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A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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30 Jul 2012, 02:20

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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?

(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. (2) Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped.

Practice Questions Question: 9 Page: 275 Difficulty: 650

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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30 Jul 2012, 02:21

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SOLUTION

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, both \(x\) and \(y\) must be integers. Question: \(x=?\)

(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) We have two distinct linear equation with two unknowns, hence we can solve for \(x\) and \(y\). Sufficient.

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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16 Nov 2013, 14:35

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Expert's post

indiheats wrote:

Bunuel wrote:

SOLUTION

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, both \(x\) and \(y\) must be integers. Question: \(x=?\)

(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) We have two distinct linear equation with two unknowns, hence we can solve for \(x\) and \(y\). Sufficient.

Answer: C.

Hi Banuel,

I get hung up on when i create the equation 15x + 18y = 38700.... When I see a equation like this, should i automatically assume that mutiple combinations of x and y are possible to satisfy the equation or are there instances where I should actually work out the math.... I spend a lot of time contemplating this, although I see the obvious answer in C of two liner equations...

thoughts?

Thanks

Generally such kind of linear equations (ax+by=c) have infinitely many solutions for x and y, and we cannot get single numerical values for the variables. But since x and y here represent the # of oranges and the # of grapefruits, then they must be non-negative integers and in this case 15x + 18y = 38700 is no longer simple linear equation, it's Diophantine equation (equations whose solutions must be integers only) and for such kind on equations there might be only one combination of x and y possible to satisfy it. When you encounter such kind of problems you must always check by trial and error whether it's the case.

In my post above there links to several such problems. _________________

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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05 Oct 2015, 07:20

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indiheats wrote:

Hi Banuel,

I get hung up on when i create the equation 15x + 18y = 38700.... When I see a equation like this, should i automatically assume that mutiple combinations of x and y are possible to satisfy the equation or are there instances where I should actually work out the math.... I spend a lot of time contemplating this, although I see the obvious answer in C of two liner equations...

thoughts?

Thanks

Your thought process should be how can we make these numbers more manageable. 15 and 18 both share 3 as a factor, but 38700 looks pretty gnarly. A quick check confirms that 3 is a factor, 3+8+7=18, which is divisible by 3

Now we have the equation in something easier to work with 5x + 6y = 12,900.

It still looks pretty daunting. So here's my thought process, what two values when added give us 12,900, in other words, we're asking what gives us 12,000 + 900

So that equation now becomes, 5x+6y = 12,000 + 900 Can we get an x such that we get 12,000 or 900. Yes. Can we get a y such that we can get 12,000 or 900. Yes. 120 is divisible by 6, and 90 is divisible by 6.

What does that mean for us? Well, we can have a case x=2,400 and G = 150 5*2400 + 6*150 = 12,900

OR We can have a case where x=180 and G= 2000

5*180 + 6*2000 = 12,900

You don't have to actually do the arithmetic. Just do a quick sense check, can we have multiple values for x and y, such that we can get 12000 + 900. Use divisibility rules, if x can give us either 900 or 12000 when multiplied by 5, both are divisible by 5, and if Y can give us either 12000 or 900 when multiplied by 6. Both are divisible by 6. So we can get different values for x and y, and still satisfy 12,900.

Hope that helped someone, I know this post is a bit dated.

A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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24 Nov 2015, 06:22

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malkadhi wrote:

indiheats wrote:

Hi Banuel,

I get hung up on when i create the equation 15x + 18y = 38700.... When I see a equation like this, should i automatically assume that mutiple combinations of x and y are possible to satisfy the equation or are there instances where I should actually work out the math.... I spend a lot of time contemplating this, although I see the obvious answer in C of two liner equations...

thoughts?

Thanks

Your thought process should be how can we make these numbers more manageable. 15 and 18 both share 3 as a factor, but 38700 looks pretty gnarly. A quick check confirms that 3 is a factor, 3+8+7=18, which is divisible by 3

Now we have the equation in something easier to work with 5x + 6y = 12,900.

