Welcome to Class 12 Applied Maths Chapter 14, where we embark on an exciting journey into the world of advanced mathematical concepts tailored for Class 11 students.” Unlock the power of applied mathematics with expert solutions crafted by professionals at AppliedMath.com. Designed to propel students towards academic success, our meticulously curated ML Aggarwal Solutions for Applied Mathematics cater to Class 11 and class 12 students seeking mastery in their examinations. Every query from the CBSE ML Aggarwal Books finds a comprehensive answer on our platform, complete with detailed explanations and step-by-step solutions presented in an easily understandable language.
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Here we provide you with Class 12 Applied Maths Chapter 14, to help you gain a comprehensive understanding of the chapter and its concepts. https://appliedmathsolution.com/wp-admin/post.php?post=6&action=edit
EXERCISE- 14.3
Q.1 A piece of equipment that costs ₹25000 has an estimated useful life of 8 years and Rs.0 scrap value. Find the annual depreciation using the linear method.
Ans. Cost = Rs.25000, useful life = 8, scrap value = 0
Annual depreciation = (Cost – Scrap value)/useful life
Therefore,
Annual depreciation = (25000 – 0)/8 = Rs.3125
Q.2 A machine costing ₹30000 is expected to have a useful life of 13 years and a final scrap value of ₹4000. Find the annual depreciation charge using the straight-line method.
Ans. Cost = Rs.30000, useful life = 13, scrap value = Rs.4000
Annual depreciation = (Cost – Scrap value)/useful life
Therefore,
Annual depreciation = (30000 – 4000)/13 = 26000/13
= Rs.2000
Q.3 An asset the life of which is estimated to be 8 years costs ₹80000. If annual depreciation is Rs.9000, find its scrap value using the linear method.
Ans. Cost = Rs.80000, useful life = 8, scrap value = ?, Dep. = Rs.9000
Annual depreciation = (Cost – Scrap value)/useful life
Therefore,
9000 = (80000 – scrap value)/8
72000 = 80000 – scrap value
Hence,
Scrap value = Rs.8000
Q.4 A vehicle costing ₹900000 has a scrap value of Rs.270000. If the annual depreciation charge is Rs.70000, find its useful life in years.
Ans. Cost = Rs.900000, useful life = ?, scrap value = Rs.270000,
Dep. = Rs.70000
Annual depreciation = (Cost – Scrap value)/useful life
Therefore,
70000 = (900000 – 270000)/useful life
Useful life = 630000/70000
= 9 years
Q.5 A machine has a scrap value of Rs.22500 after 15 years of its purchase. If the annual depreciation charge is ₹8500, find its original cost using the linear method.
Ans. Cost = ?, useful life = 15, scrap value = Rs.22500,
Dep. = Rs.8500
Annual depreciation = (Cost – Scrap value)/useful life
Therefore,
8500 = (Cost – 22500)/15
127500 + 22500 = Cost
SO,
Original Cost = Rs.150000
FAQ’s related to Class 12 Applied Maths Chapter 14 on Returns, Growth, and Depreciation :
Q.1 What are the main topics covered in Chapter 14 on Returns, Growth, and Depreciation?
Ans. Chapter 14 covers the following key topics:
- Simple Interest
- Compound Interest
- Depreciation
- Growth rates
- Returns on investments
- Applications of these concepts in real-life scenarios such as savings, loans, and investments.
Q.2 What is the difference between simple interest and compound interest?
Ans. Simple Interest: Interest calculated only on the principal amount. The formula is 𝑆𝐼=(𝑃×𝑅×𝑇)/100 where P is the principal, R is the rate of interest, and T is the time period.
Compound Interest: Interest calculated on the principal amount and also on the interest of previous periods. The formula is CI = P (1 + R/100)^T – P.
Q.6 What are some real-life applications of returns, growth, and depreciation?
Ans. Real-life applications include:
- Calculating interest on savings and loans.
- Determining the future value of investments.
- Estimating the depreciation of assets such as vehicles and machinery.
- Understanding growth rates in populations, economies, and businesses.
These are few Frequently Asked Questions relating to Class 12 Applied Maths Chapter 14
In Class 12 Applied Maths chapter 14, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 12 Applied Maths Chapter 14 promises an enriching learning experience that will set you on the path to success. Class 12 Applied Maths Chapter 14, we delve deep into advanced mathematical concepts that are crucial for understanding.
