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In the first hour of a bake sale, students sold either cho
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08 Jun 2013, 18:04
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In the first hour of a bake sale, students sold either chocolate chip cookies, which sold for $1.30, or brownies, which sold for $1.50. What was the ratio of chocolate chip cookies sold to brownies sold during the first hour of the bake sale? (1) The average price for the items sold during that hour was $1.42 (2) The total price for all the goods sold was $14.20
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Re: In the first hour of a bake sale, students sold either cho
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09 Jun 2013, 01:38




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Re: In the first hour of a bake sale, students sold either cho
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08 Jun 2013, 18:20
(1) \(\frac{1.30c+1.50b}{c+b} = 1.42\) \(1.30c+1.50b=1.42(c+b)\) \(1.30c+1.50b=1.42c+1.42b\) \(0.08b=0.12c\) \(\frac{c}{b} = \frac{0.08}{0.12}\) \(\frac{c}{b} = \frac{2}{3}\) Sufficient (2) \(1.30c+1.50b=14.20\) the only possible solution is: \(1.30(4)+1.50(6)=14.20\) Ratio is 4:6 or 2:3 Sufficient Answer is D Kudos if you like the explanation




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Re: In the first hour of a bake sale, students sold either cho
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10 Feb 2014, 05:35
Bunuel wrote: In the first hour of a bake sale, students sold either chocolate chip cookies, which sold for $1.30, or brownies, which sold for $1.50. What was the ratio of chocolate chip cookies sold to brownies sold during the first hour of the bake sale?(1) The average price for the items sold during that hour was $1.42 > \(\frac{1.30c+1.50b}{c+b}=1.42\) > \(8b=12c\) > \(\frac{c}{b} = \frac{2}{3}\). Sufficient. (2) The total price for all the goods sold was $14.20. This statement is a bit trickier: \(1.30c+1.50b=14.20\) > \(13c+15b=142\). Since c and b must be integers, then we should check whether this equation has one or more than one positive integer solutions: \(15b=14213c\) > 142 minus multiple of 13 must be a multiple of 15: only c=4 and b=6 satisfies the equation, thus \(\frac{c}{b} = \frac{4}{6}\). Sufficient. Answer: D. For more on this type of questions check: eunicesoldseveralcakesifeachcakesoldforeither109602.htmlmarthaboughtseveralpencilsifeachpencilwaseithera100204.htmlarentalcaragencypurchasesfleetvehiclesintwosizesa105682.htmljoeboughtonlytwentycentstampsandthirtycentstamps106212.htmlacertainfruitstandsoldapplesfor070eachandbananas101966.htmljoannaboughtonly015stampsand029stampshowmany101743.htmlatanamusementparktomboughtanumberofredtokensand126814.htmlcollectionsconfusedneedahelp81062.htmlHope it helps. The most important thing here is.. How do we get the numbers fast for the second statement. I remember @VeritasPrepKarishma explained the theory behind number picking for linear equations with integer constraints somewhere.. Bunuel would you please advice on this issue? It would be a huge takeaway for future problems Thanks!! Cheers J



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Re: In the first hour of a bake sale, students sold either cho
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10 Feb 2014, 23:33
St1: The average price for the items sold during that hour was $1.42.
Sufficient. We can use the allegation method to get the ratios. There is no need to calculate but it comes out as 2:3.
Down to A or D.
St2: The total price for all the goods sold was $14.20.
1.3c + 1.5b = 14.2
Multiplying by 10 across
13c + 15b = 142
After some hit and trial, we find 90 (15*6) when subtracted from 142 gives 52 which is 13*4. Hence, we get the quantities are 4 and 6 respectively. Ratio = 2:3.
This statement is also sufficient.
Answer (D).



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Re: In the first hour of a bake sale, students sold either cho
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03 Jun 2014, 08:10
Bunuel wrote: In the first hour of a bake sale, students sold either chocolate chip cookies, which sold for $1.30, or brownies, which sold for $1.50. What was the ratio of chocolate chip cookies sold to brownies sold during the first hour of the bake sale?
(1) The average price for the items sold during that hour was $1.42 > \(\frac{1.30c+1.50b}{c+b}=1.42\) > \(8b=12c\) > \(\frac{c}{b} = \frac{2}{3}\). Sufficient.
(2) The total price for all the goods sold was $14.20. This statement is a bit trickier: \(1.30c+1.50b=14.20\) > \(13c+15b=142\). Since c and b must be integers, then we should check whether this equation has one or more than one positive integer solutions: \(15b=14213c\) > 142 minus multiple of 13 must be a multiple of 15: only c=4 and b=6 satisfies the equation, thus \(\frac{c}{b} = \frac{4}{6}\). Sufficient.
Answer: D.
For more on this type of questions check:
Hope it helps. How can we use B to solve it? I agree that you have found out a ratio 4/6 but that is the ratio of total goods sold not the ratio of goods sold during first hour of the sale.



