Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Compound Interest - EXERCISE 14.3
Compound Interest - EXERCISE 14.3 | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Time (t) = 3 years
By using the formula,
CI – SI = 283.50
P [(1 + R/100)
n
- 1] – (PTR)/100 = 283.50
P [(1 + 15/100)
3
- 1] – (P(3)(15))/100 = 283.50
P[1.520 - 1] – (45P)/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
? The sum is Rs 4000
6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the
end of two years Rs. 1290 as interest compounded annually, find the sum she
borrowed.
Solution:
Given details are,
Rate = 15 % per annum
Time = 2 years
CI = Rs 1290
By using the formula,
CI = P [(1 + R/100)
n
- 1]
1290 = P [(1 + 15/100)
2
- 1]
1290 = P [0.3225]
P = 1290/0.3225
= 4000
? The sum is Rs 4000
7. The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4%
per annum. Find the period for which the compound interest is Rs. 163.20.
Solution:
Given details are,
Rate = 4 % per annum
CI = Rs 163.20
Principal (P) = Rs 2000
By using the formula,
CI = P [(1 + R/100)
n
- 1]
163.20 = 2000[(1 + 4/100)
n
- 1]
163.20 = 2000 [(1.04)
n
-1]
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Time (t) = 3 years
By using the formula,
CI – SI = 283.50
P [(1 + R/100)
n
- 1] – (PTR)/100 = 283.50
P [(1 + 15/100)
3
- 1] – (P(3)(15))/100 = 283.50
P[1.520 - 1] – (45P)/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
? The sum is Rs 4000
6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the
end of two years Rs. 1290 as interest compounded annually, find the sum she
borrowed.
Solution:
Given details are,
Rate = 15 % per annum
Time = 2 years
CI = Rs 1290
By using the formula,
CI = P [(1 + R/100)
n
- 1]
1290 = P [(1 + 15/100)
2
- 1]
1290 = P [0.3225]
P = 1290/0.3225
= 4000
? The sum is Rs 4000
7. The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4%
per annum. Find the period for which the compound interest is Rs. 163.20.
Solution:
Given details are,
Rate = 4 % per annum
CI = Rs 163.20
Principal (P) = Rs 2000
By using the formula,
CI = P [(1 + R/100)
n
- 1]
163.20 = 2000[(1 + 4/100)
n
- 1]
163.20 = 2000 [(1.04)
n
-1]
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
163.20 = 2000 × (1.04)
n
– 2000
163.20 + 2000 = 2000 × (1.04)
n
2163.2 = 2000 × (1.04)
n
(1.04)
n
= 2163.2/2000
(1.04)
n
= 1.0816
(1.04)
n
= (1.04)
2
So on comparing both the sides, n = 2
? Time required is 2years
8. In how much time would Rs. 5000 amount to Rs. 6655 at 10% per annum
compound interest?
Solution:
Given details are,
Rate = 10% per annum
A = Rs 6655
Principal (P) = Rs 5000
By using the formula,
A = P (1 + R/100)
n
6655 = 5000 (1 + 10/100)
n
6655 = 5000 (11/10)
n
(11/10)
n
= 6655/5000
(11/10)
n
= 1331/1000
(11/10)
n
= (11/10)
3
So on comparing both the sides, n = 3
? Time required is 3years
9. In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest
compounded half-yearly?
Solution:
Given details are,
Rate = 8% per annum = 8/2 = 4% (half yearly)
A = Rs 4576
Principal (P) = Rs 4400
Let n be ‘2T’
By using the formula,
A = P (1 + R/100)
n
4576 = 4400 (1 + 4/100)
2T
4576 = 4400 (104/100)
2T
(104/100)
2T
= 4576/4400
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Time (t) = 3 years
By using the formula,
CI – SI = 283.50
P [(1 + R/100)
n
- 1] – (PTR)/100 = 283.50
P [(1 + 15/100)
3
- 1] – (P(3)(15))/100 = 283.50
P[1.520 - 1] – (45P)/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
? The sum is Rs 4000
6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the
end of two years Rs. 1290 as interest compounded annually, find the sum she
borrowed.
Solution:
Given details are,
Rate = 15 % per annum
Time = 2 years
CI = Rs 1290
By using the formula,
CI = P [(1 + R/100)
n
- 1]
1290 = P [(1 + 15/100)
2
- 1]
1290 = P [0.3225]
P = 1290/0.3225
= 4000
? The sum is Rs 4000
7. The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4%
per annum. Find the period for which the compound interest is Rs. 163.20.
Solution:
Given details are,
Rate = 4 % per annum
CI = Rs 163.20
Principal (P) = Rs 2000
By using the formula,
CI = P [(1 + R/100)
n
- 1]
163.20 = 2000[(1 + 4/100)
n
- 1]
163.20 = 2000 [(1.04)
n
-1]
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
163.20 = 2000 × (1.04)
n
– 2000
163.20 + 2000 = 2000 × (1.04)
n
2163.2 = 2000 × (1.04)
n
(1.04)
n
= 2163.2/2000
(1.04)
n
= 1.0816
(1.04)
n
= (1.04)
2
So on comparing both the sides, n = 2
? Time required is 2years
8. In how much time would Rs. 5000 amount to Rs. 6655 at 10% per annum
compound interest?
