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A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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22 Feb 2017, 04:24
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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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22 Feb 2017, 04:34
Bunuel wrote: A certain store sells only jacks and marbles. A single jack costs 19 cents, and a single marble costs 25 cents. In how many different ways can one spend exactly $10 buying items at this store?
A. 0 B. 1 C. 2 D. 3 E. 4 0.19x + 0.25y = 1000; 1. 40 marbles = 1000; 2. 15 jacks = 2.75 + 29 single marbles = 7.25 3. 50 jacks = 8.5 + 6 marbles = 1.5
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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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22 Feb 2017, 04:37
RMD007 wrote: Bunuel wrote: A certain store sells only jacks and marbles. A single jack costs 19 cents, and a single marble costs 25 cents. In how many different ways can one spend exactly $10 buying items at this store?
A. 0 B. 1 C. 2 D. 3 E. 4 0.19x + 0.25y = 1000; 1. 40 marbles = 1000; 2. 15 jacks = 2.75 + 29 single marbles = 7.25 3. 50 jacks = 8.5 + 6 marbles = 1.5 15 jacks => 15*0.19 => 2.85 so this is not one of the expected outcome



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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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22 Feb 2017, 04:46
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Option D:Assume no. of jacks to be P and that of marbles to be Q, hence, 19P + 25Q = 1000. Identify number of combinations of P,Q which satisfies the aforementioned equation. :19P + 25Q = 1000 : (19P/25 + Q) = 40 i.e., P should be a multiple of 25 , possible values P can take = 0, 25, 50, 75 . . . . Lets put these back for P and check: : (19*0 + Q) = 40 ; Q = 40 : (19*1 + Q) = 40 ; Q = 21 : (19*2 + Q) = 40; Q = 2 : (19*3 + Q) = 40; Q = 17 NOT POSSIBLE. Total number of combinations = 3
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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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08 Mar 2017, 22:29
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Bunuel wrote: A certain store sells only jacks and marbles. A single jack costs 19 cents, and a single marble costs 25 cents. In how many different ways can one spend exactly $10 buying items at this store?
A. 0 B. 1 C. 2 D. 3 E. 4 Official solution from Veritas Prep. We can translate this problem algebraically as 0.19j + 0.25m = 10.00. To get rid of the decimals, multiply through by 100 and we’ll be working with the equation 19j + 25m = 1000. Note also that j and m must both be nonnegative integers. We’re now dealing with something known as a Linear Diophantine Equation. That may sound intimidating, but it’s really just an equation of the form Ax + By = C in which A, B, C, x, and y are all integers. What makes Diophantine Equations interesting (aside from the fact that they pop up from time to time on the GMAT) is that such integervalued equations can sometimes allow you to solve for more variables than you have equations. That is, you might be able to solve for both x and y using only a single equation, because of the nonnegative integer restriction implicit in the idea of jacks and marbles. Furthermore, if such an equation has multiple solutions, there is an easy and important relationship among all of the different solutions to the equation. Specifically, the key is to look at common multiples of the constants A and B (in this case A = 19 and B = 25). It turns out that each possible solution can produce all other solutions by exchanging items whose total values are common multiples of those constants. In other words, the LCM of 19 and 25 – two numbers that share no common factors besides 1 – is simply 19 * 25 = 475, so $4.75 worth of items can be produced by taking either 25 of the 19cent jacks or else 19 of the 25cent marbles. And once an initial solution is identified, we can produce all other solutions by swapping out jacks for marbles in these increments. At this point it’s worth noting that the easiest way to make it to exactly $10 is to just buy marbles, since $0.25 divides evenly (and easily) into $10.00. $10.00/$0.25 = 40, and 0 jacks/40 marbles is our first solution. Now we trade out 19 of our 25cent marbles ($4.75 worth of marbles) for 25 of the 19cent jacks ($4.75 worth of jacks), giving a second solution of 25 jacks/21 marbles. Now we trade out 19 more of our 25cent marbles for 25 more of the 19cent jacks, giving a third solution of 50 jacks, 2 marbles. Since we don’t have enough marbles to continue making trades in this manner, there are no more solutions to the equation. All told, then, there are three ways to spend exactly $10 at this store. The answer is D.
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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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10 Mar 2017, 12:43
Hi All, From the answer choices, we know that there cannot be that many ways to get a total of $10.00 when adding a multiple of $0.19 and a multiple of $0.25.... A multiple of $0.25 will always end in one of the following.... .00, .25, .50 or .75... so we really just need to determine in how many ways the $0.19 ends in a 'compliment' (re: .75, .50, .25 or .00) to that number so that the total = $10.00. To end in a 0 or a 5, we need to multiply $0.19 by a multiple of 5).... ($0.19)(0) = $0... so the 'complement' would be ($0.25)(40). This IS an option. ($0.19)(5) = $0.95... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. ($0.19)(10) = $1.90... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. ($0.19)(15) = $2.85... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. At this point, you should notice that we're just adding $0.95 to the prior sum, so doing lots of multiplication is NOT necessary.... $3.80... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. $4.75... so the 'complement' would be ($0.25)(21). This IS an option. $5.70... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. $6.65... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. $7.60... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. $8.55... but we can't get a 'complement' from a multiple of $0.25 to total $10. This is NOT an option. $9.50... so the 'complement' would be ($0.25)(2). This IS an option. Total options = 3 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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01 Jun 2017, 07:36
Hello
I am trying to understand the linear Diophantine Equation. Can anyone please elaborate on this principle?
I understand why we use the LCM but then I can't understand the following to find the next solutions: "Now we trade out 19 of our 25cent marbles ($4.75 worth of marbles) for 25 of the 19cent jacks ($4.75 worth of jacks), giving a second solution of 25 jacks/21 marbles" Maybe it would be useful to use a table to make it clearer. Thank you all



