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

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

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

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

Question 2.
Solution:
x and y are inversely proportional
xy is equal
Now,
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 2.1

Question 3.
Solution:
Let required number of days = x
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 3.1

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
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 5.1

Question 6.
Solution:
Let required time take = x hour
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 6.1
By inverse proportion
60 : x :: 75 : 5
x = \(\\ \frac { 50X5 }{ 75 } \)
Time required = 4 hours

Question 7.
Solution:
Let machines required = x
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 7.1

Question 8.
Solution:
Let 8 taken will fill in tank in x hour
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 8.1

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
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 10.1

Question 11.
Solution:
Let for x day, the food provision is sufficient for 900 + 500 = 1400 men
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 11.1

Question 12.
Solution:
Let the food will be for x days
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 12.1

Question 13.
Solution:
Let each period will be of x minutes
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B 13.1

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

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

Question 1.
Solution:
(i) \(\\ \frac { x }{ y } \) = \(\\ \frac { 3 }{ 9 } \) = \(\\ \frac { 1 }{ 3 } \)
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 1.1
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 1.2
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 1.3

Question 2.
Solution:
x and y are directly proportional
\(\\ \frac { x }{ y } \) = \(\\ \frac { 3 }{ 72 } \) = \(\\ \frac { 1 }{ 24 } \)
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 2.1

Question 3.
Solution:
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 3.1

Question 4.
Solution:
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 4.1

Question 5.
Solution:
Let distance covered = x then
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 5.1

Question 6.
Solution:
Let no. of dolls = x, then
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 6.1

Question 7.
Solution:
Let x kg of sugar will be bought
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 7.1

Question 8.
Solution:
Let cloth bought = x m
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 8.1

Question 9.
Solution:
Let length of model ship = x m
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 9.1

Question 10.
Solution:
Let x kg dust will be picked up
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 10.1

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.
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 11.1

Question 12.
Solution:
Let required distance covered = x km
Speed of man = 5 km/hr
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 12.1

Question 13.
Solution:
Let required thickness = x mm
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 13.1

Question 14.
Solution:
Let men required = x
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 14.1

Question 15.
Solution:
Let no. of words type in 8 minutes = x
RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A 15.1

 

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

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

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 1.1

Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 10% p.a.
Period (n) = 3 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 2.1

Question 3.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 12% p.a.
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 3.1

Question 4.
Solution:
Principal (P) = Rs. 4000
Rate (R) = 10% p.a.
Period (a) = 2 years 3 months
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 4.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 5.1

Question 6.
Solution:
Principal (P) = Rs. 6250
Rate (R) = 8% p.a. or 4% half yearly
Period (n) = 1 year or 2 half years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 6.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 7.1

Question 8.
Solution:
Present population (P) = 24000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 8.1

Question 9.
Solution:
3 years ago, the value of machine = Rs. 60000
Rate of depreciation (R) = 10%
Period (n) = 3 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 9.1

Question 10.
Solution:
Present value = Rs. 40000
Rate of depreciation (R) = 20% p.a.
Value of machine 2 years ago
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 10.1

Question 11.
Solution:
Rate of growth in population (R) = 10%
Present population = 33275
Population 3 years ago = A
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 11.1

Question 12.
Solution:
S.I. = Rs. 1200
Rate (R) = 5%
Period (T) = 3 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 12.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 13.1
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 13.2

Question 14.
Solution:
Amount = Rs. 4913
Rate (R) = \(6\frac { 1 }{ 4 } \) = \(\\ \frac { 25 }{ 4 } \) %
Period (n) = 3 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 14.1

Question 15.
Solution:
Sum (P) = Rs. 7500
Amount (A) = 8427
Period = 2 years
Let R be the rate of p.a., then
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D 15.1

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

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

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 1.1

Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 11% p.a.
Period (n) = 2 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 2.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 3.1

Question 4.
Solution:
Principal (P) = Rs. 10240
Rate (R) = \(12\frac { 1 }{ 2 } \)% = \(\\ \frac { 25 }{ 2 } \)% p.a.
Period (n) = 3 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 4.1

Question 5.
Solution:
Principal (P) = Rs. 62500
Rate (R) = 12% p.a.
Period (n) = 2 years 6 months
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 5.1

Question 6.
Solution:
Principal (P) = Rs. 9000
Rate (R) = 10% p.a.
Period (n) = 2 years 4 months
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 6.1

