ML Aggarwal Simple and Compound Interest Exe-8.1 Class 8 ICSE Ch-8 Maths Solutions. We Provide Step by Step Answer of Exe-8.1 Questions for Simple and Compound Interest as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.
ML Aggarwal Simple and Compound Interest Exe-8.1 Class 8 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 8th |
| Chapter-8 | Simple and Compound Interest |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-8.1 Questions |
| Edition | 2023-2024 |
Simple and Compound Interest Exe-8.1
ML Aggarwal Class 8 ICSE Maths Solutions
Page-136
Question 1. Find the simple interest on 4000 at 7.5% p.a. for 3 year 3 months. Also find the amount.
Answer :
Principal (P) = Rs 4000
Rate of interest (R) = 7.5% p.a.
= (15 / 2) % p.a.
Time (T) = 3 years 3 months
= 3(3/12)years
= 3(1/4)years
= 13 / 4 years
Simple Interest (I) = (P × R × T) / 100
= Rs {4000 × (15 / 2) × (13 / 4)} / 100
= Rs (4000 × 15 × 13) / (100 × 2 × 4)
= Rs 5 × 15 × 13
= Rs 975
Amount = P + I
= Rs 4000 + Rs 975
= RS 4, 975
Question 2. What sum of money will yield ₹170·10 as simple interest in 2 years 3 months at 6% per annum?
Answer :
I = Rs 170.10
T = 2 years 3 months
= 2(3/12)years
= 2(1/4) years
= 9 / 4 years
R = 6%
Hence,
P = (I × 100) / (R × T)
= Rs (170.10 × 100) / {6 × (9 / 4)}
On calculating further, we get,
= Rs (170.10 × 100 × 4) / (6 × 9)
= Rs (17010 × 4) / (6 × 9)
= Rs (17010 × 2) / (3 × 9)
= Rs 34020 / 27
= Rs 1260
Question 3. Find the rate of interest when ₹800 fetches ₹130 as a simple interest in 2 years 6 months.
Answer:
P = Rs 800
T = 2 years 6 months
= 2(6/12)years
= 2(1/2)years
= 5 / 2 years
R = (I × 100) / (P × T)
= (130 × 100) / {800 × (5 / 2)} % p.a.
On simplification, we get,
= (130 × 100 × 2) / (800 × 5) % p.a.
= (130 × 2) / 40 %
= 130 / 20 % p.a.
= 13 / 2 %
= 6.5% p.a.
Hence, the required rate of interest is 6.5% p.a.
Question 4. Find the time when simple interest on ₹3·3 lakhs at 6·5% per annum is ₹75075.
Answer:
P = 3.3 lakhs
= Rs 3.3 × 100000
= Rs 330000
R = 6.5% per annum
I = Rs 75075
T = (I × 100) / (P × R)
= (75075 × 100) / (330000 × 6.5) years
= (75075 × 100 × 10) / (330000 × 65) years
= (75075) / (330 × 65) years
= 1155 / 330 years
= 7 / 2 years
= 3(1/2)years
Simple and Compound Interest Exe-8.1
ML Aggarwal Class 8 ICSE Maths Solutions
Page-137
Question 5. Find the sum of money when:
(i) simple interest at 71⁄4% p.a. for years is ₹2356·25
(ii) the final amount is ₹ 11300 at 4% p.a. for 3 years 3 months.
Answer:
(i) I = Rs 2356.25
R =
% p.a.
= 29 / 4 % p.a.
T = 2(1/2)years
= 5 / 2 years
P = (I × 100) / (R × T)
= Rs (2356.25 × 100) / (29 / 4) × (5 / 2)
= Rs (2356.25 × 100 × 4 × 2) / (29 × 5)
= Rs (235625 × 8) / (29 × 5)
= Rs (47125 × 8) / 29
= Rs 1625 × 8
= Rs 13000
(ii) Amount (A) = Rs 11300
Rate (R) = 4% p.a.
Time (T) = 3 years 3 months
= 3(3/12)years
= 3(1/4)years
= 13 / 4 years
Let the principal be Rs x
S.I. = (P × R × T) / 100
= Rs (x × 4 × 13) / (100 × 4)
= Rs 13x / 100
Amount = Principal + Simple Interest
= Rs x + Rs 13x / 100
= Rs (x + 13x) / 100
= Rs (100x + 13x) / 100
= Rs (113x / 100)
But, the amount given is Rs 11300
113x / 100 = 11300
x = 11300 × 100 / 113
x = 100 × 100
x = 10000
hence, principal (P) = Rs 10000
Question 6. How long will it take a certain sum of money to triple itself at 131⁄3% per annum simple interest?
Answer:
Let the sum of money be x
Amount = 3 × Rs x
= Rs 3x
Interest = Amount – Principal
= Rs 3x – Rs x
= Rs 2x
Rate = 13(1/3)% p.a.
= 40 / 3 % p.a.
Time (T) = (I × 100) / (P × R)
= (2x × 100) / x × (40 / 3) years
= (2 × 100 × 3) / 40 years
= (100 × 3) / 20 years
= 5 × 3 years
= 15 years
Question 7. At a certain rate of simple interest ₹4050 amounts to ₹4576·50 in 2 years. At the same rate of simple interest, how much would ₹1 lakh amount to in 3 years?
Answer:
P = Rs 40000
A = Rs 4576.50
T = 2 years
Interest = Amount – Principal
= Rs 4576.50 – Rs 4050
= Rs 526.50
Let the rate of simple interest = R% per annum
R = (I × 100) / (P × T)
= (526.50 × 100) / (4050 × 2) % p.a.
= (526.50 × 10) / (405 × 2) % p.a.
= 5265 / 810 % p.a.
= 6.5% p.a.
P = Rs 1 lakh
= Rs 100000
R = 6.5% p.a.
T = 3 years
I = (P × R × T) / 100
= Rs (100000 × 6.5 × 3) / 100
= RS 1000 × 6.5 × 3
= Rs 19500
Amount = Principal + Interest
= Rs 100000 + Rs 19500
= Rs 119500
Question 8. What sum of money invested at 7.5% p.a. simple interest for 2 years produces twice as much interest as ₹9600 in 3 years 6 months at 10% p.a. simple interest?
Answer:
First Case:
Principal (P1) = Rs 9600
Rate (R1) = 10%
Period = (T) = 3 years 6 months
= 3(1/2)years = 7 / 2 years
Simple interest = (P × R × T) / 100
= (9600 × 10 × 7) / (100 × 2)
= Rs 3360
Second case:
Simple interest = Rs 3360 × 2
= Rs 6720
Rate (R) = 7.5% p.a. and
Period (T) = 2 years
Hence,
Principal = (S.I × 100) / (R × T)
= (6720 × 100) / (7.5 × 2)
= Rs (6720 × 100 × 10) / (75 × 2)
= 6720000 / 150
= Rs 44800
— : End of ML Aggarwal Simple and Compound Interest Exe-8.1 Class 8 ICSE Maths Solutions :–
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