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A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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) 14
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Originally posted by Bunuel on 30 Sep 2010, 05:08.
Last edited by Bunuel on 20 Feb 2019, 04:26, edited 2 times in total.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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30 Sep 2010, 05:22
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.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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21 Dec 2010, 16:20



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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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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15 Oct 2012, 04:51
Let the no of apples sold = a no of bananas sold = b Question is a+b=? Thus 70a + 50b = 630 7a + 5b = 63 Quick Tip In order to find out the value of a & b its better to find the value of 'a' as "63 7a" must leave a number which will end either with 0 or with 5. (Think about it for a second) Thus the only value which satisfies above equation is a=4 & b=7 a+b=11 (Other values of a & b will lie outside the answer choices) Answer B
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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15 Oct 2012, 08:39
7A + 5B = 63
5B = 63  7A
B= (63  7A)/5 => must be an integer, as we are dealing with quantities unless we can buy 1/4 of an apple or 1/3 of a banana (That could be funny :D )
so 63  7A must yield to a multiple of 5.
Now, let's pick some numbers.
if, A= 1, 637= 56 (not divisible by 5) A= 2, 6314= 49 (not divisible by 5) A= 3, 6321= 42 (not divisible by 5) A= 4, 6328= 35 (divisible by 5)
so A= 4, B= 7 A+B= 11 , Answer B



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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15 Oct 2012, 10:50
To solve this question  we can take numbers, as price of apple 7, 5 for Banana and 63 total for ease. Now we can determine quickly that total number should range between 63/7 <= N <=63/5, so ans should be between 9 and 12. Now solving the expression 7A+5B =63 first possibility with 9 apples, 0 banana we get 6.30 total amount, but question says customer purchased both, apple and banana. so not correct. So next choice, for 7A+5B =63 would come by decreasing 63 in multiple of 5 and checking divisibility of that number by 7. this way we get 4 Apples *0.70 + 7 banana *050 = 6.30 Hence total number is 7+4 =11 Ans B Bunuel wrote: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectA 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 Practice Questions Question: 64 Page: 161 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a solution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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16 Oct 2012, 03:53
so apples is 0.70 * A
Bananas 0.50 * B
then 0.70A + 0.50B = 6.30
multiply by 10 we get
7A + 5B = 63
5B = 63  7A B = 7(9A)/5
now to satisfy this equation we need 9  A = 5 only then it will be divisible by 5 therefore A is 4 and when solve we get B is 7
7(94)/5 = 7*5/5 then we need the sum of A + B = 7 + 4 = 11
Answer B



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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17 May 2016, 19:54
Attached is a visual that should help.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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09 Jan 2017, 10:33
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 We are given that apples were sold for $0.70 each and that bananas were sold for $0.50 each. We can set up variables for the number of apples sold and the number of bananas sold. b = number of bananas sold a = number of apples sold With these variables, it follows that: 0.7a + 0.5b = 6.3 We can multiply this equation by 10 to get: 7a + 5b = 63 Notice that we do not have any other information to set up a second equation, as we sometimes do for problems with two variables. So, we must use what we have. Keep in mind that variables a and b MUST be whole numbers, because you can't purchase 1.4 apples, for example. Notice also that 7 and 63 have a factor of 7 in common. Thus, we can move 7a and 63 to one side of the equation and leave 5b on the other side of the equation, and scrutinize the new equation carefully: 5b = 63 – 7a 5b = 7(9 – a) b = [7(9 – a)]/5 Remember that a and b MUST be positive whole numbers here. Thus, 5 must evenly divide into 7(9 – a). Since we know that 5 DOES NOT divide evenly into 7, it MUST divide evenly into (9 – a). We can ask the question: What must a equal so that 5 divides into 9 – a? The only value a can be is 4. We can check this: (9 – a)/5 = ? (9 – 4)/5 = ? 5/5 = 1 Since we know a = 4, we can use that to determine the value of b. b = [7(9 – 4)]/5 b = [7(5)]/5 b = 35/5 b = 7 Thus a + b = 4 + 7 = 11. Answer: B
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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07 Oct 2017, 06:43
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 Here's an approach where we test the POSSIBLE CASES. FACT #1: ( total cost of apples) + ( total cost of bananas) = 630 CENTS FACT #2: total cost of bananas is DIVISIBLE by 50, since each banana costs 50 cents. Now let's start testing POSSIBLE scenarios. Customer buys 1 apple. 1 apple costs 70 cents, which means the remaining 560 cents was spent on bananas. Since 560 is NOT divisible by 50, this scenario is IMPOSSIBLE Customer buys 2 apples. 2 apple costs 140 cents, which means the remaining 490 cents was spent on bananas. Since 490 is NOT divisible by 50, this scenario is IMPOSSIBLE Customer buys 3 apples. 3 apple costs 210 cents, which means the remaining 520 cents was spent on bananas. Since 520 is NOT divisible by 50, this scenario is IMPOSSIBLE Customer buys 4 apples. 4 apple costs 280 cents, which means the remaining 350 cents was spent on bananas. Since 350 IS divisible by 50, this scenario is POSSIBLE 350 cents buys 7 bananas. So, the customer buys 4 apples and 7 bananas for a total of 11 pieces of fruit Answer: Cheers, Brent
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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26 May 2018, 07:46
After I thought of an alternative method to solve this type of problems ( an equation with two variables and constraints ) I found this method : this equation 7x+5y= 63 represents a line draw the line (by choosing two points , the easiest is when x=0 ,then y= ? ; when y=0,then x=? ) after drawing the line , search for a point that has x ,y integers it's (4,7) However , this method is not practical because it requires precise drawing , but it may give you an indication ( for example , plug and try values of x , not y because x has fewer values )
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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15 Jan 2019, 04:39
Consider the following equation: 2x + 3y = 30. If x and y are nonnegative integers, the following solutions are possible: x=15, y=0 x=12, y=2 x=9, y=4 x=6, y=6 x=3, y=8 x=0, y=10 Notice the following: The value of x changes in increments of 3 (the coefficient for y). The value of y changes in increments of 2 (the coefficient for x). This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers. Bunuel 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 70x + 50y = 630 7x + 5y = 63 In accordance with the pattern illustrated above, we get the following nonnegative solutions for x and y: x=9, y=0 x=4, y=7Here  since apples and bananas are both purchased  x and y must both be positive. Thus, only the option in green is viable, with the result that x+y = 4+7 = 11.
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