Question 7.
Solution:
Principal (P) = Rs. 8000
Period (n) = 2 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 7.1

Question 8.
Solution:
Principal (p) = Rs. 1, 25,000
Rate of interest (r) = 8% p.a.
Period (n) = 3 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 8.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 9.1

Question 10.
Solution:
Amount of loan taken (P)
= Rs. 18000
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 10.1

Question 11.
Solution:
Amount borrowed from Bank (P) = Rs. 24000
Rate (R) = 10% p.a.
Period (n) = 2 years 3 months
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 11.1

Question 12.
Solution:
In case of Abhay,
Principal (p) = Rs. 16000
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 12.1
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 12.2

Question 13.
Solution:
Simple interest (S.I.) = Rs. 2400
Rate (R) = 8% p.a.
Period (T) = 2 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 13.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 14.1
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 14.2

Question 15.
Solution:
Let sum (p) = Rs. 100
Rate (r) 10% p.a.
Period (t) = 3 years.
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 15.1
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 15.2

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 16.1

Question 17.
Solution:
Amount (A) = Rs. 21296
Rate (r) = 10% p.a.
Period (n) = 3 years.
Let P be the sum, Then
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 17.1

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,
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 18.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 19.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 20.1

Question 21.
Solution:
Principal (P) = Rs. 6250
Amount (A) = Rs. 7290
Rate (R) = 8% p.a.
Let n be the time, then
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 21.1

Question 22.
Solution:
Present population (P) = 125000
Rate of increasing (R) = 2% p.a.
Period (n) = 3 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 22.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 23.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 24.1

Question 25.
Solution:
Initially bacteria = 500000
Increase in bacteria = 2% per hour
Period (n) = 2 hours
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 25.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 26.1

Question 27.
Solution:
Value of machine (P) = Rs. 625000
Rate of depreciation (R) = 8% p.a.
Period (n) = 2 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 27.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 28.1

Question 29.
Solution:
Cost of car = Rs. 34800
Rate of depreciation (R1) = 10% p.a. for first year
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 29.1

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B 30.1

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

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

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

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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 2.1

Question 3.
Solution:
Principal (P) = Rs. 5000
Rate (R) = 9% p.a.
Time (n) = 2 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 3.1

Question 4.
Solution:
Amount of loan (p) = Rs. 25000
Rate of interest (r) = 8% p.a.
Period (t) = 2 years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 4.1

Question 5.
Solution:
In case of Harpreet :
Amount borrowed by Harpreet (P) = Rs. 20000
Rate (r) = 12%
Period (t) = 2 Years
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 5.1

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 Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 6.1
= 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
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 7.1

Question 8.
Solution:
Principal (p) = Rs. = 16000
Rate (r) = 10% p.a. or 5% half yearly
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 8.1
RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A 8.2

 

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

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

Question 1.
Solution:
Answer = (c)
C.P. of toy Rs. = 75
S.P. = Rs. 100
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 1.1

Question 2.
Solution:
Answer = (b)
C.P. of bat = Rs. 120
S.P. = Rs. 105
Loss = Rs. 120 – Rs. 105 = Rs. 15
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 2.1

Question 3.
Solution:
Answer = (b)
S.P. of book = Rs. 100
gain = Rs. 20
C.P. = 100 – 20 = Rs. 80
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 3.1

Question 4.
Solution:
SP of an article = Rs. 48
Loss = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 4.1

Question 5.
Solution:
First time gain = 10%
Let SP – Rs. 100
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 5.1

Question 6.
Solution:
Let no. of bananas bought = 6
Now C.P. of bananas at the sale of 3
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 6.1

Question 7.
Solution:
SP of 10 pens = CP of 12 pens
= Rs. 100 (suppose)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 7.1

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)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 8.1

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. \(\\ \frac { 1 }{ 2 } \)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 9.1

Question 10.
Solution:
CP of 5 oranges = Rs. 10
SP of 6 oranges = Rs. 15
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 10.1

Question 11.
Solution:
SP of a radio = Rs. 950
Loss = 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 11.1

Question 12.
Solution:
Let CP of an article = Rs. 100
SP = \(\\ \frac { 6 }{ 5 } \) of CP = \(\\ \frac { 6 }{ 5 } \) x 100 = Rs. 120
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 12.1

Question 13.
Solution:
SP of a chair = Rs. 720
Loss = 25%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 13.1

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
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 14.1

Question 15.
Solution:
SP of first chair = Rs. 500
Gain = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 15.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 15.2

Question 16.
Solution:
Gain % SP of Rs. 625 = Loss on SP of Rs. 435
CP of an article = x
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 16.1

Question 17.
Solution:
CP of an article = Rs. 150
Overhead expenses = 10% of CP
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 17.1

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%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 18.1

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%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 19.1

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%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 20.1

Question 21.
Solution:
Price of watch including VAT = Rs. 825
VAT% = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 21.1

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RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C.

