Class 8 Maths RS Aggarwal Solutions

RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D

Tick the correct answer in each of the following

Question 1.
Solution:
Principal (P) = Rs. 5000
Rate (R) = 8% p.a.
Period (n) = 2 years

Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 10% p.a.
Period (n) = 3 years

Question 3.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 12% p.a.

Question 4.
Solution:
Principal (P) = Rs. 4000
Rate (R) = 10% p.a.
Period (a) = 2 years 3 months

Question 5.
Solution:
Principal (P) = Rs. 25000
Rate (R1) = 5% for the first year
R2 = 6% for the second year
R3 = 8% for the third year

Question 6.
Solution:
Principal (P) = Rs. 6250
Rate (R) = 8% p.a. or 4% half yearly
Period (n) = 1 year or 2 half years

Question 7.
Solution:
Principal (P) = Rs. 40000
Rate (R) = 6% p.a. \(\\ \frac { 6 }{ 4 } \) = \(\\ \frac { 3 }{ 2 } \) % quarterly
Period (n) = 6 months = 2 quarters

Question 8.
Solution:
Present population (P) = 24000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years

Question 9.
Solution:
3 years ago, the value of machine = Rs. 60000
Rate of depreciation (R) = 10%
Period (n) = 3 years

Question 10.
Solution:
Present value = Rs. 40000
Rate of depreciation (R) = 20% p.a.
Value of machine 2 years ago

Question 11.
Solution:
Rate of growth in population (R) = 10%
Present population = 33275
Population 3 years ago = A

Question 12.
Solution:
S.I. = Rs. 1200
Rate (R) = 5%
Period (T) = 3 years

Question 13.
Solution:
C.I. on a sum = Rs. 510
Rate (R) = \(12\frac { 1 }{ 2 } \) % = \(\\ \frac { 25 }{ 2 } \) % p.a.
Period (n) = 2 years

Question 14.
Solution:
Amount = Rs. 4913
Rate (R) = \(6\frac { 1 }{ 4 } \) = \(\\ \frac { 25 }{ 4 } \) %
Period (n) = 3 years

Question 15.
Solution:
Sum (P) = Rs. 7500
Amount (A) = 8427
Period = 2 years
Let R be the rate of p.a., then

Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D

Question 1.
Solution:
Principal (P) = Rs. 8000
Rate (R) = 10% p.a. or 5% half yearly
Period (n) = 1 year or 2 half years

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 2.
Solution:
Principal (P) = Rs. 31250
Rate (R) = 8% p.a. or 4% half yearly
Period (n) = \(1\frac { 1 }{ 2 } \) years or 3 half years

= Rs 3902

Question 3.
Solution:
Principal (P) = Rs 12800
Rate (R) = \(7\frac { 1 }{ 2 } \)% p.a. = \(\\ \frac { 15 }{ 2 } \)% half yearly
Period (n) = 1 year or 2 half years

Question 4.
Solution:
Principal (P) = Rs. 160000
Rate (R) = 10% p.a. or 5% half yearly
Period (n) = 2 years or 4 half years

Question 5.
Solution:
Principal (P) = Rs. 40960
Rate (R) = \(12\frac { 1 }{ 2 } \) = \(\\ \frac { 25 }{ 2 } \)% p.a. or \(\\ \frac { 25 }{ 4 } \) % half yearly

Question 6.
Solution:
Loan received for the cost of the house (P) = Rs. 125000
Rate of interest (R) = 12% p.a. or 6% half yearly

Question 7.
Solution:
Amount deposit in the bank = Rs. 20000
Rate of interest (R) = 6% p.a. or 3% half-yearly
Period (n) = 1 year or 2 half years
Amount received after 1 year
= \({ \left( 1+\frac { R }{ 100 } \right) }^{ n }\)

Question 8.
Solution:
Amount of loan = Rs. 65536
Rate of interest (R) = 12 \(12\frac { 1 }{ 2 } \) % = \(\\ \frac { 25 }{ 2 } \)%

Question 9.
Solution:
Amount deposit in the bank (P)
= Rs. 32000
Rate of interest (R) = 5% p.a.

Question 10.
Solution:
Amount taken from finance company (P) = Rs. 390625
Rate of interest (R) = 16% p.a.

Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D

By using the formula, find the amount and compound interest on :

Question 1.
Solution:
Principal (P) = Rs. 6000
Rate (R) = 9% p.a.
Period (n) = 2 years

Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 11% p.a.
Period (n) = 2 years

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 3.
Solution:
Principal (P) = Rs. 31250
Rate (R) = 8% p.a.
Period (n) = 3 years

Question 4.
Solution:
Principal (P) = Rs. 10240
Rate (R) = \(12\frac { 1 }{ 2 } \)% = \(\\ \frac { 25 }{ 2 } \)% p.a.
Period (n) = 3 years

Question 5.
Solution:
Principal (P) = Rs. 62500
Rate (R) = 12% p.a.
Period (n) = 2 years 6 months

Question 6.
Solution:
Principal (P) = Rs. 9000
Rate (R) = 10% p.a.
Period (n) = 2 years 4 months

Question 7.
Solution:
Principal (P) = Rs. 8000
Period (n) = 2 years

Question 8.
Solution:
Principal (p) = Rs. 1, 25,000
Rate of interest (r) = 8% p.a.
Period (n) = 3 years

Question 9.
Solution:
Price of a buffalo (P) = Rs. 11000
Rate of interest (R) = 10% p.a.
Period (n) = 3 years
Price of buffalo at present

Question 10.
Solution:
Amount of loan taken (P)
= Rs. 18000

Question 11.
Solution:
Amount borrowed from Bank (P) = Rs. 24000
Rate (R) = 10% p.a.
Period (n) = 2 years 3 months

Question 12.
Solution:
In case of Abhay,
Principal (p) = Rs. 16000

Question 13.
Solution:
Simple interest (S.I.) = Rs. 2400
Rate (R) = 8% p.a.
Period (T) = 2 years

Question 14.
Solution:
Difference between C.I. and S.I.
= Rs. 90
Rate (R) = 6% p.a.
Period (n) = 2 years
Let principal (P) = Rs. 100

Question 15.
Solution:
Let sum (p) = Rs. 100
Rate (r) 10% p.a.
Period (t) = 3 years.

Question 16.
Solution:
Amount (A) = Rs. 10240
Rate (r) = \(6\frac { 2 }{ 3 } \)% = \(\\ \frac { 20 }{ 3 } \)% p.a.
Period (n) = 2 years
Let sum = P, then

Question 17.
Solution:
Amount (A) = Rs. 21296
Rate (r) = 10% p.a.
Period (n) = 3 years.
Let P be the sum, Then

Question 18.
Solution:
Principal (P) = 4000
Amount (A) = Rs. 4410
Period (n) = 2 years
Let r be the rate per cent per annum
We know that,

Question 19.
Solution:
Principal (P) = Rs. 640
Amount (A) = Rs. 774.40
Period (n) = 2 years
Let r be the rate per cent per annum.
We know that

Question 20.
Solution:
Principal (P) = Rs. 1800
Amount (A) = Rs. 2178
Rate (r) = 10% p.a.
Let n be the number of years,
We know that

Question 21.
Solution:
Principal (P) = Rs. 6250
Amount (A) = Rs. 7290
Rate (R) = 8% p.a.
Let n be the time, then

Question 22.
Solution:
Present population (P) = 125000
Rate of increasing (R) = 2% p.a.
Period (n) = 3 years

Question 23.
Solution:
3 years ago, the population was = 50000
Rate of increase successively (r1, r2, r3) = 4%, 5% and 3% p.a.
Period (n) = 3 years
Present Population

Question 24.
Solution:
Population of a city in 2013 = 120000
Increase in next year = 6%
and decrease in the following year = 5%
Population in 2015

Question 25.
Solution:
Initially bacteria = 500000
Increase in bacteria = 2% per hour
Period (n) = 2 hours

Question 26.
Solution:
Growth of bacteria in a culture (R1) = 10% in first hour
Decrease in next hour (R2) = 10%
Increase in the third hour (R3) = 10%
Bacteria in the beginning = 20000
Bacteria after 3 hours

Question 27.
Solution:
Value of machine (P) = Rs. 625000
Rate of depreciation (R) = 8% p.a.
Period (n) = 2 years

Question 28.
Solution:
Value of scooter (P) = Rs. 56000
Rate of depreciation (R) = 10% p.a.
Period = 3 years
Value of scooter after 3 years

Question 29.
Solution:
Cost of car = Rs. 34800
Rate of depreciation (R1) = 10% p.a. for first year

Question 30.
Solution:
Rate of depreciation (R) = 10% p.a.
Period (n) = 3 years
Present value (A) = Rs. 291600
Value of machine 3 years ago

Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C
• RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D

Question 1.
Solution:
Principal (p) = Rs. 2500
Rate (r) = 10% p.a.
Period (t) = 2 years
Interest for the first year = \(\\ \frac { prt }{ 100 } \)
= \(\\ \frac { 2500X10X1 }{ 100 } \)
= Rs. 250
Amount at the end of first year = Rs. 2500 + Rs. 250
= Rs 2750
Principal for the second year = Rs 2750
Interest for the second year = Rs \(\\ \frac { 2750X10X1 }{ 100 } \)
= Rs. 275
Amount at the end of second year = Rs 2750 + Rs. 275
= Rs. 305
and compound interest for the 2 years = Rs. 3025 – Rs. 2500
= Rs 525 Ans.