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Re: In the first hour of a bake sale, students sold either cho
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03 Jun 2014, 08:20
rohan567 wrote: Bunuel wrote: In the first hour of a bake sale, students sold either chocolate chip cookies, which sold for $1.30, or brownies, which sold for $1.50. What was the ratio of chocolate chip cookies sold to brownies sold during the first hour of the bake sale?
(1) The average price for the items sold during that hour was $1.42 > \(\frac{1.30c+1.50b}{c+b}=1.42\) > \(8b=12c\) > \(\frac{c}{b} = \frac{2}{3}\). Sufficient.
(2) The total price for all the goods sold was $14.20. This statement is a bit trickier: \(1.30c+1.50b=14.20\) > \(13c+15b=142\). Since c and b must be integers, then we should check whether this equation has one or more than one positive integer solutions: \(15b=14213c\) > 142 minus multiple of 13 must be a multiple of 15: only c=4 and b=6 satisfies the equation, thus \(\frac{c}{b} = \frac{4}{6}\). Sufficient.
Answer: D.
For more on this type of questions check:
Hope it helps. How can we use B to solve it? I agree that you have found out a ratio 4/6 but that is the ratio of total goods sold not the ratio of goods sold during first hour of the sale. The point is that the red parts above talk about the same thing. Agree that it would have been better if the question specified that.
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Re: In the first hour of a bake sale, students sold either cho
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21 Nov 2016, 01:06
When the values of two groups are averaged, the result is a weighted average; it's called weighted average because instead of being exactly in the middle of the two groups, it is closer to the value of the larger group; there is a proportional relationship to the sizes of the groups: if group 1 is twice as big as group 2, the weighted average will be twice as close to group 1's value.
All this means that there are just 4 variables in a weighted avg question between 2 groups; the values of the individual groups, the relative weights of the groups (ratio of group sizes), and the weighted average. Knowing 3 of these unknowns is enough to discover the 4th.
Let's solve:
Statement 1: The prompt gives us 2 values up front; the values of the individual groups (price of choc chip and price of brownies). Statement 1 gives us the weighted average, so we have 3 out of 4 unknowns; that's enough to find the 4th unknown, the ratio of group sizes. SUFFICIENT
Statement 2: If we call b and c the number of brownies and cookies, we can write the equation 1.3c + 1.5b = 14.2. Simplify to: 13c + 15b = 142
In general, when you have two variables and one equation you will not have enough data to solve. There is at least one important exception:
✔ If your equation is of the form Ax + By = C where A, B and C are known, and ✔ if your unknowns are positive integers, and ✔ if C < the least common multiple of A & B, then
the equation has only 1 solution! Since our equation (13c + 15b = 142) meets all three criteria I can tell at a glance that it is SUFFICIENT.
Ans: D



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Re: In the first hour of a bake sale, students sold either cho
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23 Nov 2016, 19:27
I was stuck between A or D but I chose A because the second statement says for "ALL GOODS sold" and not "ALL GOODS sold in the first hour."



