Tickets for next month's production of The Sea Gull at the local commu

Question

Tickets for next month’s production of The Sea Gull at the local community theater went on sale yesterday. The ticket price for a person older than 12 is $A, and the ticket price for a person 12 or under is $C, where A > C. Yesterday, the sale of 33 tickets generated a total revenue of $323, and today the sale of 44 tickets generated a total revenue of $424.

In the table, select a value for A and a value for C that are jointly consistent with the given information. Make only two selections, one in each column.

In the table, select a value for  A  and a value for  C  that are jointly consistent with the given information. Make only two selections, one in each column.

ID: 700186

KarishmaB · Accepted Answer

the sale of 33 tickets generated a total revenue of $323,

From 33 tickets revenue is $323 which means that the average revenue per ticket is 9.something.
Hence, cost C should be 6/8 and cost A should be 10/11/12.

Next, to generate a revenue of $323, an odd amount, we need one price to be odd. Except 11, all other 4 options are even.
Hence A must be $11.
ANSWER

Next, assume all 33 tickets were for $11. The revenue then would be $363. But the actual revenue is $40 less (it is $323) because some of these tickets were cheaper.
If some of these tickets were for $6 i.e. $5 less than $11, then it would explain the reduced revenue of $40 (8 tickets were cheaper).
But if some of these tickets were for $8 i.e. $3 less than $11, then we cannot reduce the revenue by $40. We will not get an integer number of tickets and hence C cannot be $8.
Hence C is $6.
ANSWER

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HarshaBujji · Answer

My initial thought  
Are you sure the question didn't provide any further details, As there are 4 variables and 2 equations.

Later, while commenting got this idea. But is this really an OG question ??  
 But one solution as per my analysis is A:11, C:6

Between the 2 days the there are 11 tickets & it costed 101 rs more.

Hence these 11 tickets contributed to 101 rs.

Let x be adults whose ticket is A rs,
So 11-x children with C rs

Ax + C(11-x) = 101.

POE

Let see at the options we have for C.

(A-C)x = 101 - 11C. > 0 (As A-C is >0 )

C cannot be 12,11,10.

C has to be either 6 or 8.

Lets take C as 6.

(A-6)x = 35.

A cannot be an even number. So A has to be 11 so x=7,

7 Adults with 11 rs, 4 children with 6 rs. = 77 + 24 =101.

Now let's take C as 8.

(A-8)x = 13. x cannot be 13, & A cannot be 21.

Hence only choice is A:11 , C:6.

chetan2u · Answer

Although official solution too talks of 11 tickets costing 101, I would take it with a pinch of salt.
All tickets are not of same value, so 33 tickets and 44 tickets would surely have different combination. Say 33 tickets consisted of 22 A and 11 C while 44 tickets consisted of 20A and 24 C. In that case the difference would be 13 of C and -2 of A. 
But if 44 tickets had more of each of A and C, say 28A and 16C, then you will have 6 of A and 5 of C adding to 11.

Anyways, I would make equations and then work on number properties:
 Yesterday:  Let there be x of A and 33-x of C => x*A + (33-x) * C = 323.
(1) Surely exactly one of xA or (33-x)C would be odd to get odd+even = odd.
If x is odd, then 33-x is even and vice versa. Surely one of A and C will be odd.
 Noe, there is only one odd option, so one of A or C would be 11. 
(2) If we take all to be of C value, we will get the least sum, of tickets => 33*10 = 330, which is greater than given sum 323.
 Hence, C is less than 10. Only possibility 6 or 8. Also, A will be 11 for sure

Just from yesterday sale, we know A=11 and C is either 8 or 6.
Although, you can substitute (11,8) and (11,6) here itself to check which is correct, let use today's sale to ease our calculations. After all, efforts have been made to add the details to the question.

Today:  Let there be y of A and 44-y of C => y*A + (44-y) * C = 424.

Subtract the two equations to get (y-x)*A + 11*C -(y-x)*C = 101
(A-C) (y-x) = 101-11C
( 11,8 ): (11-8)(y-x) = 101-11*8 ........ 3(y-x) = 13 ...... y-x =  \frac{13}{3} , Not an integer.
( 11,6 ): (11-6)(y-x) = 101-11*6 ........ 5(y-x) = 35 ...... y-x = 7, Possible. Correct

Solution:  A=11 and C=6

ashutosh_73 · Answer

Hi KarishmaB I couldn't understand the yellow part in the above explanation. How can we conclude that C can be 6/8 and A can be 10/11/12? One thing i could figure out while solving was that, ''11'' will be one of the values, because total = 323 is an odd no. GMAT-Club-Forum-dvb63r0l.png

KarishmaB · Answer

When you sell 33 tickets and get $323 for it, your average price per ticket is 323/33 = 9.8 (approximately) If your tickets are actually priced at 2 different price points, then the average must be somewhere between those price points. Say you are selling tickets of $8 and $10. The average will be between 8 and 10. Say you are selling tickets of $6 and $8, the average will be between 6 and 8. If your average is 9.8, one ticket price will be less than 9.8 and the other will be more than 9.8. So C will be 6 or 8 (only two options less than 9.8) and A will be 10/11/12 (three options more than 9.8). GMAT-Club-Forum-keqvhwsl.png

Akshaynandurkar · Answer

I think faster way to solve this question would be to make equation in terms of no of tickets sold for people aged >12 or people aged < = 12
Let x = no of ticket for people aged > 12 and y = no of ticket for people aged <=12
x + y = 33 and Ax+By = 323
x = 33-y therefore A(33-y) + By =323
33A - Ay +By = 323
since A > B . it makes sense to write equation in (A-B) form
(A-B)y = 33A - 323
y = (33A-323)/(A-B)
now A should be such that 33A is greater than 323
so assume A = 10 from option
y = 7/(10-B)

