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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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02 Nov 2016, 09:30
TheLordCommander wrote: Although I got this right, it took me 10 minutes to solve it. Can anyone help me with a strategy to tackle this type of question on GMAT? There is no way im going to solve this question on the exam. I was in the same boat. glt13's solution is elegant (it uses proportions to arrive at the result). Could you follow the solution?
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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02 Nov 2016, 20:56
Dev1212 wrote: Not convinced mate... I am able to get to fractions 1/4 and 2/3.... but given the total number of items is 10, why would make the ratio of 1/4 to 2/8 and keep the ratio 2/3 as it is ? If someone can breakdown this using the same formula that would be helpful. colorblind wrote: Bunuel wrote: At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Kudos for a correct solution. \(\frac{Wa}{Wo} = \frac{(6056)}{(5640}) = \frac{1}{4}\) Apple = 2 & Orange = 8 New Avg \(\frac{Wa}{Wo} = \frac{(6052)}{(5240)} = \frac{2}{3}\) \(\frac{2}{Wo} = \frac{2}{3}\) Orange = 3
Hence, Oranges have to reduce from 8 to 3 i.e. 5 The original ratio of number of apples to number of oranges we found is 1:4. So there is 1 apple for every 4 oranges. Total number of fruits on the ratio scale is 1+4 = 5. Total number of fruits is actually 10. So actually, there must have been 2 apples and correspondingly, 8 oranges. With the new average of 52, the ratio of apples to oranges is 2:3. So for every 2 apples, there are 3 oranges. We don't know the total number of fruits after putting oranges back. Now note that number of apples doesn't change. Only oranges are put back. In original case, we found that there were 2 apples. So finally also there will be 2 apples only. Since the ratio of apples to oranges is 2:3 and number of apples actually is 2, number of oranges actually must be now 3. From 8, number of oranges has gone down to 3. So 5 oranges have been put back.
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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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13 Mar 2017, 15:28
Hi,
I don't think this is the best way to go about solving this problem, but this is how I solved this:
Since the average is 56c, the total cost of the 10 fruits is 560 cents. We need to bring it down to 52c average, by removing x number of oranges that cost 60c. So the equation becomes
(total cost in cents)  (reduced amount) = (new average cost) * (new number of fruits)
560  60x = 52 * (10x)
simplifying
560  60x = 520  52x
simplifying further
8x = 40
results in x = 5. So she needs to remove 5 oranges in order to bring the average down to 52c.
BTW, is there a way to estimate what level problem is this? i.e. is it a 500 level or 600 level problem? My apologies, I'm new at this.



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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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13 Mar 2017, 19:29
Bunuel wrote: At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 1 apple & orange pair costs $1.00 5 apple & orange pairs cost $5.00 we need to have 10 total pieces at a total price of $5.60 since oranges cost 20¢ more than apples, we need to add 3 more oranges and subtract 3 apples thus, she buys 5+3=8 oranges and 53=2 apples for $5.60 let x=number of oranges she needs to put back [2*40+(8x)60]/(10x)=52¢ x=5 E
Last edited by gracie on 23 Jul 2017, 09:20, edited 1 time in total.



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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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25 Mar 2017, 08:57
I really loved this method for weighted average problems: A is for 0,4 $ O is for 0,6 $ average is 0,56$ Let's find the ration
5640/6056=16/4=8/2 8 oranges and 2 apples
Let's find the ration after she put the fruits back 5240/6052=3/2 Note that nothing happened to apples, hence there' are 2 apples and 3 oranges
83=5 5 Oranges had to be put back Answer is E



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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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28 Jun 2017, 17:14
VeritasPrepKarishma wrote: Been looking at Karishma's responses to a lot of questions because I want to get better at using weighted averages to solve. Since it doesn't look like there are a lot of responses to this post that used weighted average to solve, so here is my attempt! Weight of Apple/Weight of orange= (orange price  overall average)/(overall average  apple price) A/O = (6056)/(5640)=4/16 = 1/4 So we know 4x+1x=10 (for quantity) x=2. 2 apples, and 8 oranges is what she originally has Then we do the weighted average equation for the new desired average: W of Apples/W of Oranges = (6052)/(5240) = 8/12 = 2/3 2 apples for every 3 oranges to get the new average since we know she already had 2 apples, we know then that she should have 3 oranges, which is 5 less than 8 the answer is 5 Am I getting the process of weighted averages right here?



