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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 CEO Joined: 12 Sep 2015 Posts: 2630 Location: Canada Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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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. 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 _________________ Brent Hanneson – Founder of gmatprepnow.com VP Joined: 07 Dec 2014 Posts: 1032 A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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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 Intern Joined: 06 Jul 2017 Posts: 3 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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22 Jun 2018, 14:32
Factorise 630 to get 7*3*3*5*2, out of these we have 7 and 5 already, so take 7*9 +5*2. Therefore total =9+2=11.
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 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 Posted from my mobile device _________________ Not a professional entity or a quant/verbal expert or anything. So take my answers with a grain of salt. Intern Joined: 11 Sep 2013 Posts: 5 Schools: CBS '19 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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06 Jul 2018, 07:56

Given:

0.70A+0.50B=6.30 - Multiply both sides by 10

7A+5B=63 - divide by 7

A+$$\frac{5}{7}$$B=9 - From here we see that B=7
(If B=14,21,28,35... then the equation A+$$\frac{5}{7}$$B=9 becomes invalid)

If B=7 then A=4.
Hence, A+B= 4+7=11

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