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09 Jan 2017, 09:33
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 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Joined: 12 Jul 2016 Posts: 18 Re: A certain fruit stand sold apples for$0.70 each and bananas  [#permalink]

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26 May 2017, 08:02
0.7A+0.5B = 6.3
Let's multiply by 10 => 7A+5B = 63
Now we enjoy some factorization
5B = 63 - 7A ==> 5B = 7(9-A)
7+4=11
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Posts: 3231
Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 07 Oct 2017, 05:43 Top Contributor 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.

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

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

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

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
So, the customer buys 4 apples and 7 bananas for a total of 11 pieces of fruit

Cheers,
Brent
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Joined: 07 Dec 2014
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A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 07 Oct 2017, 10:31 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 know the customer bought at least two apple-orange pairs
one pair=$1.20 4 pairs=$4.80
$6.30-$4.80=$1.50=3 bananas 4 apples+7 bananas=$6.30
11
B
Intern
Joined: 06 Jul 2017
Posts: 3

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23 Jun 2018, 02:01
Usually these questions will always have a ‘limiting’ number, the one that is well, not divisible by a number ending in 0. So it’ll most likely not end with a 5, 0, or similar.

If the other ends in 0, then we need to figure out a multiple of $0.70 that ends with .30. If the other number ends in 5 (such as in this case:$0.50), the we need to figure a multiple of $0.70 that ends in either .30 or .80 (0.3 + 0.5). Preferably the lowest number possible. In that case, we have 9*$0.70=$6.30, and 4*$0.70=$2.80. However the first answer is impossible, so we are left with the second. I.e. there are four$0.70. Substract $2.80 from$6.30 and we get $3.50, which translates to 7*$0.50.

Together, that’s 11, hence B.

In this case, it’d be $0.70. We just need to figure out what$0.70 multiplies with to get the ‘.10

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### Show Tags

01 Sep 2018, 02:11
you DONT need to assume values let x be number of apples and y number of bananas.
A is total number of fruits.
equation derived is 0.7x + 0.5y = 6.30
y in form of A and substituting in above equation.
0.2x + 0.5A =6.30
or
2x + 5A + 63
now just enter value of A from options 11 is going to satisfy.
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Joined: 09 Jun 2014
Posts: 215
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 17 Oct 2018, 22:35 1 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

This problem is simple but trial and error can cost on GMAT so the best approach is to do the trial and error as quickly as possible..

Now since we have the equation 7a+5b=63 (aftre decimal adjustment)
We can say a 7a= 63 - 5b
which further implies a = 9- (5/7)b

Since a and B will be integers so b has to be a multiple of 7 ..Now lets have b =7 then a = 9- 5=4

This gives us a+b =11

This indeed saves us from many trials..

Press Kudos if it helps!!
Intern
Joined: 17 Sep 2018
Posts: 10
Location: India

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25 Nov 2018, 16:02
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: $$0.7b+0.5a=6.3$$ 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... Hey, i used this method: 7a + 5b = 63 notice that 7 and 63 are common factors 5b = 63 - 7a 5b = 9(7-a) b = 9/5 (7-a) you must make the number in the parenthesis to be 5 in order to make b an integer, then a = 2 b = 9/5 (7-2) b = 9/5 (5) b = 9 ; a = 2 ; a+b= 11 Answer B Re: A certain fruit stand sold apples for$0.70 each and bananas &nbs [#permalink] 25 Nov 2018, 16:02

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