12 Days of Christmas GMAT Competition 2025 - Day 7: A team composed of

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24 Dec 2025, 00:54

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Bunuel

Alex and Maya sold souvenirs at the two performances at the local orchestra, one at the afternoon and the other at evening.

The booklet price at the afternoon was $4.

The booklet price at the evening was $6.

Based on the information provided in the question.

Alex afternoon = twice the sales of Maya evening. = 2*x

similarly, Maya afternoon is HALF of Alex Evening = (1/2)*(2y)

	Alex	Maya
A.N	2x	y
Eve	2y	x

So, Alex total = 2x+2y

Maya total = x+y

The total earning is $1800.

A.N sales + Evening sales = $1800

4*(2x+y) + 6*(2y+x) = $1800

14x + 16y = $1800

Thus, 7x + 8y = $900

Alex total = 2x+2y =?

2*(x+y) = 120

then, (x+y) =60

7x+8y = 900

can be rewritten as , 7x+7y+y = 900

7*(x+y) +y = 900

substituting here, we get 7*60 + y = 900

y= 480, and x = -420.

case 2: 2*(x+y) = 180

then , x+y = 90

substituing we get , 7*(90)+y = 900

y = 270, x = -ve.

case 3: 2*(x+y) = 240

x+y =120

7*(120)+y = 900

y= 60

x = 60 and y =60.

valid case.

Alex total = 2*(x+y) = 2*120=240

Afternoon Total = 2x + y = 2*60+ 60 = 180

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24 Dec 2025, 01:00

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afternoon
alex= 2b
maya= a/2

evening
alex= a
maya= b

in afternoon each book sold at 4$
evening each at 6$

8b + 2a + 6a + 6b = 1800
7b+4a= 900
4a = 900 - 7b
a= 225-(7b/4)
b must be divisible by 4.
after trying value,
we get, b=80, a= 120

so alex total= 120+120 =240
afternoon total= 120+60= 180

Bunuel

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24 Dec 2025, 01:10

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Let the number of booklets sold by Maya in the evening be x. Let y be the number of booklets sold by Alex in the evening be y. Then Alex sold 2x booklets in the afternoon and Maya sold y/2 booklets in the afternoon. If each booklet costs $4 int the afternoon and $6 in the evening and if team earned $1800 all the day then:

4*(2x+y/2) + 6*(x+y) =1800
7x+4y =900

Now we need to find Alex total which is 2x+y and Afternoon total which is 2x+y/2 . If you notice from the options , only one pair is consistent which is 2x+y = 240 and 2x+y/2 = 180.

So Alex total is 240 and afternoon total is 180.

Bunuel

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24 Dec 2025, 01:28

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With the information provided we can create a table something like -
A | M
Afternoon - 2a b
Evening - 2b a

And hence alex total - 2(a+b), afternoon total = (b+2a)

At this point I started to put in alex total from options, got a valid equation set from alex total = 240.

Answer should be Alex total = 240, Afternoon total = 180

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24 Dec 2025, 01:47

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Let maya evening sales be x meaning Alex afernoon will be 2x.
Let alex evening be y meaning maya evening will be 0.5y
So from the above given the prices we can create an equation
4(2x+0.5y)+6(y+x)= 1800
8x+2y+6y+6x=1800
14x+8y=1800
7x+4y=900
So y can be said to be 900-7x/4. From here we can just test cases given the choices
Remember Alex total= 2x+y
While afternoon total = 2x+0.5y. So as we test cases we should notice that the answers will differ by 0.5y meaning y could be 120 based on the difference in the answer choices. Using y as 120 we get x to be 60
So Aalex total is 240
Afternoon total is 180

Bunuel

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24 Dec 2025, 02:51

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	Afternoon($4)	Evening($6)	Total
Alex	2m	2a	2(m+a)
Maya	a	m
Total	2m+a

We need Alex's total booklets sold and Afternoon's total booklets sold.

We are given, (2m+a)*4 + (2a+m)*6 = 1800
Simplifying it, we get 7m + 8a = 900, take this as equation (1)

Now, let's test options for Alex's total and then find if we have a corresponding option for Afternoon's total.
2(m+a) = 120, we get m+a = 60, now solving this with (1), we get a high value for a(if we solve for a first), which isn't valid.
2(m+a) = 180, m+a = 90, same as above when solved with (1)
2(m+a) = 240, m+a = 120, we get a = 60 when solved with (1), which means m is 60. So the afternoon's total is 2m+a = 120+60 = 180, we have an option.