Solution:
Given details are,
Rate = 10% per annum
A = Rs 6655
Principal (P) = Rs 5000
By using the formula,
A = P (1 + R/100)
n
6655 = 5000 (1 + 10/100)
n
6655 = 5000 (11/10)
n
(11/10)
n
= 6655/5000
(11/10)
n
= 1331/1000
(11/10)
n
= (11/10)
3
So on comparing both the sides, n = 3
? Time required is 3years
9. In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest
compounded half-yearly?
Solution:
Given details are,
Rate = 8% per annum = 8/2 = 4% (half yearly)
A = Rs 4576
Principal (P) = Rs 4400
Let n be ‘2T’
By using the formula,
A = P (1 + R/100)
n
4576 = 4400 (1 + 4/100)
2T
4576 = 4400 (104/100)
2T
(104/100)
2T
= 4576/4400
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
(104/100)
2T
= 26/25
(26/25)
2T
= (26/25)
1
So on comparing both the sides, n = 2T = 1
? Time required is ½ year
10. The difference between the S.I. and C.I. on a certain sum of money for 2 years at
4% per annum is Rs. 20. Find the sum.
Solution:
Given details are,
Rate = 4 % per annum
Time = 2years
Compound Interest (CI) – Simple Interest (SI)= Rs 20
By using the formula,
CI – SI = 20
P [(1 + R/100)
n
- 1] – (PTR)/100 = 20
P [(1 + 4/100)
2
- 1] – (P(2)(4))/100 = 20
P[51/625] – (2P)/25 = 20
51/625P – 2/25P = 20
(51P-50P)/625 = 20
P = 20 × 625
P = 20/7.918
= 12500
? The sum is Rs 12500
11. In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, compound
interest?
Solution:
Given details are,
Principal = Rs 1000
Amount = Rs 1331
Rate = 10% per annum
Let time = T years
By using the formula,
A = P (1 + R/100)
n
1331 = 1000 (1 + 10/100)
T
1331 = 1000 (110/100)
T
(11/10)
T
= 1331/1000
(11/10)
T
= (11/10)
3
So on comparing both the sides, n = T = 3
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.83/5 Rating |
|
Nov 24, 2024 Last updated |
Document Description: Compound Interest - EXERCISE 14.3 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for Compound Interest - EXERCISE 14.3 have been prepared according to the Class 8 exam syllabus. Information about Compound Interest - EXERCISE 14.3 covers topics
like and Compound Interest - EXERCISE 14.3 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Compound Interest - EXERCISE 14.3.
Introduction of Compound Interest - EXERCISE 14.3 in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & Compound Interest - EXERCISE 14.3 in Hindi for Mathematics (Maths) Class 8 course.
Download more important topics related with notes, lectures and mock test series for Class 8
Exam by signing up for free. Class 8: Compound Interest - EXERCISE 14.3 | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for Compound Interest - EXERCISE 14.3 | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download
Information about Compound Interest - EXERCISE 14.3
In this doc you can find the meaning of Compound Interest - EXERCISE 14.3 defined & explained in the simplest way possible. Besides explaining types of
Compound Interest - EXERCISE 14.3 theory, EduRev gives you an ample number of questions to practice Compound Interest - EXERCISE 14.3 tests, examples and also practice Class 8
tests
|
Explore Courses for Class 8 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
video lectures
,study material
,Extra Questions
,Viva Questions
,mock tests for examination
,Compound Interest - EXERCISE 14.3 | Mathematics (Maths) Class 8
,Summary
,MCQs
,shortcuts and tricks
,Compound Interest - EXERCISE 14.3 | Mathematics (Maths) Class 8
,Previous Year Questions with Solutions
,Free
,ppt
,past year papers
,Sample Paper
,Objective type Questions
,Semester Notes
,Exam
,Compound Interest - EXERCISE 14.3 | Mathematics (Maths) Class 8
,practice quizzes
,Important questions
;
Additional Information about Compound Interest - EXERCISE 14.3 for Class 8 Preparation
Compound Interest - EXERCISE 14.3 Free PDF Download
The Compound Interest - EXERCISE 14.3 is an invaluable resource that delves deep into the core of the Class 8 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Compound Interest - EXERCISE 14.3 now and kickstart your journey towards success in the Class 8 exam.
Importance of Compound Interest - EXERCISE 14.3
The importance of Compound Interest - EXERCISE 14.3 cannot be overstated, especially for Class 8 aspirants.
This document holds the key to success in the Class 8 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Compound Interest - EXERCISE 14.3 Notes
Compound Interest - EXERCISE 14.3 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Compound Interest - EXERCISE 14.3.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Compound Interest - EXERCISE 14.3 Notes on EduRev are your ultimate resource for success.
Compound Interest - EXERCISE 14.3 Class 8 Questions
The "Compound Interest - EXERCISE 14.3 Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Compound Interest - EXERCISE 14.3 on the App
Students of Class 8 can study Compound Interest - EXERCISE 14.3 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Compound Interest - EXERCISE 14.3,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Compound Interest - EXERCISE 14.3 is prepared as per the latest Class 8 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup