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Cramya DS 1

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Cramya DS 1

by cramya » Mon Dec 22, 2008 7:57 pm
At a certain college , students can major in science , maths , history ,or linguists .If there are 1/3 as many science majors as there are history majors ,and 2/3 as many maths major as there are history majors , how many of the 2000 students major in linguistics?

1.There are as many linguistics majors as there are math majors.

2.There are 250 more maths majors than there are science major

OA : D

Takeaway IMO Knowing how to translate this algebrically: 1/3 as many science majors as there are history majors ,and 2/3 as many maths major as there are history majors and determining ratios involving relationship between more than 2 quantities.


Good luck!
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by ronniecoleman » Mon Dec 22, 2008 9:36 pm
When the question says:

1/3 as many x majors as y

then if y = k

then x = 1/3 k
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by amitabhprasad » Mon Dec 22, 2008 10:12 pm
Not sure of this is correct, but this is the way I solved it
History = 3x
==> Science = x and maths = 2x
stmt 1: Linguistic = Math = 2x
thus 3x+2x+x+2x = 2000 ==> x = 250
suff.
stmt2: Maths - Science = 250 ==> x = 250
suff.
hence "D"

Sounds simplistic as per your standard cramya, must I am missing some thing or not understanding this correctly.
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by cramya » Mon Dec 22, 2008 10:16 pm
Amit,

U got it right! Came across this problem so wanted to share with everyone interpreting 1/3 as many part and then combining ratios involving more than 2 quantities.

Good work and best of luck.

Regards,
Cramya
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by vivek.kapoor83 » Tue Dec 23, 2008 4:47 am
cramya,
i checkd it twice be4 doing and stll nto sure if i am correct coz u cnt get it rite by ur std.

History - x
sci - x/3
maths -2/3x
from 1 - l = 2/3 x
and all this =2000 , So Suff

2 - we can cal x , 2/3x -x/3 -250
x =250
So, L can b cal.
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by cramya » Tue Dec 23, 2008 7:12 am
i checkd it twice be4 doing and stll nto sure if i am correct coz u cnt get it rite by ur std.

Vivek,
Trust me there are some (I am no exception) I sit back and just wonder. This one I was able to solve but just thought of sharing it with the group. Also I wanted to see the different approaches. I did it like Amit,converted all ratios to whole numbers whereas u kept it as fractions. All the same but the approach varies a little. Also I had seen a similar ps question where people had some questions on interpreting the 1/3 as many verbage so wanted to share this one in the ds section.

By the way good work and all the best for your GMAT.

Regards,
Cramya
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by vivek.kapoor83 » Tue Dec 23, 2008 8:01 am
hey,
u took it seriously, it aws on lighther side. neways, it was also confidence booster ....thanks for wishes.
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by cramya » Tue Dec 23, 2008 10:37 pm
u took it seriously, it aws on lighther side. neways, it was also confidence booster ....thanks for wishes.
Next time I will watch out for the lighter side of Vivek.... :-)
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Re: Cramya DS 1

by Stuart@KaplanGMAT » Tue Dec 23, 2008 10:45 pm
cramya wrote:At a certain college , students can major in science , maths , history ,or linguists .If there are 1/3 as many science majors as there are history majors ,and 2/3 as many maths major as there are history majors , how many of the 2000 students major in linguistics?

1.There are as many linguistics majors as there are math majors.

2.There are 250 more maths majors than there are science major

OA : D

Takeaway IMO Knowing how to translate this algebrically: 1/3 as many science majors as there are history majors ,and 2/3 as many maths major as there are history majors and determining ratios involving relationship between more than 2 quantities.


Good luck!
I always look for opportunities to plug my favourite rule, # of equations vs # of unknowns!

From the question stem:

4 variables (s, h, m, l) and 3 distinct linear equations (s vs h; m vs h; total #). So, to solve ANYTHING related to those 4 variables, we need 1 more equation.

(1) l = m. One more distinct linear equation: sufficient.

(2) equation with m and s. One more distinct linear equation: sufficient.

