Q. Every pupil in a school was given one ticket for a concert. The school was charged a total of $6000 for these tickets, all of which were of equal value. What was the price of one ticket? If the price of each ticket had been one dollar less, the total cost would have been 1,200 less. If the price of each ticket had been $2 more, the total bill would have increased by 40%.

• $5 each ticket
  sol: let no. of people = X
  price of one ticket = Y
  a/c question,
  X*Y=6000
  X*(y-1)=4800
  X*(Y+2)=6000+6000*40/100
  =8400
  after solving
  y=5 ans.
• 10 years agoHelpfull: Yes(6) No(0)
• let no: of tickets = x; and total no: of pupil = y;
  then => xy-y(x-1) = 1200;
  then no: of pupil = 1200;
  then we get the price of each ticket as $5;
• 11 years agoHelpfull: Yes(5) No(0)
• when the price of each ticket is reduced by 1,then the saving is $1200 means $1 each student, so total student = 1200;
  =>price of each ticket = 6000/1200=$5
• 10 years agoHelpfull: Yes(1) No(0)
• 5...6000-4800=1200,and 6000/1200=5
• 10 years agoHelpfull: Yes(0) No(0)
• let no. of student=n
  price of one tkt,x=6000/n
  ((6000/n)-1)*n=6000-1200
  n=1200
  x=5
• 10 years agoHelpfull: Yes(0) No(0)
• Step 1: Define the variables
  Let:
  
  𝑝
  p be the price of one ticket.
  𝑛
  n be the number of pupils in the school.
  We know that the total cost for all the tickets is $6000, so the total cost can be written as:
  
  𝑛
  ×
  𝑝
  =
  6000
  (Equation 1)
  n×p=6000(Equation 1)
  Step 2: Analyze the situation where the price is $1 less
  If the price of each ticket were $1 less, the total cost would be $1,200 less. So, the total cost would be
  6000
  −
  1200
  =
  4800
  6000−1200=4800. This can be expressed as:
  
  𝑛
  ×
  (
  𝑝
  −
  1
  )
  =
  4800
  (Equation 2)
  n×(p−1)=4800(Equation 2)
  Step 3: Analyze the situation where the price is $2 more
  If the price of each ticket were $2 more, the total bill would have increased by 40%. So, the new total cost would be:
  
  6000
  ×
  1.40
  =
  8400
  6000×1.40=8400
  Thus, the total cost with the new price of
  𝑝
  +
  2
  p+2 is:
  
  𝑛
  ×
  (
  𝑝
  +
  2
  )
  =
  8400
  (Equation 3)
  n×(p+2)=8400(Equation 3)
  Step 4: Solve the system of equations
  We now have the following system of equations:
  
  𝑛
  ×
  𝑝
  =
  6000
  n×p=6000
  𝑛
  ×
  (
  𝑝
  −
  1
  )
  =
  4800
  n×(p−1)=4800
  𝑛
  ×
  (
  𝑝
  +
  2
  )
  =
  8400
  n×(p+2)=8400
  Solve Equation 1 and Equation 2:
  From Equation 1, we have:
  
  𝑛
  =
  6000
  𝑝
  n=
  p
  6000
  ​
  
  Substitute this expression for
  𝑛
  n into Equation 2:
  
  6000
  𝑝
  ×
  (
  𝑝
  −
  1
  )
  =
  4800
  p
  6000
  ​
  ×(p−1)=4800
  Simplify:
  
  6000
  ×
  𝑝
  −
  1
  𝑝
  =
  4800
  6000×
  p
  p−1
  ​
  =4800
  6000
  (
  𝑝
  −
  1
  )
  𝑝
  =
  4800
  p
  6000(p−1)
  ​
  =4800
  Multiply both sides by
  𝑝
  p:
  
  6000
  (
  𝑝
  −
  1
  )
  =
  4800
  𝑝
  6000(p−1)=4800p
  Expand:
  
  6000
  𝑝
  −
  6000
  =
  4800
  𝑝
  6000p−6000=4800p
  Now, move all terms involving
  𝑝
  p to one side:
  
  6000
  𝑝
  −
  4800
  𝑝
  =
  6000
  6000p−4800p=6000
  Simplify:
  
  1200
  𝑝
  =
  6000
  1200p=6000
  Solve for
  𝑝
  p:
  
  𝑝
  =
  6000
  1200
  =
  5
  p=
  1200
  6000
  ​
  =5
  So, the price of each ticket is $5.
  
  Step 5: Verify the solution
  We can verify this by checking the other conditions:
  
  Total cost with
  𝑝
  =
  5
  p=5:
  
  𝑛
  ×
  𝑝
  =
  6000
  5
  =
  1200
  n×p=
  5
  6000
  ​
  =1200
  There are 1200 pupils, and the total cost is indeed $6000.
  
  If the price were $1 less:
  
  𝑛
  ×
  (
  𝑝
  −
  1
  )
  =
  1200
  ×
  4
  =
  4800
  n×(p−1)=1200×4=4800
  This is correct, as the total cost would have been $4800.
  
  If the price were $2 more:
  
  𝑛
  ×
  (
  𝑝
  +
  2
  )
  =
  1200
  ×
  7
  =
  8400
  n×(p+2)=1200×7=8400
  This is correct, as the total cost would have increased by 40% to $8400.
  
  Thus, the price of one ticket is $5.
• 14 Days agoHelpfull: Yes(0) No(0)