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A citrus fruit grower receives $15 for each crate of oranges [#permalink] ### Show Tags 18 Jul 2009, 17:49 4 This post was BOOKMARKED 00:00 Difficulty: (N/A) Question Stats: 91% (02:23) correct 9% (00:00) wrong based on 113 sessions ### HideShow timer Statistics Guys, i found 2 problems following are in the same pattern, but the OAs are different, making me very confused and time-wasted. Can you tell me the logic disguided behind them? Thanks 1. A citrus fruit grower receives$15 for each crate of oranges shipped and $18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week? a. Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped. b. Last week the grower received a total of$38,700 from the crates of oranges and grapefruit shipped.
The OA is C

2.Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? a. She bought$4.40 worth of stamps
b. She bought an equal number of $0.15 stamps and$0.29 stamps.
But the OA is A
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Re: Collections confused-need a help [#permalink]

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18 Jul 2009, 20:36
IMHO both the answers are C. Where did you find this question and the answer for the second one ? Sometimes the docs that fly around in the forums have wrong answers.
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Re: Collections confused-need a help [#permalink]

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19 Jul 2009, 09:25
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1. A citrus fruit grower receives $15 for each crate of oranges shipped and$18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week?

a. Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped.
b. Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped. The OA is C obviously C 2.Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?

a. She bought $4.40 worth of stamps b. She bought an equal number of$0.15 stamps and $0.29 stamps. But the OA is A[/quote] 0.15X+0.29Y = 4.4 IE: 15X+29Y=440 15* anything will always give an intiger ending in either 5 or 0 ie 15x can = 15,30,45..etc 29*anything can give an intiger ending in 9,8,7,6,5,4,3,2,1,0 thus y has to be a multiple of 5 ie y = 5b rephrase 15x+145b = 440 try b = 1 or 3 , x becomes a fraction , only if b = 2 x is a whole number, thus 145b = 290 thus x = 10 Senior Manager Joined: 21 Jul 2009 Posts: 265 Location: New York, NY Followers: 3 Kudos [?]: 110 [0], given: 23 Re: Collections confused-need a help [#permalink] ### Show Tags 14 Oct 2009, 16:13 I saw this question on two different "sets" of the pdf's that are floating around the internet; one had OA as A, and the other said C. Exteremely unreliable. I think the answer is A, because there is only one combination of 29 and 15 that fit into 440: ten and fifteen. In order for there to be another combination there would need to be a second common multiple of the two numbers. Senior Manager Joined: 31 Aug 2009 Posts: 419 Location: Sydney, Australia Followers: 9 Kudos [?]: 280 [6] , given: 20 Re: Collections confused-need a help [#permalink] ### Show Tags 14 Oct 2009, 16:53 6 This post received KUDOS 3 This post was BOOKMARKED The questions are similar but not the same. The OAs I believe are correct in both cases. Just because the logic is the same to derive both answers does not imply that the answers should be the same in both cases. First Question Statement 1 only tells us the ratio oranges shipped is 20 more than twice gfruits. The number could be 20 times anything. Insuf. Statement 2 This gives us a formula. Orange x something + gfruit x something =$38,700
But this could be any combination of the Oranges and Grapefruits for example
Oranges 15 x 1380 crates = 20700
GFruit 18 x 1000 crates = 18000
This works but so does this:
Oranges 15 x 1620 crates = 24300
GFruit 18 x 800 crates = 14400

We need to determine the number of Orange crates to grapefruit crates to determine. Ie. We need statement 1. Hence ANS = C
(In case you’re curious solving the equations gives you the second set above, ie. 1620 crates of oranges)

Second Question
Statement 1 tells us the total. Similar to the first question we do not know the ratio of $0.15 to$0.29 stamps. However, unlike the first question there are only a few possibilities. The total figure, $4.40, ends in a 0. This would only be possible if the number of$0.29 is a multiple of 5 (or obviously 10). Quickly testing the only possible 3 cases

5 stamps x 0.29 = $1.45$4.40 – $1.45 =$2.95 (not divisible by 15, quick way to check this is not divisible by 3. You can use the fact that 2+9+5=16 which is the quick way to check divisibility by 3).

