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Martha bought several pencils. If each pencil was either a

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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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New post 22 Sep 2017, 06:42
One good way to quickly find values of variables i.e x and y in such scenario's is carry information in statement 1 to 2 (Just for these type of scenario's) and make use of the concept: 'Statements NEVER contradict each other in DS !'...

Once you have the value of x and y forget about statement 1 and check if there is another possible pair... by using the method recommended by Bunuel :

It's quite easy to check whether 10F+9C=66 has one or more solutions. 9C=66-10F so 66 minus multiple of 10 must be multiple of 9: 66 is not multiple of 9; 56 is not; 46 is not; 36 IS MULTIPLE OF 9 (F=3 and C=4); 26 is not; 16 is not and 6 is not. So only one combination of F and C satisfies equation 10F+9C=66, namely F=3 and C=4. Sufficient.
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New post 02 Apr 2018, 18:54
ketanniper wrote:
I am approaching the problem from divisibility angle.

Since 23x+21y=130 and 130 is not divisible by 23 or 21. Martha can not only buy one type of pencil.
Now, If she buys both I need to know how many of 23 and 21 number pencil she bought. i.e. 23+21 = 44 => 130/44 = 2*44 + 42. So she can buy 2 -2 pencil of each number.

Now, reminder 42 is divisible by only 21 i.e. 21*2. so she bought additional two 21 number pencil.

Net we found the values that satisfies the conditions. the statement is sufficient - The answer is B.


Bunuel, Can we generalize the approach as suggested by ketanniper to all the problems that you have shared in your response earlier in this post?
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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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New post 08 Aug 2018, 09:54
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Bull78 wrote:
Martha bought several pencils. If each pencil was either a 23 cent pencil or a 21 pencil, how many 23 cent pencils did martha buy?

(1) Martha bought a total of 6 pencils

(2) The total value of the pencils Martha bought was 130 cents.


Target question: How many 23 cent pencils did Martha buy?

Given: Martha bought several pencils. Each pencil was either a 23-cent pencil or a 21-cent pencil.

Statement 1: Martha bought a total of 6 pencils
Clearly, statement 1 is NOT SUFFICIENT

Statement 2: The total value of the pencils Martha bought was 130 cents.
This statement APPEARS to be insufficient. However, since the pencils cost 21 cents and 23 cents, we can't buy many for 130 cents.
So, let's LIST THE POSSIBLE CASES

Case a: Martha bought ZERO 23-cent pencils (for a total of 0 cents).
This means she spent all 130 cents on the 21-cent pencils. However, 21 cents does not evenly divide into 130 cents
So, it cannot be the case that Martha bought ZERO 23-cent pencils

Case b: Martha bought ONE 23-cent pencil.
This means she spent the remaining 107 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 107 cents.
So, it cannot be the case that Martha bought ONE 23-cent pencil

Case c: Martha bought TWO 23-cent pencils (for a total of 46 cents).
This means she spent the remaining 84 cents on the 21-cent pencils. 21 cents divides into 84 cents exactly FOUR TIMES.
So, it's possible that Martha bought TWO 23-cent pencils and FOUR 21-cent pencils
In this case, the answer to the target question is Martha bought TWO 23-cent pencils

Case d: Martha bought THREE 23-cent pencils (for a total of 69 cents).
This means she spent the remaining 61 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 61 cents.
So, it cannot be the case that Martha bought THREE 23-cent pencils

Case e: Martha bought FOUR 23-cent pencils (for a total of 92 cents).
This means she spent the remaining 38 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 38 cents.
So, it cannot be the case that Martha bought FOUR 23-cent pencils

Case f: Martha bought FIVE 23-cent pencils (for a total of 15 cents).
This means she spent the remaining 15 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 15 cents.
So, it cannot be the case that Martha bought FIVE 23-cent pencils

Now that we've examined all of the cases, the only possible solution is the one described in case c
So, it MUST be the case that Martha bought TWO 23-cent pencils
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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New post 08 Aug 2018, 13:28
Is it fair to assume that even though statement 1 may be insufficient, that the statement is still correct?
Because if I assume so, then i can use it to make my trial and error considerably easier when evaluating if statement B alone is sufficient - as I know that the sum is 6 and the solutions are hence limited to :
1,5
2,4
3,3
4,2
5,1

So i can very quickly see that 2,4 will satisfy and determine that statement B alone is sufficient.

I guess my point is, a D.S question may give a statement that is insufficient but would it ever give a statement that's outright false?
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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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New post 31 Jan 2019, 06:21
To check if Statement (2) alone was sufficient, I created two columns with the multiples of the numbers 23 and 21 side by side, until they get to 130, so I could check how many combinations of these two columns would reach 130 together.

In this case:

23 21
46 42
69 63
92 84
115 108
- 129

So, I stated eliminating the numbers from the "column 23", starting with the number 23 itself: is there a number in the "column 21" that ends with 7? Because we need to reach a combination that sums 130. If there were, It would de needed to sum and check if it gets to 130. After eliminating each number of the first column checking it with all from the second column, I could see that there is only one possible combination: 46+84=130. Therefore, there is only one value for X and one for Y so the equation sums 130.

This occurs only in some cases where the numbers are positive integers.
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Re: Martha bought several pencils. If each pencil was either a   [#permalink] 31 Jan 2019, 06:21

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