Total=x+1 envelopes

Let number of 2C envelopes=a

Let number of 5C envelopes=b

a+b=x+1

a=b+3

b+3+b=x+1

2b+3=x+1

b=(x-2)/2

Total cost=2a+5b

=2(b+3)+5b=7b+6

=7(x-2)/2+6

=(7x-2)/2

Pick B

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

x+1 envelopes.
a+b = x+1
a-b = 3
2a= x+4
a = x/2 + 2
b = x/2 -1

Total cost = 2a+5b
2(x/2 +2) + 5(x/2 -1)
=x+4 + 5x/2 -5
=7x/2 -1

pick B

Aside from the 3 extra 2-cent envelopes, there are (x-2)/2 2-cent envelopes, and the same number of 5-cent envelopes. Adding those together, plus the 6 cents for the 3 extra 2-cent envelopes, the total cost is:

[(x-2)/2][5+2] + 6

=[(x-2)/2]*7 + 6

=(7x - 14 + 12)/2

=(7x - 2)/2

Pick B

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.

Plugging in is a good idea but don't use 4 as A also gives you 13.

If you want to do it algebraically:

Total envelopes (x+1)

a = total 2 cent envelopes (there are 3 more of these)
b = total 5 cent envelopes

2a + 5b = total cost

Calculate the three extra 2 cent envelopes separately so that you have an equal number of 2 cent and 5 cent envelopes. Note that it is a total of 6 cents.


((x-2)/2) * 7 + 6)

(7x-14)/2 +6
(7x -14 +12)/2
(7x-2)/2

And so we choose B.

Hope this helps!

nicolezl wrote:Aside from the 3 extra 2-cent envelopes, there are (x-2)/2 2-cent envelopes, and the same number of 5-cent envelopes. Adding those together, plus the 6 cents for the 3 extra 2-cent envelopes, the total cost is:

[(x-2)/2][5+2] + 6

=[(x-2)/2]*7 + 6

=(7x - 14 + 12)/2

=(7x - 2)/2

Pick B

thank u very much..........i understood how to solve it....

Picking #'s and plugging in here is probably the best and easiest way to do it. Doing the algebra can be confusing and can possibly lead to trap answers in other problems like this one.

T1 is the number of 2 cents envelopes

T2 is the number of 5 cents envelopes

T1+ T2 = X+1

T1 - T2 = 3

solving both equations T1 = X+4/2

substituting T1 in T1 - T2 = 3 , you will get T2 = X-2/2

Total cost = 2(X+4/2) = 5 (x-2/2) = 7x-2/2

IMO: B

total envelopes = x+1

assume number of 2 cent envelope = e
so number of 5 cent envelopes = e -3

e+e-3 = x+1
e = x/2 + 2
e - 3 = x/2 -1

cost = 2 (e) + 5 (e-3)
= 2. x/2 + 2.2 + 5. x/2 - 5
= 7.x/2 -1 = (7x-2)/2

choice B

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

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

Total=x+1
let # of $2=n
then $5 = n-3
Also n+n-3=x+1 -> n=x/2 + 2
Thus cost = 2n + 5(n-3)
=7n-15 = 7x/2 + 14-15 = 7x/2 -1
IMO B

total evnelop = x+1
if difference is 3 then the total no of 2c env. = (x+1)/2 + 3/2
total cost of 2c env. = (x+4)*2/2 = x+4

total no of 5c env. = (x+1)/2 - 3/2
total cost of 2c env. = (x-2)*5/2

total cost = x+4+(5x/2)-5 = (7x - 2)/2

hence B.

I solved this by picking a number for X, solving for the how many 2cent and 5cent stamps adding them to come up with a total order cost. then plug in my number for X in each answer choice... luckily, B was the right answer!

Pick B and move on.