Other Exercises

Question 1.
Solution:
x2 + 8x + 16
= (x)2 + 2 × x × 4 + (4)2
= (x + 4)2

Question 2.
Solution:
x2 + 14x + 49
= (x)2 + 2 × x × 7 + (7)2
= (x + 7)2 Ans.

Question 3.
Solution:
1 + 2x + x2
= (1)2 + 2 × 1 × x + (x)2
= (1 + x)2 Ans.

Question 4.
Solution:
9 + 6z + z2
= (3)2 + 2 x 3 x z + (z)2
= (3 + z)2 Ans.

Question 5.
Solution:
x2 + 6ax + 9a2
= (x)2 + 2 × x × 3a + (3a)2
= (x + 3a)2 Ans.

Question 6.
Solution:
4y2 + 20y + 25
= (2y)2 + 2 x 2y x 5 + (5)2
= (2y2 + 5)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 7.
Solution:
36a2 + 36a + 9
= 9 [4a2 + 4a + 1]
= 9 [(2a)2 + 2 x 2a x 1 + (1)2]
= 9 [2a + 1]2

Question 8.
Solution:
9m2 + 24m + 16
= (3m)2 + 2 x 3m x 4 + (4)2
= (3m + 4)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 9.
Solution:
z2 + z + \(\\ \frac { 1 }{ 4 } \)
= (z)2 + 2 x z x \(\\ \frac { 1 }{ 2 } \) + \({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)
= \({ \left( z+\frac { 1 }{ 2 } \right) }^{ 2 }\)

Question 10.
Solution:
49a2 + 84ab + 36b2
= (7a)2 + 2 x 7a x 6b + (6b)2
{ ∵ a2 + 2ab + b2 = (a + b)2}
= (7a + 6b)2

Question 11.
Solution:
p2 – 10p + 25
= (p)2 – 2 x p x 5 + (5)2
= (p – 5)2
{ ∵ a2 – 2ab + b2 = (a – b)2}

Question 12.
Solution:
121a2 – 88ab + 16b2
= (11a)2 – 2 x 11a x 4b + 4(b)2
= (11a – 4b)2

Question 13.
Solution:
1 – 6x + 9x2
= (1)2 – 2 x 1 x 3x + (3x)2
= (1 – 3x)2
{ ∵ a2 – 2ab + b2 = (a – b)2}

Question 14.
Solution:
9y2 – 12y + 4
= (3y)2 – 2 x 3y x 2 + (2)2
{ ∵ a2 – 2ab + b2 = (a – b)2}
= (3y – 2)2

Question 15.
Solution:
16x2 – 24x + 9
= (4x)2 – 2 x 4x x 3 + (3)2
= (4x – 3)2 Ans.

Question 16.
Solution:
m2 – 4mn + 4n2
= (m)2 -2 x m x 2n + (2n)2
= (m – 2n)2 Ans.

Question 17.
Solution:
a2b2 – 6abc + 9c2
= (ab)2 – 2 x ab x 3c + (3c)2
= (ab – 3c)2 Ans.

Question 18.
Solution:
m4 + 2m2n2 + n4
= (m2)2 + 2m2n2 + (n2)2
= (m2 + n2)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 19.
Solution:
(l + m)2 – 4lm
= l2 + m2 + 2lm – 4lm
= l2 + m2 – 2lm
= l2 – 2lm + m2
= (l – m)2
{ ∵ a2 – 2ab + b2 = (a – b)}

 

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C 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 9 Percentage Ex 9B

RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B.

