In a hospital there is a fixed registration charge and charge per day based on the number of days a patient gets admitted. A patient admitted for 5 days had to pay Rs. 4575 while another patient admitted for 7 days had to pay Rs. 6225. Find the amount to be paid by a patient staying for ten days.
Answers
Sᴏʟᴜᴛɪᴏɴ :-
Let us assume that, fixed registration charge is Rs.x and charge per day is Rs.y .
Than,
→ charge for 5 days = fixed charge + 5*charge per day.
→ x + 5y = 4575 --------------- Eqn(1) .
Similarly,
→ charge for 7 days = fixed charge + 7*charge per day.
→ x + 7y = 6225 --------------- Eqn(2) .
Subtracting Eqn(1) from Eqn(2) , we get,
→ (x + 7y) - (x + 5y) = 6225 - 4575
→ x - x + 7y - 5y = 1650
→ 2y = 1650
→ y = Rs.825
Putting value of y in Eqn.(1) Now,
→ x + 5*825 = 4575
→ x + 4125 = 4575
→ x = 4575 - 4125
→ x = Rs.450
Therefore ,
→ charge for 10 days = fixed charge + 10*charge per day.
→ x + 10y
→ 450 + 10*825
→ 450 + 8250
→ Rs.8700 (Ans.)
Hence, the amount to be paid by a patient staying for ten days is Rs.8700.
GIVEN:
- In a hospital there is a fixed registration charge and charge per day based on the number of days a patient gets admitted.
- A patient admitted for 5 days had to pay Rs. 4575.
- while another patient admitted for 7 days had to pay Rs.6225
TO FIND:
- Amount to be paid by a patient staying for ten days.
SOLUTION:
Let, the fixed registration charge be x and per day charge be y.
A.T.Q
=> 5 days charge = 4575.
=> fixed charge + 5 × daily charge =4575
=> x + 5y = 4575••••••••••••••• (1)
Similarly,
=> 7 days charge = 6225
=> fixed charge + 7 × daily charge = 6225
=> x + 7y = 6225•••••••••••••••(2)
Now subtracting (1) From (2),
=> (x + 7y) - ( x +5y) = 6225 - 4575
=> x - x + 7y - 5y = 1650
=> y = 1650/2
=> y = 825
Now ,
=> x + 5 × 825 = 4575
=> x + 4125 = 4575
=> x = 450
Now,
=> 10 days charge = fixed charge + 10 × daily charge
=> x + 10y
=> 450 + 10 × 825
=> 450 + 8250