Hotel chanakya in chanakyapuri has a fixed monthly cost of ₹1000000. The advertising cost is ₹10000 per month. It has 5 A/C rooms, which cost ? 600
per day and 10 non-A/C rooms, which cost 350
per day. Direct costs are * 100 per day for an AIC
room, and 50 for a non-A/C room. In the month of
April 2020, the occupancy rate of A/C rooms is 50%
while that of non-A/C rooms is 45%. Find the profit
of the hotel in rupee terms for the month of April
2020.
(a) 33,600
((32,000) Loss
(b) 28,800
(d) (17,750) Loss
Answers
Solution :-
→ Fixed monthly cost of hotel = ₹1,00,000
→ Advertising cost per month = ₹10,000
→ Maintenance cost of AC room = ₹100 / day .
→ Maintenance cost of non - AC room = ₹50 / day .
so,
→ Total monthly cost = 1,00,000 + 10,000 + 5*100*30 + 10*50*30 = 1,00,000 + 10,000 + 15000 + 15000 = ₹ 1,40,000 .
now,
→ Amount recieved = 50% of AC rooms * 600 * 30 + 45% of non AC room * 350 * 30 = (2.5 * 18000) + (4.5 * 10500) = 45000 + 47250 = ₹ 92250 .
then,
→ Loss of the hotel = 140000 - 92250 = ₹ 47750 (Ans.)
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Answer:
32000 Loss
Step-by-step explanation:
Monthly Cost: 1,00,000
Ad Cost: 10000
Total monthly Cost on the hotel:
100000+10000+(50/100)*500*30 + 500*30*(45/100) = 124250
Total Cost recovered: (50/100)*5*600*30 + (45/100)*10*350*30 = 92250
Loss of
124250-92250 = 320000