So Alex's total = 240, Afternoon's total = 180

Bunuel

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24 Dec 2025, 03:35

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Bunuel

afternoon performance sold at = 4 dollar

evening performance sold at = 6 dollar

Alex afternoon booklet sold number (AA) = 2X : X= ME - Maya evening sold number

AE = 2Y : Y = MA

so, 4 * (2x) + 6* (X) + 6*(2Y) + 4*(Y) = 1800

=8X + 6X+12Y+4Y = 1800
=7X + 8Y = 900 -------- 1

Now here is tricky part we got two variable equation and I was stuck here, then I thought of trying options, which really helped me.

AA + AE = 240 ( took 240 from question as a possible outcome)
2x + 2y = 240
x + y = 240 ----- 2

solved 1 & 2 and got x =60 and y = 60

So I got the answer Alex total book sold number = 240 & total book sold during afternoon = 180

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24 Dec 2025, 03:52

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This one is all about being quick to find the right expression that'll get you the answer... trial and error won't work until you want to be sitting with this for a 1/3rd of your allotted QR time

Alex sells souvenirs after the afternoon show. Maya sells souvenirs after the evening show.
Alex sells souvenirs after the afternoon show. Maya sells souvenirs after the evening shows.

Thus, we have 4 instances of sales, and the relationship between these will help us find the answer - that how many tickets Alex sells in total (Alex's Afternoon + Alex's Evening) and how many tickets of the afternoon show were sold (Alex's Afternoon + Maya's Afternoon).

It'll help to see the relationships between the two. Also, there's a common link between the two: Alex's Afternoon booklets figure in both equations.

More key info:

Each afternoon booklet is for $4 and each evening booklet's for $6.
The total earnings from the two is $1,800.

Cool, now, let's see the relationships between the variables:

Alex sold twice as many booklets in the afternoon as Maya sold in the evening.
Alex sold twice as many booklets in the evening as Maya sold in the morning.

In other words, whatever Alex sold was twice as much as Maya sold. How's this? Imagine Alex sold 2 booklets in the afternoon - then Maya sold 1 in the evening. Let's say she sold 800 tickets in the evening. Then Maya sold 400.

That's a total sale of 802 for Alex, and 401 for Maya.

Take any other value: Alex sold 1500 booklets in the afternoon, then Maya sold 750 in the evening. Alex sold 2 tickets in the evening, then Maya sold 1 in the morning.

It's always a 2x correlation.

But this won't help us solve the question on its own, as we aren't asked to compare Alex's total with Maya's. We need Alex's total booklets sold with the total sold in the afternoon.

Alex's tickets sold = Twice of Maya's tickets

If Maya's tickets is x, Alex's is 2x.

Their total is 3x.

From the sale of 3x tickets, they made $1,800. This includes Alex's $4 tickets and $6 tickets, as well as Maya's.

This also means, 3x = $1800

or x = $600, or the total value of Alex's tickets is $1200, and Maya's is $600.

Hence, 4y + 6z = $600 (y & z = respective number of tickets Maya sold in the afternoon and evening shows, at respective rates)

And 4a + 6b = $1200 (a & b = respective number of tickets Alex sold in the afternoon and evening shows)

We also know that, given the relationships between each variable, Alex sold twice as many booklets in the afternoon as Maya in the evening; and twice as many in the evening as Maya in the afternoon).

So, what Alex sold in the afternoon - 4a - is twice of what Maya sold in the evening - 6z. Hence, 4a = 2*6z = 12z, or a = 3z. Hence, we can replace 6z in the first equation with 2a.

Similarly, what Alex sold in the evening - 6b - is twice as what Maya sold in the afternoon - 4y. Hence, 6b = 2*4y, or 6b = 8y or 3b = 4y. Hence, we can replace the 4y in the first equation with 3b.

With this substitution, we can add both the equations to get ourselves:

2a + 3b = $600
4a + 6b = $1200
--------------------
6a + 9b = $1800
--------------------

This just means the same number of tickets were sold in the morning and evening.

But the tickets in the evening are prices 1.5x the tickets in the morning, so the sales from the evening ticket will be 1.5x the sales from the morning tickets.

Let's now take x = sales from the morning tickets. Then 1.5x will be the sales from the evening ticket.

Hence:

x + 1.5x = 1800

2.5x = 1800

x = 720.

With that, $720 is the sales in the morning, and $1080 is sales in the evening.

We also know that in both cases, Alex sold twice as much as Maya.

