Class 8 Maths RS Aggarwal Solutions

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

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 10C.

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:
List price of refrigerator = Rs. 14650
Sales tax = 6%

Question 2.
Solution:
(i) Lost of tie = Rs. 250
ST = 6%

Question 3.
Solution:
Price of watch including VAT = Rs. 1980
Rate of VAT = 10%

Question 4.
Solution:
Price of shirt including VAT = Rs. 133750
Rate of VAT = 7%
∴ Original price of the shirt

Question 5.
Solution:
Sale price of 10 g gold including VAT = Rs. 15756
Rate of VAT = 1%

Question 6.
Solution:
Sale price of computer including VAT = Rs. 37960
Rate of VAT = 4%
∴ Original price of computer

Question 7.
Solution:
Sale price of car parts including VAT = Rs. 20776
Rate of VAT = 12%
∴ Original price of car parts

Question 8.
Solution:
Sale price of TV set including VAT = Rs. 27000
Rate of VAT = 8%

Question 9.
Solution:
Sale price of shoes including VAT = Rs. 882
Original price = Rs 840

Question 10.
Solution:
Sale price of VCR including VAT = Rs. 19980
Original price = Rs. 18500
∴Amount of VAT

Question 11.
Solution:
Sale price of car including VAT = Rs. 382500
Basic price of the car = Rs. 340000
Amount of VAT = Rs. 382500 – 340000
= Rs. 42500

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

• 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:
Marked price of cooler = Rs 4650
Rate of discount = 18%
Selling price

Question 2.
Solution:
Marked price = Rs 960
Selling price = Rs 816
Total Discount = M.P. – S.P.
= Rs 960 – Rs 816
= Rs 144

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

Question 4.
Solution:
S.P. of toy = Rs 216.20
Discount = 8%

Question 5.
Solution:
S.P. of tea set = Rs 528
Rate of discount = 12%

Question 6.
Solution:
Let C.P. of goods = Rs 100
Marked price = Rs 100 + 35
= Rs 135
Rate of discount = 20%

Question 7.
Solution:
Let C.P. of phone = Rs 100
.’. Marked price = Rs 100 + 40
= Rs 140
Rate of discount = 30%

Question 8.
Solution:
C.P. of fan = Rs. 1080
Gain = 25%

Question 9.
Solution:
C.P of refrigerator = Rs. 11515
and gain % = 20%
S.P. of refrigerator

Question 10.
Solution:
C.P. of ring = Rs. 1190
Gain = 20%

Question 11.
Solution:
Let Marked price = Rs. 100
Discount = 10%

Question 12.
Solution:
Let C.P. = Rs. 100
Gain = 8%

Question 13.
Solution:
Marked price of TV = Rs. 18500
Series of two successive discounts = 20% and 5%

Question 14.
Solution:
Let M.P. = Rs. 100
First discount = 20%
and second discount = 5%

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.

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

• 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:
(i) C.P. = Rs. 620
S.P. = Rs. 713
Gain = S.P. – C.P. = Rs. 713 – Rs. 620
= Rs. 93

Question 2.
Solution:
(i) C.P. = Rs. 1650
Gain% = 4%

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%

Question 4.
Solution:
C.P. of iron safe = Rs 12160
Paid for its transportation = Rs 340
Total cost = Rs 12160 + Rs 340
= Rs 12500

Question 5.
Solution:
C.P. of car = Rs 73500
Overhead charges = Rs 10300 + 2600
= Rs 12900

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.

Question 7.
Solution:
Ratio of the mixture = 5 : 2
Let 5 kg of coffee is mixed with 2 kg of chicory

Question 8.
Solution:
Let CR of 17 water bottles = Rs. 17
Then C.P. of one bottle

Question 9.
Solution:
Let C.P. of 12 candIes =Rs. 12
C.P of 1 candle = Re. 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

Question 11.
Solution:
S.P. of 45 lemons = C.P. of 45 lemons – loss

Question 12.
Solution:
CP. of 6 oranges = Rs 20
and C.P of 1 orange = Rs \(\\ \frac { 20 }{ 6 } \) = Rs \(\\ \frac { 10 }{ 3 } \)

Question 13.
Solution:
C.P. of 12 bananas = Rs 40
and C.P of 1 banana = Rs \(\\ \frac { 40 }{ 12 } \) = Rs \(\\ \frac { 10 }{ 3 } \)

Question 14.
Solution:
C.P. of 10 apples = Rs 75
and C.P of 1 apple = Rs \(\\ \frac { 75 }{ 10 } \) = Rs \(\\ \frac { 15 }{ 2 } \)

Question 15.
Solution:
Let eggs purchased were = 3 x 16 = 48

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

Question 17.
Solution:
S.P. of a pen = Rs 54
Loss = \(\\ \frac { 1 }{ 10 } \) of her outlay

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%

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%

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

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

Question 22.
Solution:
C.P. of first bat = Rs 840
Gain% = 15%

Question 23.
Solution:
C.P. of first jean = Rs 1450
Gain% = 8%

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%

Question 25.
Solution:
Let C.P. = Rs 100
then S.P. = \(\\ \frac { 6 }{ 5 } \) of Rs 100

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 \(\\ \frac { 500 }{ 6 } \)

Question 27.
Solution:
S.P. of a bouquet = Rs 322
Gain = 15%

Question 28.
Solution:
In first case,
S.P. of an umbrella = Rs 336
Loss% = 4%

Question 29.
Solution:
S.P. of Radio = Rs 3120
Loss% = 4%

Question 30.
Solution:
S.P. Of first sarees = Rs 1980
Loss% = 10%

Question 31.
Solution:
S.P. of first fan = Rs 1140
Gain% = 14%

Question 32.
Solution:
C.P. for Manoj = Rs 3990
or S.P. for Arun = Rs 3990
loss% = 5%
C.P for Arun
\(=\frac { S.P\times 100 }{ 100-losspercent} =\frac { 3990\times 100 }{ 100-5 } \)

