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

These kinds of questions (Variables in the Answer Choices - VIACs) can be answered algebraically or using the INPUT-OUTPUT approach.
Enchanting has solved the question using the INPUT-OUTPUT approach, so let's solve it algebraically..

3 more $4 tickets than $7 tickets were purchased
Let x = the NUMBER of $7 tickets purchased
So, x + 4 = the NUMBER of $4 tickets purchased
So, the total NUMBER of tickets purchased = x + (x + 3) = 2x + 3

The question tells us that 2t + 1 tickets were purchased.
So, we can write: 2x + 3 = 2t + 1
Subtract 1 from both sides of the equation to get: 2x + 2 = 2t
Divide both sides by 2 to get: x + 1 = t
Subtract 1 from both sides to get: x = t - 1 we'll use this information later on!

Which of the following expresses the total COST, in dollars, of the 2t + 1 tickets?
Let's first answer this question using the variables we assigned earlier.
The cost of x tickets costing $7 each = 7x dollars
The cost of x + 3 tickets costing $4 each = 4(x + 3) dollars
So, the TOTAL cost = 7x + 4(x + 3)
= 7x + 4x + 12
= 11x + 12
We've now expressed the total cost in terms of x.
In order to express the total cost in terms of t, we'll use the fact that x = t - 1

Take: 11x + 12
Substitute to get: 11(t - 1) + 12
Expand: 11t - 11 + 12
Simplify: 11t + 1

Answer: A

Cheers,
Brent
_________________

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.

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)
_________________

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?

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
_________________

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

\(4(y+3) + 7y\)

\(4y + 12 + 7y\)

\(11y + 12\)

\(11(t-1) + 12\)

\(11t - 11 + 12 = 11t + 1\)
_________________

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)

The answer is A.

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.
Add (3x4) to 11t -11

Therefore total cost = 11t -11 + 12 = 11t +1

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

Answer A
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________