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 10 Profit and Loss Ex 10C

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

Question 1.
Solution:
List price of refrigerator = Rs. 14650
Sales tax = 6%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 1.1

Question 2.
Solution:
(i) Lost of tie = Rs. 250
ST = 6%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 2.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 2.2

Question 3.
Solution:
Price of watch including VAT = Rs. 1980
Rate of VAT = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 3.1

Question 4.
Solution:
Price of shirt including VAT = Rs. 133750
Rate of VAT = 7%
∴ Original price of the shirt
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 4.1

Question 5.
Solution:
Sale price of 10 g gold including VAT = Rs. 15756
Rate of VAT = 1%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 5.1

Question 6.
Solution:
Sale price of computer including VAT = Rs. 37960
Rate of VAT = 4%
∴ Original price of computer
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 6.1

Question 7.
Solution:
Sale price of car parts including VAT = Rs. 20776
Rate of VAT = 12%
∴ Original price of car parts
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 7.1

Question 8.
Solution:
Sale price of TV set including VAT = Rs. 27000
Rate of VAT = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 8.1

Question 9.
Solution:
Sale price of shoes including VAT = Rs. 882
Original price = Rs 840
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 9.1

Question 10.
Solution:
Sale price of VCR including VAT = Rs. 19980
Original price = Rs. 18500
∴Amount of VAT
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 10.1

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
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 11.1

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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

 

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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 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 9 Percentage Ex 9A

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

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:
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 3.1

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 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:
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 6.1
We see that 15% or \(\\ \frac { 3 }{ 20 } \) is the largest

Question 7.
Solution:
(i) Let x% of 150
= 96
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 7.1
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 7.2

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%
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 10.1

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
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 12.1

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
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 19.1

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
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 21.1
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 21.2

Question 22.
Solution:
24-carat gold is 100% pure
22 parts out of 24 part in 22-carat gold
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 22.1

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.

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

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

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

Other Exercises

Objective Questions :
Tick the correct answer in each of the following:

Question 1.
Solution:
2x – 3 = x + 2
=> 2x – x
= 2 + 3
= 5 (c)

Question 2.
Solution:
5x + \(\\ \frac { 7 }{ 2 } \) = \(\\ \frac { 3 }{ 2 } \) x – 14
=> 5x – \(\\ \frac { 3 }{ 2 } \) x = – 14 – \(\\ \frac { 7 }{ 2 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8C 2.1

Question 3.
Solution:
z = \(\\ \frac { 4 }{ 5 } \)(z + 10)
=> 5z = 4z + 40
=> 5z – 4z = 40
=> z = 40 (a)

Question 4.
Solution:
3m = 5m – \(\\ \frac { 8 }{ 5 } \)
=> 3m – 5m = \(\\ \frac { -8 }{ 5 } \)
=> – 2m = \(\\ \frac { -8 }{ 5 } \)
=> m = \(\\ \frac { -8 }{ -5×2 } \) = \(\\ \frac { 4 }{ 5 } \) (c)

Question 5.
Solution:
5t – 3 = 3t, – 5
=> 5t – 3t = – 5 + 3
=> 2t = – 2
=> t = \(\\ \frac { -2 }{ 2 } \) = – 1 (b)

Question 6.
Solution:
2y + \(\\ \frac { 5 }{ 3 } \) = \(\\ \frac { 26 }{ 3 } \) – y
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8C 6.1

Question 7.
Solution:
\(\\ \frac { 6x+1 }{ 3 } \) +1 = \(\\ \frac { x-3 }{ 6 } \)
\(\\ \frac { 12x+2+6=x-3 }{ 6 } \)
12x – x = – 3 – 2 – 6
11x – 11
=> x = \(\\ \frac { -11 }{ 11 } \) = – 1 (b)

Question 8.
Solution:
\(\\ \frac { n }{ 2 } \) – \(\\ \frac { 3n }{ 4 } \) + \(\\ \frac { 5n }{ 6 } \) = 21
\(\\ \frac { 6n-9n+10n= 252 }{ 12 } \)
LCM of 2, 4, 6 = 12
16n – 9n = 252
=> 7n = 252
=> n = \(\\ \frac { 252 }{ 7 } \) = 36 (c)

Question 9.
Solution:
\(\\ \frac { x+1 }{ 2x+3 } \) = \(\\ \frac { 3 }{ 8 } \)
=> 8 (x + 1) = 3 (2x + 3)
(By cross multiplication)
x + 8x + 9 = 8x – 6x = 9 – 8
=> 2x = 1
=> x = \(\\ \frac { 1 }{ 2 } \)
x = \(\\ \frac { 1 }{ 2 } \) (d)

