# Samacheer Kalvi 8th Maths Book Solutions Term 2 Chapter 1 | Life Mathematics Answers For Tamil Nadu Board

Samacheer Kalvi 8th Maths Book Solutions Term 2 Chapter 1 – Life Mathematics is available here. The Samacheer Kalvi 8th Maths book answers of Term 2 Chapter 1, available at BYJU’S, contain step by step explanations designed by our Mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on Samacheer Kalvi 8th Maths solutions. Find here, in this article below, the compiled list of Samacheer Kalvi 8th Class Maths Textbook Term 2 Chapter 1 Questions and Solutions for the students to practice.

## Samacheer Kalvi Class 8 Maths Book Term 2 Chapter 1 Questions with Solutions

Term 2 Chapter 1 of the Samacheer Kalvi 8th Maths guide will help the students to solve problems related to the applications of percentage in problems, profit and loss, discount, overhead expenses and GST, compound interest.

### Samacheer Kalvi 8th Maths Term 2 Chapter 1: Life Mathematics Book Exercise 1.1 Questions and Solutions

Question 1: A bank pays ₹240 as interest for 2 years for a sum of ₹3000 deposited as savings. Find the rate of interest given by the bank.

Solution:

The formula for simple interest is given by:

SI = PTR / 100 where P is the principal amount, T is the time, R is the rate of interest

Here, P = 3000, T = 2 years, R = ?, I = 240

240 = [3000 * 2 * R] / 100

240 * 100 = 6000R

R = 24000 / 6000

R = 4%

Question 2: What is 25% of 30% of 400?

Solution:

30% of 400

= [30 / 100] * 400

= 120

25% of the above is

= [25 / 100] * 120

= 30

Question 3: A number when decreased by 20% gives 80. Find the number.

Solution:

Let the number be x.

When it is decreased by 20%, the number obtained is 80.

x – [20 / 100] * x = 80

[100x – 20x] / 100 = 80

80x / 100 = 80

80x = 80 * 100 = 8000

x = 8000 / 80

x = 100

### Samacheer Kalvi 8th Maths Term 2 Chapter 1: Life Mathematics Book Exercise 1.2 Questions and Solutions

Question 1: If selling an article for ₹820 causes a 10% loss on the selling price, find its cost price.

Solution:

Selling price = 820

Loss percentage = 10%

Selling price = Cost price * [(100 – Loss %) / 100]

820 = CP *[(100 – 10) / 100]

820 = CP * [90 / 100]

820 / 0.9 = CP

CP = 911.11

Question 2: A shopkeeper buys goods at 4 / 5 of its marked price and sells them at 7 / 5 of the marked price to find his profit percentage.

Solution:

Let the marked price be MP.

He buys goods at 4 / 5 of its marked price.

Cost price = (4 / 5) * MP

Selling price = (7 / 5) * MP

Profit = Selling price – Cost price

= (7 / 5) * MP – (4 / 5) * MP = (3 / 5) * MP

Profit percentage = [Profit / Marked price] * 100

= {[(3 / 5) MP] / MP} * 100

= (3 / 5) * 100

= 300 / 5

= 60%

Question 3: If a mattress is marked for ₹7500 and is available at two successive discounts of 10% and 20%, find the amount to be paid by the customer.

Solution:

The marked price of the mattress = 7500

Discount d1 = 10%

Discount d2 = 20%

Price after discount d1 = MP * ([100 – d1] / 100)

= 7500 * ([100 – 10] / 100)

= 7500 * ([90 / 100]

= 6750

Price after discount d2 = MP * ([100 – d2] / 100)

= 6750 * ([100 – 20] / 100)

= 6750 * ([80 / 100]

= 5400

### Samacheer Kalvi 8th Maths Term 2 Chapter 1: Life Mathematics Book Exercise 1.3 Questions and Solutions

Question 1: Find the compound interest on ₹3200 at 2.5% p.a for 2 years, compounded annually.

Solution:

Principal = 3200

R = 2.5%

Time = 2 years, compounded annually

Amount = A = P [1 + (r / 100)]n

= 3200 * [1 + (2.5 / 100)]2

= 3200 * [1.050625]

= 3362

Compound interest = Amount – Principal = 3362 – 3200 = 162 rupees

Question 2: What is the difference in simple interest and compound interest on ₹15000 for 2 years at 6% per annum compounded annually.

Solution:

Principal = 15000

T = 2 years

Rate of interest = 6% per annum compounded annually

Difference between simple interest and compound interest is

CI – SI = P (r / 100)n

= 15000 (6 / 100)2

= 15000 (0.06)2

= 54 rupees

Question 3: The number of conversion periods, if the interest on a principal is compounded every two months is___________.

Solution:

The conversion period is the time period after which the interest is added to the principal. If the principal is compounded every two months, then, in a year, there will be 6 (12 / 2) conversion periods.

### Samacheer Kalvi 8th Maths Term 2 Chapter 1: Life Mathematics Book Exercise 1.4 Questions and Solutions

Question 1: P’s income is 25% more than that of Q. By what percentage is Q’s income less than P’s?

Solution:

Let Q’s income be Rs.100.

P’s income is 25% more than Q.

P’s income = 100 + (25 / 100) * 100 = 125

Q’s income is 25% less than P.

In percentage terms, Q’s income is less than P’s with respect to P’s income is

[(P – Q) / P] * 100

= ([125 – 100] / 125) * 100

= [25 / 100] * 100

= 25%

Question 2: If the first number is 20% less than the second number and the second number is 25% more than 100, then find the first number.

Solution:

The second number is 25% more than 100.

The second number is 100 + (25 / 100) * 100

= 125

The first number is 20% less than the second number

The first number is = (125) – (20 / 100) * 125

= 125 – 25

= 100

The first number is 100.

Question 3: The simple interest on a certain principal for 3 years at 10 % p.a is ₹300. Find the compound interest accrued in 3 years.

Solution:

Let the principal be P.

Rate of interest is 10% per annum.

Time period = 3 years

Principal = (300 * 100) / (10 * 3) = 1000

Compound interest = CI = A – P

Amount = P (1 + (r / 100))n

= 1000 (1 + [10 / 100])3

= 1000 * (110 / 100)3

= 1331

Compound interest = Amount – Principal

= 1331 – 1000

= 331 rupees