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A shopkeeper sells apples for $7 each and strawberries for $15.00 each

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A shopkeeper sells apples for $7 each and strawberries for $15.00 each  [#permalink]

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New post 20 Mar 2018, 13:16
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A shopkeeper sells apples for $7 each and strawberries for $15.00 each. How many apples did the shopkeeper sell today?

(1) The number of apples sold today is 1 more than twice the number of strawberries sold.
(2) Today the shopkeeper received a total of $65 from the sale of both apples and strawberries.

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A shopkeeper sells apples for $7 each and strawberries for $15.00 each  [#permalink]

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New post 20 Mar 2018, 13:54
itisSheldon wrote:
A shopkeeper sells apples for $7 each and strawberries for $15.00 each. How many apples did the shopkeeper sell today?

(1) The number of apples sold today is 1 more than twice the number of strawberries sold.
(2) Today the shopkeeper received a total of $65 from the sale of both apples and strawberries.



Let A be the number of apples and S the number of strawberries

1. A = 2S + 1

If S = 1, A = 2+1 = 3
If S = 2, A = 4+1 = 5
We cannot find a unique value for the number of apples sold.(Insuffiicient)

2. 7A + 15S = 65

This is only possible when A = 5, S = 2 since both
apples and strawberries are sold (Sufficient - Option B)
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Re: A shopkeeper sells apples for $7 each and strawberries for $15.00 each  [#permalink]

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New post 20 Mar 2018, 18:16
St1: no of apple = 2* number of strawberry +1. No other info . Hence insufficient.
St2:7x+5y=65. Since the number of fruits can only be POSITIVE INTEGER, hence finding integer solution, we get apple =5, and strawberry=2 is the only INTEGER solution.sufficient.

hence answer B

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Re: A shopkeeper sells apples for $7 each and strawberries for $15.00 each  [#permalink]

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New post 21 Mar 2018, 10:07
1
Let x = # of apples sold
Let y = # of strawberries sold

Statement 1, represented as an equation is
\(x = y + 1\)
Clearly not sufficient, since we have 1 linear equation with 2 unknowns.

Statement 2, represented as an equation is
\(7x + 15y = 65\)

It's very tempting to immediately choose Answer C since we have 2 equations with 2 unknowns, but this is a trap. Let's see what possible values of x and y would fit the equation...
Note that we can't have x=0 or y=0 because neither 7 is factor of 65, nor 15 is a factor of 65. And since we're dealing with # of fruit, x and y must be some positive integers.

Let's take \(7x\) first. It's obvious that \(65\) is a multiple of 5 and \(15y\) is a multiple of 5. Thus, \(7x\) must be a multiple of 5. So x can equal {5,10,15,etc}. However if x=10, then 7x=70, which means we won't have a positive value for y. So x=5 is the only possible value for x. Substituting x=5 into the equation and solving for y we get:
\(7x + 15y = 65\)
\(7(5) + 15y = 65\)
\(15y = 30\)
\(y = 2\)

Thus, selling 5 apples and 2 strawberries is the only way the shopkeeper could have received $65.
Sufficient.

Answer: B
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A shopkeeper sells apples for $7 each and strawberries for $15.00 each  [#permalink]

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New post Updated on: 24 Mar 2018, 17:32

Solution


Let the number of apples sold today be ‘\(a\)’ and the number of strawberries sold today be ‘\(b\)’.

Statement-1The number of apples sold today is \(1\)more than twice the number of strawberries sold “.

    • \(a = 2b+1\)

The above expression does not give any information about the value of the number of apples sold today.

Hence, Statement 1 alone is not sufficient to answer the question.

Statement-2Today the shopkeeper received a total of $\(65\) from the sale of both apples and strawberries.

    • \(7a+15b=65\)
    • \(15b\) ends either with a zero or 5.
    •Thus, for \(7a+15b\) to end with a \(5\), \(7a\) must end either with the \(5\) or \(0\).
    • The only possible value of ‘\(a\)’ and ‘\(b\)’ is \(5\) and \(2\) respectively.

Hence, Statement 2 alone is sufficient to answer the question.

Answer: B
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Originally posted by EgmatQuantExpert on 24 Mar 2018, 14:55.
Last edited by EgmatQuantExpert on 24 Mar 2018, 17:32, edited 3 times in total.
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Re: A shopkeeper sells apples for $7 each and strawberries for $15.00 each  [#permalink]

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New post 24 Mar 2018, 15:07
EgmatQuantExpert wrote:

Solution


Let the number of apples sold today be ‘\(a\)’ and the number of strawberries sold today be ‘\(b\)’.

Statement-1The number of apples sold today is \(1\)more than twice the number of strawberries sold “.

    • \(a = 2b+1\)

The above expression does not give any information about the value of the number of apples sold today.

Hence, Statement 1 alone is not sufficient to answer the question.

Statement-2Today the shopkeeper received a total of $\(65\) from the sale of both apples and strawberries.

    • \(7a+10b=65\)
    • \(10b\) always ends with a zero.
    •Thus, for \(7a+10b\) to end with a \(5\), \(7a\) must end with the \(5\).
    • The only possible value of ‘\(a\)’ and ‘\(b\)’ is \(5\) and \(3\) respectively.

Hence, Statement 2 alone is sufficient to answer the question.

Answer: B


Re statement 2, strawberries cost $15 each, not $10 each.
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Re: A shopkeeper sells apples for $7 each and strawberries for $15.00 each &nbs [#permalink] 24 Mar 2018, 15:07
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