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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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28 Jun 2017, 19:05
Bunuel wrote: At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Kudos for a correct solution. 1. Total cost is 560 cents 2. If she puts 1 orange back , average cost is 500/9. Similarly for 2,3,4, and 5 oranges back it is 440/8, 380/7, 320/6, 260/5 We see 260/5 gives 52 cents average cost.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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29 Jun 2017, 00:36
brianne5 wrote: VeritasPrepKarishma wrote: Been looking at Karishma's responses to a lot of questions because I want to get better at using weighted averages to solve. Since it doesn't look like there are a lot of responses to this post that used weighted average to solve, so here is my attempt! Weight of Apple/Weight of orange= (orange price  overall average)/(overall average  apple price) A/O = (6056)/(5640)=4/16 = 1/4 So we know 4x+1x=10 (for quantity) x=2. 2 apples, and 8 oranges is what she originally has Then we do the weighted average equation for the new desired average: W of Apples/W of Oranges = (6052)/(5240) = 8/12 = 2/3 2 apples for every 3 oranges to get the new average since we know she already had 2 apples, we know then that she should have 3 oranges, which is 5 less than 8 the answer is 5 Am I getting the process of weighted averages right here? Yes, it is correct. Good job!
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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22 Jul 2017, 19:14
VeritasPrepKarishma wrote: karishma, for this type of questions, do we need to maintain the initial quantity. if we have to maintain 10 fruits and achieve 52. it is still possible. 4 : 6. so, so i thought number of fruits to put back was 2. please let me know how to go about this questions!!



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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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22 Jul 2017, 21:27
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Bunuel wrote: At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Kudos for a correct solution. Let the number of Apples and Oranges be \(x\) and \(y\), respectively. \(40x + 60y = 560\) \(x + y = 10\) Solving the above equations, we get \(x = 2\) and \(y = 8\). Let the number of Oranges that Mary must keep for the average to be 52 cents be \(k\). Then, \(\frac{(40*2) + 60k}{2+k} = 52\) Solving the above equation gives \(k = 3\). So, the number of Oranges Mary must put back = \(83 = 5\). Ans  E.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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24 Jul 2017, 23:11
Avinash_R1 wrote: VeritasPrepKarishma wrote: karishma, for this type of questions, do we need to maintain the initial quantity. if we have to maintain 10 fruits and achieve 52. it is still possible. 4 : 6. so, so i thought number of fruits to put back was 2. please let me know how to go about this questions!! The question tells you that initial quantity is not maintained. It doesn't talk about picking more fruits/replacing fruits etc. If you want to maintain the number of fruits to 10 and achieve 52 cents as the average price, you know that w1/w2 = (60  52)/(52  40) = 8/12 = 2/3 So there should be 2 apples for every 3 oranges. This means there should be 4 apples and 6 oranges. But actually she has 2 apples and 8 oranges. So she needs to put back 2 oranges and pick 2 more apples.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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31 Oct 2017, 00:35
For the first step, since 56 cents is closer to 60 cents, you must infer that number of oranges is more than that of apples. Hence compute the average price for 4 apples & 6 oranges > 52 cents 3 apples & 7 oranges > 54 cents 2 apples & 8 oranges > 56 cents You know know that initially there were 2 apples and 8 oranges. Then you answer the question ..........
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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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16 Nov 2017, 17:00
Please see the attached snapshot for a weighted average (teetertotter or seesaw view of the problem)
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teet.PNG [ 82.59 KiB  Viewed 183 times ]
Last edited by FANewJersey on 03 Dec 2017, 20:23, edited 1 time in total.



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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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03 Dec 2017, 16:55
colorblind wrote: Bunuel wrote: At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Kudos for a correct solution. \(\frac{Wa}{Wo} = \frac{(6056)}{(5640}) = \frac{1}{4}\) Apple = 2 & Orange = 8 New Avg \(\frac{Wa}{Wo} = \frac{(6052)}{(5240)} = \frac{2}{3}\) \(\frac{2}{Wo} = \frac{2}{3}\) Orange = 3
Hence, Oranges have to reduce from 8 to 3 i.e. 5 I think this is the most efficient way to solve this problem. I was able to reduce my solving time by 2/3 using the weighted average approach for this problem. I'm just not sure how I can determine that this method will work on a problem similar to this one!



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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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03 Dec 2017, 20:29
hdavies wrote: colorblind wrote: Bunuel wrote: At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Kudos for a correct solution. \(\frac{Wa}{Wo} = \frac{(6056)}{(5640}) = \frac{1}{4}\) Apple = 2 & Orange = 8 New Avg \(\frac{Wa}{Wo} = \frac{(6052)}{(5240)} = \frac{2}{3}\) \(\frac{2}{Wo} = \frac{2}{3}\) Orange = 3
Hence, Oranges have to reduce from 8 to 3 i.e. 5 I think this is the most efficient way to solve this problem. I was able to reduce my solving time by 2/3 using the weighted average approach for this problem. I'm just not sure how I can determine that this method will work on a problem similar to this one! It will work as long as you practice the concept of weighted average over and over again. Here is a tip for you: Teeter Tooter approach to solving weighted average works on questions, in which we have two averages and then the average of those two averages. Question for you: Why do you think that 40 cents is average for the price of Apples and that 60 cents is the average of the price of Oranges. If you know the answer to this question then it is one step in the right direction to discern if this method of applying weighted average concept could help bring benefits.




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