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by Abdulla » Thu Dec 25, 2008 3:21 pm
Marta bought several pencils, if each pencil was either a 23 cent pencil or a 21 cent pencil, how many 23 cent pencils did Marta buy?

1. Marta bought a total of 6 pencils

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

Thoughts:

B(23) = Value => value/23 = B .. The question asked for B ?

Given that A(21) + B(23) = Value

Statement 1 : A+B = 6 Insufficient still we need to know the value of each variable alone.

Statement 2 : A(21)+B(23)=130 .. My question is when we can say that there is only one unique way to get 130 ?? because this what makes the question tricky. On the other hand there are other similar questions we chose C as the answer, but in here the OA is (B) .
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by amitabhprasad » Thu Dec 25, 2008 3:39 pm
If you look at equation formed by statement 2
23x+21y = 130
only one combination is possible
23*2+21*4 = 130
Hence select "B"

I got this same question yesterday while taking GMATPrep,I selected "B"
as I don't see any other way how cost of the pencil can be 130. I took hint of "A" but that is to save time and see if 6 can be broken such that some combination results in 130 that helped me narrow down number trials. I am sure there must be some better alternatives
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by Abdulla » Thu Dec 25, 2008 4:05 pm
So lets say we know nothing about x and y and the total value was 12,534
How to solve it without knowing the value of at least one of them ( x or y) ??
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by amitabhprasad » Thu Dec 25, 2008 4:39 pm
Abdulla wrote:So lets say we know nothing about x and y and the total value was 12,534
How to solve it without knowing the value of at least one of them ( x or y) ??
I am not sure if this will e easy enough to solve in 2 min, unless number is such that it can be factored out easily. Most of the GMATPrep questions are structured logically.
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by cramya » Thu Dec 25, 2008 8:12 pm
Abduall refer here:
https://www.beatthegmat.com/cramya-ds-1-t26252.html

Stuart has provided some explanation for rare situations like these. There is one other problem like this for which I have provided the link to in the post above(follow through the post above and u will see).

Hope this also helps!
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by ronniecoleman » Thu Dec 25, 2008 10:54 pm
Marta bought several pencils, if each pencil was either a 23 cent pencil or a 21 cent pencil, how many 23 cent pencils did Marta buy?

1. Marta bought a total of 6 pencils

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

very nice question:
Infact tricky question


normal course of actino would be to have OA as C

Two equations and two variables, but we need to be careful .

b says total cost 130

Unit digit is 0
So 23* x + 21* k = should have a unit digit as 0

doing hit and trial ..
23*2 = 46
so we require 4 in the units place more..
21*4= 84

46+ 84 = 130 Bingo

But you mail ask for other options:

23*1 = 23
21*7 = 147 not allowed

23*3= 69
21*1= 21
sum = 90 not allowed

like thsi you will see other options getting eliminated.
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by Abdulla » Fri Dec 26, 2008 1:19 am
cramya wrote:Abduall refer here:
https://www.beatthegmat.com/cramya-ds-1-t26252.html

Stuart has provided some explanation for rare situations like these. There is one other problem like this for which I have provided the link to in the post above(follow through the post above and u will see).

Hope this also helps!
Cramya.. He did not explain the way that he solved the problem rather than using the number of equations = the number of Variables logic..
Did you mean this approach ?
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by cramya » Fri Dec 26, 2008 1:23 am
Did you mean this approach ?
What I meant is in general to solve for all n variables u need n distinct equations. But sometimes u may need to watch out for situations like Stuart desribed in the post(u may not need n distinct equations if u r solving for less than n variables or trying to find a relation between variables).

Example Stuart gave:
2a+3b = 5

a=1 b=1 are the only solutions given a and b are positive

For example in the problem u posted here there was only 1 equation but 2 disticnt variables but it still could be solved (since both numbers were positive and the only combination was 23*2+21*4 = 130 like Amit explained in his post.

Hope this helps!
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by Ian Stewart » Fri Dec 26, 2008 10:08 am
In this question, if she bought x pencils, it must have cost between 21x and 23x cents. So from 2), we quickly discover x must be 6, and then it must be that only one combination of 21 and 23 cent pencils will give us the right total price.

