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A citrus fruit grower receives $15 for each crate of oranges [#permalink]
18 Jul 2009, 17:49

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

86% (02:08) correct
14% (00:00) wrong based on 36 sessions

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 _________________

Re: Collections confused-need a help [#permalink]
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.

Re: Collections confused-need a help [#permalink]
19 Jul 2009, 09:25

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

Re: Collections confused-need a help [#permalink]
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.

Re: Collections confused-need a help [#permalink]
14 Oct 2009, 16:53

1

This post received KUDOS

2

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)

15 stamps x 0.29 = $4.45 (no need to calculate just add the above two) Leaving 5 cents which is not divisible by 15c.

Hence there is only one possible solution and A is sufficient. B is insufficient because it just tells us the ratio of A:B is 1:1. Does not tell us total or anything else. ANS = A

With the equation that you construct 23a+21b=130 if you use trial and error you will arrive at one unique combination where the sum is 130 the important thing to be noted is a and b will be whole nos. say for a=1 solving for b doesn't yield a whole no. for a=2 b = 4

Hence using only statement 2 gives us a definite answer.

Unfortunately such questions are seen in DS only test setters are keen to lay a trap where 2 variable and 2 equation would lead to option C

when solutions have to be whole numbers (as in this case) always be wary of options C, D and E if you have a linear equation whose solutions must be whole numbers then you must test with plugging in numbers in some sort of organized way, there's really no better way to do problems like this.

Answer is B. You have to utilize 130 cents in such a way that you can buy both 21 cents and 23 cents pencil.

x - 21 cents pencils y - 23 cents pencils

Equation - 21x + 23y = 130

x is multiple of 21 and y multiple of 23. x and y should be such that it satisfies the above equation. By putting values of x and y, the only possible values are - x=4 and y=2.

Hence B is sufficient. There could be no other possible integer values of x and y that can satisfy the above equation.

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)

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.

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.

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

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. _________________

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. _________________

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.

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.

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? _________________

Re: Marta bought several pencils. if each pencil was either 23 [#permalink]
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:

Re: Marta bought several pencils. if each pencil was either 23 [#permalink]
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

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 \(15x+29y=440\)?

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