Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Word Problem DS: Simple Equations [#permalink]

Show Tags

18 Feb 2011, 04:28

1

This post received KUDOS

3

This post was BOOKMARKED

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 = 140-0 = 140; not a multiple of 19. x=1; 19y = 140-17=123; not a multiple of 19. x=2; 19y = 140-34=106; not a multiple of 19. x=3; 19y = 140-51=89; not a multiple of 19. x=4; 19y = 140-68=72; not a multiple of 19 x=5; 19y = 140-95=55; not a multiple of 19 x=6; 19y = 140-112=38; is a multiple of 19 x=7; 19y = 140-129=11; not a multiple of 19 x=8; 19y = 140-146=-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!!!
_________________

Re: Word Problem DS: Simple Equations [#permalink]

Show Tags

18 Feb 2011, 04:37

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 = 140-0 = 140; not a multiple of 19. x=1; 19y = 140-17=123; not a multiple of 19. x=2; 19y = 140-34=106; not a multiple of 19. x=3; 19y = 140-51=89; not a multiple of 19. x=4; 19y = 140-68=72; not a multiple of 19 x=5; 19y = 140-95=55; not a multiple of 19 x=6; 19y = 140-112=38; is a multiple of 19 x=7; 19y = 140-129=11; not a multiple of 19 x=8; 19y = 140-146=-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 G-Day, I'd rather solve it..
_________________

My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

Re: Word Problem DS: Simple Equations [#permalink]

Show Tags

18 Feb 2011, 05:31

4

This post received KUDOS

4

This post was BOOKMARKED

The fact that its 8 cakes doesn't give sufficient information. So A is not sufficient.

Since total revenue is 140 and average price is 18, it can only happen if total cakes are ~8 (18*8 = 144) - it cant be very different than 8 as price difference is not so high. Since 140 is 4 less than 144, the lower priced cake has to be certain number of units greater than higher priced one it can be 7,1 or 6,2 or 5,3. Quick plugging would tell us that 6 cakes of 17 and 2 cakes of 19 are the only possibility and hence statement 2 is sufficient.

Re: Word Problem DS: Simple Equations [#permalink]

Show Tags

12 Mar 2011, 09:18

Nowadays, whenever I see this question pattern, I immediately select option B. However, on the D Day, I'm sure we wont have enough time to do trial and error

Re: Word Problem DS: Simple Equations [#permalink]

Show Tags

12 Mar 2011, 09:35

2

This post received KUDOS

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 ?
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Re: Word Problem DS: Simple Equations [#permalink]

Show Tags

12 Mar 2011, 22:05

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 ?

Re: Eunice sold several cakes. If each cake sold for either [#permalink]

Show Tags

15 Nov 2012, 21:25

Easier way to solve instead of trial and error. First of all when we smell its going to 'B'...Choose values as below:

from first statement X+Y=8 from second statment 17x+19y=140

solve for x,y, you get X=6 Y=2...we dont need to bother much duplicate values as 2 eq's will have unique value if a1/a2!= b1/b2!=c1/c2 for 2 eq's a1x+b1y+c1=0 a2x+b2y+c2=0

Dont waste time in D-day using trial and error....

Not only will we have to find the correct value which satisfies the equation, we would also have to check all the others to make sure they DO NOT satisfy.

Is it safe to just jump onto B. (or A, depending on the question).
_________________

Re: Eunice sold several cakes. If each cake sold for either [#permalink]

Show Tags

26 Dec 2013, 04:27

you can make it quicker by changing Fluke's strategy slightly. As we need to find number of cakes that sell for 19 dollars. Set y=0 first and then go through the approach. With this you need to find y0, y1, and y3.

