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is linear eqn with 2 variables sufficient

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is linear eqn with 2 variables sufficient

by vishal.pathak » Sat Nov 19, 2011 6:06 am
Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.

In this case statement 2 is sufficient

But

2 digit number TU = ?
(1) TU^2 = 196
(2) 3T + 5U = 47

Here statement 2 is not.

How do we decide if linear equation with 2 variables is sufficient or not
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by user123321 » Sat Nov 19, 2011 8:29 am
vishal.pathak wrote:Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.

In this case statement 2 is sufficient

But

2 digit number TU = ?
(1) TU^2 = 196
(2) 3T + 5U = 47

Here statement 2 is not.

How do we decide if linear equation with 2 variables is sufficient or not
question 1)
A)this is clearly insufficient.
B)23x+21y = 130
since x,y are +ve integers
we need to verify whether we are going to get multiple x,y pairs or not.
by trail and error we know x = 2,y=4 clearly satisfies & nothing else.
hence B is sufficient.

question 2)
A)T.U^2 = 196
give T is ten's digit & U unit's digit
the only possible value is when T.U^2 = 4.7^2
in rest of the cases either T or U is having more one digit. hence sufficient.

B)3T + 5U = 47
check the cases like we did for first question
T U TU
9 4 94
4 9 49
here two cases are possible. hence insufficient.

user123321
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by vishal.pathak » Sat Nov 19, 2011 8:54 am
user123321 wrote:
vishal.pathak wrote:Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.

In this case statement 2 is sufficient

But

2 digit number TU = ?
(1) TU^2 = 196
(2) 3T + 5U = 47

Here statement 2 is not.

How do we decide if linear equation with 2 variables is sufficient or not
question 1)
A)this is clearly insufficient.
B)23x+21y = 130
since x,y are +ve integers
we need to verify whether we are going to get multiple x,y pairs or not.
by trail and error we know x = 2,y=4 clearly satisfies & nothing else.
hence B is sufficient.

question 2)
A)T.U^2 = 196
give T is ten's digit & U unit's digit
the only possible value is when T.U^2 = 4.7^2
in rest of the cases either T or U is having more one digit. hence sufficient.

B)3T + 5U = 47
check the cases like we did for first question
T U TU
9 4 94
4 9 49
here two cases are possible. hence insufficient.

user123321
Trying every case is a lot of hit and trial. Is there a way to determine whether a particular linear equation has more than 1 solution. Like in the 1st example here, how did you know that more solution do not exist. Did you check every option. If yes, will you do the same in the exam
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by user123321 » Sat Nov 19, 2011 9:45 am
you dont need to find all solutions...just need to check whether it is possible to get more than one. that is enough. In exam, all questions wont be tough. The time we save by answering easy questions quickly, will help us invest more time for answering tough questions.

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by vishal.pathak » Sat Nov 19, 2011 10:15 am
user123321 wrote:you dont need to find all solutions...just need to check whether it is possible to get more than one. that is enough. In exam, all questions wont be tough. The time we save by answering easy questions quickly, will help us invest more time for answering tough questions.

user123321
buddy that is true, but I guess in hit and try one normally tries 2-3 options and time does not permit to try more. So to try as long as we get al alternate answer is not wise IMO

EXPERTS PLEASE HELP
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by user123321 » Sat Nov 19, 2011 10:27 am
vishal.pathak wrote:
user123321 wrote:you dont need to find all solutions...just need to check whether it is possible to get more than one. that is enough. In exam, all questions wont be tough. The time we save by answering easy questions quickly, will help us invest more time for answering tough questions.

user123321
buddy that is true, but I guess in hit and try one normally tries 2-3 options and time does not permit to try more. So to try as long as we get al alternate answer is not wise IMO

EXPERTS PLEASE HELP
I think you just started preparation like me. you better should learn hit & trail if you need to get past DS. because when you start reading solving inequalities, you need to assume lot of values to get into conclusions. The more you practice, the faster you will see the patterns and without noting the values you can recognize whether multiple values are possible or not. I am no expert but it's just a simple suggestion :)

