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

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

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New post 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
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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 Oct 2017, 06:43
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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
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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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

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New post 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]

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

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New post 06 Jul 2018, 07:56
Question asks for A+B=?

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


Answer B. 11
Re: A certain fruit stand sold apples for $0.70 each and bananas   [#permalink] 06 Jul 2018, 07:56

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