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Jose purchased several large and small dressers at a flea market to ad

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Jose purchased several large and small dressers at a flea market to ad [#permalink]

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New post 15 Sep 2014, 01:25
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Jose purchased several large and small dressers at a flea market to add to his antique furniture
collection. The large dressers cost $31 apiece, while the small dressers cost $25 apiece. If Jose did not
purchase any other items at the flea market, how many small dressers did he buy?

1) Jose purchased as many small dressers as large dressers.

2) In total, Jose spent $280 at the flea market.
[Reveal] Spoiler: OA

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Re: Jose purchased several large and small dressers at a flea market to ad [#permalink]

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New post 15 Sep 2014, 11:53
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clipea12 :

Statement 1 : It doesn't say any information about how many small dressers did jose bought. Apart from saying,

Number of small= number of large. So insufficient.

Statement 2: It says total amount is 280 dollars.

let, L=large & S=small.

31L + 25S= 280.

It may look insufficient in the beginning. But if you look closely, and plugging values you can find.

If S=10 and L=1. We get 281. So the total number of large and small dressers should be less 11.

So if you continue doing that, we get only one value gives us 280, that is 5.

31*5 + 25*5= 155+ 125=280.

Other method is that when you add the unit digit of multiple of 31 and multiple of 25 should give us 0 and it should be 280 as well.

When you multiply 25 with any number the unit digit will either be 0 or 5. Never other numbers so we can eliminate- 1 to 9(apart from 5).

And if unit digit is 0. then unit digit for multiple of 31 also be 0. But we get only in 310. So unit digit of 25 cannot be 0.

you can go about like this and find that only when we multiple 5 to both 31 and 25. We get a sum of 280.

Statement 2 is sufficient.

How to check, whether 5 will be only number?. Well, we can now look at statement-1(NOTE: Just to verify, not while solving). In GMAT, the both statements never contradicts each other.

St.1 says Number of large dresser=Number of small dresser. So 5 is the number of small dresser.
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Re: Jose purchased several large and small dressers at a flea market to ad [#permalink]

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New post 15 Sep 2014, 23:04
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clipea12 wrote:
Jose purchased several large and small dressers at a flea market to add to his antique furniture
collection. The large dressers cost $31 apiece, while the small dressers cost $25 apiece. If Jose did not
purchase any other items at the flea market, how many small dressers did he buy?

1) Jose purchased as many small dressers as large dressers.

2) In total, Jose spent $280 at the flea market.



Let number of small dressers be "S"
Number of large dressers be "L"

Here the idea is to understand that the dressers have to take a positive integer value.

Statement 1)

the statement only tells us that S = L and nothing else.

Say L=S= 6, then 25*S + 31*L = 25*6 + 31*6 = 336
If L=S= 5, then 25*S + 31*L = 25*5 + 31*5 = 280

So, nothing can be concluded from this statement. Hence, Statement 1 is "Insufficient"


Statement 2)

This statement tells us that 25*S + 31*L = 280

Now we know that the number of dressers is an integer value and that total cost of both dressers is 280

So there is only one value that satisfies this condition when S=5 & L=5

Hence Statement 2 is " Sufficient "

Therefore answer B.

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Re: Jose purchased several large and small dressers at a flea market to ad [#permalink]

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New post 10 Aug 2017, 04:37
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Jose purchased several large and small dressers at a flea market to ad [#permalink]

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New post 25 Aug 2017, 20:10
Let number of small dresser is x, number of large dresser is y
We need to find x

Statement 1: x=y. INSUFF, x and y could be any number

Statement 2: We know the price of each unit so we can set up the equation as

25x + 31y = 280.
or x = \(\frac{(280-31y)}{25}\)

In order for x to be an interger, 280-31y must be divisible by 25.
All multiple of 25 has 5 or 0 as its last digit, to make (280 - 31y) has 5 or 0 as the last digit 31y must have 0 or 5 as the last digit. Therefore y can be 0, 5, 10, 15, etc..
y can't be 0, and if y = 10 then 31y > 280. Therefore only y= 5 is possible.
y =5 then x = 5.

SUFF

Answer is B.

Kudos [?]: 8 [0], given: 17

Re: Jose purchased several large and small dressers at a flea market to ad   [#permalink] 25 Aug 2017, 20:10
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Jose purchased several large and small dressers at a flea market to ad

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