Other Exercises

Tick the correct answer in each of the following:

Question 1.
Solution:
\(\\ \frac { 3 }{ 5 } \)
= \(\\ \frac { 3X20 }{ 5X20 } \)
= \(\\ \frac { 60 }{ 100 } \)
= 60% (d)

Question 2.
Solution:
0.8%
= \(\\ \frac { 0.8 }{ 100 } \)
= \(\\ \frac { 8 }{ 10X100 } \)
= \(\\ \frac { 8 }{ 1000 } \)
= 0.008 (b)

Question 3.
Solution:
6 : 5
= \(\\ \frac { 6 }{ 5 } \)
= \(\\ \frac { 6X20 }{ 5X20 } \)
= \(\\ \frac { 120 }{ 100 } \)
= 120% (c)

Question 4.
Solution:
5% of a number is 9
Number = \(\\ \frac { 9X100 }{ 5 } \)
= 180 (d)

Question 5.
Solution:
Let x% of 90 = 120
=> \(\\ \frac { x }{ 100 } \) x 120 = 90
=> x% = \(\\ \frac { 120X100 }{ 90 } \)
= \(133\frac { 1 }{ 3 } %\) (c)

Question 6.
Solution:
Let x% 10 kg = 250 g
\(\\ \frac { x }{ 100 } \) x 10 kg
= \(\\ \frac { 250 }{ 1000 } \) kg
x = \(\\ \frac { 250X100 }{ 1000X10 } \)
= \(\\ \frac { 25 }{ 10 } \)%
= 2.5% (d)

Question 7.
Solution:
40% of x = 240
=> x = \(\\ \frac { 240 }{ 40 } \) x 100
= 600 (b)

Question 8.
Solution:
?% of 400 = 60
=> x% of 400 = 60
\(\\ \frac { x }{ 100 } \) x 400 = 60
x = \(\\ \frac { 60X100 }{ 400 } \)
= 15 (c)

Question 9.
Solution:
(180% of ?)÷2 = 504
\(\left( \frac { 180 }{ 100 } \times x \right) \div 2\) = 504
\(\frac { 180 }{ 100 } x\) = 504 x 2
\(x=\frac { 504\times 2\times 100 }{ 180 }\)
= 560 (d)

Question 10.
Solution:
20% of Rs. 800
= \(\\ \frac { 20 }{ 100 } \) x 800
= Rs 160 (a)

Question 11.
Solution:
Nitin gets = 98 marks
and it is 56% of total marks
Total-marks = \(\\ \frac { 98X100 }{ 56 } \)
= 175 (c)

Question 12.
Solution:
Let a number be = 1000
Then increase = 10%
Increased number = \(\\ \frac { 100X110 }{ 100 } \) = 110
Now decrease = 1%
Decreased number = \(\\ \frac { 110X90 }{ 100 } \) = 99
Difference = 100 – 99 = 1
% decrease = 1% (b)

Question 13.
Solution:
4 hours 30 min = \(4\frac { 1 }{ 2 } %\)
= \(\\ \frac { 9 }{ 2 } \) hours
% of a day = \(\\ \frac { 9 }{ 2 } \) x \(\\ \frac { 100 }{ 24 } \)%
= \(\\ \frac { 75 }{ 4 } \)%
= \(18\frac { 3 }{ 4 } %\) (a)

Question 14.
Solution:
Let total number of examinees = 100
Passed = 65
Failed = 100 – 65 = 35
Now 35% of total examinees = 420
Total examinees = \(\\ \frac { 420X100 }{ 35 } \)
= 1200 (c)

Question 15.
Solution:
Let number = x
Then x = \(\\ \frac { xX20 }{ 100 } \) = 40
=> 100x – 200x = 4000
80x = 4000
=> x = \(\\ \frac { 4000 }{ 80 } \) = 50
Number = 50 (a)

Question 16.
Solution:
Rate of decrease = \(27\frac { 1 }{ 2 } %\)
Let number = x
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B 16.1

Question 17.
Solution:
Let x% of 20 = 0.05
\(\\ \frac { x }{ 100 } \) x 20 = 0.05
x = \(\\ \frac { 0.05X100 }{ 20 } \)
= 0.25% (c)

Question 18.
Solution:
\(\\ \frac { 1 }{ 3 } \) = 1206 = 402
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B 18.1

Question 19.
Solution:
x% of y is y% = \(\\ \frac { xy }{ 100 } \)
\(\\ \frac { y }{ 100 } \) × x
= y% of x (a)

Question 20.
Solution:
Let x% of \(\\ \frac { 2 }{ 7 } \) = \(\\ \frac { 1 }{ 35 } \)
=>\(\\ \frac { x }{ 100 } \) x \(\\ \frac { 2 }{ 7 } \) = \(\\ \frac { 1 }{ 35 } \)
x = \(\\ \frac { 1X100X7 }{ 35X2 } \)
= 10% (b)

Hope given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B 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 10A

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

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 10A.

