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# A certain store sells only jacks and marbles. A single jack costs 19

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A certain store sells only jacks and marbles. A single jack costs 19  [#permalink]

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18 Feb 2017, 01:43
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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 ##### Most Helpful Expert Reply Magoosh GMAT Instructor Joined: 28 Dec 2011 Posts: 4472 Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink] ### Show Tags 22 Feb 2017, 10:39 4 4 manishtank1988 wrote: I do know that these type of problems can be solved by interchanging the constant values of a equation. Also, because of value of marbles 25 cents i got - C - 2 as the answer but the answer is D - 3. I wasted a lot of time on this question still got it wrong. Can someone suggest a simpler or straight forward method to solve this and these types of questions? Thanks Dear manishtank1988, I'm happy to respond. I think of problems of this sort in terms of number sense. That blog article will recommend ways to develop this important skill. Think about the numbers. The number$0.25 is a nice number that divides evenly into $10. The number$0.19 is a prime number of cents that doesn't divide evenly into $10. We can easily build$10 out of a lot of $0.25's only. If we want to use some of the$0.19 marbles, we will need to find some multiple of 19 that is divisible by 25, so that we can fill the remaining space with the $0.25 marbles. Again, since 19 is a prime number, the only way we will get a multiple of 25 is if we multiply 19*25. 20*25 = 500, so 19*25 = (20 - 1)*25 = 500 - 25 = 475 1) We could build$4.75 from the $0.19 marbles, and fill the remaining$5.25 with $0.25 marbles. 2) We could build 2($4.75) = $9.50 from the$0.19 marbles, and fill the remaining $0.50 with two$0.25 marbles.
3) We could build all $10 from only the$0.25 marbles.
Those are the three possibilities.

Does all this make sense?
Mike
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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
26
9
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 afore-mentioned 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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18 Feb 2017, 02:43
1
1
vienbuuchau 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 Hi j - # of jacks, m - # of marbles 0.19j + 0.25m = 10 19j + 25m = 1000 19j = 1000 - 25m m = 40 - 19n j = 25n where n is integer >=0, because we are not told that at lest one of each needs to be bought. Hence only possible options fo n are 0, 1 and 2. Answer D. Intern Joined: 07 Oct 2016 Posts: 4 Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink] ### Show Tags 18 Feb 2017, 09:02 vitaliyGMAT wrote: vienbuuchau 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

Hi

j - # of jacks, m - # of marbles

0.19j + 0.25m = 10

19j + 25m = 1000

19j = 1000 - 25m

m = 40 - 19n

j = 25n

where n is integer >=0, because we are not told that at lest one of each needs to be bought.

Hence only possible options fo n are 0, 1 and 2.

Hi, can you please explain how you got these two equations:

m = 40 - 19n

j = 25n
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Re: A certain store sells only jacks and marbles. A single jack costs 19  [#permalink]

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18 Feb 2017, 09:50
2
mohdtaha wrote:
vitaliyGMAT wrote:
vienbuuchau 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 Hi j - # of jacks, m - # of marbles 0.19j + 0.25m = 10 19j + 25m = 1000 19j = 1000 - 25m m = 40 - 19n j = 25n where n is integer >=0, because we are not told that at lest one of each needs to be bought. Hence only possible options fo n are 0, 1 and 2. Answer D. Hi, can you please explain how you got these two equations: m = 40 - 19n j = 25n Hi 19j = 1000 - 25m 19j = 25(40 - m) j is a multiple of 25 and (40 - m) is a multiple of 19 -----> j=25n and 40 - m = 19n or m = 40 - 19n starting from n=3 our expression 40 - 19n < 0, hence we have only 3 possibilities for n = 0, 1 and 2. Hope this helps Regards Math Expert Joined: 02 Aug 2009 Posts: 7984 Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink] ### Show Tags 18 Feb 2017, 10:16 1 1 vienbuuchau 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

Hi,

The equation is 19x+25y=1000...
Since 25y and 1000 are multiple of 25, 19x is also multiple of 25..
What all values can 19x or 19*25*z take..
Z can be 0,1 or 2.. so THREE values..
When z=3, 19*25*3>1000..

D
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Re: A certain store sells only jacks and marbles. A single jack costs 19  [#permalink]

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18 Feb 2017, 18:07
1
19x+25y=1000

We need multiples of 25 since 19 does not divide evenly into 1000 ($10). One obvious choice is x=0. The LCM of 19 and 25 is 475. So 19*25=475 which means in this instance y=21 (19(25)+25(21)=1000). So x can be 25 to complete the equation. Now 19*25*2=950 is another multiple of 25 that is less than 1000. So x can be 0, 1 or 2 since 19*25*3=1475 which is more than the$10 spent at the store.

