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To find the ratio of adults to children

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To find the ratio of adults to children

by gmattesttaker2 » Wed Dec 25, 2013 6:07 pm
Hello,

Can you please assist with this problem. I was not able to solve Statement 2:

If at a school trip each adult was charged $24 and each child was charged $16, what is
the ratio of adults to children at the school trip?

1) The ratio of the total revenue from children to the total revenue from
adults was 8 to 3.

2) Total revenue was $2,200

OA: A


I tried to solve as follows:

Let a = number of adults
c = number of children

a/c = ?

Statement 1:
24a / 16c = 3 / 8

Solving we get a/c = 1/4

Sufficient.


Statement 2:

24a + 16c = 2200
=> 3a + 2c = 275

I was wondering how to proceed from here?

Thanks for your help.

Regards,
Sri
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by Uva@90 » Wed Dec 25, 2013 8:48 pm
gmattesttaker2 wrote:Hello,

Can you please assist with this problem. I was not able to solve Statement 2:

If at a school trip each adult was charged $24 and each child was charged $16, what is
the ratio of adults to children at the school trip?

1) The ratio of the total revenue from children to the total revenue from
adults was 8 to 3.

2) Total revenue was $2,200

OA: A


I tried to solve as follows:

Let a = number of adults
c = number of children

a/c = ?

Statement 1:
24a / 16c = 3 / 8

Solving we get a/c = 1/4

Sufficient.


Statement 2:

24a + 16c = 2200
=> 3a + 2c = 275

I was wondering how to proceed from here?

Thanks for your help.

Regards,
Sri
Sri,
3a + 2c = 275
3a =275 -2c => a =(275-2c)/3
When c =1::a =91 and when c=4::a =89
we are getting different ratios.
Hence Statement 2 is insufficient.

You found statement 1 is sufficient.

Hence Ans is A

Regards,
Uva.
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by sanju09 » Thu Dec 26, 2013 12:11 am
gmattesttaker2 wrote:Hello,

Can you please assist with this problem. I was not able to solve Statement 2:

If at a school trip each adult was charged $24 and each child was charged $16, what is
the ratio of adults to children at the school trip?

1) The ratio of the total revenue from children to the total revenue from
adults was 8 to 3.

2) Total revenue was $2,200

OA: A


I tried to solve as follows:

Let a = number of adults
c = number of children

a/c = ?

Statement 1:
24a / 16c = 3 / 8

Solving we get a/c = 1/4

Sufficient.


Statement 2:

24a + 16c = 2200
=> 3a + 2c = 275

I was wondering how to proceed from here?

Thanks for your help.

Regards,
Sri
(1) Once a:c can be answered uniquely, it's sufficient! AD On BCE Gone!

(2) This gives us $(24a + 16c) = $2,200. If reduced to 3a + 2c = 275, still it gives us many possible pairs (a, c) to answer a:c differently. For example, if a = 25, then c = 100, that in turn gives a;c equal to 1:4, but if a = 45 then c = 69, and this in turn gives a;c equal to 15:23, a different answer; hence [spoiler]insufficient. D Died A Alive!

So pick A[/spoiler]
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by ceilidh.erickson » Sun Dec 29, 2013 10:36 am
There are times when a single equation is sufficient to solve for two unknowns - take for example OG DS #132:

Joanna bought only $0.15 and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.

Here, statement 1 is sufficient to answer the question. Even though we only have a single equation:
0.15x + 0.29y = 4.40
... we actually have enough information to answer the question, because there are implied integer constraints on the question. We can only have an integer number of stamps. When we try various combinations, we'll find that only one scenario yields $4.40: 10 stamps of each value. Because 15 and 29 do not share any factors, there is no overlap in possible dollar amounts. Multiples of 15 have to end in a units digit of 5 or 0, so to add up to a sum with a units digit of 0, the 29 cent stamps would have to be in a multiple of 5 or 10. Those are the only values we have to test. Five 29 cent stamps would yield $1.45, meaning that the 15 cent stamps would have to add to $2.95. Since this is not a multiple of 15, this doesn't work. Next, we can try ten of the 29 cent stamps, and we get $2.90, which would leave $1.50 for the 15 cent stamps. This works! Thus, there is only one possible solution: $1.50 + $2.90.