Re: Eunice sold several cakes. If each cake sold for either [#permalink]

Show Tags

16 Aug 2014, 12:16

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 = 140-0 = 140; not a multiple of 19. x=1; 19y = 140-17=123; not a multiple of 19. x=2; 19y = 140-34=106; not a multiple of 19. x=3; 19y = 140-51=89; not a multiple of 19. x=4; 19y = 140-68=72; not a multiple of 19 x=5; 19y = 140-95=55; not a multiple of 19 x=6; 19y = 140-112=38; is a multiple of 19 x=7; 19y = 140-129=11; not a multiple of 19 x=8; 19y = 140-146=-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!!!

Hi,

Is there a quicker way to do this without going through every single trial? I followed the same approach but was at 3+ when I was done with it.

Re: Eunice sold several cakes. If each cake sold for either [#permalink]

Show Tags

17 Aug 2014, 16:00

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 ?

By checking the unit's place: I have 19*5 ( unit digit 5) and 17*5 ( unit didgit 5) or 19* 4 ( 6) and 17*2 ( 4) .... In the real test, I think I will choose B without trying and error --> Save more time for other questions
_________________

......................................................................... +1 Kudos please, if you like my post

Re: Eunice sold several cakes. If each cake sold for either [#permalink]

Show Tags

06 Oct 2015, 02:39

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 = 140-0 = 140; not a multiple of 19. x=1; 19y = 140-17=123; not a multiple of 19. x=2; 19y = 140-34=106; not a multiple of 19. x=3; 19y = 140-51=89; not a multiple of 19. x=4; 19y = 140-68=72; not a multiple of 19 x=5; 19y = 140-95=55; not a multiple of 19 x=6; 19y = 140-112=38; is a multiple of 19 x=7; 19y = 140-129=11; not a multiple of 19 x=8; 19y = 140-146=-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!!!

Thanks Fluke. Not a great finding but it might confuse others. 17*6=102 and not 112.

Re: Eunice sold several cakes. If each cake sold for either [#permalink]

Show Tags

09 Jul 2016, 08:49

1

This post received KUDOS

I have used a simple approach however I am not sure whether the argumentation is valid:

Statement 2 says: 140 = 17x + 19 y

We know that there has to be one solution for x and y. The question is whether it is possible to receive another solution. If we modify x_solution and y_solution 140 still has to be the result. Hence I have looked for the LCM of 17 and 19 because if I reduce the number of 17s I have to increase the number of 19s, receiving the same result. Because 17 and 19 are both prime numbers their LCM is 17*19 > 140. Therefore there cannot be another solution for this equation.

=> Statement 2 sufficient.

Is this argumentation valid? Thank you for your help.

Eunice sold several cakes. If each cake sold for either [#permalink]

Show Tags

08 Sep 2016, 03:32

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 = 140-0 = 140; not a multiple of 19. x=1; 19y = 140-17=123; not a multiple of 19. x=2; 19y = 140-34=106; not a multiple of 19. x=3; 19y = 140-51=89; not a multiple of 19. x=4; 19y = 140-68=72; not a multiple of 19 x=5; 19y = 140-95=55; not a multiple of 19 x=6; 19y = 140-112=38; is a multiple of 19 x=7; 19y = 140-129=11; not a multiple of 19 x=8; 19y = 140-146=-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!!!

Fluke, I cant think of a different approach, but I can suggest a modification to your method to save a couple of minutes. You have carried out the subtraction 9 times. Let us try to reduce the number of subtractions.

You have to subtract the sum of 17 and 19 i.e, 36, successively from 140 and check whether 17 or 19 is a factor of the resultant.

140-36=104 (not a multiple of either 17 or 19) 104-36= 68 ( IS a multiple of 17, NOT a multiple of 19)

Stop here.

y= number of times successive subtraction was carried out = 2. x= number of times successive subtraction was carried out + 4 =6. (As you have to count the number of 17's in 64 as well as the number of times 17 was subtracted from 140).

gmatclubot

Eunice sold several cakes. If each cake sold for either
[#permalink]
08 Sep 2016, 03:32

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...