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If Company M ordered a total of 50 computers and printers an

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If Company M ordered a total of 50 computers and printers an [#permalink]

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If Company M ordered a total of 50 computers and printers and Company N ordered a total of 60 computers and printers, how many printers did company M order?

(1) Company M and Company N ordered the same number of computers
(2) Company N ordered 10 more printers than Company M

[Reveal] Spoiler:
I get stuck in this one! Why is not C the right answer??? :shock:

Thanks!!
[Reveal] Spoiler: OA

Last edited by Engr2012 on 21 Jan 2016, 05:34, edited 2 times in total.
Formatted the question.
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 29 Jan 2013, 12:54
mjg2110 wrote:
Hi

I get stuck in this one! Why is not C the right answer??? :shock:
Thanks!!


If company M ordered a total of 50 computers and printers and company N order a total of 60 computers and printers. How many computers does company M order?

1) Company M and N order the same number of computers
2) Company N order 10 computers more than M.


I found this q weird.
For M, C(M) + P(M) = 50
For N, C(N)+ P(N) = 60

A does not say anything about printers for M and N. It can be 10 and 10, 20 and 20, 30 and 30 or any such combi
B same issue as A
On Combining, the info in both options is ambiguous. M and N computers are same where B says N has 10 more than M. How both of them can be true?
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 29 Jan 2013, 17:06
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 30 Jan 2013, 04:45
Expert's post
mjg2110 wrote:
If Company M ordered a total of 50 computers and printers and Company N ordered a total of 60 computers and printers, how many printers did company M order?

(1) Company M and Company N ordered the same number of computers
(2) Company N ordered 10 more printers than Company M

I get stuck in this one! Why is not C the right answer??? :shock:

Thanks!!


It's straight E. Consider the following cases:

#1: M ordered 5 computers and 45 printers, N ordered 5 computers and 55 printers;
#2: M ordered 10 computers and 40 printers, N ordered 10 computers and 50 printers;

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 08 Aug 2014, 02:51
This is absolute killer.
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 23 Oct 2015, 08:21
This looks straight forward at first. From the question, two equations are introduced and four variables are introduced.

Then (1) and (2) each introduce what appears to be an additional equation so we have four equations and four variables.

At this point I selected answer C without solving assuming that our variable could be solved by rules of solvability with 4 equations. The problem is that the equation that (2) introduces is in fact identical to the information you already have. In essence it is not a new equation and thus you are still left with 4 variables and only 3 equations.

Does anyone have a more succinct way to explain this or identify this pattern in future problems?
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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jbburf wrote:
This looks straight forward at first. From the question, two equations are introduced and four variables are introduced.

Then (1) and (2) each introduce what appears to be an additional equation so we have four equations and four variables.

At this point I selected answer C without solving assuming that our variable could be solved by rules of solvability with 4 equations. The problem is that the equation that (2) introduces is in fact identical to the information you already have. In essence it is not a new equation and thus you are still left with 4 variables and only 3 equations.

Does anyone have a more succinct way to explain this or identify this pattern in future problems?


The rule is that you need to have n "distinct" equations to solve to "n" variables. This is especially true for DS questions. Do not mark C or E in DS questions without actually loooking at the equations you get either from the question stem and the statements. This is the "pattern" you are talking about.

You get the same equations of \(C_m + P_m = 50\) from the question stem+2 statements combined.

Thus, be very careful in DS questions when you are given 'n' variables and 'n' equations. These equations must be distinct to give any unique value.

Hope this helps.
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 23 Oct 2015, 13:00
Engr2012 wrote:
jbburf wrote:
This looks straight forward at first. From the question, two equations are introduced and four variables are introduced.

Then (1) and (2) each introduce what appears to be an additional equation so we have four equations and four variables.

At this point I selected answer C without solving assuming that our variable could be solved by rules of solvability with 4 equations. The problem is that the equation that (2) introduces is in fact identical to the information you already have. In essence it is not a new equation and thus you are still left with 4 variables and only 3 equations.

Does anyone have a more succinct way to explain this or identify this pattern in future problems?


The rule is that you need to have n "distinct" equations to solve to "n" variables. This is especially true for DS questions. Do not mark C or E in DS questions without actually loooking at the equations you get either from the question stem and the statements. This is the "pattern" you are talking about.

You get the same equations of \(C_m + P_m = 50\) from the question stem+2 statements combined.

Thus, be very careful in DS questions when you are given 'n' variables and 'n' equations. These equations must be distinct to give any unique value.

Hope this helps.


Well said, and thank you for the response. The problem here was that an additional equation was introduced but it was not distinct.
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 21 Jan 2016, 02:02
mjg2110 wrote:
If Company M ordered a total of 50 computers and printers and Company N ordered a total of 60 computers and printers, how many printers did company M order?

