Danny purchased a number of grease pumps of only two possible variants

Question

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.

MarmikUpadhyay · Answer

Let the number of $5 pumps = 13x and number of $25 pumps = 7x, where x is a positive integer.

Total purchase price = $5 * 13x + $25 * 7x = $240x
=> Total purchase price will be a multiple of 240.

13x =?

Statement 1: 
Total purchase price < $400
=> Total purchase price = $240
=> x = 1 and 13x = 13
Statement 1 is Sufficient.

Statement 2:
Total purchase price > $200
=> Total purchase price can be any value from {240, 480, 720 and so on}
=> x can be any value from {1, 2, 3 and so on}
We do not get a definite answer for 13x.
Statement 2 is Not Sufficient.

So, correct answer is option  A .

Bunuel · Answer

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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rye · Answer

Bunuel  If \frac{no. of $5 pumps}{no. of $25 pumps} = \frac{13}{7}

then,  WHY can't we put   no. of $25 pumps = \frac{7 x no. of $5 pumps}{13}

In that case, the equation using Condition A will be:

5 x (no. of $5 pumps) + 25 x (no. of $25 pumps) < 400

5 x (no. of $5 pumps) + 25 x \frac{7 x no. of $5 pumps}{13} < 400

(no. of $5 pumps) < \frac{65}{3}

(no. of $5 pumps) < 21.something

Thus, (no. of $5 pumps)  <  21

Where am I incorrect? Why do I only have to put (no. of $5 pumps) and (no. of $25 pumps) as 13 x and 7x

SatvikVedala · Answer

Hi  Bunuel

Regarding the Question statement, Consider

$5 pumps = x
$ 25 pumps = y

x/y = 13/7 (given)

this implies, 7x = 13y

Since 7 is prime & 13 is prime --> only way we get 7x = 13y is when x = 13 & y = 7

The above can be inferred without Option (1) and Option (2)

If this is the case, Option D would be suitable since 5(13)+25(7) gives 240 & $200< $ 240 <$ 400

Request your opinion

SaquibHGMATWhiz · Answer

Hey,

A couple of things here:

The first trigger point for you should be that you cannot get the asnwer just from the question statement itself. This is not possible in GMAT DS question.  
 Secondly, when you say ' only way we get 7x = 13y is when x = 13 & y = 7 ' are you sure? 
  What if x = 26 and y = 14? Think about it.    
 The correct inference to draw from 7x = 13y is that the  minimum possible values of x and y are 13 and 7 respectively

Danny purchased a number of grease pumps of only two possible variants

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