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A certain fruit stand sold apples for $0.70 each and bananas

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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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 \(b=7\) --> \(a=4\) --> \(a+b=4+7=11\).

Answer: B.
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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
A+B= 11 , Answer B
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 15 Oct 2012, 04:51
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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 [#permalink]

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New post 15 Oct 2012, 09:13
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0.7x + 0.5y = 6.3
=> 0.2x +0.5(x+y) = 6.3
x+y -> is the total number of fruits bought..... substituting values only option B, i.e 11 gives a integral value for x, i.e x = 4

Please let me know if the logic seems to be flawed in any aspect....
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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

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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

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Question: 64
Page: 161
Difficulty: 600


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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 16 Oct 2012, 03:53
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1
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

Answer B
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 10 Jan 2014, 01:26
1
Reframe the question to give:
7a + 5b= 63...this to avoid dealing with decimals.
Again adjust the eqn to get:

b=(63 - 7a)/5

It is obvious from the above eqn that 'a' has to be less than 9 (since he purchased both type of fruits)
Then look for a value from 1 to 8 that will give you either '5' or '0' in the units place, this so that 'b' has to be a multiple of '5' and hence an integer.

Only value is a=4 which gives b=7
Therefore, total fruits = 4+7=11

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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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

Answer = B
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 11 Sep 2014, 04:31
Another approach is to look at the unit digits.

Multiple of 7 + Multiple of 5 = 6.3. that is unit digit of multiple of 7 + unit digit of multiple of 5 = 3

That will leave with two options, since unit digit of 5 will 0 or 5. (unit of multiple of 7 are all different)

Option 1: 5 multiples Unit digit is 0. then unit digit of 7 multiple need to be 3. That is 9*7 = 63
9*0.70 + 0*0.50 = 6.3.
Since it was said both apple and banana this is not possible.

Option 2: 5 multiples Unit digit is 5. then unit digit of 7 multiple need to be 8. that is 4*7 = 28
4*0.70 + 7*0.50 = 6.3

So 4 apples and 7 bananas. total 11.
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 11 Dec 2015, 04:30
7x+5y=63

just test every multiple of 5 less than 63 to find multiple of 7

only 35+28=63, so 7+4=11

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 17 May 2016, 19:54
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Attached is a visual that should help.
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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 22 May 2016, 11:20
This is probably the easiest approach you can take in this situation:
we are given two numbers and the result of the sum of their multiples.

So we have \(63\) resulting from \(x\) times \(7\) and \(y\) times \(5\).

The fastest way to solve it is by asking ourselves how can we get a \(3\) unit from \(7\) and \(5\)?
The main culpable has to be the \(7\) as the \(5\) can only give us \(0\) or \(5\).

So how can the \(7\) gives us \(3\) as a unit digit? Well every \(9\) numbers we get \(63\) that would be the answer if the problem didn't tell us "If a customer purchased both apples and bananas". Given that we have also to use the five we understand that we need to find a multiple of \(7\) that gives us \(8\), so that \(8+5 = ...3\) and in our case \(7\) results in \(8\) as \(4*7=28\). So we get \(28+5=33\), and as the total we are given at the beginning is \(63\) then we just need to add \(30\) or \(5*6\), to sum it up we did the following calculations \(7*4 + 5*1 + 5*6\) or -> \(4+6+1=11\)
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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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 :cry:
4 apple-banana pairs=4.80; 6.30-4.80=1.50, enough for 3 more bananas :-D
4 apples+7 bananas=6.30=11 total fruits
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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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 \(b=7\) --> \(a=4\) --> \(a+b=4+7=11\).

Answer: B.


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 :)
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 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

answer b
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 08 Apr 2018, 00:08
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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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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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New post 26 May 2018, 07:46
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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   [#permalink] 26 May 2018, 07:46
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