PS algebraic formula

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wayofjungle: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Fri Mar 11, 2011 8:55 am; Thanked: 12 times

by wayofjungle » Mon Mar 28, 2011 12:33 pm

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?

3x + 1
(7x - 2) / 2
11x + 31
(7x - 6) / 2
(13x + 3) / 2

[spoiler] Plug in. If x = 10, then there are seven 2-cent envelopes (costing a total of 14 cents) and four 5-cent envelopes (for a total of 20 cents). The total cost is 14 + 20 = 34. The answer choice B equals 34 using an x of 10: = 34. [/spoiler]

Source: Beat The GMAT — Quantitative Reasoning | Mon Mar 28, 2011 12:33 pm

6983manish: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Mon Mar 07, 2011 12:27 am; Thanked: 6 times

by 6983manish » Tue Mar 29, 2011 3:23 am

wayofjungle 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?

3x + 1
(7x - 2) / 2
11x + 31
(7x - 6) / 2
(13x + 3) / 2

[spoiler] Plug in. If x = 10, then there are seven 2-cent envelopes (costing a total of 14 cents) and four 5-cent envelopes (for a total of 20 cents). The total cost is 14 + 20 = 34. The answer choice B equals 34 using an x of 10: = 34. [/spoiler]

Lets solve it using algebra.

Let the number of 5-cent envelopes is "n"
Then number of 2-cent envelopes would be n+3.

Its given that n+n+3 = x+1

=> 2n + 3 = x+1
=> n = (x-2)/2
n is number of 5-cent envelopes.
n+3 = (x-2)/2 +3 is number of 2-cent envelopes.

Total Cost = 2 [(x-2)/2 + 3 ] +5(x-2)/2 = (7X-2)/2

Hence , answer would be B.

wayofjungle: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Fri Mar 11, 2011 8:55 am; Thanked: 12 times

by wayofjungle » Tue Mar 29, 2011 1:02 pm

Thanks, I tried to solve by plugging in, but the x+1 kept throwing me off

jaymw: Master | Next Rank: 500 Posts; Posts: 144; Joined: Sun Aug 29, 2010 9:17 amFollowed by: 4 members; Thanked: 40 times; GMAT Score:760

by jaymw » Fri Apr 01, 2011 1:34 am

Plugging in works just fine here. I believe it is the easier method of getting to the correct answer, too.

You need numbers such that t(two cent envelopes) is 3 more than f(five cent envolopes).

t=5
f=2

The above numbers fulfil those requirements and are quite easy to deal with. We now have t+f=7 envelopes, therefore x must be 6.

The total cost when plugging in 6 for x should be 5*2+2*5=20.

Let's try:

a) 3*6+1=19 FALSE
b) (7*6-2)/2=20 CORRECT
c) 11*6+31=too much FALSE
d) (7*6-6)/2=18 FALSE
e) (13*6+2)/2=way too much FALSE

Only b works, therefore it must be the correct answer.

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