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# A person purchased a total of 2t + 1 tickets. Some of the tickets cost

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Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink]
I took a lot of time to think on how to proceed with this question. Then thought why "+1", why not "2t" or just "t".

Then realized, 2t+1 is always odd and qtn also states that number of $4 tickets is 3 more than$7 tickets.

so tried t =1==> tickets = 3; it implies all $4 tickets; price = 12 i.e. 2t+1 t=2 ==> tickets = 5;tickets = 2 + 3; one each for$4 and $7 and 3 tickets extra for$4. Price = 1 tickets for $11 and 3 extra tickets for$12; total price = 23 i.e. (2t+1)
t=3 ==>tickets = 7; tickets = 4 + 3; two each for $4 and$7 and 3 tickets extra for $4. Price = 2 tickets for$11 and 3 extra tickets for $12 ; total price = 34 i.e. (2t+1) I combined price of$4 and $7 =$11 and it means 2 tickets combined for $11 each ($4+$7) Intern Joined: 27 Jan 2020 Posts: 12 Own Kudos [?]: 2 [0] Given Kudos: 1029 Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink] Enchanting wrote: Bunuel wrote: A person purchased a total of 2t + 1 tickets. Some of the tickets cost$4 each and the remaining tickets cost $7 each. If 3 more$4 tickets than $7 tickets were purchased, which of the following expresses the total cost, in dollars, of the 2t + 1 tickets? A. 11t + 1 B. 11t + 12 C. 22t – 10 D. 22t + 11 E. 22t + 23 PS20417 We will solve this by substitution. If t = 1, then number of tickets = 3 Number of$ 4 tickets = 3
Cost of $4 tickets = 12 Number of$ 7 tickets = 0
Cost of $7 tickets =0 Total cost =12 = 11*1+1 = 11t+1 We are getting 11t + 1 for any value of t. Answer: Option A. Why did you considered all the 3 tickets as of 4$ and 0 for 7$? I didn't get the highlighted part, can you elaborate please? e-GMAT Representative Joined: 04 Jan 2015 Posts: 3710 Own Kudos [?]: 17353 [2] Given Kudos: 165 Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink] 2 Bookmarks Expert Reply Solution Given In this question, we are given that • A person purchased a total of 2t + 1 tickets. • Some of the tickets cost$4 each and the remaining tickets cost $7 each To find We need to determine • If 3 more$4 tickets than $7 tickets were purchased, which of the following expresses the total cost, in dollars, of the 2t + 1 tickets Approach and Working out Let the number of$4 tickets be x, and the number of $7 tickets be y • x + y = 2t + 1 • Cost = 4x + 7y Now, if x = y + 3 • Cost = 4(y + 3) + 7(y) = 11y + 12 • And, 2y + 3 = 2t + 1 o Implies, y = t - 1 • Thus, Cost = 11(t – 1) + 12 = 11t + 1 Thus, option A is the correct answer. Correct Answer: Option A Director Joined: 09 Jan 2020 Posts: 953 Own Kudos [?]: 235 [1] Given Kudos: 432 Location: United States Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink] 1 Kudos Bunuel wrote: A person purchased a total of 2t + 1 tickets. Some of the tickets cost$4 each and the remaining tickets cost $7 each. If 3 more$4 tickets than $7 tickets were purchased, which of the following expresses the total cost, in dollars, of the 2t + 1 tickets? A. 11t + 1 B. 11t + 12 C. 22t – 10 D. 22t + 11 E. 22t + 23 PS20417 x = 4 dollar tickets y = 7 dollar tickets $$x = y + 3$$ $$x + y = 2y+ 3$$ $$2y + 3 = 2t + 1$$ $$2y = 2t - 2$$ $$y = t - 1$$ $$4x+7y$$ = total cost $$4(y+3) + 7y$$ $$4y + 12 + 7y$$ $$11y + 12$$ $$11(t-1) + 12$$ $$11t - 11 + 12 = 11t + 1$$ Senior Manager Joined: 18 Jun 2018 Posts: 334 Own Kudos [?]: 201 [1] Given Kudos: 1283 Concentration: Finance, Healthcare A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink] 1 Bookmarks Let t = 2, which means that we have 2(2)+1 = 5 tickets in total. Let the number of$4 tickets = 4 and the number of $7 tickets = 1 (i.e. number of$4 tickets = number of $7 ticket(s) + 1) Total$4 ticket = 4 x $4 =$16 and total of $7 ticket = 1 x$7 = $7 for a combined total of$23 (so we have to look for this in the answer choices)

Now test the options:

A: 11(2) + 1 = $23 (keep) B: 11(2) + 12 =$34 (you can skip this one outrightly since it is $11 greater than option A) C: 22(2) - 10 =$34 (not what we are looking for)
D: 22(2) + 11 = $55 (not what we are looking for; this one can also be skipped since we can tell that it is more than$23 w/o doing the math)
E: 22(2) + 23 = $67 (not what we are looking for; this one can also be skipped since we can tell that it is more than$23 w/o doing the math)

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Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink]
Given the total number of tickets = 2t+1 (odd number of tickets)
4 dollar tickets + 7 dollar tickets = 2t +1

Given that 4 dollar ticket is 3 more than 7 dollar ticket

Therefore , subtracting 3 from 2t+1 will give the number of equal 7 and 4 dollar tickets

2t +1 -3 = 2t -2 = 2 (t-1)

we have 2 types of tickets , equally dividing 2 (t-1) i.e 2 (t-1)/2 we get t-1 tickets in 7 dollars and t-1 tickets in 4 dollars
Multiplying the cost of each type, we get
7(t-1) + 4(t-1) = 11t -11.

Remember, we subtracted 3 - 4 dollar tickets from the total number of tickets.

Therefore total cost = 11t -11 + 12 = 11t +1
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Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink]
Put t = 2 so total tickets purchased are: 5

The number of tickets of $4: 4 [cost 4 * 4 = 16] The number of tickets of$7: 1 [cost 7 * 1 = 7]

So, 5 tickets when t = 2, will have a total cost of \$23.

2t + 1 tickets then will have total cost as in when t = 2 gives us 23: 11(t) + 1

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Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink]
gmatophobia ScottTargetTestPrep

I was able to recognize that "2t+1" indicates an odd number of tickets but was stumped beyond that point.

I'm not sure why there's a need to introduce additional variables x and y (as others have) when t clearly stands for the number of tickets and the answer choices are also in terms of t.

What might be the simplest way to approach this question?
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Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink]
1
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achloes wrote:
gmatophobia ScottTargetTestPrep

I was able to recognize that "2t+1" indicates an odd number of tickets but was stumped beyond that point.

I'm not sure why there's a need to introduce additional variables x and y (as others have) when t clearly stands for the number of tickets and the answer choices are also in terms of t.

What might be the simplest way to approach this question?

I think the simplest approach is to test answer choices. Just pick some value for t and calculate the cost of 2t + 1 tickets according to the information in the question stem. Then, substitute the value of t in the answer choices. Eliminate any answer choice that does not give you the same value as the cost you calculated. By picking different values for t if necessary (it won't be), repeat the process until all but one of the answer choices are eliminated.

For your question about using variables other than t, if you can somehow come up with the expressions t - 1 and t + 2, you can calculate 7(t - 1) + 4(t + 2) to obtain the total cost directly. However, it is not easy to come up with two expressions that add up to 2t + 1 where one of the expressions is 3 greater than the other. We simply use the variables x and y (or x and x + 3) to determine these two expressions. If you can determine these two expressions mentally, then by all means, go for it!
Re: A person purchased a total of 2t + 1 tickets. Some of the tickets cost [#permalink]
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