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Re: Sara went to the store and bought both German chocolate bars and Swiss [#permalink]
Hoehenheim wrote:
We are given two variables say S and G to be quantified as per the equation xS + yG = 35

Quote:
(1) The Swiss chocolate bars cost $3.50 per bar, and the German chocolate bars cost $2.50 per bar.

now we know it is actually 3.5x + 2.5y = 35. On initial analysis it might seem that this is not enough as it is one equation with two unknown variables, however do know that we are to find positive integer values for both and that decreases the probability of there being many different solutions especially for smaller numbers. What if, we can arrive at a single unique solution using this equation by trial and error?
Let's see:
we know that Sara purchased both, so neither value can be zero. That discards 10x3.5 = 35, i.e 10S as the answer, so number of S has to be less than 10. Working our way backwards, it will be important to establish a number where say 3.5a=2.5b. This is only true for when 3.5x5=17.5=2.5x7
Since 5S and 7G are the only possible solutions for 3.5x + 2.5y = 35 as x has to lie between 1 and 10, we have arrived at our unique solution. Note that the only scenario that has led to this solution is the fact that the shop is selling whole numbered- chocolate items.
(1) is sufficient.

Quote:
(2) Sara purchased 2 more German chocolate bars than she did Swiss chocolate bars.

We do not know how much they cost, it could be 10S and 50G or even 1S and 1G.
Not sufficient.

Answer: A

One trick to solving this quickly is to sneak a peek onto Statement (2) only for the values and check if (C) i.e, both statements together stand, it they do, try to arrive at those values using only Statement 1 and if at all there are alternative solutions. That is how you decide between (A) and (C)

These kind of questions are particularly to be expected when you are scoring well, so it is advisable to be on the lookout for prime number values and easily multiplicative numbers for such DS questions.

­Hello, 

In the solution are we making an assumption that Sara invested the same amount to buy both chocolates as suggested by 3.5a=2.5b?
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Re: Sara went to the store and bought both German chocolate bars and Swiss [#permalink]
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vik.gupta007 wrote:
In the solution are we making an assumption that Sara invested the same amount to buy both chocolates as suggested by 3.5a=2.5b?

­No, we can't assume anything like that unless we're told. It just happens that there's no other way for the numbers to total up to $35. Here's one way to think about it: start with all $3.50 bars, so 10 bars for a total of $35. As Hoehenheim pointed out, the question doesn't allow this case, so let's adjust and look for another one. If we don't have 10 $3.50 bars, we need to take some out, but the remaining quantity must be divisible by $2.50. So what happens when we start putting bars back? When we buy 9 bars for $3.50, there is now $3.50 left to spend, but $2.50 won't go into $3.50. What if we put another bar back? Now we have $7 remaining, but $2.50 won't go into that, either. Keep on going, and you'll have this list of possible amounts to spend on $2.50 bars: $3.50, $7, $10.50, $14, $17.50, $21, $24.50, $28, $31.50. How can we tell which of these $2.50 might go into? A multiple of $5 would work, but the first one of those we'll see is $35, so that's out. So we need something that is $2.50 more than a multiple of $5. Only $17.50 works.

 
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Re: Sara went to the store and bought both German chocolate bars and Swiss [#permalink]
DmitryFarber wrote:
vik.gupta007 wrote:
In the solution are we making an assumption that Sara invested the same amount to buy both chocolates as suggested by 3.5a=2.5b?

­No, we can't assume anything like that unless we're told. It just happens that there's no other way for the numbers to total up to $35. Here's one way to think about it: start with all $3.50 bars, so 10 bars for a total of $35. As Hoehenheim pointed out, the question doesn't allow this case, so let's adjust and look for another one. If we don't have 10 $3.50 bars, we need to take some out, but the remaining quantity must be divisible by $2.50. So what happens when we start putting bars back? When we buy 9 bars for $3.50, there is now $3.50 left to spend, but $2.50 won't go into $3.50. What if we put another bar back? Now we have $7 remaining, but $2.50 won't go into that, either. Keep on going, and you'll have this list of possible amounts to spend on $2.50 bars: $3.50, $7, $10.50, $14, $17.50, $21, $24.50, $28, $31.50. How can we tell which of these $2.50 might go into? A multiple of $5 would work, but the first one of those we'll see is $35, so that's out. So we need something that is $2.50 more than a multiple of $5. Only $17.50 works.


 

­Hello Dmitry, 

Thank you for the detailed response and clarifying my query. 

This is much appreciation and very helpful. 

Thanks!
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Re: Sara went to the store and bought both German chocolate bars and Swiss [#permalink]
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