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

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18 Feb 2011, 04:28

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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]

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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.. _________________

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

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18 Feb 2011, 05:31

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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]

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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]

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12 Mar 2011, 09:35

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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 ? _________________

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

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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]

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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]

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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]

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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]

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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 _________________

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

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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]

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09 Jul 2016, 08:49

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.

gmatclubot

Re: Eunice sold several cakes. If each cake sold for either
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09 Jul 2016, 08:49

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