Each statement is sufficient on its own: choose (D).

One of the beautiful things about using this rule is that we don't actually have to translate any of the equations, we just have to recognize how many we have, what variables are involved and if any of the equations are non-linear (e.g. x^2 = 25 is non-linear).
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by cramya » Tue Dec 23, 2008 10:50 pm
I always look for opportunities to plug my favourite rule, # of equations vs # of unknowns!
Stuart,
I agree. Is the problem below an exception to this rule(this is the only problem I have seen so far ):

https://www.beatthegmat.com/word-transla ... 15309.html

After solving something similar I try to convince myself each time that 1 equation wiht 2 unknowns in insuffucient by plugging in a few different values rather than convincing myself on a theoritical basis .

Please advice on what the difference is,in the problem posted above(1 equation 2 unknowns but answer is A).


Regards,
Cramya
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by Stuart@KaplanGMAT » Wed Dec 24, 2008 2:59 am
cramya wrote:
I always look for opportunities to plug my favourite rule, # of equations vs # of unknowns!
Stuart,
I agree. Is the problem below an exception to this rule(this is the only problem I have seen so far ):

https://www.beatthegmat.com/word-transla ... 15309.html

After solving something similar I try to convince myself each time that 1 equation wiht 2 unknowns in insuffucient by plugging in a few different values rather than convincing myself on a theoritical basis .

Please advice on what the difference is,in the problem posted above(1 equation 2 unknowns but answer is A).


Regards,
Cramya
That question is different because we have an extra piece of information: the two variables have to be positive integers.

Here's the exact wording of the rule:
In order to solve for a system of n variables, one needs n distinct linear equations.
However, there are many occasions on which we can get away with fewer than n variables. Here are the two most common "special" situations:

1) if you're only solving for some of the variables in the system, you may not require all n equations.

For example:
If a + b + c + d = 6, what's the value of d?

(1) a + b + c = 4
Statement (1) gets rid of (a + b + c) all in one shot, so even though we only have 2 equations for our 4 variables, we can solve for d.

2) if you're solving for a relationship among variables instead of the actual value of the variables, you may not require all n equations.

For example:
What's the value of 3x + 2y?

(1) 10 - 4y = 6x
Statement (1) can be rewritten as 6x + 4y = 10; we then divide both sides by 2 to get 3x + 2y = 5, answering the question for us. Note, we have no clue what the value of x or y is (and no way to solve for them), but that's not what the question is asking.

The situation in the question you linked is rarer, but another exception to the general rule. When the types of numbers available are limited, you may not need n distinct equations to solve the system.

Here's another example similar to the one you linked:
If a and b are positive integers, what's the value of b?

(1) 2a + 3b = 5
Well, the only way we can satisfy statement (1) is if both a and b are 1, so it's sufficient to answer the question.
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by cramya » Wed Dec 24, 2008 11:23 am
Wow, thank you so much Stewart for a very detailed explanation. Now it makes sense.
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by logitech » Fri Dec 26, 2008 3:46 pm
Stuart Kovinsky - you math genius you!

Thanks for the explanation!
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by Ian Stewart » Fri Dec 26, 2008 5:40 pm
Stuart Kovinsky wrote: Here's the exact wording of the rule:
In order to solve for a system of n variables, one needs n distinct linear equations.
That word distinct deserves some attention as well, because it's the basis of some n equations/n unknowns traps on the GMAT. Sometimes it can appear that you have, say, three distinct equations, three unknowns, but you can't solve for any of your unknowns. The following question, for example, might look like an obvious 'C', but it's an 'E', because the three equations are not independent (fundamentally distinct):

If a + 2b + c = 8, what is the value of c?
1) a + b = 5
2) b + c = 3

(Try adding the two equations in the statements; you'll arrive at the equation given in the question).

If you do know precisely what to look for when you have n equations/n unknowns, you can solve a lot of DS questions very quickly, but you have to be very careful about the exceptions or you'll likely fall into some traps.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by cramya » Fri Dec 26, 2008 5:59 pm
Thanks Ian!
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