A flight has equal numbers of seats in Classes A, B, and C,

Question

A flight has equal numbers of seats in Classes A, B, and C, and all tickets in each class are sold for the same price. On Monday, 80% of maximum revenue was earned, as \(\frac{4}{5}\), \(\frac{2}{3}\), and 100% of all seats in Classes A, B, and C, respectively, were sold. On Tuesday, 85% of maximum revenue was earned, as 100%, \(\frac{1}{3}\), and \(\frac{1}{2}\) of all seats in Classes A, B, and C, respectively, were sold.

When every seat on the flight is sold, the seats in Class A account for what percent of the total revenue?

(A) \(60\)%
(B) \(62\frac{1}{2}%\)%
(C)\( 65\)%
(D) \(66\frac{2}{3}%\)%
(E) \(75\)%

When every seat on the flight is sold, the seats in Class A account for what percent of the total revenue?

(A)  60 %
(B)  62\frac{1}{2}% %
(C)  65 %
(D)  66\frac{2}{3}% %
(E)  75 %

ankushsambare · Answer

Let the total revenue from fully selling Classes A, B, and C be A, B, and C.
So, total revenue when all seats are sold = A + B + C.
 From Monday: 
(4/5)A + (2/3)B + C = (4/5)(A + B + C)
Expanding the right side:
(4/5)A + (2/3)B + C = (4/5)A + (4/5)B + (4/5)C
Cancel (4/5)A from both sides:
(2/3)B + C = (4/5)B + (4/5)C
Rearranging:
(1/5)C = (2/15)B
Therefore,
C = (2/3)B

From Tuesday: 
A + (1/3)B + (1/2)C = (17/20)(A + B + C)
Substitute C = (2/3)B:
A + (1/3)B + (1/3)B = (17/20)(A + (5/3)B)
A + (2/3)B = (17/20)A + (17/12)B
Rearranging:
(3/20)A = (3/4)B
So,
A = 5B
Thus,
A : B : C = 5 : 1 : 2/3
Multiplying by 3:
A : B : C = 15 : 3 : 2
Total revenue ratio = 15 + 3 + 2 = 20
Class A's share of total revenue = 15/20 = 3/4 = 75%
 Answer: E . 75%

vasu1104 · Answer

lets consider numbers of seats in each class= P

now it says, all tickets in each class are sold for the same price. means revenue is same for all classes.
lets assume revenue for A,B and C = A, B and C respectively.

it doesnt mean price for price seat in all classes is same.

maximun revenue is when it sells all seats from all classes.
so revenue = PA + PB + PC

Monday

seats sold ,
A= 4/5p
B = 2/3p
C =p

revenue = (4/5)PA + (2/3)PB + PC = (4/5) (PA+PB+PC)

upon solving this, we get C =(2/3)B

Tuesday

seats sold,
A= P
B= (1/3)P
C= (1/2)P

revenue = PA +(1/3)PB + (1/2)PC = 85/100 (PA+PB+PC)
A + B/3 + 2B/6 = 85/100 (A+ B + 2B/3)

upon solving this, we get A = 5B

PA / PA +PB+PC

5PB / 5PB+ PB + (2/3)PB

15/20

75%

option E

BongBideshini · Answer

Let the full revenue from Classes A, B, C be proportional to A, B, C.
Monday: 80% earned ⇒ 20% lost
Unsold seats:
 
 A: 1/5 unsold 
 B: 1/3 unsold 
 C: 0 unsold  
Therefore:
(1/5)A + (1/3)B = 20% of total revenue.
Multiply by 15:
3A + 5B = 3(A+B+C)
⇒ 5B = 3B + 3C
⇒ 2B = 3C
So,  B : C = 3 : 2

Tuesday: 85% earned ⇒ 15% lost
Unsold seats:
 
 A: 0 
 B: 2/3 unsold 
 C: 1/2 unsold  
Thus:
(2/3)B + (1/2)C = 15% of total revenue.
Let B = 3k and C = 2k.
Then:
(2/3)(3k) + (1/2)(2k) = 15% (A+B+C)
=> 2k + k = (3/20)(A+B+C)
=> A + B + C = 20k
Total revenue = 20k.
Therefore:
A + B + C = 20k
A + 3k + 2k = 20k
A = 15k
Hence:
A : B : C = 15 : 3 : 2
So Class A's share of total revenue is:
15/(15+3+2) = 15/20 = 75%.
 Ans D

A flight has equal numbers of seats in Classes A, B, and C,

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A flight has equal numbers of seats in Classes A, B, and C,

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