Eunice sold several cakes. If each cake sold for either exactly 17 or

Consider the following equation:
2x + 3y = 30.

If x and y are nonnegative integers, the following solutions are possible:
x=15, y=0
x=12, y=2
x=9, y=4
x=6, y=6
x=3, y=8
x=0, y=10

Notice the following:
The value of x changes in increments of 3 (the coefficient for y).
The value of y changes in increments of 2 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.



Let x = the number of $17 cakes and y = the number of $19 cakes.

Statement 1:
x+y = 8
Here, x and y can be any nonnegative values that sum to 8.
INSUFFICIENT.

Statement 2:
17x+19y = 140

Solve for x and y when x+y=8, as required in Statement 1.
Multiplying x+y = 8 by 17, we get:
17x + 17y = 136
Subtracting the blue equation from the red equation, we get:
2y = 4
y=2, implying in the blue equation that x=6.

Thus, one solution for 17x+19y=140 is as follows:
x=6, y=2
In accordance with the rule discussed above, the value of x may be altered only in INCREMENTS OF 19 (the coefficient for y), while the value of y may be altered only in INCREMENTS OF 17 (the coefficient for x).
Not possible:
If x increases by 19 and y decreases by 17, then y will be negative.
If x decreases by 19 and y increases by 17, then x will be negative.
Implication:
The only nonnegative integral solution for 17x+19y=140 is x=6 and y=2.
SUFFICIENT.

Truth be told, we can determine WITHOUT VIRTUALLY NO WORK AT ALL that Statement 2 is sufficient.
From my earlier post:



Statement 1 is clearly INSUFFICIENT.

Statement 2 implies the following:
17x + 19y = 140

Clearly, this equation must yield at least one viable integer combination for x and y.
As noted above:
Once a viable option for x has been determined, the value of x may change only in increments of 19 (the coefficient for y).
BUT SUCH A CHANGE IS NOT POSSIBLE.
If the value of x increases by 19, then the value of 17x will exceed the sum of 140.
If the value of x decreases by 19, then the value of 17x will be negative.
Implication:
ONLY ONE INTEGER COMBINATION for x and y will satisfy the equation.
SUFFICIENT.



Using this line of reasoning, we can determine the correct answer with virtually no work at all.

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..

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

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 ?

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?

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!

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....

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.

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).

Hello Friends,
My 2 cents...

In such scenarios i cheat: i use S-1 info in S-2 and try to solve - if i am able to solve viola we have to answer otherwise NOT

For example, here - we know that a + b = 8 and we need to find b therefore a = 8 -b in other equation:
a = 17 cents
b = 19 cents
17a+19b = 140
17*(8-b) + 19b = 140
136 - 17b + 19b = 140
2b = 4
b = 2 | we have our answer!!!

PLEASE NOTE: i ONLY do it in such scenarios as usually we SHOULD NOT carry data from S-1 to S-2.

I used following approach:

140 is divisible by 5 and 10 and is even.
19 is obviously a prime. How can we transfer the unit digit into a 0? E.g. by adding +1
How do we get +1 from 17? Multiply it by 3.
Hence, 19 + 17*3 = 19 + 51 = 70. Stop here. This answer is sufficient.

Dear GMATGuruNY

1- Are any answer of x=15, y=0 or x=0, y=10 correct in DS questions? Is it considered solution or logic to say that a person bought zero of x or y?

2- If the equation is altered to be (multiply by 2):

4x + 6y = 60

then the increment process above won't work
x=15, y=0

x=9, y=4

But such solution like x=12, y=2 won't appear if we take increments. should the equation be reduced to the last common factor before taking increments?


Thanks

Unless we are told -- or something in the prompt implies -- that x and y must be positive, we should assume that x=0 and y=0 are viable options.



The increment approach requires that the coefficients for x and y be in their most reduced form.

Hi, VeritasKarishma, is there any method wherein you know when to stop testing answers? I got (2,6) as a correct pair giving 17*6+19*2 as 140; but how do I know with certainty that there can be no other pair to satisfy the conditions?

Yes, only one solution needs to be found using hit and trial. After that, you can figure out all the rest.
Check here: https://anaprep.com/algebra-integer-sol ... variables/

Video solution from Quant Reasoning starts at 0:25
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