The document RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 5), Class 8, Maths Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

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**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.

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- RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 4), Class 8, Maths
- RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 3), Class 8, Maths
- RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 2), Class 8, Maths
- RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 1), Class 8, Maths