Other Exercises

Question 1.
Solution:
(i) C.P. = Rs. 620
S.P. = Rs. 713
Gain = S.P. – C.P. = Rs. 713 – Rs. 620
= Rs. 93
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 1.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 1.2

Question 2.
Solution:
(i) C.P. = Rs. 1650
Gain% = 4%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 2.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 2.2
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 2.3

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:
(i)S.P.= Rs 1596
Gain % = 12%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 3.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 3.2

Question 4.
Solution:
C.P. of iron safe = Rs 12160
Paid for its transportation = Rs 340
Total cost = Rs 12160 + Rs 340
= Rs 12500
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 4.1

Question 5.
Solution:
C.P. of car = Rs 73500
Overhead charges = Rs 10300 + 2600
= Rs 12900
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 5.1

Question 6.
Solution:
C.P. of 20 kg @ Rs 36 per kg.
= 20 x 36 = Rs 720
C.P. of 25 kg @ Rs32 per kg.
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 6.1

Question 7.
Solution:
Ratio of the mixture = 5 : 2
Let 5 kg of coffee is mixed with 2 kg of chicory
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 7.1

Question 8.
Solution:
Let CR of 17 water bottles = Rs. 17
Then C.P. of one bottle
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 8.1

Question 9.
Solution:
Let C.P. of 12 candIes =Rs. 12
C.P of 1 candle = Re. 1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 9.1

Question 10.
Solution:
S.P. of 125 cassettes – gain = C.P. of 125 cassettes
=> S.P. of 125 cassettes – S.P. of 5 cassettes = C.P. of 125 cassettes
=> S.P. of 120 cassettes = C.P. of 125 cassettes
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 10.1

Question 11.
Solution:
S.P. of 45 lemons = C.P. of 45 lemons – loss
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 11.1

Question 12.
Solution:
CP. of 6 oranges = Rs 20
and C.P of 1 orange = Rs \(\\ \frac { 20 }{ 6 } \) = Rs \(\\ \frac { 10 }{ 3 } \)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 12.1

Question 13.
Solution:
C.P. of 12 bananas = Rs 40
and C.P of 1 banana = Rs \(\\ \frac { 40 }{ 12 } \) = Rs \(\\ \frac { 10 }{ 3 } \)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 13.1

Question 14.
Solution:
C.P. of 10 apples = Rs 75
and C.P of 1 apple = Rs \(\\ \frac { 75 }{ 10 } \) = Rs \(\\ \frac { 15 }{ 2 } \)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 14.1

Question 15.
Solution:
Let eggs purchased were = 3 x 16 = 48
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 15.1

Question 16.
Solution:
S.P. of camera = Rs 1080
C.P. of camera + gain = S.P. of camera
=> C.P. of camera + \(\\ \frac { 1 }{ 8 } \) of C.P. = S.P. of camera
= \(\\ \frac { 9 }{ 8 } \) x C.P. of camera = S.P. of camera = Rs 1080
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 16.1

Question 17.
Solution:
S.P. of a pen = Rs 54
Loss = \(\\ \frac { 1 }{ 10 } \) of her outlay
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 17.1

Question 18.
Solution:
Let C.P. of table = Rs 100
In first case, loss = 10%
S.P. = Rs 100 – 10 = Rs 90
and in second case, gain = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 18.1

Question 19.
Solution:
Let C.P. of chair = Rs 100
In first case, gain = 15%
then S.P. = Rs 100 + 15 = Rs 115
and in second case, gain = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 19.1

Question 20.
Solution:
Let the C.P. of cycle = Rs 100
In first case, gain = 10%
then S.P. = Rs 100 + 10 = Rs 110
In second case, gain = 14%
Difference between their S.P.s = Rs 114 – Rs 110 = Rs 4
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 20.1

Question 21.
Solution:
Cost price of 40 kg @ Rs 12.50 per kg.
= 40 x 12.50 = Rs 500
and cost price of 30 kg @ of Rs 14 per kg
= 30 x 14 = Rs 420
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 21.1

