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Cost of Envelope

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by dhiren8182 » Wed Feb 15, 2012 10:23 pm
Can anyone explain by picking numbers.
Dhiren
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by kul512 » Mon Feb 20, 2012 9:56 am
needthis wrote:First problem for me in the morning and I guess my brain is short-circuited.
I solved like like this:

2*((x+1)/2+3)+5*((x+1)/2-3)
which eventually simplifies to (7x-2)/2
i think you have missed a point here-
you divided two bunches into (x+1)/2+3 and (x+1)/2-3 but if you subtract the later one(5 cent bunch) from former one(2 cent bunch), you get 6, which is not the given information. Bunches will be ((x+1)+3)/2 and ((x+1)-3)/2, solving this will give the correct answer.
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by Mike@Magoosh » Wed Feb 22, 2012 10:39 am
dhiren8182 wrote:Can anyone explain by picking numbers.
Dhiren
Hi, there, I'm happy to help with this. :)

First of all, I would say: getting comfortable with algebraic manipulations is very important for the GMAT, so for future problems like this, give the algebraic approach an earnest attempt. Nevertheless, here is a solution by picking numbers.

I'll start by picking x = 10. Then, there are 11 envelopes total --- four of the 5 cent envelopes, and seven of the 2 cent envelopes.

Cost = 4*5 + 2*7 = 20 + 14 = 34 cents

Plug x = 10 into the answers

A)3x+1 = 3*10 + 1 = 31 :(
B)(7x-2)/2 = (70 - 2)/2 = 68/2 = 34 this works :)
C)11x+31 = 11*10 + 31 = 141 :(
D)(7x-6)/2 = (70 - 6)/2 = 64/2 = 32 :(
E)(13x+3)/2 = (133)/2 = 66.5 cents (bizarre and wrong!):(

So, as it happens, we got lucky, and on the first possible choice for x, we were able eliminate four answers and narrow the choices down to one. That will not always happen --- sometimes, we have to make one choice for x, eliminate 2 or 3 answers, and then make another choice for x to eliminate more answers.

Does that make sense?

Here for free is another problem to try, with variables in the answer choices.

https://gmat.magoosh.com/questions/342

The question at that link should be followed by a free video solution when you submit your answer.

Please let me know if you have any questions about this.

Mike :)
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by ronnie1985 » Thu Mar 29, 2012 8:02 am
(B) QED
Follow your passion, Success as perceived by others shall follow you
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by singhmanj » Wed Apr 25, 2012 11:39 pm
Hi Mike,

Through the question posted on Magoosh, you have conveyed the importance of knowing algebra or else one tends to make mistakes.

Unless we put the 5th item, value of which is given in the question stem, in proper algebraic format, we will tend to ignore that 4 is not negative but sign changes due to minus outside.

Thanks.

Mike@Magoosh wrote:
dhiren8182 wrote:Can anyone explain by picking numbers.
Dhiren
Hi, there, I'm happy to help with this. :)

First of all, I would say: getting comfortable with algebraic manipulations is very important for the GMAT, so for future problems like this, give the algebraic approach an earnest attempt. Nevertheless, here is a solution by picking numbers.

I'll start by picking x = 10. Then, there are 11 envelopes total --- four of the 5 cent envelopes, and seven of the 2 cent envelopes.

Cost = 4*5 + 2*7 = 20 + 14 = 34 cents

Plug x = 10 into the answers

A)3x+1 = 3*10 + 1 = 31 :(
B)(7x-2)/2 = (70 - 2)/2 = 68/2 = 34 this works :)
C)11x+31 = 11*10 + 31 = 141 :(
D)(7x-6)/2 = (70 - 6)/2 = 64/2 = 32 :(
E)(13x+3)/2 = (133)/2 = 66.5 cents (bizarre and wrong!):(

So, as it happens, we got lucky, and on the first possible choice for x, we were able eliminate four answers and narrow the choices down to one. That will not always happen --- sometimes, we have to make one choice for x, eliminate 2 or 3 answers, and then make another choice for x to eliminate more answers.

Does that make sense?

Here for free is another problem to try, with variables in the answer choices.

https://gmat.magoosh.com/questions/342

The question at that link should be followed by a free video solution when you submit your answer.

Please let me know if you have any questions about this.

Mike :)
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by heymayank08 » Sat Apr 28, 2012 11:25 pm
easy..pick B
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by Mike@Magoosh » Mon Apr 30, 2012 10:29 am
singhmanj wrote:Hi Mike, Through the question posted on Magoosh, you have conveyed the importance of knowing algebra or else one tends to make mistakes.
Unless we put the 5th item, value of which is given in the question stem, in proper algebraic format, we will tend to ignore that 4 is not negative but sign changes due to minus outside.
I'm not sure I understand the point you are making here. If one plugs in numbers ---- say, x = 3, then the average is 3(3)+4 = 13, which means the sum of the five numbers is 65, and the fifth number is 7(3)-4 = 17. When we leave it in algebraic form, we have to worry about the sign reversal (subtracting a minus sign), but when we put everything into numbers, the subtraction of 4 is simply part of the calculation and included in the value of the fifth number, 17. Then, we would just subtract 17 from the sum. We don't have to worry about the negative-of-a-negative problem when we use a numerical approach.

If I haven't allayed your concern, then please elaborate.

Mike :-)
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by Lifetron » Sat Aug 25, 2012 7:18 pm
It is 'B' !
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by chris558 » Thu Sep 20, 2012 6:07 am
Let T=two cent envelopes and
F=five cent evenelopes

T+F=x+1
T=F+3-->T-f=3

add these equations together...
T+F=x+1
+T-F=3
----------
2T=x+4
T=(x+4)/2

Total cost= 2T+5(x+1-T)
=2((x+4)/2) + 5(x+1-(x+4)/2)
=(2x+8)/2 + 5[(2x+2-x-4)/2)
=(2x+8)/2 + 5(x-2)/2
=(2x+8+5x-10)/2
=(7x-2)/2

Answer is B
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by chibapawan » Wed Dec 19, 2012 3:38 am
kvcpk wrote:
rashmi.kaushal wrote:A company hired a printer to produce a total of x + 1 envelopes. The job consisted of two types of envelopes, 2¢ envelopes and 5¢ envelopes. If the company requested 3 more 2¢ envelopes than 5¢ envelopes, which of the following expressions denotes the cost, in cents, of the total x + 1 envelopes ?

A)3X+1

B)7x-2/2

C)11x+31

D)7x-6/2

E)13x+3/2
Another easy way of solving this is assuming x to be 4.
This means there are 4 2cents envelopes and 1 5cents envelope
Total cost =8+5 = 13
By substituting in options, only B satisfies.
Am i doing something wrong here?

Will x=4 not result in 13 for option 1 as it does for option 2?
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by rajeshsinghgmat » Mon Apr 01, 2013 3:03 am
(B) (7x - 2)/2

Let, the no. 5 cent envelopes =a

Let, the no. 2 cent envelopes =b

b = a + 3

a+b = x+1

a + a + 3 = x + 1

2a = x - 2

a = (x-2)/2

Cost of envelopes = 5a + 2b
= 5a + 2(a+3)
= 7a + 6
= 7(x-2)/2 +6
= (7x -14 + 12)/2
= (7x -2)/2
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by Pratiti » Tue May 14, 2013 1:12 am
OA is B

Take any value say total number of envelopes=09
therefore we get total number of envelopes will be 3+6
cost will be (3*5)+(2*6)=27
now verify the answers which give 27 as the answer
that is B
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by Java_85 » Sat Oct 26, 2013 9:42 am
IMO B
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