In the ratio question that we have here, we also have integer constraints (adults and children have to be whole numbers), but $24 and $16 share factors. This means that there will be multiple combinations that yield $2,200. Sanju and Uva both gave good examples of these.

There really is no algebraic shortcut to determine sufficiency in these cases. In each question, we had a single equation with two variables. You have to test numbers to see if there is more than one set of values that would yield the same sum. You can use the idea of shared factors as a clue, but it's advisable to just do the math on these.
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by sanju09 » Sun Dec 29, 2013 11:32 pm
I'd like to add a bit to this nice post by ceilidh.erickson. Firstly, this type of single linear equation in two variables is called a "Diophantine Equation", in which the biggest constraint is that the variables must be positive integers (not just integers) and there exists a unique pair of solution for the two variables, which makes the single linear equation in two variables sufficient by itself.

A Diophantine Equation normally has the coefficients of the variables as co-primes. But, great care must be taken because Simultaneous Equations appear in DS only, where sufficiency test is the ultimate task.
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by gmattesttaker2 » Wed Jan 08, 2014 10:25 pm
ceilidh.erickson wrote:There are times when a single equation is sufficient to solve for two unknowns - take for example OG DS #132:

Joanna bought only $0.15 and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.

Here, statement 1 is sufficient to answer the question. Even though we only have a single equation:
0.15x + 0.29y = 4.40
... we actually have enough information to answer the question, because there are implied integer constraints on the question. We can only have an integer number of stamps. When we try various combinations, we'll find that only one scenario yields $4.40: 10 stamps of each value. Because 15 and 29 do not share any factors, there is no overlap in possible dollar amounts. Multiples of 15 have to end in a units digit of 5 or 0, so to add up to a sum with a units digit of 0, the 29 cent stamps would have to be in a multiple of 5 or 10. Those are the only values we have to test. Five 29 cent stamps would yield $1.45, meaning that the 15 cent stamps would have to add to $2.95. Since this is not a multiple of 15, this doesn't work. Next, we can try ten of the 29 cent stamps, and we get $2.90, which would leave $1.50 for the 15 cent stamps. This works! Thus, there is only one possible solution: $1.50 + $2.90.

In the ratio question that we have here, we also have integer constraints (adults and children have to be whole numbers), but $24 and $16 share factors. This means that there will be multiple combinations that yield $2,200. Sanju and Uva both gave good examples of these.

There really is no algebraic shortcut to determine sufficiency in these cases. In each question, we had a single equation with two variables. You have to test numbers to see if there is more than one set of values that would yield the same sum. You can use the idea of shared factors as a clue, but it's advisable to just do the math on these.

Hello Ceilidh,

Thanks a lot for your excellent post.

Best Regards,
Sri
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by GMATGuruNY » Thu Jan 09, 2014 6:56 am
gmattesttaker2 wrote:Hello,

Can you please assist with this problem. I was not able to solve Statement 2:

If at a school trip each adult was charged $24 and each child was charged $16, what is
the ratio of adults to children at the school trip?

1) The ratio of the total revenue from children to the total revenue from
adults was 8 to 3.

2) Total revenue was $2,200
Another way to evaluate statement 2:

Let $C = the total revenue from the children and $A = the total revenue from the adults.

Since the two statements cannot contradict each other, it must be possible in statement 2 that the $2200 in total revenue is divided according to the ratio indicated in statement 1:
$C : $A = 8:3 = 800:300 = 1600:600, implying that $C = $1600 and that $A = $600, for a total of $2200 in revenue.
Here, the number of children = 1600/16 = 100, while the number of adults = 600/24 = 25.

Check whether OTHER revenue ratios are also possible.
Since the ticket values of $16 and $24 are both factors of $48, we can add $48 to $C, while subtracting $48 from $A:
$C = $1648 and $A = $552, for a total of $2200 in revenue.
Here, the number of children = 1648/16 = 103, while the number of adults = 552/24 = 23.