Question 22.
Solution:
C.P. of first bat = Rs 840
Gain% = 15%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 22.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 22.2

Question 23.
Solution:
C.P. of first jean = Rs 1450
Gain% = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 23.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 23.2

Question 24.
Solution:
Total quantity of rice = 200 kg.
C.P. of 200 kg @ Rs 25 per kg.
= Rs 200 x 25 = Rs 5000
Gain% on total = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 24.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 24.2

Question 25.
Solution:
Let C.P. = Rs 100
then S.P. = \(\\ \frac { 6 }{ 5 } \) of Rs 100
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 25.1

Question 26.
Solution:
Let C.P. of flower vase = Rs 100
then S.P. \(\\ \frac { 5 }{ 6 } \) of C.P. = \(\\ \frac { 5 }{ 6 } \) of Rs 100
= Rs \(\\ \frac { 500 }{ 6 } \)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 26.1

Question 27.
Solution:
S.P. of a bouquet = Rs 322
Gain = 15%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 27.1

Question 28.
Solution:
In first case,
S.P. of an umbrella = Rs 336
Loss% = 4%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 28.1

Question 29.
Solution:
S.P. of Radio = Rs 3120
Loss% = 4%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 29.1

Question 30.
Solution:
S.P. Of first sarees = Rs 1980
Loss% = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 30.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 30.2

Question 31.
Solution:
S.P. of first fan = Rs 1140
Gain% = 14%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 31.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 31.2

Question 32.
Solution:
C.P. for Manoj = Rs 3990
or S.P. for Arun = Rs 3990
loss% = 5%
C.P for Arun
\(=\frac { S.P\times 100 }{ 100-losspercent} =\frac { 3990\times 100 }{ 100-5 } \)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 32.1

Question 33.
Solution:
C.P. of plot of land = Rs 480000
C.P. of \(\\ \frac { 2 }{ 5 } \)th part = Rs 480000 x \(\\ \frac { 2 }{ 5 } \)
= Rs 192000
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 33.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 33.2

Question 34.
Solution:
C.P. of sugar = Rs 4500
C.P. of \(\\ \frac { 1 }{ 3 } \) of sugar
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 34.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A 34.2

 

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10A 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 8 Linear Equations Ex 8A

RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A.

Other Exercises

Solve :

Question 1.
Solution:
8x + 3 = 27 + 2x
=> 8x – 2x
=> 27 – 3
=> 6x = 24
=> x = \(\\ \frac { 24 }{ 6 } \) = 4
x = 4

Question 2.
Solution:
5x + 7 = 2x – 8
=> 5x – 2x = – 8 – 7
=> 3x = – 15
=> x = \(\\ \frac { -15 }{ 3 } \) = – 5
x = – 5

Question 3.
Solution:
2z – 1 = 14 – z
=> 2z + z = 14 + 1
=> 3z = 15
=> z = \(\\ \frac { 15 }{ 3 } \) = 5
z = 5

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 4.
Solution:
9x + 5 = 4(x – 2) +8
=> 9x + 5 = 4x – 8 + 8
=> 9x – 4x = – 8 + 8 – 5
=> 5x = – 5
=> x = \(\\ \frac { -5 }{ 5 } \) = – 1
x = – 1

Question 5.
Solution:
\(\\ \frac { 7y }{ 5 } \) = y – 4
Multiplying both sides by 5,
\(\\ \frac { 7y }{ 5 } \) x 5 = 5(y – 4)
=> 5 (y-4)
=> 7y = 5y – 20
=> 7y – 5y = – 20
=> 2y = – 20
=> y = \(\\ \frac { -20 }{ 2 } \) = – 10
Hence y = – 10 Ans.

Question 6.
Solution:
3x + \(\\ \frac { 2 }{ 3 } \) = 2x + 1
=> 3x – 2x = 1 – \(\\ \frac { 2 }{ 3 } \)
=> x = \(\\ \frac { 3-2 }{ 3 } \) = \(\\ \frac { 1 }{ 3 } \)
Hence x = \(\\ \frac { 1 }{ 3 } \) Ans.

