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15 Oct 2012, 04:33
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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 Practice Questions Question: 64 Page: 161 Difficulty: 600 [Reveal] Spoiler: OA _________________ Math Expert Joined: 02 Sep 2009 Posts: 44609 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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15 Oct 2012, 04:34
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SOLUTION

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 Given: $$0.7b+0.5a=6.3$$ Question: $$a+b=?$$ $$0.7a+0.5b=6.3$$ --> $$7a+5b=63$$ --> $$5b=63-7a$$ --> $$5b=7(9-a)$$ --> $$5b$$ 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 >13 (as $$5b$$ in this case would be more than$6.30), so $$b=7$$ --> $$a=4$$ --> $$a+b=4+7=11$$.

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15 Oct 2012, 08:39
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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, 63-7= 56 (not divisible by 5)
A= 2, 63-14= 49 (not divisible by 5)
A= 3, 63-21= 42 (not divisible by 5)
A= 4, 63-28= 35 (divisible by 5)

so A= 4, B= 7
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15 Oct 2012, 10:50
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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 Project

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 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 Project Each 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! _________________ Lets Kudos!!! Black Friday Debrief Manager Joined: 21 Sep 2012 Posts: 228 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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16 Oct 2012, 03:53
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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(9-A)/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(9-4)/5 = 7*5/5 then we need the sum of A + B = 7 + 4 = 11

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24 Aug 2014, 23:18
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Getting rid of decimals

70x + 50y = 630

630 is divisible by 70; so 50y should also be divisible by 70 & 50

LCM of 50 & 70 = 350

y $$= \frac{350}{50} = 7$$

x $$= \frac{280}{70} = 4$$

Total = x+y = 7+4 = 11

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03 Dec 2014, 20:58
Honestly, I hemmed and hawed over this easy question before I found the right answer.

How I reached the original answer: I knew that 6.3 had to have some a multiple of 0.5 in it so I added and subtracted 0.5 to see if the result was a multiple of 7.

How I would do it now: I would break down 7a + 5b= 63.
b= 63 - 7a / 5
b = 7 (9 - a) / 5

We know that 9-a has to equal a multiple of 5 and you can't have a negative of a banana.
Set 9-a=5
a=4
Solve the rest of the equation and dominate the world
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17 May 2016, 19:54
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Attached is a visual that should help.
Attachments

Screen Shot 2016-05-17 at 7.43.35 PM.png [ 103.29 KiB | Viewed 4738 times ]

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15 Jun 2017, 21:43
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 1 apple-banana pair=1.20 5 apple-banana pairs=6.00; 6.30-6.00=.30, not enough for another banana 4 apple-banana pairs=4.80; 6.30-4.80=1.50, enough for 3 more bananas 4 apples+7 bananas=6.30=11 total fruits B Senior Manager Joined: 09 Mar 2016 Posts: 441 A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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07 Apr 2018, 07:33
Bunuel wrote:
SOLUTION

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 Given: $$0.7b+0.5a=6.3$$ Question: $$a+b=?$$ $$0.7a+0.5b=6.3$$ --> $$7a+5b=63$$ --> $$5b=63-7a$$ --> $$5b=7(9-a)$$ --> $$5b$$ 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 >13 (as $$5b$$ in this case would be more than$6.30), so $$b=7$$ --> $$a=4$$ --> $$a+b=4+7=11$$.

Hello pushpitkc , i am kindof confused by Bunuel`s problem solving method, hence i have questions can you please explain:-)

I understand that we factored and got $$5b=7(9-a)$$ but after this I didn't understand the reasoning/ explanation ...

1. i understand that $$5b$$ must be multiple of 7 , but how can $$b$$ be the multiple of 7 .... $$5b$$ is one whole number

2. can you explain why do we conclude that if both apples and bananas are bought then total quantity should be less than >13

3. (as $$5b$$ in this case would be more than $6.30), so $$b=7$$ ...cant understand 4... and how we conclude that total number is 11.... thank you and have great weekend Manager Joined: 08 Sep 2016 Posts: 62 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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07 Apr 2018, 09:52
The way I approached this problem was by making a list of multiples of 7 and multiples of 5. I then compared both list to see which values add up to 63.

70x + 50y = 630
7x+5y= 63

x: 7,14,21,28,35,42,49,56
7: 5,10,15,20,25,30,35

28+35 = 63

28 = 7*4
35= 5*7

7+4=11

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] Show Tags 08 Apr 2018, 00:08 1 This post received KUDOS 1 This post was BOOKMARKED Hey dave13 Please find my answers inline 1. I understand that $$5b$$ must be multiple of 7, but how can $$b$$ be the multiple of 7 .... $$5b$$ is one whole number The only possibility for 5*b to be a multiple to 7 and as 5 is not divisible by 7, b has to be divisible by 7 2. can you explain why do we conclude that if both apples and bananas are bought then total quantity should be less than >13 3. (as $$5b$$ in this case would be more than$6.30), so $$b=7$$ ...cant understand
We know that 7a + 5b = 63
(Worst case scenario: you can have around 12 bananas each costing 5 dollars
As soon as you have 13 bananas - the cost will become 65 which is greater than 63)

4... and how we conclude that total number is 11
Since $$5b = 63 - 7a$$ -> $$b=\frac{(63 - 7a)}{5}$$
a = 9, b = 0 -> Not possible because both type of fruits can be bought.
a = 4, b = 7 is our answer

Hope this helps you!
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