10 stamps x 0.29 = $2.95$4.40 – $2.90 =$1.50 (obviously this leaves 10stamps x 0.15c)

2) Alfredo would have paid $10 if he had bought 16 rollers & 12 doughnuts I've noticed that 2 statements are similar but my question is why each statement alone is not sufficient. Thanks, i would appreciate if u can clarify the difference between this problem (doughnut) and the previous one (pencil) Math Expert Joined: 02 Sep 2009 Posts: 37148 Followers: 7274 Kudos [?]: 96815 [6] , given: 10786 Re: Gmat prep2 [#permalink] ### Show Tags 19 Apr 2010, 07:28 6 This post received KUDOS Expert's post 9 This post was BOOKMARKED sandranjeim wrote: Hi, Can anyone help me on this, i am a little bit confused. I have here a somehow similar problem and both statements are not sufficient to get to the answer. So what is exactly the difference betwen the below problem and the problem of pencils??? Thanks Now find the problem: At a certain bakery, each roll costs r cents and each doughnut cost of cents. If Alfredo bought rolls & doughnuts at the bakery, how many cents did he pay for each roll? 1) Alfredo paid$5 for 8 rolls and 6 doughnuts
2) Alfredo would have paid $10 if he had bought 16 rollers & 12 doughnuts I've noticed that 2 statements are similar but my question is why each statement alone is not sufficient. Thanks, i would appreciate if u can clarify the difference between this problem (doughnut) and the previous one (pencil) At a certain bakery, each roll costs r cents and each doughnut costs d cents. If Alfredo bought rolls and doughnuts at the bakery, how many cents did he pay for each roll? Let $$r$$ be the price of rolls in cents and $$d$$ be the price of doughnuts in cents. Note that $$r$$ and $$d$$ must be an integers. Q: $$r=?$$ (1) Alfredo paid$5.00 for 8 rolls and 6 doughnuts --> $$8r+6d=500$$ --> $$4r+3d=250$$. Multiple solutions are possible, for instance: $$r=25$$ and $$d=50$$ OR $$r=10$$ and $$d=70$$. Not sufficient.

(2) Alfredo would have paid $10.00 if he had bought 16 rolls and 12 doughnuts --> $$16r+12d=1000$$ --> $$4r+3d=250$$. The same. Not sufficient. (1)+(2) No new info. Not sufficient. Answer: E. Marta bought several pencils. if each pencil was either 23 cents pencil or a 21 cents pencil. How many 23 cents pencils did Marta buy? Let $$x$$ be the # of 23 cent pencils and $$y$$ be the # of 21 cent pencils. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) Marta bought a total of 6 pencils --> $$x+y=6$$. Clearly not sufficient. (2) The total value of the pencils Marta bought was 130 cents --> $$23x+21y=130$$. Now x and y must be an integers (as they represent the # of pencils). The only integer solution for $$23x+21y=130$$ is when $$x=2$$ and $$y=4$$. Sufficient. Answer: B. Similar problems: A citrus fruit grower receives$15 for each crate of oranges shipped and $18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week? Let $$x$$ be the # of oranges and $$y$$ the # of grapefruits. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped --> $$x=2y+20$$. Not sufficient to calculate $$x$$ (2) Last week the grower received a total of$38,700 from the crates of oranges and grapefruit shipped --> $$15x+18y=38700$$ --> $$5x+6y=12900$$. Multiple values are possible, for istance: $$x=180$$ and $$y=2000$$ OR $$x=60$$ and $$y=2100$$.

(1)+(2) Two unknowns, two different linear equations --> We can calculate unique value of $$x$$. Sufficient.

Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? Let $$x$$ be the # of$0.15 stamps and $$y$$ the # of $0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient.

(2) She bought an equal number of $0.15 stamps and$0.29 stamps --> $$x=y$$. Not sufficient.

So when we have equation of a type $$ax+by=c$$ and we know that x and y are integers, there can be multiple solutions possible for x and y (eg $$5x+6y=12900$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it's sufficient.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.
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19 Apr 2010, 07:52
Thanks for the lovely explanation,

But one last question, what is the quickest way to know if ax + by = c is sufficient or not...

Is it by trial and error??

Many thanks
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19 Apr 2010, 08:02
I don't know if it is the right explanation to my previous question (what is the quickest way to know if it is sufficient)

Maybe if numbers do not have common factors the solution is sufficient as it the case of 23x + 21y = 130

Otherwise (when we have common factors consequently we will have many combinations) and the statement will be insufficient as it is the case of
8r + 6d = 500

Does it sound logic??

Thanks a lot
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19 Apr 2010, 08:10
sandranjeim wrote:
Marta bought several pencils. if each pencil was either 23 cents pencil or a 21 cents pencil. How many 23 cents pencils did Marta buy?