Hence, $720 = 3x, or x = $240 (what Alex sold in the morning is 2x or $480 / $4 = 120 tickets).

And $1080 = 3x or x = $360 (what Alex sold in the evening is 2x or $720 / $6 = 120 tickets).

Alex, hence, has sold 240 tickets.

In the afternoon, $480 of the tickets were hers, hence $240 of the tickets were Maya's, or Maya sold 60 tickets.

hence, 120 + 60 = 180 is the total number of tickets both sold in the afternoon.

Bunuel

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24 Dec 2025, 05:57

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let AA=alex afternoon
AE=alex evening
MA=maya afternoon
ME= maya evening
afternoon 4$ Evening 6$
alex to maya evening AA=2ME
maya to alex evening MA=I/2AE
Total revenue 4(AA+MA)+6(AE+ME)=1800
ME=x, AE=y, AA=2x, MA=1/2y
4(2x+y/2)+6(y+x)=14x+8y=1800
7x+4y=900
alex total AA+AE=2x+y=2x+2k=2(x+k)
afternoon total AA+MA=2X+Y/2=2X+k
Alex total =240
Afternoon total=180

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24 Dec 2025, 06:33

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Bunuel

Maya sold y booklets in the afternoon show and x booklets in the evening show. Therefore, Alex sold 2x booklet in the afternoon show and 2y booklet in the evening show.

14x + 16y = 1800

7x + 8y = 900

Alex Total = 2(x+y)

A) 2(x+y) = 120

x + y = 60

8x + 8y = 480

This gets us negative y, eliminate.

B) 2(x+y) = 180

x+y = 90

8x + 8y = 720

Again this gets y negative, eliminate

C) 2(x+y) = 240

x + y = 120

8x + 8 y = 960

y = 60

TOtal books sold in the afternoon = 60 + 120 = 180

Number of books Alex Sold = 2(60+60) = 240
Booklets sold in the afternoon = 180

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24 Dec 2025, 06:39

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based on the given info : Aa = 2Me and Ma=1/2Ae => Ae=2Ma
Afternoon cost = 4, evening cost = 6

4(Aa+Ma)+6(Ae+Me) = 1800 => 4(Aa+1/2Ae) + 6 (Ae+1/2Aa) = 1800 => 4Aa+2Ae+6Ae+3Aa = 1800 => 7Aa+8Ae = 1800
=> 7(At)+ Ae = 1800 (At = Alex total)

Using the values, when Alex total = 240
=> Ae = 1800 - (7*240) => 120
=> Ma = 120/2 = 60, Aa = 240-120 = 120 Afternoon total = Aa +Ma = 120+60 = 180

=> When Alex total = 240, afternoon total = 180, consistent

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24 Dec 2025, 06:42

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Let's translate into equations the info provided:

Let Aa the number of books Alex sold in the afternoon, and Ae in the evening, same for Maya.

I) Aa = 2*Me
II) Ae = 2*Ma
III) 4(Aa + Ma) + 6*(Ae + Me) = 1800

we want ( Aa +Ma) and ( Aa + Ae).

plugging the I in the II we get ==> Aa + Ae = 2 (Me + Ma) ==> Alex sold the double of books as Maya
from the III ==> 7*(Me +Ma) = 900 - Ma

Plugging the number of Alex sold books we have directly the number of Maya total, then we have Ma and everything follows.

Plugging Aa + Ae = 240 we get Me + Ma = 120
and Ma = 900 - 840 = 60 ==> Me = 60 and Aa = 120
Hence Aa + Ma = 120 + 60 = 180

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24 Dec 2025, 06:48

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Having denoted the relative sales on different shows as X and Y in a table (attached), it's clear that the values we need to find are ${Alex}=2x+2y$ and ${Afternoon}=2x+y.$
Knowing that the total is 1800, we can calculate that $4(2x+y)+6(2y+x)=14x+16y=1800$, meaning $7x+8y=900$

From there, we can see that $4*{Alex}=8x+8y=900+x$, so Alex must've sold at least $900/4=225.$

Then let's check if she earned 240. Then, $900+x=240*4=960, x=60 => y=(240-2x)/2=(240-120)/2=60$, and the Afternoon total is equal 180. This fits, hence the answer is 240 for Alex and 180 for the afternoon.

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24 Dec 2025, 07:13

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A team composed of Alex and Maya sold souvenir booklets at two performances of a local orchestra on the same day: one performance took place in the afternoon and the other took place in the evening. Each booklet sold for $4 at the afternoon performance and $6 at the evening performance. Alex sold twice as many booklets in the afternoon performance as Maya sold in the evening performance, and Maya sold half as many booklets in the afternoon performance as Alex sold in the evening performance. Altogether, the team earned $1,800 from booklet sales that day.