(1) Company M and Company N ordered the same number of computers
(2) Company N ordered 10 more printers than Company M

I get stuck in this one! Why is not C the right answer??? :shock:

Thanks!!


Admins can you please edit this question take this part out "I get stuck in this one!Why is not C the right answer??? :shock: " sothat one cannot see the answer. Thanks.

Company M: a+b=50, Company N: x+y=60, b=?

(1) a=x, ok, y-b=10, not sufficient
(2) y=b+10 same information as above, not sufficient
(1)+(2) We have twice the same info, so still not sufficient.

Answer E
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 23 Jun 2016, 03:36
Bunuel wrote:
mjg2110 wrote:
If Company M ordered a total of 50 computers and printers and Company N ordered a total of 60 computers and printers, how many printers did company M order?

(1) Company M and Company N ordered the same number of computers
(2) Company N ordered 10 more printers than Company M

I get stuck in this one! Why is not C the right answer??? :shock:

Thanks!!


It's straight E. Consider the following cases:

#1: M ordered 5 computers and 45 printers, N ordered 5 computers and 55 printers;
#2: M ordered 10 computers and 40 printers, N ordered 10 computers and 50 printers;

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html



Bunuel

Hey!
I am also stuck with this question. The above statements that you mentioned didn't give us a unique answer on their own. But when i am combining them, i am using this approach and reaching to an answer.

Let (M computers) MC = x
therefore, (M Printers) MP = 50-x .......(eq1)
On other hand NC = y
and NP= 60-y

Now statement 1 says : x=y ie. Company M and N ordered same number of computer.
Thus NP becomes 60-x (because NP WAS "60-y" and now x=y)
so now, NP= 60-x .........(eq2)

statement 2 says: NP=10MP

Now combining 1 and 2 gives us:
ie (60-x) = 10(50-x)

Please help
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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 23 Jun 2016, 03:42
Expert's post
ashutoshsh wrote:
Bunuel wrote:
mjg2110 wrote:
If Company M ordered a total of 50 computers and printers and Company N ordered a total of 60 computers and printers, how many printers did company M order?

(1) Company M and Company N ordered the same number of computers
(2) Company N ordered 10 more printers than Company M

I get stuck in this one! Why is not C the right answer??? :shock:

Thanks!!


It's straight E. Consider the following cases:

#1: M ordered 5 computers and 45 printers, N ordered 5 computers and 55 printers;
#2: M ordered 10 computers and 40 printers, N ordered 10 computers and 50 printers;

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html



Bunuel

Hey!
I am also stuck with this question. The above statements that you mentioned didn't give us a unique answer on their own. But when i am combining them, i am using this approach and reaching to an answer.

Let (M computers) MC = x
therefore, (M Printers) MP = 50-x .......(eq1)
On other hand NC = y
and NP= 60-y

Now statement 1 says : x=y ie. Company M and N ordered same number of computer.
Thus NP becomes 60-x (because NP WAS "60-y" and now x=y)
so now, NP= 60-x .........(eq2)

statement 2 says: NP=10MP

Now combining 1 and 2 gives us:
ie (60-x) = 10(50-x)

Please help


(2) says: Company N ordered 10 more printers than Company M, not 10 times as many.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If Company M ordered a total of 50 computers and printers an [#permalink]

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New post 23 Jun 2016, 09:28
Straight E
(1) M and N bought same number of computer = {Both 20 computers} or {30 computers} or {40 computers} ..etc etc. Cannot find no. of printers by this information INSUFFICIENT
(2) N bought 10 more printers than N= {M-30 N-40} or {M-44 N-54} or {M-1 N-11} No. of printers is again variable INSUFFICIENT

COMBINE
MC & NC=25 MP=25 NP=35 [MP+MC= 50 NP+NC=60]
or
MC & NC=10 MP=40 NP=50 [MP+MC= 50 NP+NC=60]
or
MC & NC=1 MP=49 NP=59 [MP+MC= 50 NP+NC=60]
or
MC & NC=0 MP=50 NP=60 [MP+MC= 50 NP+NC=60]

Even afer combining the two statements number of printers that company M bought can be anything from 1 to 50

HENCE 0

DO NOT BE AFRAID TO CHOOSE E WHEN CALCULATIONS OBVIOUSLY POINTS THAT THE OPTION IS INDEED E

mjg2110 wrote:
If Company M ordered a total of 50 computers and printers and Company N ordered a total of 60 computers and printers, how many printers did company M order?

(1) Company M and Company N ordered the same number of computers
(2) Company N ordered 10 more printers than Company M

[Reveal] Spoiler:
I get stuck in this one! Why is not C the right answer??? :shock:

Thanks!!

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Re: If Company M ordered a total of 50 computers and printers an   [#permalink] 23 Jun 2016, 09:28
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