An automobile rental agency purchases new cars from the manufacturer

Question

An automobile rental agency purchases new cars from the manufacturer in lots of 5 and releases the oldest cars to an auctioneer in lots of 3. Last year the number of cars owned by the agency increased by 14. If a purchase of a lot and a release of a lot each count as one transaction, which of the following is the least number of transactions that could have occurred last year?

A. Three

B. Four

C. Five

D. Six

E. Seven

A. Three

B. Four

C. Five

D. Six

E. Seven

Fish181 · Accepted Answer

Wow this problem tricked me into thinking that a full purchase and release counted as one transaction. If I had known that they were independent of each other then 6 definitely makes sense.

KarishmaB · Answer

Recall Integer Solutions to equation in 2 variables discussed here: https://anaprep.com/algebra-integer-sol ... variables/ If x lots were brought in and y were offloaded (x and y are both integers), 5x - 3y = 14 x has to be at least 3 because 15 cars were certainly brought in but then we have no integer value for y. If x = 4, y = 2 works. Hence there were 6 transactions. Answer (D) Method 2: Alternatively, since numbers are small, we can use quick logic: Each purchase adds 5 cars. A 'purchase + release' adds 2 cars. As much as possible, we should maximum independent purchases to have fewer transactions. 2 purchases will add 10 cars. 4 transactions of '2 purchases + 2 releases' will add 4 cars. Hence 6 total transactions.

gmatophobia · Answer

Let the number of lots in which a purchase was made = x

Let the number of lots in which a release was made = y

5*x - 3*y = 14

3y = 5x + 14

3y = 3x + 12 + 2 + 2x

y = \frac{3x + 12 + 2 + 2x}{3}

y = x+ 4 + \frac{2 + 2x}{3}

x = -1 ; y = -1 + 4 = 3

One pair of (x,y) that satisfies the equation is (-1, 3)

Other pairs (x,y) = (4,2), (7,7) ...

When x = 4, and y = 2 , the number of transactions = 6

This is the minimum number of transactions possible.

Option D

Anki111 · Answer

If you think logically. The least is possible only when {5x4} is subtracted by (3x2) to give 14 remaining cars in the agency.
There is no other way possible which is less than 6 transactions.

There can not be 3x5 = 15 or less subtracted by any multiple of 3 to give 14.

Half battle is won if you understand the question.

rey1009 · Answer

5x-3y =14
x=4, y=2
total 6

mauricioalcan · Answer

Let the number of new cars purchased in one lot = x

Let the number of old cars purchased in one lot = y

5*x - 3*y = 14

3y = 5x + 14

How did you get from 5x-3y = 14 to 3y = 5x + 14

It is not 5x = 14 + 3y --> 5x - 14 = 3y ?

curiousPope · Answer

what I understood was. Agency  buys 5 cars and releases 3 car  and this combined makes one transaction. (which is obviously wrong).  gmatophobia  can you please help me understand the question and How  5x-3y=14  is making one transaction?

gmatophobia · Answer

curiousPope

5x - 3y = 14 is the translation of "  Last year the number of cars owned by the agency increased by 14.  " We can have multiple transactions leading to the increase. Hence, that equation doesn't represent one transaction.

A key point to note in this question is that x, and y can only be integers as they represent a lot in an auction.

Yes2GMAT · Answer

Fish181  The wording is terrible. I got confused as well.

DanTheGMATMan · Answer

Perhaps it's easiest to just eyeball which combination of 5x - 3y will equal 14. Can also isolate for one variable, taking advantage of the fact that x and y have to be integers:

https://www.youtube.com/watch?v=h25MSRwvgsg

ayushiii02 · Answer

. If a purchase of a lot and a release of a lot each count as one transaction

What does this mean?

Posted from my mobile device

sarthak1701 · Answer

Haha, same here. I took the equation as 14 = 5x + 3y while it should be 14 = 5x - 3y

Purnank · Answer

This problem was pain. I spent unnecessary time on this it was just so easy

like 5x coming in and 3y going out.

Therefore 5x - 3y = 14

so we need to find minimum number for this

so lets take 3y on RHS 5x = 14+3y

minimum y will need to make RHS multiple of 5 is 2

y=2 and x=4 total 6 transactions!

Answer is D.

Fullridedreams · Answer

This problem also counts as a critical reasoning problem! At least to me.

If you miss "If a purchase of a lot and a release of a lot  each  count as one transaction" in the stem, this problem becomes a lot harder than it needs to be.

I did a practice exam and quant has always been my weak point. I got this one right though and got a Q78 for the first time! Thank you GMAT Club!

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=83637 Harsha GMAT-Club-Forum-vmy88kmm.png

Dream009 · Answer

How I solved was 
1) + 5 - 3 = + 2
2) +2 + 5 - 3= 4
3) + 4 + 5- 3 = 6
4) + 6 + 5 - 3 = 8
5) + 8 + 5 - 3 = 10
6) + 10 + 5 - Now this is where we get 14

nihilre · Answer

I thought the same. 1 purchase + 1 sle = 1 complete transaction

soniasw16 · Answer

Same here, how to avoid such misunderstandings?

An automobile rental agency purchases new cars from the manufacturer

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An automobile rental agency purchases new cars from the manufacturer

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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