Question 7.
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0
=> 15y – 2y + 5y = 60 – 18 – 30
=> 18y = 12
=> y = \(\\ \frac { 12 }{ 18 } \) = \(\\ \frac { 2 }{ 3 } \)
=> y = \(\\ \frac { 2 }{ 3 } \)

Question 8.
Solution:
3(5x – 7) – 2(9x – 11) = 4(8x – 13) – 17
=> 15x – 21 – 18x + 22 = 32x – 52 – 17
=> 15x – 18x – 32x = – 52 – 17 + 21 – 22
=> 15x – 50x = – 70
=> – 35x = – 70
=> x = \(\\ \frac { -70 }{ -35 } \) = 2
x = 2

Question 9.
Solution:
\(\\ \frac { x-5 }{ 2 } \) – \(\\ \frac { x-3 }{ 5 } \) = \(\\ \frac { 1 }{ 2 } \)
Multiplying each term by 10, the L.C.M. of 2 and 5
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 9.1

Question 10.
Solution:
\(\\ \frac { 3t-2 }{ 4 } \) – \(\\ \frac { 2t+3 }{ 3 } \) = \(\\ \frac { 2 }{ 3 } \) – t
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 10.1

Question 11.
Solution:
\(\frac { 2x+7 }{ 5 } +\frac { 3x+11 }{ 2 } =\frac { 2x+8 }{ 3 } -5\)
Multiplying by 30, the L.C.M. of 5, 2 and 3.
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 11.1

Question 12.
Solution:
\(\frac { 5x-4 }{ 6 } =4x+1-\frac { 3x+10 }{ 2 } \)
Multiplying by 6, the L.C.M. of 6 and 2
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 12.1

Question 13.
Solution:
\(5x-\frac { 1 }{ 3 } \left( x+1 \right) =6\left( x+\frac { 1 }{ 30 } \right) \)
=> 5x – \(\\ \frac { x+1 }{ 3 } \) = 6x + \(\\ \frac { 1 }{ 5 } \)
Multiplying by 15, the L.C.M. of 3 and 5
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 13.1

Question 14.
Solution:
\(4-\frac { 2\left( z-4 \right) }{ 3 } =\frac { 1 }{ 2 } \left( 2z+5 \right)\)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 14.1

Question 15.
Solution:
\(\frac { 3\left( y-5 \right) }{ 4 } -4y=3-\frac { \left( y-3 \right) }{ 2 } \)
Multiplying by 4, the L.C.M. of 4 and 2
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 15.1

Question 16.
Solution:
\(\\ \frac { 8x-3 }{ 3x } \) = \(\\ \frac { 2 }{ 1 } \)
By cross multiplication,
8x – 3 = 6x
=> 8x – 6x = 3
=> 2x = 3
=> x = \(\\ \frac { 3 }{ 2 } \)
x = \(\\ \frac { 3 }{ 2 } \)

Question 17.
Solution:
\(\\ \frac { 9x }{ 7-6x } \) = \(\\ \frac { 15 }{ 1 } \)
By cross multiplication,
9x = 105 – 90x
=> 9x + 90x = 105
=> 99x = 105
=> x = \(\\ \frac { 105 }{ 99 } \) = \(\\ \frac { 35 }{ 33 } \)
x = \(\\ \frac { 35 }{ 33 } \)

Question 18.
Solution:
\(\\ \frac { 3x }{ 5x+2 } \) = \(\\ \frac { -4 }{ 1 } \)
By cross multiplication,
3x × 1 = – 4×(5x + 2)
=> 3x = – 20x – 8
=> 3x + 20x = – 8
=> 23x = – 8
=> x = \(\\ \frac { -8 }{ 23 } \)
Hence x = \(\\ \frac { -8 }{ 23 } \)

Question 19.
Solution:
\(\\ \frac { 6y-5 }{ 2y } \) = \(\\ \frac { 7 }{ 9 } \)
By cross multiplication,
9(6y – 5) = 7 × 2y
=> 54y – 45 = 14y
=> 54y – 14y = 45
=> 40y = 45
=> y = \(\\ \frac { 45 }{ 40 } \) = \(\\ \frac { 9 }{ 8 } \)
Hence y = \(\\ \frac { 9 }{ 8 } \) Ans.

Question 20.
Solution:
\(\\ \frac { 2-9z }{ 17-4z } \) = \(\\ \frac { 4 }{ 5 } \)
By cross multiplication,
5 (2 – 9z) = 4(17 – 4z)
=> 10 – 45z = 68 – 16z
=> – 45z + 16z = 68 – 10
=> – 29 = 58
=> z = \(\\ \frac { 58 }{ -29 } \) = – 2
Hence z = – 2 Ans.

