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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
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Bunuel
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.

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

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SatvikVedala
Bunuel
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.

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

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
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