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# A county fair sold 1,750 tickets, each of which was either an adult or

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A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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09 Aug 2017, 23:39
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Difficulty:

75% (hard)

Question Stats:

61% (03:29) correct 39% (03:09) wrong based on 59 sessions

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A county fair sold 1,750 tickets, each of which was either an adult or children's ticket, and earned a total of \$27,000. The fair earned 25% more from adult tickets than from children's tickets, but sold 25% fewer adult tickets than children's tickets. How much did a children's ticket cost?

A. \$10
B. \$12
C. \$15
D. \$20
E. \$25

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A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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10 Aug 2017, 00:26
Bunuel wrote:
A county fair sold 1,750 tickets, each of which was either an adult or children's ticket, and earned a total of \$27,000. The fair earned 25% more from adult tickets than from children's tickets, but sold 25% fewer adult tickets than children's tickets. How much did a children's ticket cost?

A. \$10
B. \$12
C. \$15
D. \$20
E. \$25

Let the number of children tickets be c => c+0.75c = 1750
c = 1000 , adult tickets will be 750
price of children,adult ticket can be (10,12.5), (12,15),(15,75/4) (20,25), (25,125/4) from the given options
10*1000 + 12.5*750 = 19375
12*1000 + 15*750 = 23250
15*1000 + 75/4*750 = ~29000
20000 + 25*750 = 38750

Please correct me.. i think some 25% more than means 20% less logic is in play but i am not able to get it
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Re: A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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10 Aug 2017, 01:40
Let c=No. of children's ticket sold, so (1750-c)= No. of Adult's ticket sold.
Now, Let p1= Cost of children's ticket sold and p2= Cost of Adult's ticket sold.
So,
cp1 + (1750-c)p2 = 27000..................(1)

Now, the fair earned 25% more from adult tickets than from children's tickets.
So cp1 = 1.25(1750-c)p2....................(2)

Using (1) and (2), we get
(1750-c)p2 (1+1.25) = 27000..................(3)

Also, Fair sold 25% fewer adult tickets than children's tickets.
So, c=.75(1750-c)...................(4)

Solving (4) we get c = 750. Using this and (3) we get p2 = 12 and using p2 value and (2), we get p1 = 20.

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Re: A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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10 Aug 2017, 01:47
Luckisnoexcuse wrote:
Bunuel wrote:
A county fair sold 1,750 tickets, each of which was either an adult or children's ticket, and earned a total of \$27,000. The fair earned 25% more from adult tickets than from children's tickets, but sold 25% fewer adult tickets than children's tickets. How much did a children's ticket cost?

A. \$10
B. \$12
C. \$15
D. \$20
E. \$25

Let the number of children tickets be c => c+0.75c = 1750
c = 1000 , adult tickets will be 750
price of children,adult ticket can be (10,12.5), (12,15),(15,75/4) (20,25), (25,125/4) from the given options
10*1000 + 12.5*750 = 19375
12*1000 + 15*750 = 23250
15*1000 + 75/4*750 = ~29000
20000 + 25*750 = 38750

Please correct me.. i think some 25% more than means 20% less logic is in play but i am not able to get it

I think i got it after shivam2506 post
I misread "The fair earned 25% more from adult tickets than from children's tickets"

Revenue from Adult tickets : Revenue from children tickets = 5:4
that means Revenue from children tickets = 12000 (4/9 * 27000)
Let no. of children tickets = c
c + 0.75c = 1750
c= 1000
or no of children tickets: no of adult tickets = 4:3
no. of children tickets will be = 1000 (4/7 * 1750)
Child Ticket's cost = 12000/1000 = 12
B
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Re: A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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10 Aug 2017, 01:50
If 25% less adult tickets were sold, 1000 children tickets and 750 adult tickets were sold. (c+0.75c=1750)
in total 750 adult tickets earned 25% more than 1000 children tickets. 1000c+1.25c=27000 --> c=12
B is correct
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A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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10 Aug 2017, 03:14
Total number of tickets sold : 1750
Total amount earned : 27000

If the number of tickets sold to children is x, adults were sold .75x tickets.
Therefore 1.75x = 1750 or x = 1000.
There are 750 tickets adult's tickets sold and 1000 children's tickets sold.

If the amount made by selling adult tickets is 25% more than children,
the amount made by selling adult tickets is 1.25y when y is the amount made by selling children's ticket.
Therefore, 1.25y + y = 27000 or y = 27000/2.25 = 12000.
The amount made by selling children's ticket is 12000 and that made by selling adult's ticket is 15000(1.25y)

Therefore the children's ticket cost 12000/1000 = 12\$(Option B)
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Re: A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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15 Aug 2017, 10:56
Bunuel wrote:
A county fair sold 1,750 tickets, each of which was either an adult or children's ticket, and earned a total of \$27,000. The fair earned 25% more from adult tickets than from children's tickets, but sold 25% fewer adult tickets than children's tickets. How much did a children's ticket cost?

A. \$10
B. \$12
C. \$15
D. \$20
E. \$25

We can let the number of children's tickets = c; thus, the number of adults’ tickets = 0.75c. We can create the following equation:

c + 0.75c = 1,750

1.75c = 1,750

c = 1,000

Next, we can let the revenue from children’s tickets = p; thus, the revenue from adults’ tickets = 1.25p. We can create the following equation:

p + 1.25p = 27,000

2.25p = 27,000

p = 12,000

Since the total revenue from children’s tickets is \$12,000 and 1,000 children’s tickets were sold, each children’s ticket cost \$12,000/1,000 = \$12.

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Re: A county fair sold 1,750 tickets, each of which was either an adult or  [#permalink]

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16 Sep 2018, 14:15
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Re: A county fair sold 1,750 tickets, each of which was either an adult or   [#permalink] 16 Sep 2018, 14:15
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