Question 21.
Solution:
\(\\ \frac { 4x+7}{ 9-3x } \) = \(\\ \frac { 1 }{ 4 } \)
By cross multiplication,
4(4x + 7) = 1 (9 – 3x)
=> 16x + 28 = 9 – 3x
=> 16x + 3x = 9 -28
=> 19x = – 19
=> x = \(\\ \frac { -19 }{ 19 } \) = – 1
Hence x = – 1 Ans.

Question 22.
Solution:
\(\\ \frac { 7y+4}{ y+2 } \) = \(\\ \frac { -4 }{ 3 } \)
By cross multiplication,
3 (7y + 4) = – 4 (y + 2)
=> 21y + 12 = – 4y – 8
=> 21y + 4y = – 8 – 12
=> 25y = – 20
=> y = \(\\ \frac { -20 }{ 25 } \) = \(\\ \frac { -4 }{ 5 } \)
y = \(\\ \frac { -4 }{ 5 } \)

Question 23.
Solution:
\(\\ \frac { 15(2-y)-5(y+6) }{ 1-3y } \) = \(\\ \frac { 10 }{ 1 } \)
By cross multiplication,
15 (2 – y) – 5(y + 6) = 10 (1 – 3y)
=> 30 – 15y – 5y – 30 = 10 – 30y
=> – 15y – 5y + 30y = 10 – 30 + 30
=> 30y – 20y = 10
=> 10y = 10
y = \(\\ \frac { 10 }{ 10 } \) = 1
Hence y = 1 Ans.

Question 24.
Solution:
\(\\ \frac { 2x-(7-5x) }{ 9x-(3x+4x) } \) = \(\\ \frac { 7 }{ 6 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 24.1

Question 25.
Solution:
\(m-\frac { \left( m-1 \right) }{ 2 } =1-\frac { \left( m-2 \right) }{ 3 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 25.1

Question 26.
Solution:
\(\\ \frac { 3x+5 }{ 4x+2 } \) = \(\\ \frac { 3x+4 }{ 4x+7 } \)
By cross multiplication,
(3x + 5)(4x + 7) = (3x + 4)(4x + 2)
=> 12x² + 21x + 20x + 35
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 26.1

Question 27.
Solution:
\(\\ \frac { 9x-7 }{ 3x+5 } \) = \(\\ \frac { 3x-4 }{ x+6 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 27.1

Question 28.
Solution:
\(\\ \frac { 2-7x }{ 1-5x } \) = \(\\ \frac { 3+7x }{ 4+5x } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 28.1

 

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A are helpful to complete your math homework.

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

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

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 10B.

Other Exercises

Question 1.
Solution:
Marked price of cooler = Rs 4650
Rate of discount = 18%
Selling price
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 1.1

Question 2.
Solution:
Marked price = Rs 960
Selling price = Rs 816
Total Discount = M.P. – S.P.
= Rs 960 – Rs 816
= Rs 144
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 2.1

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:
S.P. of shirt = Rs 1092
Discount = Rs 208
M.P. of shirt = S.P. + discount
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 3.1

Question 4.
Solution:
S.P. of toy = Rs 216.20
Discount = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 4.4

Question 5.
Solution:
S.P. of tea set = Rs 528
Rate of discount = 12%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 5.1

Question 6.
Solution:
Let C.P. of goods = Rs 100
Marked price = Rs 100 + 35
= Rs 135
Rate of discount = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 6.1

Question 7.
Solution:
Let C.P. of phone = Rs 100
.’. Marked price = Rs 100 + 40
= Rs 140
Rate of discount = 30%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 7.1

Question 8.
Solution:
C.P. of fan = Rs. 1080
Gain = 25%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 8.1

Question 9.
Solution:
C.P of refrigerator = Rs. 11515
and gain % = 20%
S.P. of refrigerator
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 9.1

Question 10.
Solution:
C.P. of ring = Rs. 1190
Gain = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 10.1

Question 11.
Solution:
Let Marked price = Rs. 100
Discount = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 11.1

Question 12.
Solution:
Let C.P. = Rs. 100
Gain = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 12.1

Question 13.
Solution:
Marked price of TV = Rs. 18500
Series of two successive discounts = 20% and 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 13.1

Question 14.
Solution:
Let M.P. = Rs. 100
First discount = 20%
and second discount = 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 14.1

 

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10B 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.