1) Marta bought a total of 6 pencils

2) The total value of the pencils Marta bought was 130 cents

OA = B

Why isn't C? how can we get tp the answer with only the second information.

Thanks

OA = B is correct.
At first look C looks close contender but if you take a careful look then you can find that 23 is a prime number while 21 is not.
1. We dont know the total value of pencils purchased. [Insuff]
2. Say x is the # of 23 cents pencils and y is for 21 cents pencils. Making an equation now:
23x + 21y = 130 >>>>> x=(130-21y)/23

23 has multiple: 23, 46, 69, 92 and 115. If you check for these values then only 46 survives. Sufficient.
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19 Apr 2010, 08:25
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sandranjeim wrote:
Thanks for the lovely explanation,

But one last question, what is the quickest way to know if ax + by = c is sufficient or not...

Is it by trial and error??

Many thanks

Yes, by trial and error plus some logic and knowledge of basics of number properties. Just be aware that C might be a trap answer for such questions.
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04 May 2011, 11:27
Bunuel wrote:
sandranjeim wrote:
Hi,

Can anyone help me on this, i am a little bit confused. I have here a somehow similar problem and both statements are not sufficient to get to the answer. So what is exactly the difference betwen the below problem and the problem of pencils???

Thanks

Now find the problem:

At a certain bakery, each roll costs r cents and each doughnut cost of cents. If Alfredo bought rolls & doughnuts at the bakery, how many cents did he pay for each roll?

1) Alfredo paid $5 for 8 rolls and 6 doughnuts 2) Alfredo would have paid$10 if he had bought 16 rollers & 12 doughnuts

I've noticed that 2 statements are similar but my question is why each statement alone is not sufficient.

Thanks, i would appreciate if u can clarify the difference between this problem (doughnut) and the previous one (pencil)

At a certain bakery, each roll costs r cents and each doughnut costs d cents. If Alfredo bought rolls and doughnuts at the bakery, how many cents did he pay for each roll?

Let $$r$$ be the price of rolls in cents and $$d$$ be the price of doughnuts in cents. Note that $$r$$ and $$d$$ must be an integers. Q: $$r=?$$

(1) Alfredo paid $5.00 for 8 rolls and 6 doughnuts --> $$8r+6d=500$$ --> $$4r+3d=250$$. Multiple solutions are possible, for instance: $$r=25$$ and $$d=50$$ OR $$r=10$$ and $$d=70$$. Not sufficient. (2) Alfredo would have paid$ 10.00 if he had bought 16 rolls and 12 doughnuts --> $$16r+12d=1000$$ --> $$4r+3d=250$$. The same. Not sufficient.

(1)+(2) No new info. Not sufficient.

Marta bought several pencils. if each pencil was either 23 cents pencil or a 21 cents pencil. How many 23 cents pencils did Marta buy?

Let $$x$$ be the # of 23 cent pencils and $$y$$ be the # of 21 cent pencils. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$

(1) Marta bought a total of 6 pencils --> $$x+y=6$$. Clearly not sufficient.

(2) The total value of the pencils Marta bought was 130 cents --> $$23x+21y=130$$. Now x and y must be an integers (as they represent the # of pencils). The only integer solution for $$23x+21y=130$$ is when $$x=2$$ and $$y=4$$. Sufficient.

Similar problems:

A citrus fruit grower receives $15 for each crate of oranges shipped and$18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week?

Let $$x$$ be the # of oranges and $$y$$ the # of grapefruits. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$

(1) Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped --> $$x=2y+20$$. Not sufficient to calculate $$x$$

(2) Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped --> $$15x+18y=38700$$ --> $$5x+6y=12900$$. Multiple values are possible, for istance: $$x=180$$ and $$y=2000$$ OR $$x=60$$ and $$y=2100$$. (1)+(2) Two unknowns, two different linear equations --> We can calculate unique value of $$x$$. Sufficient. Answer: C. Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?

