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Eunice sold several cakes. If each cake sold for either

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

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New post 18 Dec 2016, 02:42
gmat1220 wrote:
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 ?


I would use the same "Unit Digit" approach.
Though, without taking A into consideration, how can you assume the total number of cakes are 8 though?

Or are we essentially taking info from A and just to try if there is only 1 combination for condition B?
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New post 23 Mar 2017, 22:59
1
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.
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New post 23 Mar 2017, 23:00
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!!!



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.
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New post 29 Mar 2017, 18:23
The way I looked at it -

1) Not enough but true
2) we need 17x + 19y = 140.

I was going to check for (x*7)+(x*9) = 0, only one combination worked. Tried it out and it worked. So hence B.
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New post 10 Nov 2017, 08:37
this is never an official problem

in og, we meet similar but different problem, in which oa is b but

it is a very simple math to find that there is one couple x,y which meet choice B.

we never have to do a lot of computation like in this problem.
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New post 15 Dec 2017, 11:16
gmatpapa wrote:
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..




Not true always, if a question offers several sets of possibilites then answer wont always be B. it would be C or E instead.

here is an example of such question.

https://gmatclub.com/forum/joe-bought-o ... 06212.html
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New post 15 Dec 2017, 11:53
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!!!



this would take so long to do!
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New post 07 Mar 2018, 12:55
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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.
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New post 21 Mar 2018, 22:56
1
There is an easier way. A bit shorter.

Now, we have 17x + 19y = 140

Notice that 140 has a 0 at its units digit, we just need to find which combination of units digit of 17x and 19y would give us a 0.

17x ----> 7, 4, 1, 8, 5....
19y ----> 9, 8, 7, 6, 5....

We see that 1 of 17x and 9 of 19y make up 0, and also both the 5s, and 4 & 6.

Let's take them.

(x = 3, y = 1)

17*3 + 19*1 = 70 !! (Voila! 70 is exactly the half of 140) (So, double them)

(17*3 + 19*1)*2 = 140

Which shows that x = 6 and y = 2

6 & 2 are the answers, and we can guess that no other integers would satisfy this equation. Also, we can cross-check this with option A that 6+2 = 8. And we cannot say that the other case of 5 & 5 would be 8. So we can ignore this.

To check for 6 & 4, try it 17*2 + 19*4 = 110 (Not close enough)

So, solely by checking units digit, and by doing some educated leaps, we can conclude that the answer is B.

:D


gmatpapa wrote:
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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New post 25 May 2019, 03:25
1
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.

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?

(1) Eunice sold a total of 8 cakes.
(2) Eunice made 140 dollars in total revenue from her cakes.


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.


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New post 16 Jun 2019, 13:34
GMATGuruNY wrote:
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


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
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New post 17 Jun 2019, 03:43
1
Mo2men wrote:
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?


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.

Quote:
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?


The increment approach requires that the coefficients for x and y be in their most reduced form.
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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 17 Jun 2019, 08:24
Given either exact 17 $ or 19 $ is price of cake

1. total 8 cakes ,,,,not sufficient
2. 140 $ total = 17 x + 19 y = 140
put x =0,1,2,3,4,5,6,7,8,9 randomly and check value of y

at X =6 , Y =2

B is sufficient
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Re: Eunice sold several cakes. If each cake sold for either   [#permalink] 17 Jun 2019, 08:24

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