Question 2.
Solution:
Principal (P) = Rs. 15625 Rate
(R) = 12% p.a.
Period (n) = 3 years

Question 3.
Solution:
Principal (P) = Rs. 5000
Rate (R) = 9% p.a.
Time (n) = 2 years

Question 4.
Solution:
Amount of loan (p) = Rs. 25000
Rate of interest (r) = 8% p.a.
Period (t) = 2 years

Question 5.
Solution:
In case of Harpreet :
Amount borrowed by Harpreet (P) = Rs. 20000
Rate (r) = 12%
Period (t) = 2 Years

Question 6.
Solution:
Principal (p) = Rs. 64000
Rate (r) = \(7\frac { 1 }{ 2 } \) = \(\\ \frac { 15 }{ 2 } \)%
Period (t) = 3 years
Interest for the first year = \(\\ \frac { prt }{ 100 } \)
= Rs \(\\ \frac { 64000X15X1 }{ 100X2 } \)
= Rs 4800

= Rs 79507

Question 7.
Solution:
Principal (p) = Rs 6250
Rate (r) 8% p.a. or 4% half yearly
Period (t) = 1 year = 2 half years

Question 8.
Solution:
Principal (p) = Rs. = 16000
Rate (r) = 10% p.a. or 5% half yearly

Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10A
• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10B
• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10C
• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10D

Question 1.
Solution:
Answer = (c)
C.P. of toy Rs. = 75
S.P. = Rs. 100

Question 2.
Solution:
Answer = (b)
C.P. of bat = Rs. 120
S.P. = Rs. 105
Loss = Rs. 120 – Rs. 105 = Rs. 15

Question 3.
Solution:
Answer = (b)
S.P. of book = Rs. 100
gain = Rs. 20
C.P. = 100 – 20 = Rs. 80

Question 4.
Solution:
SP of an article = Rs. 48
Loss = 20%

Question 5.
Solution:
First time gain = 10%
Let SP – Rs. 100

Question 6.
Solution:
Let no. of bananas bought = 6
Now C.P. of bananas at the sale of 3

Question 7.
Solution:
SP of 10 pens = CP of 12 pens
= Rs. 100 (suppose)

Question 8.
Solution:
Gain on 100 pencils = SP of 20 pencils
SP of 100 pencils gains = CP of 100 pencils
=> SP of 100 pencils – SP of 20 pencils = CP of 100 pencils
=> SP of 80 pencils – CP of 100 pencils = Rs 100 (suppose)

Question 9.
Solution:
SP of 5 toffees = Re. 1
SP of 2 toffees = Re. 1
Now CP of 1 toffee = Rs. \(\\ \frac { 1 }{ 5 } \)
and SP of 1 toffee = Rs. \(\\ \frac { 1 }{ 2 } \)

Question 10.
Solution:
CP of 5 oranges = Rs. 10
SP of 6 oranges = Rs. 15

Question 11.
Solution:
SP of a radio = Rs. 950
Loss = 5%

Question 12.
Solution:
Let CP of an article = Rs. 100
SP = \(\\ \frac { 6 }{ 5 } \) of CP = \(\\ \frac { 6 }{ 5 } \) x 100 = Rs. 120

Question 13.
Solution:
SP of a chair = Rs. 720
Loss = 25%

Question 14.
Solution:
Ratio in CP and SP = 20 : 21
Let CP = Rs. 20
and SP = Rs. 21
Gain = SP – CP = Rs. 21 – 20 = Re. 1

Question 15.
Solution:
SP of first chair = Rs. 500
Gain = 20%

Question 16.
Solution:
Gain % SP of Rs. 625 = Loss on SP of Rs. 435
CP of an article = x

Question 17.
Solution:
CP of an article = Rs. 150
Overhead expenses = 10% of CP

Question 18.
Solution:
In first case, gain = 5%
and in second case, loss = 5%
and difference = Rs. 5 more
But difference in % = 5 + 5 = 10%

Question 19.
Solution:
Let CP of an article = Rs. 100
List price = Rs. 100 + 20% of Rs. 100
= Rs. 100 + 20 = Rs. 120
Discount = 10%

Question 20.
Solution:
Let CP of an article = Rs. 100
Then Marked price
= Rs. 100 + 10% of 100
= Rs. 100 + 10 = Rs. 110
Discount = 10%

Question 21.
Solution:
Price of watch including VAT = Rs. 825
VAT% = 10%

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.