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Re: In the first hour of a bake sale, students sold either cho
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24 Nov 2016, 09:25
nrl.amin wrote: In general, when you have two variables and one equation you will not have enough data to solve. There is at least one important exception:
✔ If your equation is of the form Ax + By = C where A, B and C are known, and ✔ if your unknowns are positive integers, and ✔ if C < the least common multiple of A & B, then
the equation has only 1 solution! Since our equation (13c + 15b = 142) meets all three criteria I can tell at a glance that it is SUFFICIENT.
Ans: D Are you sure if any equation fulfill these 3 criteria, the equation will have only one possible integral solution? Can you explain it, how?
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Re: In the first hour of a bake sale, students sold either cho
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21 Mar 2017, 21:57
emmak wrote: In the first hour of a bake sale, students sold either chocolate chip cookies, which sold for $1.30, or brownies, which sold for $1.50. What was the ratio of chocolate chip cookies sold to brownies sold during the first hour of the bake sale?
(1) The average price for the items sold during that hour was $1.42 (2) The total price for all the goods sold was $14.20 OFFICIAL SOLUTION Correct Answer: D In statement (1), if the average price of the two items sold is given, then that information gives you the exact ratio of the two items. For instance, if it was learned that average price was $1.40 it would be clear that an equal number of each were sold. Since the average price is $1.42 (closer to $1.50) then it is clear that more brownies were sold and the ratio of cookies to brownies is the ration of the inverted distances from the average to each individual price or 8:12, which is reduced to 2:3 (see problem solving book for a review of this method). This can also be shown algebraically with c = # of cc cookies and b = # of brownies. $1.30(c) + $1.50(b) = $1.42(c + b). Removing parentheses the equation becomes 1.3c + 1.5b = 1.42c + 1.42b. Combining like terms the equation becomes .12c = .8b and the ratio of c:b = 8:12 = 2:3. Statement (1) is then sufficient. Statement (2) appears to be insufficient because only one equation can be created as follows: $1.30c + $1.50b = $14.20. However because c and b must be whole numbers and because of the properties of the numbers involved, this equation actually gives the exact number of each that were sold and thus the ratio. To see this, first simplify the equation to 13c + 15b = 142. Looking at the equation it is clear that it is not possible that more than 11 items were sold (11 of the cheapest item is more than 142) and it is not possible that less than 9 items were sold (9 of the most expensive items is still less than 142). With these limits, this equation also tells you that b + c = 10 and it is possible to solve and see that b = 6 and c = 4 and the ratio of c:b is 2:3. The correct answer is D, each statement alone is sufficient.
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Re: In the first hour of a bake sale, students sold either cho
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27 Apr 2017, 10:22
jlgdr wrote: Bunuel wrote: In the first hour of a bake sale, students sold either chocolate chip cookies, which sold for $1.30, or brownies, which sold for $1.50. What was the ratio of chocolate chip cookies sold to brownies sold during the first hour of the bake sale?(1) The average price for the items sold during that hour was $1.42 > \(\frac{1.30c+1.50b}{c+b}=1.42\) > \(8b=12c\) > \(\frac{c}{b} = \frac{2}{3}\). Sufficient. (2) The total price for all the goods sold was $14.20. This statement is a bit trickier: \(1.30c+1.50b=14.20\) > \(13c+15b=142\). Since c and b must be integers, then we should check whether this equation has one or more than one positive integer solutions: \(15b=14213c\) > 142 minus multiple of 13 must be a multiple of 15: only c=4 and b=6 satisfies the equation, thus \(\frac{c}{b} = \frac{4}{6}\). Sufficient. Answer: D. For more on this type of questions check: http://gmatclub.com/forum/eunicesolds ... 09602.htmlhttp://gmatclub.com/forum/marthabought ... 00204.htmlhttp://gmatclub.com/forum/arentalcar ... 05682.htmlhttp://gmatclub.com/forum/joeboughton ... 06212.htmlhttp://gmatclub.com/forum/acertainfru ... 01966.htmlhttp://gmatclub.com/forum/joannabought ... 01743.htmlhttp://gmatclub.com/forum/atanamuseme ... 26814.htmlhttp://gmatclub.com/forum/collectionsc ... 81062.htmlHope it helps. The most important thing here is.. How do we get the numbers fast for the second statement. I remember VeritasPrepKarishma explained the theory behind number picking for linear equations with integer constraints somewhere.. Bunuel would you please advice on this issue? It would be a huge takeaway for future problems Thanks!! Cheers J 13c+15b=142 or in other words, 15b=14213c. This last expression says that RHS will be a number which has either 0 or 5 as its units digit (since it is a multiple of 15). Which means 13c has 2 or 7 as its units digit. Hence 'c' can be 4 or 9 (Note that, c >10 not possible, so no need to check beyond 9). c=9 gives you b= \(\frac{25}{15},\) so c=4 is the only solution.



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Re: In the first hour of a bake sale, students sold either cho
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22 Sep 2018, 11:38
ryuken101 wrote: I was stuck between A or D but I chose A because the second statement says for "ALL GOODS sold" and not "ALL GOODS sold in the first hour." Statement (2) appears to be insufficient because only one equation can be created as follows: $1.30c + $1.50b = $14.20. However because c and b must be whole numbers and because of the properties of the numbers involved, this equation actually gives the exact number of each that were sold and thus the ratio. To see this, first simplify the equation to 13c + 15b = 142. Looking at the equation it is clear that it is not possible that more than 11 items were sold (11 of the cheapest item is more than 142) and it is not possible that less than 9 items were sold (9 of the most expensive items is still less than 142). With these limits, this equation also tells you that b + c = 10 and it is possible to solve and see that b = 6 and c = 4 and the ratio of c:b is 2:3. The correct answer is D, each statement alone is sufficient.



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Re: In the first hour of a bake sale, students sold either cho
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22 Sep 2018, 12:50
HimanshuW11 wrote: ryuken101 wrote: I was stuck between A or D but I chose A because the second statement says for "ALL GOODS sold" and not "ALL GOODS sold in the first hour." Statement (2) appears to be insufficient because only one equation can be created as follows: $1.30c + $1.50b = $14.20. However because c and b must be whole numbers and because of the properties of the numbers involved, this equation actually gives the exact number of each that were sold and thus the ratio. To see this, first simplify the equation to 13c + 15b = 142. Looking at the equation it is clear that it is not possible that more than 11 items were sold (11 of the cheapest item is more than 142) and it is not possible that less than 9 items were sold (9 of the most expensive items is still less than 142). With these limits, this equation also tells you that b + c = 10 and it is possible to solve and see that b = 6 and c = 4 and the ratio of c:b is 2:3. The correct answer is D, each statement alone is sufficient. thanks for the explanation




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