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GMAT 1: 710 Q48 V40 GMAT 2: 740 Q49 V42 Eunice sold several cakes. If each cake sold for either  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 69% (01:57) correct 31% (01:36) wrong based on 1654 sessions

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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.

This one is a typical problem!

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

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

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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

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!!!
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GMAT 1: 710 Q48 V40 GMAT 2: 740 Q49 V42 Re: Word Problem DS: Simple Equations  [#permalink]

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1
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

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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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?

A. Eunice sold a total of 8 cakes.
B. Eunice made 140 dollars in total revenue from her cakes.

This one is a typical problem!

For more on the questions with Diophantine equation (equations whose solutions must be integers only) check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html
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Re: Word Problem DS: Simple Equations  [#permalink]

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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 Retired Moderator B
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Re: Word Problem DS: Simple Equations  [#permalink]

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2

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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1
I am trying to reduce the number of test cases for Statement 2). Pls verify this reasoning -

17a + 19b = 140
2 cases
odd + odd = even
even + even = even

a,b - both even
a,b - both odd

5 sets of (a,b) values
(2,6)
(3,5)
(4,4)
(5,3)
(6,2)

I can delete 4,4 but that means 4 cases are still left. How to reduce the cases?
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Re: Word Problem DS: Simple Equations  [#permalink]

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2

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:

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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1
I have to guess the answer B in the test. The trial and error calculation is too long.
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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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2
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....
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Re: Word Problem DS: Simple Equations  [#permalink]

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Bunuel wrote:
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?

A. Eunice sold a total of 8 cakes.
B. Eunice made 140 dollars in total revenue from her cakes.

This one is a typical problem!

For more on the questions with Diophantine equation (equations whose solutions must be integers only) check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Well, ESP on Test-Day,

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

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

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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

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.

There has to be an easier way?
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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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subhashghosh wrote:

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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How come C isn't the answer? I mean, both statements together are more than sufficient to give an exact answer.
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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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Ooh.. got it. Statement A does not say she sold total of 8 cakes of 17 and 19 dollars, but just total cakes.
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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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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

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

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

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1
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

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

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