Frank ICSE Maths Class 10 Solutions 2023-24

• Chapter 1 – Compound Interest

• Chapter 2 – Sales Tax and Value Added Tax

• Chapter 3 – Banking

• Chapter 4 – Shares and Dividends

• Chapter 5 – Linear Inequalities

• Chapter 6 – Quadratic Equations

• Chapter 7 – Problems Based On Quadratic Equations

• Chapter 8 – Reflections

• Chapter 9 – Ratio and Proportion

• Chapter 10 – Remainder and Factor Theorems

• Chapter 11 – Matrices

• Chapter 12 – Distance and Section Formulae

• Chapter 13 – Equation of a Straight Line

• Chapter 14 – Symmetry

• Chapter 14 – Symmetry

• Chapter 15 – Similarity

• Chapter 16 – Loci

• Chapter 17 – Circles

• Chapter 18 – Constructions

• Chapter 19 – Mensuration I

• Chapter 20 – Mensuration II

• Chapter 21 – Trigonometric Identities

• Chapter 22 – Heights and Distances

• Chapter 23 – Graphical Representations

• Chapter 24 – Measures of Central Tendency

• Chapter 25 – Probability

Every question in these resources relates to a subject covered in the Class 10 ICSE syllabus and was created by our specialists to offer the best solutions to issues that students encounter.

Students' ability to think logically and rationally can be aided with Frank Maths Solutions. During thorough explanations, practise questions, solved examples, and more, it assists in the promotion of conceptual clarity. Daily problem-solving using the Frank textbook helps students learn the ideas taught in each chapter.

Students are aware of the value of solving problems to improve conceptual understanding and its applications. Due to this, Vedantu offers students an all-inclusive Frank Solutions for Maths package.

A Few Key Points of Benefits of Frank Solutions From Vedantu

1. Our knowledgeable instructors have created these solutions to help you study for your exams and get high grades in Maths.

2. Chapter-by-chapter solutions are provided in PDF format, which is free to download and offline use both.

3. Problems are resolved step-by-step with thorough justifications for easier comprehension.

4. These answers not only eliminate doubts but also provide wide information about the relevant subjects.

Significance of Frank Solution of Class 10 Maths

Students who use Frank Textbook Solutions for ICSE Class 10 Mathematics will succeed in their test because they cover all the material. Our topic specialists, who have taught ICSE Class 10 for more than ten years, solve textbook questions. One of the easiest ways to comprehend all the concepts and perform better on the final exam is to use the Set of Frank Textbook Solutions.

Important Question from Frank Solution of Class 10 for Maths

1. Calculate the amount and the compound interest for each of the following:  (year 2020,21)

a. Rs7,500 at 12%p.a. In 3 years

Solution:From the question it is given that, 

Principal, P= Rs 7,500, Rate, r= 12%p.a, Time, t= 3 years

For the first year, t = 1 year

We know that, S.I. = (P*r*t)/100

= (7,500* 12* 1)/100

= 900

Then, A = P+S.I.

= 7,500 + 900

= 8,400

Therefore, the new principal is Rs 8,400

Now, for the second year, t = 1year & P = 8,400

S.I. = (P*r*t)/100

= (8,400* 12*1)/100

= 1,008

Therefore, the new principal is Rs1,008

Then, A = P+S.I.

= 8,400 + 1,008

= 9,408

Therefore, the new principal is Rs 9,408.

Now, for the third year, t = 1year & P = 9,408

S.I.= (P*r*t)/100

= (9,408*12*1)/100

= 1,128.96

Then, A = P+S.I.

= 9,408 + 1,128

= 10,536

We know 

C.I. = interest in first year + interest in second year + interest in third year

= Rs(900+1,008+1,128.96)

= Rs 3,036.96

2. Manjot deposited Rs 27,500 in a deposit scheme paying 12%p.a. Compound interest. If the duration of the deposit in 3 year, calculate:

1. The amount received by him at the end of three years.

2. The compound interest received by him.

3. The amount received by him had hr chosen the duration of the deposit to the 2 years.

Solution: From the question it is given that Manjot deposited Rs 27,500 in a scheme paying 12% p.a.

Time t = 3 years

1. C1 = (P*r*t)/100

= ( 27,500* 12*1)/100

= Rs 3,300

Then , P1 = 27,500 + 3,300

= Rs 30,800

C2= (P*r*t)/100

= ( 30,800*12*1)/100

=  Rs 3,696

Then , P2 = 30,800 + 3,696

= Rs 34,496

C3 = (P*r*t)/100

= ( 34,496* 12*1)/100

= Rs 4139.52

Then, P3 = 4,139.52 + 34,496

= Rs 38,636

2. Then  compound interest received by him = 3,300 + 3,696 + 4,139.52 = 11,135.52

3. The amount received by him had he chosen the duration of deposits to be 2 years P2 = Rs 34,496 

Conclusion

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