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by vishal.pathak » Sat Nov 19, 2011 11:28 am
user123321 wrote: I think you just started preparation like me. you better should learn hit & trail if you need to get past DS. because when you start reading solving inequalities, you need to assume lot of values to get into conclusions. The more you practice, the faster you will see the patterns and without noting the values you can recognize whether multiple values are possible or not. I am no expert but it's just a simple suggestion :)

user123321
appreciate ur suggestion buddy, practice is definitely the key n as u said buddy when hit and trial is done, it has to be based on some pattern and my hope here is that some expert can tell us a method to find patterns in case of linear equations with 2 variables
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by vishal.pathak » Sun Nov 20, 2011 3:24 am
Experts, please help. Just a confirmation of user123321's claims will be enough.

I just want to know whether there is a way to find the number of possible solutions of a given linear equation with 2 variables

Regards,
Vishal
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by shankar.ashwin » Sun Nov 20, 2011 3:48 am
I am not sure if there's a method to doing this. But you get a general idea here.

You have a equation, 23X+21Y = 130.

Now if I assume X=Y, we have X (23+21) = 130

(now) look at the LHS, you have 44X.

List out multiples of 44 close to the needed range (44,88,132)

When X = 3 we have 132, which is 2 more than the required value. Since the difference of our numbers is 2, we can increase X by 1 and decrease Y by 1,

so we have 4X+2Y which is = 130.

In general, if we have an equation of the form ax + by = c (Writing it in its simplest form)

We write it as, x = (c - by)/a , now we substitute values of y=0,1,2.. and find the corresponding value of 'x'. While I agree this is not the easiest thing to do, I believe the GMAT would never ask you to calculate values which exceed substituting y=5. Though cumbersome, it shouldn't take you more than 3 mins.

Try doing this for the second question. Our equation here is 3X+5Y=47

X = (47-5Y)/3

For X to be an integer, (47-5Y) must be divisible by 3.

47 leaves a remainder of 2 when divided by 3, so when 5Y also leaves a remainder of 2 and we subtract the 2, we get a number which is divisible by 3.

For eg, (47 - 2) = 45 which is divisible by 3.

Possible values of 5Y are (5,10,15,20,25,30,35,40)

Now by looking at the numbers you can see (5,20,35) leave a remainder of 2, these correspond to Y=1,4 and 7 respectively. So we have 3 possible cases here.
Last edited by shankar.ashwin on Sun Nov 20, 2011 5:48 am, edited 1 time in total.
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by pemdas » Sun Nov 20, 2011 4:22 am
Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.
st(1) a+b=6 and 23a+21b=? we are missing data for producing two linear equations with 2 variables Not Sufficient
st(2) find integer values for a,b to satisfy for 23a+21b=130 -> 23 is prime so start with it, 130/23=6*23 + 2/23. Click away a=6 and start with a=5 -> 130-(23*5) is divisible by 21? No it isn't. Next, 130-(23*4) is divisible by 21? No. Continue till a=2, 130-(23*2) is divisible by 21? Yes and b=4. Sufficient

b

2 digit number TU = ?
(1) TU^2 = 196
(2) 3T + 5U = 47
note both T and U must be whole numbers
st(1) TU^2=196 takes us to prime factorization of 196
196/2/98
98/2/49
49/7/7
7/7/1 -> (2*7)^2 or T=4 and U=7 Sufficient

st(2) 3T+5U=47 both coefficients are primes hence 3,5,47, hence 3T/47 +5U/47 which returns 1 is only possible if T and U are selected as pairs amongst single whole numbers. By trial method we have T=4 and U=7
T cannot be 0
T=1 -> missing 44 which cannot be factored by 5, 44/5 is non-integer
T=2 -> missing 41 which cannot be factored -----
T=3 -> missing 38, the same
T=4 -> missing 35, bingo U=7
T=5 -> missing 32, ---
T=6 -> missing 29, ---
T=7 -> missing 26, ---
T=8 -> missing 23, ---
T=9 -> missing 20, bingo U=4

Two bingos will make this Not Sufficient (TU), (47) and (94)

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by pemdas » Sun Nov 20, 2011 4:48 am
i'm not the GMAT expert, but would disagree about hit and try as the key for such questions

please note every question is wrapped around some concept, I haven't looked into previous posts not to get direction for solution but rather tried on my own. You can recognize there at least one difference which is the true key for many number property elements involved into questions and these may be embedded into equations, inequalities even into good share of geometry questions in GMAT. This key I meant is 'Perform prime factorization of the numbers given, when you're in loss and want to see what may/should come out of this"

vishal.pathak wrote:Experts, please help. Just a confirmation of user123321's claims will be enough.

I just want to know whether there is a way to find the number of possible solutions of a given linear equation with 2 variables

Regards,
Vishal
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by mehrasa » Sun Nov 20, 2011 11:33 am
vishal.pathak wrote:Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.

In this case statement 2 is sufficient

stat1) X+Y=6 not sufficient bcuz we need to have total cost of these pencils
stat2) 23x+21y=130 since we have a hidden constrain for this Q which is that x and y need to be integer.. we need to plug in numbers to find x and Y
or say x=(130-21y)/23.... sufficient ==> answer is B

2 digit number TU = ?
(1) TU^2 = 196
(2) 3T + 5U = 47

Here statement 2 is not B

stat1) tu^2=196 ==> the only two digit number which its square is 196 is 14 => TU is 14
stat2)3t+4u=47 here also the x and y should be integers.. the only pair is t=9 and u=4 ==> tu=94

while seems both reach us to a unique answer, the answer should be E bcuz the answers of statements are different




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by pemdas » Sun Nov 20, 2011 12:09 pm
@meh, you read previous posts after solving on yours?
there are two digits and not one, TU^2 and not (TU)^2 so discard 14^2
st(1) TU^2=196 takes us to prime factorization of 196
196/2/98
98/2/49
49/7/7
7/7/1 -> (2*7)^2 or T=4 and U=7 Sufficient
also st(2) doesn't offer unique answer , there are two answers

mehrasa wrote:
vishal.pathak wrote:Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.

In this case statement 2 is sufficient

stat1) X+Y=6 not sufficient bcuz we need to have total cost of these pencils
stat2) 23x+21y=130 since we have a hidden constrain for this Q which is that x and y need to be integer.. we need to plug in numbers to find x and Y
or say x=(130-21y)/23.... sufficient ==> answer is B

2 digit number TU = ?
(1) TU^2 = 196
(2) 3T + 5U = 47

Here statement 2 is not B

stat1) tu^2=196 ==> the only two digit number which its square is 196 is 14 => TU is 14
stat2)3t+4u=47 here also the x and y should be integers.. the only pair is t=9 and u=4 ==> tu=94

while seems both reach us to a unique answer, the answer should be E bcuz the answers of statements are different




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by mehrasa » Sun Nov 20, 2011 6:59 pm
hmmm. I got the point
good Q...
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by Testtrainer » Tue Nov 29, 2011 10:20 am
Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.

I think the key to answering this very difficult question is to know two things: 1) Pencils are by definition positive integers and 2) The two statements will always be consistent with one another (at least on the actual GMAT, not necessarily from other material). So one can use statement (1) to see that the 23 cent and 21 cent pencils must be limited to: 0,6; 1,5; 2,4; or 3,3. Plug each of these values into statement (2) to see whether only one set of values works. As mentioned previously, only 2,4 work, so statement (2) is sufficient by itself.

This is very tricky because although we are using statement (1), we don't need the data from statement (1) to answer the question. We just use it because we know how this stupid test works![/i][/quote]
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