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A certain fruit stand sold apples for $0.70 each and bananas
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Updated on: 03 Mar 2012, 23:43
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ? A. 10 B. 11 C. 12 D. 13 E. 15
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Originally posted by pzazz12 on 30 Sep 2010, 05:08.
Last edited by Bunuel on 03 Mar 2012, 23:43, edited 1 time in total.
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Re: help me to solve it.....
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30 Sep 2010, 05:22




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Re: help me to solve it.....
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30 Sep 2010, 19:49
pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 Without calculating anything in paper you could approach this problem. Know  Some multiple of 7 + Some multiple of 5 should yield 63. To get to a some multiple of 5, we should ensure that a 3 or 8 (5+3) should be a multiple of 7. 63 is a direct multiple of 7, however in this case there won't be any bananas. Hence the next option is to look for a multiple of 7 that has 8 as the unit digit. 28 satisfies this hence no. of apples is 4 and no of bananas is 7  Answer 11 (B).  35 seconds straight.
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Re: help me to solve it.....
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01 Oct 2010, 05:06
Bunuel wrote: pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 Given: \(0.7b+0.5a=6.3\) Question: \(a+b=?\) \(0.7a+0.5b=6.3\) > \(7a+5b=63\). After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: \(a=4\) and \(b=7\) > \(a+b=11\). Answer: B. thank you, but can you explain me how this (9,0) and (4,7) to be solve...



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Re: help me to solve it.....
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01 Oct 2010, 05:49
pzazz12 wrote: Bunuel wrote: pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 Given: \(0.7a+0.5b=6.3\) Question: \(a+b=?\) \(0.7a+0.5b=6.3\) > \(7a+5b=63\). After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: \(a=4\) and \(b=7\) > \(a+b=11\). Answer: B. thank you, but can you explain me how this (9,0) and (4,7) to be solve... Trial and error would be good for it, but here is another way: \(7a+5b=63\) > \(5b=637a\) > \(5b=7(9a)\) > \(5b\) must be multiple of 7 > \(b\) must be multiple of 7 > \(b\) can not be 0 (as "a customer purchased both apples and bananas") or >14 (as \(5b\) in this case would be more than $6.30), so \(b=7\) > \(a=4\). Hope it's clear.
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Re: help me to solve it.....
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10 Dec 2010, 20:14
ezhilkumarank wrote: pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 Without calculating anything in paper you could approach this problem. Know  Some multiple of 7 + Some multiple of 5 should yield 63. To get to a some multiple of 5, we should ensure that a 3 or 8 (5+3) should be a multiple of 7. 63 is a direct multiple of 7, however in this case there won't be any bananas. Hence the next option is to look for a multiple of 7 that has 8 as the unit digit. 28 satisfies this hence no. of apples is 4 and no of bananas is 7  Answer 11 (B).  35 seconds straight. i get some multipule of 5 and 7 make 63... but why the multiple of 7 with a 3 or 8?



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Re: help me to solve it.....
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11 Dec 2010, 00:01
mmcooley33 wrote: ezhilkumarank wrote: pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 Without calculating anything in paper you could approach this problem. Know  Some multiple of 7 + Some multiple of 5 should yield 63. To get to a some multiple of 5, we should ensure that a 3 or 8 (5+3) should be a multiple of 7. 63 is a direct multiple of 7, however in this case there won't be any bananas. Hence the next option is to look for a multiple of 7 that has 8 as the unit digit. 28 satisfies this hence no. of apples is 4 and no of bananas is 7  Answer 11 (B).  35 seconds straight. i get some multipule of 5 and 7 make 63... but why the multiple of 7 with a 3 or 8? ezhilkumarank means that as multiple of 5 ends with 5 or 0 then multiple of 7 must end with 8 or 3 in order their sum to end with 3 (63). There is another approach in my previous post. Hope it's clear.
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Re: help me to solve it.....
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11 Dec 2010, 08:20
haha crystal as usual.



