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A shopkeeper sells apples for $7 each and strawberries for $15.00 each [#permalink]
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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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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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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 Posted from my mobile devicePosted from my mobile device
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Re: A shopkeeper sells apples for $7 each and strawberries for $15.00 each [#permalink]
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21 Mar 2018, 10:07
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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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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 ‘\(b\)’. Statement1 “ The number of apples sold today is \(1\)more than twice the number of strawberries sold “. 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.Statement2 “ Today 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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Re: A shopkeeper sells apples for $7 each and strawberries for $15.00 each [#permalink]
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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 ‘\(b\)’. Statement1 “ The number of apples sold today is \(1\)more than twice the number of strawberries sold “. 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.Statement2 “ Today 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: BRe statement 2, strawberries cost $15 each, not $10 each.




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