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Re: Danny purchased a number of grease pumps of only two possible variants [#permalink]
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imo A

Danny purchased a number of grease pumps of only two possible variants: $5 pumps and $25 pumps. If the ratio between the number of $5 pumps purchased and the number of $25 pumps purchased is 13:7, How many $5 grease pumps did Danny buy?

(1) The total purchase price of the grease pumps Danny bought was less than $400 .


Number of $5 pumps purchased = 13x,
Number of $25 pumps purchased = 7x

Total purchase price= 13x*5+7x*25 = 240x

240x <400

Since x is a positive integer, x must be equal to 1.

Sufficient

(2) The total purchase price of the grease pumps Danny bought was greater than $200.


Number of $5 pumps purchased = 13x,
Number of $25 pumps purchased = 7x

Total purchase price= 13x*5+7x*25 = 240x

240x >400

Since x is a positive integer, x can be any value > 1.

Insufficient
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Danny purchased a number of grease pumps of only two possible variants [#permalink]
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Bunuel wrote:
Danny purchased a number of grease pumps of only two possible variants: $5 pumps and $25 pumps. If the ratio between the number of $5 pumps purchased and the number of $25 pumps purchased is 13:7, How many $5 grease pumps did Danny buy?

(1) The total purchase price of the grease pumps Danny bought was less than $400 .
(2) The total purchase price of the grease pumps Danny bought was greater than $200.
 

Danny purchased a number of grease pumps of only two possible variants: $5 pumps and $25 pumps. If the ratio between the number of $5 pumps purchased and the number of $25 pumps purchased is 13:7, How many $5 grease pumps did Danny buy?

Given that \(\frac{({$}5 \ pumps)}{({$}25 \ pumps)}=\frac{13x}{7x}\), for some positive integer \(x\).

(1) The total purchase price of the grease pumps Danny bought was less than $400 --> \(5*13x+25*7x<400\) --> \(x<\frac{5}{3}\). Since \(x\) is an integer then \(x=1\) --> \(({$}5 \ pumps)=13x=13\). Sufficient.

(2) The total purchase price of the grease pumps Danny bought was greater than $200 --> \(5*13x+25*7x>200\) --> \(x>\frac{5}{6}\) --> \(x\) can be any integer more than or equal to 1. Not sufficient.

Answer: A.­
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Re: Danny purchased a number of grease pumps of only two possible variants [#permalink]
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Re: Danny purchased a number of grease pumps of only two possible variants [#permalink]
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