问题1. Cherian女士以卢比的价格购买了一条船。 16000。如果船的总成本以每年5％的速度贬值，则在2年后计算其价值。
解决方案：
We have,
Price of a boat is = Rs 16000
Depreciation rate = 5% per annum
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
= P (1 + R/100)2
Since it is depreciation we use P (1 – R/100)n
= 16000 (1 – 5/100) (1 – 5/100)
= 16000 (95/100) (95/100)
= 16000 (0.95) (0.95)
= 14440
Therefore,
Value of the boat after two years is Rs 14440.
问题2。一台机器的价值每年贬值10％。如果现在的价值是100000卢比，那么2年后的价值是多少?另外，找到此期间的总折旧。
解决方案：
We have,
Present value of machine is = Rs 100000
Rate of depreciation = 10% per annum
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
= 100000 (1 – 10/100) (1 – 10/100)
= 100000 (90/100) (90/100)
= 100000 (0.9) (0.9)
= 81000
Value of machine after two years will be Rs 81000
Therefore,
Total depreciation during this period is Rs (100000 – 81000) = Rs 19000.
问题3. Pritam花了Rs买了一块地。 640000。它的值每六个月增加5％。 2年后该地块的价值是多少?
解决方案：
We have,
Price of land is = Rs 640000
Rate of increase = 5% in every six month
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
= 640000 (1 + 5/100) (1 + 5/100) (1 + 5/100) (1 + 5/100)
= 640000 (105/100) (105/100) (105/100) (105/100)
= 640000 (1.025) (1.025) (1.025) (1.025)
= 706440.25
Therefore,
The value of the plot after two years will be Rs 706440.25.
问题4. Mohan花费Rs买了房子。 30000，其价值每年以25％的速度贬值。 3年后找到房子的价值。
解决方案：
We have,
Price of house is = Rs 30000
Depreciation rate is = 25% per year
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
= 30000 (1 – 25/100) (1- 25/100) (1 – 25/100)
= 30000 (75/100) (75/100) (75/100)
= 30000 (0.75) (0.75) (0.75)
= 12656.25
Therefore,
The value of the house after 3 years is Rs 12656.25
问题5.机器的价值每年贬值10％。它是3年前购买的。如果其当前值为Rs。 43740，找到其购买价格。
解决方案：
We have,
Present value of machine is = Rs 43740
Depreciation rate of machine is = 10% per annum
Let the purchase price 3 years ago be = Rs x
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
43740 = x (1 – 10/100) (1 – 10/100) (1 – 10/100)
43740 = x (90/100) (90/100) (90/100)
43740 = x (0.9) (0.9) (0.9)
43740 = 0.729x
x = 43740/0.729
= 60000
Therefore,
The purchase price is Rs 60000.
问题6. 2年前购买的冰箱的价值每年贬值12％。如果其当前值为Rs。 9680，购买了多少?
解决方案：
We have,
Present value of refrigerator is = Rs 9680
Depreciation rate is = 12%
Let the price of refrigerator 2 years ago be = Rs x
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
9680 = x (1 – 12/100) (1 – 12/100)
9680 = x (88/100) (88/100)
9680 = x (0.88) (0.88)
9680 = 0.7744x
x = 9680/0.7744
= 12500
Therefore,
The refrigerator was purchased for Rs 12500.
问题7：一台电视机的价格引述为卢比。 1999年初价格为17000。2000年初价格上涨了5％。由于需求减少，2001年初成本降低了4％。2001年电视机的成本是多少?
解决方案：
We have,
Cost of T.V at beginning of 1999 is = Rs 17000
Hiked in price in the year 2000 is = 5%
Depreciation rate in the year 2001 is = 4%
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
= 17000 (1 + 5/100) (1 – 4/100)
= 17000 (105/100) (96/100)
= 17000 (1.05) (0.96)
= 17136
Therefore,
The cost of TV set in the year 2001 is Rs 17136.
问题8. Ashish最初的投资为Rs，开始了业务。 500000。第一年，他蒙受了4％的损失。然而，在第二年，他获得了5％的利润，第三年又上升到10％。计算整个3年的净利润。
解决方案：
We have,
Initial investment by Ashish is = Rs 500000
Incurred loss in the first year is = 4%
Profit in 2nd year is = 5 %
Profit in 3rd year is = 10%
By using the formula,
A = P (1 + R/100)t
Substituting the values, we have
= 500000 (1 – 4/100) (1 + 5/100) (1 + 10/100)
= 500000 (96/100) (105/100) (110/100)
= 500000 (0.96) (1.05) (1.1)
= 554400
Therefore,
The net profit for the entire period of 3 years is Rs 554400.