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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, 01:42
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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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 23 Mar 2017, 21:59
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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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 23 Mar 2017, 22:00
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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Joined: 27 Aug 2014
Posts: 56
Location: Canada
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GMAT 1: 660 Q45 V35
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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 29 Mar 2017, 17:23
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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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 10 Nov 2017, 07:37
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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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 15 Dec 2017, 10:16
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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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 15 Dec 2017, 10:53
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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Re: Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 07 Mar 2018, 11:55
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.
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Eunice sold several cakes. If each cake sold for either  [#permalink]

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New post 21 Mar 2018, 21:56
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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Eunice sold several cakes. If each cake sold for either &nbs [#permalink] 21 Mar 2018, 21:56

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