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Re: A certain fruit stand sold apples for $0.70
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12 Dec 2010, 05:30
ajit257 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase? A. 10 B. 11 C. 12 D. 13 E. 14
Is there a faster way to do these problems other than brute force ? The first solution will invariably involve some brute force. But (9, 0) is easy to get since 63 is a multiple of 7. Check out this post for clarification on these type of questions: http://gmatquant.blogspot.com/2010/11/integralsolutionsofaxbyc.html
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Re: Alternate approaches to OG 12 PS #65?
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21 Dec 2010, 16:20



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Re: help me to solve it.....
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21 Dec 2010, 16:58
Maybe it could be of any help to realise that 0.7y + 0.5x = 6.3 is a straight line, we want the 'y' on the left side, rewriting gives: 7y + 5x = 63 y = 9  (5/7)x Almost instantly you should see that the right term (5/7) can only be an integer if X is either 7, 14, 28 and so forth. Ruling out fourteen we only have 7 left which gives 95 = 4 +7 fruits.
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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17 Dec 2013, 03:46
pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 He buys both, and the only combination of multiples of 5 and 7 that end in the single digit 3 is the single digits of the respective multiples ending in 8 and 5, which means 28 = 4*7 for bananas, and 7*5 = 35 for apples.. From here you can simply just add a decimal inbetween so that the restriction is upheld. 4 bananas and 7 apples gives us 11 fruits.



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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24 Nov 2014, 10:29
I just started by thinking ok at least 5*7 and 7*5 is an easy approach to start with to find multiples. If you look at the answer choices that would be 12 fruits, so somewhere in the middle of your options. In total that would be 70 (or $7), so clearly "one seven less" is 63, so that's 4*7 + 7*5 = fruits 4+7=11.



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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21 Dec 2014, 13:38
from the question we get the following equation : 0.7A+0.5B = 6.3 this can be written as : 7A+5B = 63 Now rather than randomly plugging in values , we can do the following : 5B = 63  7A ==> 5B = 7(9A)
this tells us that B should be a multiple of 7. it cannot be 14 as then the value will be greater than 63. hence B = 7 and A = 4 Ans is 11



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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22 Dec 2014, 00:51
7x+5y=63
multiple of 5 gives numbers only ending 0 or 5, so multiple of 7 should end 3 or 8 respectively. 7*9=63 (eliminate  no option for 5) 7*8=56 7*7=49 7*6=42 7*5=35 7*4=28 7*3=21 7*2=14 7*1=7
it is 4+(6328)/5=4+7=11
B



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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05 Jan 2015, 08:02
A = Apples B = Bananas
0.7A + 0.5B = 6.3 7A + 5B = 63 5B = 63  7A 5B = 7 (9  A)
Therefore, B has to be = 7 And, 5 = (9  A) i.e. A = 4
A + B = 7 + 4 = 11
Answer: B



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08 Oct 2016, 20:18
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A certain fruit stand sold apples for $0.70 each and bananas
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Updated on: 09 Jan 2017, 15:33
The algebraic explanations in this thread are valid but needlessly complicated. Plus, it is unlikely that you will be able to come up with a similar approach to such a problem on your own. There is a much simpler and more broadly applicable approach that doesn't require trial and error. Because the numbers of apples and bananas have to be integers, the easiest thing to do here is start with the total of 6.3 and subtract off .7 until you arrive at a multiple of .5. 6.3  .7 = 5.6 5.6  .7 = 4.9 4.2  .7 = 4.2 4.2  .7 = 3.5 We are now at a multiple of .5, having subtracted off 4 apples. Because the bananas are 50 cents each, this gives us 7 bananas, for a total of 11 pieces of fruit.
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Originally posted by dannythor6911 on 05 Jan 2017, 15:24.
Last edited by dannythor6911 on 09 Jan 2017, 15:33, edited 1 time in total.



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A certain fruit stand sold apples for $0.70 each and bananas
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05 Jan 2017, 17:48
My approach: A>Apples:0.70 B>Bananas: 0.50 so.. Bought Spent 1A+1B= 1.20 2A+2B= 2.40 3A+3B= 3.60 4A+4B= 4.80 5A+5B= 6.00 > 5+5 =10 but still have 0,30 that were spent so it cant be (A), the next possible value is 11 according to the answers (B) I hope that works!



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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06 Jan 2017, 18:34
0.7a + 0.5 b = 6.30$ Assume a = 5, b = 5 3.5 + 2.5 = 6.0$ more 0.3$ more 10 11 B
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