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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 _________________ 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, 09:02
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
CEO
Joined: 12 Sep 2015
Posts: 2878
Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 07 Oct 2017, 06:43 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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Brent Hanneson – GMATPrepNow.com

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A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 07 Oct 2017, 11:31 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
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Joined: 06 Jul 2017
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23 Jun 2018, 03:01
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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Joined: 11 Sep 2013
Posts: 20
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01 Sep 2018, 03:11
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.
Re: A certain fruit stand sold apples for $0.70 each and bananas &nbs [#permalink] 01 Sep 2018, 03:11 Go to page Previous 1 2 [ 28 posts ] Display posts from previous: Sort by # A certain fruit stand sold apples for$0.70 each and bananas

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