Now there is no option for B <10 that would give y in integer form
Now take A = 11
y = (363-323)/(11-B)
only option suitable for B is 6 since taking B = 10 gives y = 40 which is not possible since x+y = 33
and taking B = 8 gives 11-B = 3 which cannot divide 40
so take B = 6
Answers A = 11 and B = 6

Gemmie · Answer

Ax + Cy = 323
x + y = 33

Am + Cn = 424
m + n = 44

(i) A > C
=> A * (x + y) > Ax + Cy = 323
=> A > $\frac{323}{33} = 9.78$
=> A can be 10, 11 or 12

(ii) A > C
=> C * (x+y) < Ax + Cy = 323
=> C < $\frac{323}{33} = 9.78$
=> C can be 6 or 8
=> C even

(iii) Ax + Cy = 323
323 odd, Cy even => Ax odd => A = 11

(iv) Ax + Cy = 323
=> 11x + C (33 - x) = 323
=> (11 - C) * x + 33C = 323 (P)

Similarly, (11 - C) * m + 44C = 424 (Q)

(Q) - (P) => (11 - C) * (m - x) + 11C = 101

If C = 6
=> (11 - 6) * (m - x) = 101 - 66 = 65
=> m - x = $\frac{65}{5} = 13$

If C = 8
=> (11 - 8) * (m - x) = 101 - 88 = 13
=> m - x = $\frac{13}{5}$ => not an integer

=> A = 11; C = 6

Oppenheimer1945 · Answer

All options are even except 11, Hence A=11, test C=10,18,6:

11x+10(33-x)=323 (not possible)
11x+8(33-x)=323 =>x=(323-33*8)/3 (not integer)
11x+6(33-x)=323 => x=(323-33*6)/5=25 (possible)

check with 44 tickets:
11y+(44-y)*6=424 => y=(424-44*6)/5 =>y=32 (possible)

(A,C)=(11,6)

rish_dutton · Answer

Official solution

bb · Answer

That's an insane official solution. This should not be a DI Question. This should be a calculation/arithmetic/calculator usage test, but not a Data Insights one

rish_dutton · Answer

bb  totally agree. I hope such questions don't show up on the actual test.

Dbrunik · Answer

its fair i think.

the OG solution is just going through and proving out why all the other options dont work.

once you know 11 is the difference, one of the numbers has to be odd. therefore if a>c the only solution in which C is odd is 11, in which case 12 would have to be A. So, this is really the only one you need to disprove. the rest is just algebra.

Dbrunik · Answer

the official solution is terribly messy. here is a very straighforward and easy way to see what is going on:

44-33=11 and 424-323=101

this difference must be a combination of A and C, where A>C.

since the dollar amount is odd, one of the dollar amounts (the answer choices) must be odd.

your equation then becomes: (Quantity of A)(price of A) + (11- Quantity of A)(price of C) = 101 --------- where 11-quantity of A is equal to quantity of C.

at this point you just need to test values of A and C to see what works. start at the lowest and move up.

Arzeyone · Answer

A truly overwhelming algebra for 2 minutes and i could not understand logic of assigning odd/even property to A as properties for C & number of tickets are variable also assuming incremental sales to split evenly among A&C may not be feasible:

A short approach can be:
As per 1st condition, where sales is $323. A cannot be 6/8 as 33*6 & 33*8 are both less than 323.
If A is 10; 10*33=330; is 7 more than total sales; to reduce sales by 7, number of tickets that may be sold at C,say x, should be such that x(A-C)=7; since 7 is prime A-C must be 7 and thus C has to be 3

If A is 12; 12*33=354; so $31 more(prime again), thus for A-C to be 31, C has to be negative.

Now, if A is 11; 11*33=363; 40$ more, not prime. So tickets sold at C, say x, need to reduce cost by 40$, x(A-C)=40=8*5=5*8=4*10=10*4, A-C is 5/8/10/4, thus c is 6/3/1/7. Thus C can be 6

A=11 & C=6.

anushree01 · Answer

Good question, i like it, nice solution too

agrasan · Answer

Great explanation  KarishmaB  
I spent ~5 mins to come up with an algebra equation based solution but it didn't come to my mind that we can simply leverage even-odd theory based concepts here to quickly solve this. Thanks much.

AnujSingh0147 · Answer

A very nice question considering TPA:

Assume x1 is the sale of tickets above 12 years and y1 for below 12 years similarly x2 and y2 for the next day

x1+y1 = 33
x2 + y2 = 44
A(x1) + C(y1) = 323
A(x2) + C(y2) = 424

(x2-x1) + (y2-y1) = 11
similarly 
A(x2-x1) + C(y2-y1) = 101

Assume (x2-x1) = d
(y2-y1) = 11-d
A(d) + C(11-d) = 101 
Ad + 11C - Cd = 101 
(A - C)d + 11C = 101
(This gives a final equation where we can check the inequalities from the options)

Assume the largest value for A = 12 (from the option) and C =11
(12 - 11)d = 121 = 101
d = -21 (Hence, these scenarios are not possible including 10)

A = 12/11/10 and C = 8/6

We need to make 101 therefore it is evident that one even and one odd digit is required to satisfy the situation.
A = 11 and C = 8
(11-8)d + 88 = 101
3d = 13 (No whole solution)

A = 11 and C = 6
(11-6)d + 66 = 101
5d = 35
d = 7

Therefore, through iteration A = 11 and C = 6

Tickets for next month's production of The Sea Gull at the local commu

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Tickets for next month's production of The Sea Gull at the local commu

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