PAGE NO 14.31:
Question 1:
Ms. Cherian purchased a boat for Rs 16000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
ANSWER:
Value of the boat after two years = P(1 − R/100)^{n}
⇒ 16,000(1 − 5/100)²
= 16,000(0.95)²
= 14,440
Thus, the value of the boat after two years will be Rs 14,440.
PAGE NO 14.31:
Question 2:
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 100000 ? Also, find the total depreciation during this period.
ANSWER:
Value of the machine after two years = P(1 − R/100)^{n}
⇒ 100,000(1 − 10/100)²
= 100,000(0.90)² = 81,000
Thus, the value of the machine after two years will be Rs 81,000.
Depreciation = Rs 100,000 − Rs 81,000
= Rs 19,000
PAGE NO 14.31:
Question 3:
Pritam bought a plot of land for Rs 640000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years?
ANSWER:
Given: P = Rs 64,000
R = 5% for every six monthsValue of the plot after two years = P(1 + R/100)^{n}
⇒ 64,000(1 + 5/200)^{4 }
= 64,000(1.025)^{4 }
= 706,440.25
Thus, the value of the plot after two years will be Rs 706,440.25.
PAGE NO 14.31:
Question 4:
Mohan purchased a house for Rs 30000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
ANSWER:
Value of the house after three years = P(1 − R/100)^{n}
⇒ 30,000(1 − 25/100)³
= 30,000(0.75)³
= 12,656.25
Thus, the value of the house after three years will be Rs 12,656.25.
PAGE NO 14.31:
Question 5:
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43740, find its purchase price.
ANSWER:
Purchase price = P(1 − R/100) ^{− n}
⇒ 43,740(1 − 10/100)^{ − 3 }
= 43,740(0.90) ^{− 3}
= 60,000
Thus, the purchase price of the machine was Rs 60,000.
PAGE NO 14.31:
Question 6:
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9680, for how much was it purchased?
ANSWER:
Purchase price = P(1 − R/100)^{− n}
⇒ 9,680(1 − 12/100)^{− 2 }
= 9,680(0.88)^{−2 }
= 12,500
Thus, the purchase price of the refrigerator was Rs 12,500.
PAGE NO 14.31:
Question 7:
The cost of a T.V. set was quoted Rs 17000 at the beginning of 1999. In the beginning of 2000 the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What was the cost of the T.V. set in 2001?
ANSWER:
Cost of the TV = P(1 + R/100)(1 − R/100)
⇒ 17,000(1 + 5/100)(1 − 4/100)
= 17,000(1.05)(0.96)
= 17,136
Thus, the cost the TV in 2001 was Rs 17,136.
PAGE NO 14.31:
Question 8:
Ashish started the business with an initial investment of Rs 500000. In the first year he incurred a loss of 4%. However during the second year he earned a profit of 5% which in third year rose to 10%. Calculate the net profit for the entire period of 3 years.
ANSWER:
Profit for three years = P(1 − R_{1}/100)(1 + R_{2}/100)(1 + R_{3}/100)
⇒ 500,000(1 − 4/100)(1 + 5/100)(1 + 10/100)
= 500,000(0.96)(1.05)(1.10)
= 554,400
Thus, the net profit is Rs 554,400.
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 