It still looks pretty daunting. So here's my thought process, what two values when added give us 12,900, in other words, we're asking what gives us 12,000 + 900

So that equation now becomes, 5x+6y = 12,000 + 900 Can we get an x such that we get 12,000 or 900. Yes. Can we get a y such that we can get 12,000 or 900. Yes. 120 is divisible by 6, and 90 is divisible by 6.

What does that mean for us? Well, we can have a case x=2,400 and G = 150 5*2400 + 6*150 = 12,900

OR We can have a case where x=180 and G= 2000

5*180 + 6*2000 = 12,900

You don't have to actually do the arithmetic. Just do a quick sense check, can we have multiple values for x and y, such that we can get 12000 + 900. Use divisibility rules, if x can give us either 900 or 12000 when multiplied by 5, both are divisible by 5, and if Y can give us either 12000 or 900 when multiplied by 6. Both are divisible by 6. So we can get different values for x and y, and still satisfy 12,900.

Hope that helped someone, I know this post is a bit dated.

Here are two approaches I'd take: 1) 2 sec approach though unscientific - a quick look at the equation will tell you that 12900 is a very large number compared to 5 and 6. Hence it is quite likely to fit in multiple combinations of 5 and 6. INSUF 2) 10 sec approach - let x=0, 12900 is divisible by 6(digits of 12900 add up to 3 and is an even number) so y will be an integer. This will give us one combination. Now let y=0, 12900 is obviously divisible by 5--> second combination. INSUF _________________

Please consider giving 'kudos' if you like my post and want to thank

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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01 Aug 2012, 00:08

First what comes to my mind is that it is C, so combining two statements we can figure out: let say oranges - x and grapefruit -y, combining two statements we have (2y+20)*15+18*20=3870, y=800 and x=1620

Bunuel can you please clarify: How do we know that 800 and 1620 is not the only combination? if it is the only compbination possible then the answer should be B, but how to calculate from the statement 2 alone that there is only one possible solution. I have tried to pick numbers but after few attempts, looking at the watch i said it should be C (just a good feel). _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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01 Aug 2012, 03:35

ziko wrote:

First what comes to my mind is that it is C, so combining two statements we can figure out: let say oranges - x and grapefruit -y, combining two statements we have (2y+20)*15+18*20=3870, y=800 and x=1620

Bunuel can you please clarify: How do we know that 800 and 1620 is not the only combination? if it is the only compbination possible then the answer should be B, but how to calculate from the statement 2 alone that there is only one possible solution. I have tried to pick numbers but after few attempts, looking at the watch i said it should be C (just a good feel).

if one costs 15 dollars and the other costs 18 dollars you said one solution is 800 and 1620 then at least you know that 18 crates of orange cost the same price (18*15) than 15 crates of grapefruit (15*18). So for instance 818 (800+18) and 1605 (1620-15) must be another solution. And there are many others

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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03 Aug 2012, 06:00

Expert's post

1

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SOLUTION

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, both \(x\) and \(y\) must be integers. Question: \(x=?\)

(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) We have two distinct linear equation with two unknowns, hence we can solve for \(x\) and \(y\). Sufficient.

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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20 Apr 2013, 02:25

Bunuel wrote:

SOLUTION

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, both \(x\) and \(y\) must be integers. Question: \(x=?\)

(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) We have two distinct linear equation with two unknowns, hence we can solve for \(x\) and \(y\). Sufficient.

Answer: C.

Hi Bunnel,

I marked this one as B, as i thought that by prime factorization we can get the number of multiples of 15 and 18. However i later did the prime factorization and now know that their is no way of knowing how many times 15 or 18 goes in to 38,700.

I remembered this technique as I had used it in Problem Solving, so want to know whether this technique can be used in DS questions.