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

Question 34.
Solution:
C.P. of sugar = Rs 4500
C.P. of \(\\ \frac { 1 }{ 3 } \) of sugar

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 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

• RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9A
• RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B

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

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

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 9 Percentage Ex 9A

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

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9A
• RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B

Question 1.
Solution:
(i) 48% = \(\\ \frac { 48 }{ 100 } \) = \(\\ \frac { 12 }{ 25 } \)
(ii) 220% = \(\\ \frac { 220 }{ 100 } \) = \(\\ \frac { 11 }{ 5 } \)
(iii) 2.5% = \(\\ \frac { 2.5 }{ 100 } \) = \(\\ \frac { 25 }{ 10X100 } \) = \(\\ \frac { 1 }{ 40 } \)

Question 2.
Solution:
(i) 6% = \(\\ \frac { 6 }{ 100 } \) = 0.06
(ii) 72% = \(\\ \frac { 72 }{ 100 } \) = 0.72
(iii) 125% = \(\\ \frac { 125 }{ 100 } \) = 1.25

Question 3.
Solution:

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:
Ratio is 4:5 or \(\\ \frac { 4 }{ 5 } \)
\(\\ \frac { 4 }{ 5 } \)
= \(\\ \frac { 4X20 }{ 5X20 } \)
= \(\\ \frac { 80 }{ 100 } \)
= 80%

Question 5.
Solution:
125% = \(\\ \frac { 125 }{ 100 } \)
= \(\\ \frac { 5 }{ 4 } \) (dividing by 25)
Ratio = 5:4

Question 6.
Solution:

We see that 15% or \(\\ \frac { 3 }{ 20 } \) is the largest

Question 7.
Solution:
(i) Let x% of 150
= 96

Question 8.
Solution:
\(6\frac { 2 }{ 3 } %\) of Rs 3600
= Rs \(\\ \frac { 9 }{ 2 } \) x \(\\ \frac { 3600 }{ 100 } \)
= Rs 162

Question 9.
Solution:
16% of a number = 72
Number = \(\\ \frac { 72 }{ 16 }\)%
= \(\\ \frac { 72X100 }{ 16 } \)
= 450

Question 10.
Solution:
Let monthly income = Rs x
Savings = 18%

Question 11.
Solution:
Let total games = x
Then game which a team wins = 73
and it is 35% of total games
35% of x = 7
=> x = \(\\ \frac { 7X100 }{ 35 } \) = 20
Number of total games = 20

Question 12.
Solution:
Let salary = Rs x
Increment = 20%
Total salary = x + 20% of x

Question 13.
Solution:
No. of days, Sonal attended = 204 days
and her attendance = 85% of total days
85% of total days = 204
Total number of days = \(\\ \frac { 204X100 }{ 85 } \)
= 240 days

Question 14.
Solution:
Let B’s income = Rs. 100
Then A’s income = 20% less than B’s
= 100 – 20 = Rs. 80
Difference = 100 – 80 = 20
B’s more income that A’s = \(\\ \frac { 20 }{ 80 } \) x 100
= 25%

Question 15.
Solution:
Increase in price of petrol = 10%
Let first price = Rs. 100 p.l.
Increased price = 100 + 10 = Rs. 110
Now reduction in consumption = 110 – 100= 10
Percentage reduced consumption
= \(\\ \frac { 10X100 }{ 110 } \)
= \(\\ \frac { 100 }{ 11 } \)
= \(9\frac { 1 }{ 11 } %\)

Question 16.
Solution:
Present population of a town = 54000
Rate of increase = 8% annually
Population a year ago = \(\\ \frac { 54000X100 }{ 100+8 } \)
= \(\\ \frac { 54000X100 }{ 108 } \)
= 50000

Question 17.
Solution:
Depreciation in the value of machine = 20%
Present value = Rs. 160000
Then value of machine one year ago
= \(\\ \frac { 160000X100 }{ 100-20 } \)
= \(\\ \frac { 160000X100 }{ 80 } \)
= Rs 200000

Question 18.
Solution:
In an alloy,
Copper = 40%
Nickel = 32%
Zinc = 100 – (40% + 32%)
= 100 – 72 = 28%
Then mass of zinc in one kg of alloy
= 1 kg X 28%
= \(\\ \frac { 1000X28 }{ 100 } \) g
= 280 gm

Question 19.
Solution:
In balance diet,
Protein = 12%
Fats = 25%
Carbohydrates = 63%
Total number of dories = 2600

Question 20.
Solution:
In gunpowder,
Nitre = 75%
Sulphur = 10%
(i) Amount of gunpowder if nitre is 9 kg
= \(\\ \frac { 100 }{ 75 } \) x 9
= 12 kg
(ii) Amount of gunpowder if sulphur is 2.5kg 100
= \(\\ \frac { 100 }{ 10 } \) x 2.5
= 25 kg

Question 21.
Solution:
Let C get = Rs. x

Question 22.
Solution:
24-carat gold is 100% pure
22 parts out of 24 part in 22-carat gold

Question 23.
Solution:
Let present salary = Rs. 100
Increase = 25%
Increased salary = Rs. 100 + 25
= Rs. 125
To receive the original salary, amount to be decreased = Rs. 125 – 100 = Rs. 25
∴ % decrease = \(\\ \frac { 25X100 }{ 125 } \) = 20%

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