by adding +1 How do we get +1 from 17? Multiply it by 3. Hence, 19 + 17*3 = 19 + 51 = 70. Stop here. This answer is sufficient.



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Eunice sold several cakes. If each cake sold for either
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21 Mar 2018, 22:56
There is an easier way. A bit shorter. Now, we have 17x + 19y = 140 Notice that 140 has a 0 at its units digit, we just need to find which combination of units digit of 17x and 19y would give us a 0. 17x > 7, 4, 1, 8, 5.... 19y > 9, 8, 7, 6, 5.... We see that 1 of 17x and 9 of 19y make up 0, and also both the 5s, and 4 & 6. Let's take them. (x = 3, y = 1) 17*3 + 19*1 = 70 !! (Voila! 70 is exactly the half of 140) (So, double them) (17*3 + 19*1)*2 = 140 Which shows that x = 6 and y = 2 6 & 2 are the answers, and we can guess that no other integers would satisfy this equation. Also, we can crosscheck this with option A that 6+2 = 8. And we cannot say that the other case of 5 & 5 would be 8. So we can ignore this. To check for 6 & 4, try it 17*2 + 19*4 = 110 (Not close enough) So, solely by checking units digit, and by doing some educated leaps, we can conclude that the answer is B. :D gmatpapa wrote: fluke wrote: Eunice sold x $17 cakes Eunice sold y $19 cakes
A. Eunice sold 8 cakes. Those cakes can be all 17$ cakes or all $19 cakes. Not sufficient.
B. 17x+19y = 140
Whenever we get such type of questions and we know that x and y can only be integers, it is a must we try all integral values for x and y that may satisy the equation.
Write multiples of 19 that are less than 140; 0,19,38,57,76,95,114,133
Start with x=0; x=0; 19y = 1400 = 140; not a multiple of 19. x=1; 19y = 14017=123; not a multiple of 19. x=2; 19y = 14034=106; not a multiple of 19. x=3; 19y = 14051=89; not a multiple of 19. x=4; 19y = 14068=72; not a multiple of 19 x=5; 19y = 14095=55; not a multiple of 19 x=6; 19y = 140112=38; is a multiple of 19 x=7; 19y = 140129=11; not a multiple of 19 x=8; 19y = 140146=6; not a multiple of 19 and < 0. We can stop here.
We see that the above equation satisfies only for x=6; y=38/19=2; Sufficient.
Ans: "B"
I wonder if there is an easier way to find this one!!! Not in my knowledge. Even I use trial and error method to solve such problems. However, now whenever i see such problems I instantly pick B because all such questions I have seen have the same structure and very predictable answer choice!! :D But I'm sure if I see one on GDay, I'd rather solve it..



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Re: Eunice sold several cakes. If each cake sold for either
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25 May 2019, 03:25
Consider the following equation: 2x + 3y = 30. If x and y are nonnegative integers, the following solutions are possible: x=15, y=0 x=12, y=2 x=9, y=4 x=6, y=6 x=3, y=8 x=0, y=10 Notice the following: The value of x changes in increments of 3 (the coefficient for y). The value of y changes in increments of 2 (the coefficient for x). This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers. gmatpapa wrote: Eunice sold several cakes. If each cake sold for either exactly 17 or exactly 19 dollars, how many 19 dollar cakes did Eunice sell?
(1) Eunice sold a total of 8 cakes. (2) Eunice made 140 dollars in total revenue from her cakes. Let x = the number of $17 cakes and y = the number of $19 cakes. Statement 1: x+y = 8 Here, x and y can be any nonnegative values that sum to 8. INSUFFICIENT. Statement 2: 17x+19y = 140Solve for x and y when x+y=8, as required in Statement 1. Multiplying x+y = 8 by 17, we get: 17x + 17y = 136 Subtracting the blue equation from the red equation, we get: 2y = 4 y=2, implying in the blue equation that x=6. Thus, one solution for 17x+19y=140 is as follows: x=6, y=2 In accordance with the rule discussed above, the value of x may be altered only in INCREMENTS OF 19 (the coefficient for y), while the value of y may be altered only in INCREMENTS OF 17 (the coefficient for x). Not possible: If x increases by 19 and y decreases by 17, then y will be negative. If x decreases by 19 and y increases by 17, then x will be negative. Implication: The only nonnegative integral solution for 17x+19y=140 is x=6 and y=2. SUFFICIENT.
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Eunice sold several cakes. If each cake sold for either
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16 Jun 2019, 13:34
GMATGuruNY wrote: Consider the following equation: 2x + 3y = 30.
If x and y are nonnegative integers, the following solutions are possible: x=15, y=0 x=12, y=2 x=9, y=4 x=6, y=6 x=3, y=8 x=0, y=10
Dear GMATGuruNY1 Are any answer of x=15, y=0 or x=0, y=10 correct in DS questions? Is it considered solution or logic to say that a person bought zero of x or y? 2 If the equation is altered to be (multiply by 2): 4x + 6y = 60 then the increment process above won't work x=15, y=0 x=9, y=4 But such solution like x=12, y=2 won't appear if we take increments. should the equation be reduced to the last common factor before taking increments? Thanks



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Re: Eunice sold several cakes. If each cake sold for either
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17 Jun 2019, 03:43
Mo2men wrote: Dear GMATGuruNY1 Are any answer of x=15, y=0 or x=0, y=10 correct in DS questions? Is it considered solution or logic to say that a person bought zero of x or y? Unless we are told  or something in the prompt implies  that x and y must be positive, we should assume that x=0 and y=0 are viable options. Quote: 2 If the equation is altered to be (multiply by 2):
4x + 6y = 60
then the increment process above won't work x=15, y=0
x=9, y=4
But such solution like x=12, y=2 won't appear if we take increments. should the equation be reduced to the last common factor before taking increments? The increment approach requires that the coefficients for x and y be in their most reduced form.
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Re: Eunice sold several cakes. If each cake sold for either
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17 Jun 2019, 08:24
Given either exact 17 $ or 19 $ is price of cake
1. total 8 cakes ,,,,not sufficient 2. 140 $ total = 17 x + 19 y = 140 put x =0,1,2,3,4,5,6,7,8,9 randomly and check value of y
at X =6 , Y =2
B is sufficient




Re: Eunice sold several cakes. If each cake sold for either
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