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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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03 Jul 2017, 02:38
Bunuel wrote: A certain store sells only jacks and marbles. A single jack costs 19 cents, and a single marble costs 25 cents. In how many different ways can one spend exactly $10 buying items at this store?
A. 0 B. 1 C. 2 D. 3 E. 4 Don't they give out conversion too? Or they assume we are familiar with cent and dollar conversion?



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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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04 Jul 2017, 04:13
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Bunuel wrote: A certain store sells only jacks and marbles. A single jack costs 19 cents, and a single marble costs 25 cents. In how many different ways can one spend exactly $10 buying items at this store?
A. 0 B. 1 C. 2 D. 3 E. 4 Let X is number of Jack and Y is number of Marbel . X and Y both has to an +ve Integer . 19X+25Y=1000 25Y=100019X Y=40(19/25)*X> For Y to be integer , X must be multiple of 25 => (X=0 Y=40),(X=25,Y=21),(X=50,Y=2) are the three options we have .



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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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28 Aug 2017, 08:33
gmat4varun wrote: Bunuel wrote: A certain store sells only jacks and marbles. A single jack costs 19 cents, and a single marble costs 25 cents. In how many different ways can one spend exactly $10 buying items at this store?
A. 0 B. 1 C. 2 D. 3 E. 4 Let X is number of Jack and Y is number of Marbel . X and Y both has to an +ve Integer . 19X+25Y=1000 25Y=100019X Y=40(19/25)*X> For Y to be integer , X must be multiple of 25 => (X=0 Y=40),(X=25,Y=21),(X=50,Y=2) are the three options we have . I cannot figure out where the 40 is coming from. Please help me understand this. Thanks!



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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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28 Aug 2017, 16:30
Hi rnz, We're told that single marbles cost $0.25 each. Thus, if you spend ALL $10 on just marbles, then you could purchase exactly 40 marbles (and 0 jacks). GMAT assassins aren't born, they're made, Rich
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Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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28 Aug 2017, 18:14
Bunuel wrote: A certain store sells only jacks and marbles. A single jack costs 19 cents, and a single marble costs 25 cents. In how many different ways can one spend exactly $10 buying items at this store?
A. 0 B. 1 C. 2 D. 3 E. 4 Here's my take, please correct I am wrong. 19 is 1 less than 20, which is a multiple of 1000. To offset this, we need 19 in multiples of 25, as the other integer we have is 25. 1 all marbles no jack 2 19*25=475, marbles for balance value 3 19*50=950, marbles for balance value 4th scenario is not possible as 19*75 exceeds 1000



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A certain store sells only jacks and marbles. A single jack costs 19 [#permalink]
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28 Aug 2017, 18:27
It's start with 19J+25M = 1,000
Where J and M are integers
(1,000  19J)/25 > must be integer.
J = 0 yes J = (19*25) < 1,000 yes J = (19*25*2) < 1,000 yes
D
In short, this question wanted test taker to understand 1) the number property I.e. common factors of 19 and 25 that less than 1,000 and 2) the magic number 0 should be consider
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A certain store sells only jacks and marbles. A single jack costs 19
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