Since the ratio of children to adults can be different values, INSUFFICIENT.
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by GMATGuruNY » Thu Jan 09, 2014 7:40 am
The strategy employed in my post above can also be used to evaluate statement 1 in OG DS92:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy ?

(1) She bought $4.40 worth of stamps.

(2) She bought an equal number of $0.15 stamps and $0.29 stamps.
Let $F = the total revenue from the 15¢ stamps and $T = the total revenue from the 29¢ stamps.

Since the two statements cannot contradict each other, it must be possible in statement 1 that the 440¢ in total revenue is divided according to the ratio indicated in statement 2.
Statement 2 indicates that the ratio of 15¢ stamps to 29¢ stamps = 1:1.
If an equal number of each type of stamp is purchased, we get:
$F : $T = 15:29 = 150:290, for a total of 440¢ in revenue.
Here, the number of 15¢ stamps = 150/15 = 10, and the number of 29¢ stamps = 290/29 = 10.

Check whether OTHER revenue ratios are also possible.
Since the stamp values are 15¢ and 29¢, the revenue ratio can be altered only by adding a multiple of 15 and 29 to $F or $T, while subtracting this same multiple from the remaining value.
The LCM of 15 and 29 = 15*29 = 435.
If 435¢ is added to either 150¢ or 290¢, the sum will exceed 440¢.
Thus, the revenue ratio CANNOT be altered, implying that only ONE revenue ratio will satisfy statement 1:
$F = 150¢ and $T = 290¢.
Thus, the number of 15¢ stamps = 10.
SUFFICIENT.
Last edited by GMATGuruNY on Wed Jan 13, 2016 3:24 am, edited 2 times in total.
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by Brent@GMATPrepNow » Thu Jan 09, 2014 8:04 am
A key takeaway in all of this is that we cannot necessarily conclude that 1 equation with 2 variables does not provide sufficient information in a Data Sufficiency question. This is a common myth that the GMAT likes to test.

Other myths include:
- 2 equations with 2 variables provide sufficient information
- quadratic equations do not provide sufficient information to determine the given variable

For more information about these common myths (and others), watch our free videos:
- Common GMAT Data Sufficiency Myths - Part I: https://www.gmatprepnow.com/module/gmat- ... cy?id=1106
- Common GMAT Data Sufficiency Myths - Part II: https://www.gmatprepnow.com/module/gmat- ... cy?id=1107

Cheers,
Brent
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by 2mist » Fri Jan 10, 2014 5:57 am
GMATGuruNY wrote:The strategy employed in my post above can also be used to evaluate statement 1 in OG DS92:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy ?

(1) She bought $4.40 worth of stamps.

(2) She bought an equal number of $0.15 stamps and $0.29 stamps.
Let $F = the total revenue from the 15¢ stamps and $T = the total revenue from the 29¢ stamps.

Since the two statements cannot contradict each other, it must be possible in statement 1 that the 440¢ in total revenue is divided according to the ratio indicated in statement 2.
Statement 2 indicates that the ratio of 15¢ stamps to 29¢ stamps = 1:1.
If an equal number of each type of stamp is purchased, we get:
$F : $T = 15:29 = 150:290, for a total of 440¢ in revenue.
Here, the number of 15¢ stamps = 150/15 = 10, and the number of 29¢ stamps = 290/29 = 10.

Check whether OTHER revenue ratios are also possible.
Since the stamp values are 15¢ and 29¢, the revenue ratio can be altered only by adding a multiple of 15 and 29 to $F or $T, while subtracting from this same multiple from the remaining value.
The LCM of 15 and 29 = 15*29 = 435.
If 435¢ is added to either 150¢ or 290¢, the sum will exceed 440¢.
Thus, the revenue ratio CANNOT be altered, implying that only ONE revenue ratio will satisfy statement 1:
$F = 150¢ and $T = 290¢.
Thus, the number of 15¢ stamps = 10.
SUFFICIENT
Mitch,

This post made my day! I was on a lookout for a method other than hit and try to check whether one equation with two variables has unique solution.

#yourock

Regards,
Mist
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