CBSE Class 8

RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12A
• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12B
• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12C

Question 1.
Solution:
∵ x and y are inversely proportional
Then xy are equal
(i) xy = 6 x 9 = 54
= 10 x 15 = 150
= 14 x 21 = 294
= 16 x 24 = 384
∵ xy in each case is not equal
So, x and y are not inversely proportional
(ii) xy = 5 x 18 = 90
= 9 x 10 = 90
= 15 x 6 = 90
= 3 x 30 = 90
= 45 x 2 = 90
∵ xy in each case is equal
x and y are inversely proportional
(iii) xy = 9 x 4 = 36
= 3 x 12 = 36
= 6 x 9 = 54
= 36 x 1 = 36
∵ xy in each is not equal
x and y are not inversely proportional

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Question 2.
Solution:
x and y are inversely proportional
xy is equal
Now,

Question 3.
Solution:
Let required number of days = x

Question 4.
Solution:
A pond is! dug in 8 days by = 12 men
It can be dug in 1 day by = 12 x 8 men (Less days, more men)
and it can be dug in 6 days by = \(\\ \frac { 12X8 }{ 6 } \)
= 16 men Ans. (more days, less men)

Question 5.
Solution:
Let 14 cows can graze in x days

Question 6.
Solution:
Let required time take = x hour

By inverse proportion
60 : x :: 75 : 5
x = \(\\ \frac { 50X5 }{ 75 } \)
Time required = 4 hours

Question 7.
Solution:
Let machines required = x

Question 8.
Solution:
Let 8 taken will fill in tank in x hour

Question 9.
Solution:
8 taps can fill tank in = 27 minutes
1 tap can fill that tank in = 27 x 8 minutes (less tap, more time)
8 – 2 = 6 taps can fill that tank in
= \(\\ \frac { 27X8 }{ 6 } \) minutes
= 36 minutes

Question 10.
Solution:
Let total animals can be feed with food in x days

Question 11.
Solution:
Let for x day, the food provision is sufficient for 900 + 500 = 1400 men

Question 12.
Solution:
Let the food will be for x days

Question 13.
Solution:
Let each period will be of x minutes

Question 14.
Solution:
x and y are inversely
and x = 15, y = 6
Then xy = 15 x 6 = 90
Now if x = 9, then y = \(\\ \frac { 90 }{ 9 } \)
= 10

Question 15.
Solution:
x and y are inversely and x = 18, y = 8
xy = 18 x 8 = 144
Now if y = 16,
then x = \(\\ \frac { 144 }{ 16 } \)
= 9

Hope given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12A
• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12B
• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12C

Question 1.
Solution:
(i) \(\\ \frac { x }{ y } \) = \(\\ \frac { 3 }{ 9 } \) = \(\\ \frac { 1 }{ 3 } \)

Question 2.
Solution:
x and y are directly proportional
\(\\ \frac { x }{ y } \) = \(\\ \frac { 3 }{ 72 } \) = \(\\ \frac { 1 }{ 24 } \)

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Let distance covered = x then

Question 6.
Solution:
Let no. of dolls = x, then

Question 7.
Solution:
Let x kg of sugar will be bought

Question 8.
Solution:
Let cloth bought = x m

Question 9.
Solution:
Let length of model ship = x m

Question 10.
Solution:
Let x kg dust will be picked up

Question 11.
Solution:
A speed of car = 50 km/hr
Distance travelled in 1 hr. = 5 m
Let required distance travelled in 1 hr. 12 min.

Question 12.
Solution:
Let required distance covered = x km
Speed of man = 5 km/hr

Question 13.
Solution:
Let required thickness = x mm

Question 14.
Solution:
Let men required = x

Question 15.
Solution:
Let no. of words type in 8 minutes = x

Hope given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12A are helpful to complete your math homework.

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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.