Let $$x$$ be the # of $0.15 stamps and $$y$$ the # of$0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$

(1) She bought $4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient. (2) She bought an equal number of$0.15 stamps and $0.29 stamps --> $$x=y$$. Not sufficient. Answer: A. So when we have equation of a type $$ax+by=c$$ and we know that x and y are integers, there can be multiple solutions possible for x and y (eg $$5x+6y=12900$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it's sufficient. Hope it helps. So basically it means that we have to check if there is only one solution or more than one. i am just wondering if the equations are little more complex, do we have time to do that.. or i guess its easy to figure when we form the equations, whether it will have one solution or more. Math Forum Moderator Joined: 20 Dec 2010 Posts: 2021 Followers: 162 Kudos [?]: 1743 [0], given: 376 Re: Gmat prep2 [#permalink] ### Show Tags 04 May 2011, 12:13 agdimple333 wrote: So basically it means that we have to check if there is only one solution or more than one. i am just wondering if the equations are little more complex, do we have time to do that.. or i guess its easy to figure when we form the equations, whether it will have one solution or more. GMAT won't give you tedious equations to deal with. You can easily work out all possible values for the equation. And I feel it is somewhat tedious to find out whether an equation with 2 variables has one or more integral solutions. Anyone? _________________ Director Joined: 23 Apr 2010 Posts: 584 Followers: 2 Kudos [?]: 80 [0], given: 7 Re: Marta bought several pencils. if each pencil was either 23 [#permalink] ### Show Tags 12 Nov 2011, 08:36 Quote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?

Let x be the # of $0.15 stamps and y the # of$0.29 stamps. Note that x and y must be an integers. Q: x=?

(1) She bought $4.40 worth of stamps --> 15x+29y=440. Only one integer combination of x and y is possible to satisfy 15x+29y=440: x=10 and y=10. Sufficient So the only way to solve these types of questions is trial and error? This particular question is from the OG 12th edition (question 123). How can we quickly arrive at x=10 and y = 10? My feeling, even now reviewing the problem, is that it is rather tedious and time consuming. The OG's explanation is quite esoteric (at least to me). I would like to ask our math gurus to explain. Thanks a lot! PS I've just found this great and fast solution by lagomez: joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-87449.html#p657359 Quote: My quick way, 15x will end in a 5 or 0 so 29y must end in a 0 or 5 as well to get 440. That means y has to be 5 or 10. Testing both only 10 will work Manager Status: Bell the GMAT!!! Affiliations: Aidha Joined: 16 Aug 2011 Posts: 183 Location: Singapore Concentration: Finance, General Management GMAT 1: 680 Q46 V37 GMAT 2: 620 Q49 V27 GMAT 3: 700 Q49 V36 WE: Other (Other) Followers: 6 Kudos [?]: 74 [1] , given: 43 Re: Marta bought several pencils. if each pencil was either 23 [#permalink] ### Show Tags 13 Nov 2011, 02:35 1 This post received KUDOS An interesting article at Karishma's (from Veritas) old blog to solve such equations: http://gmatquant.blogspot.com/search?up ... -results=7 Hope it will help someone as it helped me _________________ If my post did a dance in your mind, send me the steps through kudos :) My MBA journey at http://mbadilemma.wordpress.com/ Manager Joined: 08 Sep 2011 Posts: 75 Concentration: Finance, Strategy Followers: 3 Kudos [?]: 2 [0], given: 5 Re: Marta bought several pencils. if each pencil was either 23 [#permalink] ### Show Tags 16 Nov 2011, 09:23 The way i did this was, to make a combination where the units digits = 0 (since 23 and 21 have to = 130 or you have to have x number of 3's + y number of 1's end in 0) So you can have one 3 (one 23) + seven 1 (seven 21). This cant work because 7 *21 bust stmt 2 another option is three 3 (three 23) + one 1 (one 21). This cant work because this option = 90 which would make stmt 2 false and all stmts provided are true last option is two 3 ( two 23) + four 1 (four 21). = 130 so you know how many of each is need Manager Joined: 30 May 2008 Posts: 76 Followers: 1 Kudos [?]: 96 [0], given: 26 Re: Gmat prep2 [#permalink] ### Show Tags 21 Apr 2012, 06:47 Bunuel wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?

Let $$x$$ be the # of $0.15 stamps and $$y$$ the # of$0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$

(1) She bought $4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient. (2) She bought an equal number of$0.15 stamps and $0.29 stamps --> $$x=y$$. Not sufficient. Answer: A. So when we have equation of a type $$ax+by=c$$ and we know that x and y are integers, there can be multiple solutions possible for x and y (eg $$5x+6y=12900$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it's sufficient. Hope it helps. How can one identify one or multiple solution for $$ax+by=c$$? (i.e. how did you arrive at the conclusion that only one integer combo satisfy $$15x+29y=440$$? Re: Gmat prep2 [#permalink] 21 Apr 2012, 06:47 Go to page 1 2 Next [ 22 posts ] Similar topics Replies Last post Similar Topics: A fruit stand sells apples, pears, and oranges. If oranges cost$ 0.50 2 13 Dec 2015, 04:55
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