Let the number of booklets Alex sold in the evening= a
And the number of booklets Maya sold in the evening= b

It is also given that
Maya evening = b
Then, alex afternoon = 2b

Alex evening = a
Then, maya afternoon= a/2

Afternoon price = 4 and Afternoon total = 2b+a/2
Evening price= 6 and Evening total = a+b

Overall, 4(2b+(a/2))+6(a+b)=1800
8b +2a+6a+6b=1800
8a+14b=1800
4a+7b=900

Given afternoon total (2b+a/2) choices are 120,180,240,260,300
If afternoon total is 180, then a/2 +2b=180 or a=360-4b

Also, 4a+7b=900
4(360-4b)+7b=900 or b=60
Then, a=360-4(60) =360-240 =120

Alex evening total = a=120
And Alex afternoon total= 2b =2*60=120
Alex overall total =120+120=240
Afternoon total= 180

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24 Dec 2025, 07:35

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Let afternoon and evening tickets sold by Alex = Aa and Ae and
Maya = Ma and Me

Given, Aa = 2Me
and Ma = 0.5*Ae => Ae = 2Ma

Total Revenue = 4(Aa+Ma) + 6(Ae+Me) = 1800

Substitute: 4(2Me+0.5Ae) + 6(Ae+Me) = 1800

Simplify: 7Me+4Ae = 900

Analyze option choices which fits,

Ae=120 and Me= 60

Bunuel

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24 Dec 2025, 07:43

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assuming variable P and Q and acc. to question
A | M
afternoon @4 2Q P
evening @6 2P Q
8Q + 4P + 12P + 6Q = 1800 => 7Q + 8P = 900 --(i)
1st selection = 2(P + Q) and 2nd selection = 2Q + P
substituting values for selections and equating with (i) will give the answer.
P = 60 and Q = 60 satisfies => 1st selection 240 and 2nd selection 180

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24 Dec 2025, 07:44

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Table:
A M
a 2x y -> $4
e 2y x -> $6

4(2x+y)+6(2y+x) = 1800
14x + 16y = 1800 -> x=(1800-16y)/14

a = Afternoon Total = 2x+y = 2*(1800-16y)/14 + y = (1800-16y+7y)/7 = (1800-9y)/7 -> y=(1800-7a)/9
A = Alex Total = 2x+2y = 2*(1800-16y)/14 + 2y = (1800-16y+14y)/7 = (1800-2y)/7 = (1800-2(1800-7a)/9)/7 = (9*1800-2*1800+14a)/63 = (7*1800+14a)/63 = (1800+2a)/9 = 200 + 2a/9

A = 200 + 2a/9

Checking values only 180 is divisible by 9 -> a=180 and A=240

Alex Total=240
Afternoon Total=180

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24 Dec 2025, 07:44

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Bunuel

	Aft	Eve
Al	2x	2y
Ma	y	x

4*(2x+y) + 6*(2y+x) = 1800

2*(2x+y) + 3*(2y+x) = 900

4x + 2y + 6y + 3x = 900

7x + 8y = 900

total Alex = 2(x+y)
total Afternoon = 2x + y

Alex > Aft
Try options 2(x+y) = 180
x+y = 90
630+y = 900
y = 270
x < 0, not possible

2(x+y) = 240
x+y = 120
840 + y = 900
y = 60
x = 60

2x+y = 180 possible

Alex Total = 240
Afternoon Total = 180

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24 Dec 2025, 08:00

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Let Aa Ae Ma Me be afternoon and evening sales of Alex, afternoon and evening sales of Maya respectively

Given
Aa = 2Me
Ma = 1/2 Ae

Writing revenue equation
4(Aa+Ma)+6(Ae+Me) = 1800
4(2Me+1/2Ae)+6(Ae+Me) = 1800
14Me+8Ae = 1800
7Me+4Ae = 900

By trial method
If afternoon total = 180
Aa+Ma = 2Me+1/2Ae = 180
4Me+Ae = 360
Ae=360-4Me
Putting in equation
7Me + 4(360-4Me) = 900
7Me +1440 - 16Me = 900
Me = 60
Ae = 360-4Me = 120

Alex total = Ae + Aa = 2Me+Ae = 240

Hence afternoon total = 180
Alex total = 240
Both are available in the options

Bunuel