In general, it can be risky to apply the 'if you have n unknowns, you need n equations to solve' guideline that is sometimes casually repeated in test prep materials. There are many, many exceptions to this guideline on the GMAT, as this question illustrates. In particular, if the quantities in your equations must be positive integers, as is the case in many word problems, that is a severe restriction, and it means you often need much less information than you first expect when you glance at the algebra.
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by vittalgmat » Fri Dec 26, 2008 10:32 am
Ian Stewart wrote:In this question, if she bought x pencils, it must have cost between 21x and 23x cents. So from 2), we quickly discover x must be 6, and then it must be that only one combination of 21 and 23 cent pencils will give us the right total price.

In general, it can be risky to apply the 'if you have n unknowns, you need n equations to solve' guideline that is sometimes casually repeated in test prep materials. There are many, many exceptions to this guideline on the GMAT, as this question illustrates. In particular, if the quantities in your equations must be positive integers, as is the case in many word problems, that is a severe restriction, and it means you often need much less information than you first expect when you glance at the algebra.
Ian,

How would u identify an equation such as the above have unique solutions. Any signs/clues ?
Here are some of my random thoughts. Looking the two equations:
1) 21x + 23y = 130
2) 2a +3b =5

I can deduce the following:
a) atleast one of the coefficients is a prime number (23 and 2,3)
b) This implies LCM is the product of the two coefficients and GCD is 1 (as the two are coprime).

Is there anything else that can help identify such equations??

thanks
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by rajataga » Fri Dec 26, 2008 10:42 am
@vittal :- more than the prime number, a big hint to identify such types of equations is when the answer to the variables HAS TO BE whole numbers......

like in 2a + 3b = 5....here the answer has to be whole numbers, only then will you get a unique solution...
otherwise many more values of 'a' and 'b' could exist.....
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by vittalgmat » Fri Dec 26, 2008 11:31 am
Thanks Rajataga,
I agree with u that whole numbers/integers makes it restrictive. I believe that whole number is the only restriction that I have seen in the many problems that I have encountered. But I was wondering whether there is something else..

Thanks a lot
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by Ian Stewart » Sat Dec 27, 2008 11:50 am
vittalgmat wrote: Ian,

How would u identify an equation such as the above have unique solutions. Any signs/clues ?
Here are some of my random thoughts. Looking the two equations:
1) 21x + 23y = 130
2) 2a +3b =5

I can deduce the following:
a) atleast one of the coefficients is a prime number (23 and 2,3)
b) This implies LCM is the product of the two coefficients and GCD is 1 (as the two are coprime).

Is there anything else that can help identify such equations??

thanks
Your observations are certainly correct, but of course we're only looking at two different questions here, and it's actually just coincidence that the numbers were coprime in the two examples. To recognize this trap, and similar traps in, for example, ratio problems (see the men/women/children sightseeing tour question from GMATPrep for a related trap in a ratio question), you simply need to notice when the variables in a question are restricted to certain values. In the most common situation, your unknowns are constrained to be integers, which is a severe restriction, and you might be able to solve with much less information than normal.
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by vittalgmat » Sat Dec 27, 2008 5:32 pm
Thanks Ian, Cramya, Rajataga for the good discussion.

In summary, as most of you have observed, watch for restrictions on the variables. They would be Integers and often +ve integers (eg. ppl, # of stamps, # of packages, any real life situations/things/ppl).
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by cramya » Sat Dec 27, 2008 6:22 pm
In summary, as most of you have observed, watch for restrictions on the variables. They would be Integers and often +ve integers (eg. ppl, # of stamps, # of packages, any real life situations/things/ppl
Here's another one of these(making up the numbers just to illustrate the point):

Given x+y+z =1000

To find : x

Stmt I

x = .67y

INSUFF

However stmt II is

.40x = .60(y+z)

We would think smt II is insuff so its C) but here's the catch


Given : x+y+z = 1000 can be rewritten as 1000-x = y+z

.40x/.60 = y+z

We can equate
1000-x = .40x/.60

We can find x.

Stmt II looks insufficient but is indeed sufficient

I think this is tricky also :-)

Three distinct varaibles only 2 equations Can be solved since we have to find only one of the quantities and that quantity can be isolated in statement 2 to get us our 2nd equation




Moral of the story:

Try rearranging/playing with the equations (see if something cancels etcc..) given (dont have to spend too much time) to see if we have what we need to solve for something. Just a quick check helps! For some rate problems we really dont need all of the 3 components if we can establish a realtion between 2 or more of the quantities

Hope this helps also! Good luck!
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