D.
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Re: A certain store sells only jacks and marbles. A single jack costs 19  [#permalink]

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21 Feb 2017, 23:13
I do know that these type of problems can be solved by interchanging the constant values of a equation. Also, because of value of marbles 25 cents i got - C - 2 as the answer but the answer is D - 3. I wasted a lot of time on this question still got it wrong. Can someone suggest a simpler or straight forward method to solve this and these types of questions?
Thanks
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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, 00:35
7
1
manishtank1988 wrote:
Hello abhimahna, Skywalker18, mikemcgarry, msk0657, Bunuel, Vyshak, Engr2012, VeritasPrepKarishma, Abhishek009 and other forum members

I do know that these type of problems can be solved by interchanging the constant values of a equation. Also, because of value of marbles 25 cents i got - C - 2 as the answer but the answer is D - 3. I wasted a lot of time on this question still got it wrong. Can someone suggest a simpler or straight forward method to solve this and these types of questions?
Thanks

Hi,

I usually solve these questions as per following approach:

We are given 19j+25m=1000.

I can write this equation as:

m = $$1000/25$$ - $$19j/25$$

Now the number of Jacks and marbles must be integral. So, we can substitute values for j to get m as a non negative integer.(Note: I am including 0 also here.)

Only 3 values of j will work here. ( j=0,25 and 50) for which we will get (m = 40, 21 and 2 respectively.).

I hope it is clear now. It won't take more than 30 secs of your time to solve these question.
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Re: A certain store sells only jacks and marbles. A single jack costs 19  [#permalink]

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23 Feb 2017, 05:10
2
manishtank1988 wrote:
Hello abhimahna, Skywalker18, mikemcgarry, msk0657, Bunuel, Vyshak, Engr2012, VeritasPrepKarishma, Abhishek009 and other forum members

I do know that these type of problems can be solved by interchanging the constant values of a equation. Also, because of value of marbles 25 cents i got - C - 2 as the answer but the answer is D - 3. I wasted a lot of time on this question still got it wrong. Can someone suggest a simpler or straight forward method to solve this and these types of questions?
Thanks

This is how we solve this type of questions. I would suggest you to practice this method a bit. It doesn't take more than 1-2 mins.

https://www.veritasprep.com/blog/2011/0 ... -of-thumb/

If there are any doubts, let me know.
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Re: A certain store sells only jacks and marbles. A single jack costs 19  [#permalink]

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24 Feb 2017, 12:25
1
Hi manishtank1988,

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

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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
2
3
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 non-negative 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 integer-valued 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 non-negative 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 19-cent jacks or else 19 of the 25-cent 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 25-cent marbles ($4.75 worth of marbles) for 25 of the 19-cent jacks ($4.75 worth of jacks), giving a second solution of 25 jacks/21 marbles. Now we trade out 19 more of our 25-cent marbles for 25 more of the 19-cent 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
1
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

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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, 02:43
warriorguy 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 Don't they give out conversion too? Or they assume we are familiar with cent and dollar conversion? The question itself will supply the relative conversions. Though you should have a few basic ones memorized: 1 hour = 60 minutes, ... I guess if$1 = 100 cents is not a common knowledge, then this also will be provided.
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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
1
1
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=1000-19X 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 . Intern Joined: 26 Dec 2016 Posts: 28 Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink] ### Show Tags 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=1000-19X
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, 16:30
1
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).

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A certain store sells only jacks and marbles. A single jack costs 19  [#permalink]

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Updated on: 22 Mar 2018, 19:54
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 cost per pair=44 22 pairs=968; remainder=32 no 21 pairs=924; remainder=76 yes 76/19=4 jacks 25 jacks+21 marbles=1000 adding 25 jacks, 50 jacks+2 marbles=1000 subtracting 25 jacks, 40 marbles=1000 3 ways D Originally posted by gracie on 22 Mar 2018, 10:14. Last edited by gracie on 22 Mar 2018, 19:54, edited 1 time in total. Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2815 Re: A certain store sells only jacks and marbles. A single jack costs 19 [#permalink] ### Show Tags 22 Mar 2018, 15:48 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

We can let the number of jacks purchased = x and the number of marbles purchased = y. We are given that a jack costs 19 cents and a marble costs 25 cents. We need to determine how many different ways one can purchase 10 dollars’ worth, or 1,000 cents’ worth, of jacks and marbles.
Let’s create an equation:

19x + 25y = 1000

19x = 1000 - 25y

19x = 25(40 - y)

x = [25(40 - y)]/19

We see that 25(40 - y) must be a multiple of 19. However, since 19 does not evenly divide 25, we see that 19 must divide (40 - y). Thus, the possible values of y are 2, 21, and 40, and, thus, there are 3 ways one can spend exactly 10 dollars (or 1000 cents) at the store.

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Re: A certain store sells only jacks and marbles. A single jack costs 19   [#permalink] 22 Mar 2018, 15:48

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