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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20 Apr 2013, 05:20

Expert's post

cumulonimbus wrote:

Bunuel wrote:

SOLUTION

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, both \(x\) and \(y\) must be integers. Question: \(x=?\)

(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) We have two distinct linear equation with two unknowns, hence we can solve for \(x\) and \(y\). Sufficient.

Answer: C.

Hi Bunnel,

I marked this one as B, as i thought that by prime factorization we can get the number of multiples of 15 and 18. However i later did the prime factorization and now know that their is no way of knowing how many times 15 or 18 goes in to 38,700.

I remembered this technique as I had used it in Problem Solving, so want to know whether this technique can be used in DS questions.

What technique are you talking about? Can you please also give PS question for which you've used it? _________________

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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13 Oct 2013, 14:55

Bunuel wrote:

cumulonimbus wrote:

Bunuel wrote:

SOLUTION

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, both \(x\) and \(y\) must be integers. Question: \(x=?\)

(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) We have two distinct linear equation with two unknowns, hence we can solve for \(x\) and \(y\). Sufficient.

Answer: C.

Hi Bunnel,

I marked this one as B, as i thought that by prime factorization we can get the number of multiples of 15 and 18. However i later did the prime factorization and now know that their is no way of knowing how many times 15 or 18 goes in to 38,700.

I remembered this technique as I had used it in Problem Solving, so want to know whether this technique can be used in DS questions.

What technique are you talking about? Can you please also give PS question for which you've used it?

I did the same thing and marked B. How can we tell quickly that there are multiple answers for 5x + 6y = 12900?

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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14 Oct 2013, 05:16

Expert's post

1

This post was BOOKMARKED

runningguy wrote:

Bunuel wrote:

cumulonimbus wrote:

Hi Bunnel,

I marked this one as B, as i thought that by prime factorization we can get the number of multiples of 15 and 18. However i later did the prime factorization and now know that their is no way of knowing how many times 15 or 18 goes in to 38,700.

I remembered this technique as I had used it in Problem Solving, so want to know whether this technique can be used in DS questions.

What technique are you talking about? Can you please also give PS question for which you've used it?

I did the same thing and marked B. How can we tell quickly that there are multiple answers for 5x + 6y = 12900?

Trial and error plus some logic and knowledge of basics of number properties should help you to identify this.

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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16 Nov 2013, 14:29

Bunuel wrote:

SOLUTION

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, both \(x\) and \(y\) must be integers. Question: \(x=?\)

(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) We have two distinct linear equation with two unknowns, hence we can solve for \(x\) and \(y\). Sufficient.

Answer: C.

Hi Banuel,

I get hung up on when i create the equation 15x + 18y = 38700.... When I see a equation like this, should i automatically assume that mutiple combinations of x and y are possible to satisfy the equation or are there instances where I should actually work out the math.... I spend a lot of time contemplating this, although I see the obvious answer in C of two liner equations...

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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10 Jan 2014, 07:49

Bunuel wrote:

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?

(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. (2) Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped.

Practice Questions Question: 9 Page: 275 Difficulty: 650

We are given this equation: 15*O + 18*6 = Total Revenue, so we have 3 variables.

Statement 1 solves one of our variables, but it's insufficient because we have 2 more. Statement 2 solves another of our variables, but still on itself it's insufficient.

But if we combine the two statements, we have one equation and one variable, so we can solve for the last variable. The answer is C.

Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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10 Feb 2015, 11:25

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Re: A citrus fruit grower receives $15 for each crate of oranges [#permalink]

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26 Nov 2015, 07:49

Expert's post

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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?

(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. (2) Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped.

We get a "2by2" table as below:

Attachment:

GCDS Bunuel A citrus fruit grower receives (20151125).jpg [ 24.76 KiB | Viewed 1145 times ]

There are 2 variables (a,b) and 2 equations are given by the 2 conditions, so there is high chance (C) will be the answer. If we look at the conditions together,

from a=2b+20, 15a+18b=38,700, we can get the values of a and b, so this is sufficient, and the answer becomes (C).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________

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