Question 10.
Solution:
\(\\ \frac { 4x+8 }{ 5x+8 } \) = \(\\ \frac { 5 }{ 6 } \)
6(4x + 8) = 5(5x + 8)
(By cross multiplication)
24x + 48 = 25x + 40
=> 24x – 25x = 40 – 48
=> – x = – 8
=> x = 8 (c)

Question 11.
Solution:
\(\\ \frac { n }{ n+15 } \) = \(\\ \frac { 4 }{ 9 } \)
9n = 4n + 60
(By cross multiplication)
9n – 4n = 60
=> 5n = 60
=> n = \(\\ \frac { 60 }{ 5 } \) = 12
n = 12 (d)

Question 12.
Solution:
3(t – 3) = 5 (2t + 1)
3t – 9 = 10t + 5
=> 3t – 10t = 5 + 9
=> – 7t = 14
=> t = \(\\ \frac { 14 }{ -7 } \) = – 2
t = – 2 (a)

Question 13.
Solution:
Let number = x
Then \(\\ \frac { 4 }{ 5 } \)x = \(\\ \frac { 3 }{ 4 } \)x + 4
=> \(\\ \frac { 16x=15x+80 }{ 20 } \)
16x – 15x = 80
=> x = 80
:. Number = 80 (c)

Question 14.
Solution:
Ages of A : B = 5 : 7
Let A’s age = 5x
Then B’s age = 7x
After 4 years
A’s age = 5x + 4
and B’s age = 7x + 4
\(\\ \frac { 5x+4 }{ 7x+5 } \) = \(\\ \frac { 3 }{ 4 } \)
=> 3(7x + 4) = 4(5x + 4)
21x + 12 = 20x + 16
=>21x – 20x = 16 – 12
x = 4
B’s age = 7x
= 7 x 4
= 28 years (b)

Question 15.
Solution:
Perimeter of an isosceles triangle = 16 cm
and base = 6 cm
Let each equal side = x cm
x + x + 6 = 16
=> 2x = 16 – 6 = 10
=> x = \(\\ \frac { 10 }{ 2 } \) = 5
Each equal side = 5 cm (b)

Question 16.
Solution:
Let first number = x
Then second number = x + 1
and third number = x+ 2
x + x + 1 + x + 2 = 51
=> 3x + 3 = 51
=> 3x = 51 – 3 = 48
=> x = \(\\ \frac { 48 }{ 3 } \) = 16
Middle number = x + 1 = 16 + 1 = 17 (b)

Question 17.
Solution:
Let first number = x
Then second number = x + 15
x + x + 15 = 95
=> 2x = 95 – 15 = 80
=> x= \(\\ \frac { 80 }{ 2 } \) = 40
=> Smaller number = 40

Question 18.
Solution:
Ratio in boys and girls in a class = 7:5
Let no. of boys = 7x
Then no. of girls = 5x
7x – 5x = 8
=> 2x = 8
x = \(\\ \frac { 8 }{ 2 } \) = 4
Total strength = 7x + 5x = 12x
= 12 x 4
= 48 (c)

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C 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 8B

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

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

Other Exercises

Question 1.
Solution:
The numbers be 8x and 3x (∵ Ratio is 8 : 3)
Sum = 143
According to the condition
8x + 3x = 143
=> 11x = 143
=> x = \(\\ \frac { 143 }{ 11 } \) = 13
First number = 8x = 8 x 13 = 104
and second number = 3x = 3 x 13 = 39 Ans.

Question 2.
Solution:
Let the number = x
According to the condition,
x – \(\frac { 2 }{ 3 } x\) = 20
=> \(\\ \frac { 3x-2x }{ 3 } \) = 20
=> \(\\ \frac { x }{ 3 } \) = 20
=> x = 20 x 3 = 60
Hence original number = 60 Ans.

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:
Let the number = x
According to the condition
\(\frac { 4 }{ 5 } x-\frac { 2 }{ 3 } x\) = 10
=> \(\\ \frac { 12x-10x }{ 15 } \) = 10
=> \(\\ \frac { 2x }{ 15 } \) = 10
=> 2x = 10 x 15
x = \(\\ \frac { 10 x 15 }{ 2 } \) = 75
Hence number = 75 Ans.

Question 4.
Solution:
Let first part = x
then second part = 24 – x (∵ Sum = 24)
According to the condition,
7x + 5 (24 – x) = 146
=> 7x + 120 – 5x = 146
=> 2x = 146 – 120 = 26
=> x = \(\\ \frac { 26 }{ 2 } \) = 13
First part = 13
and second part = 24 – 13 = 11 Ans.

Question 5.
Solution:
Let number = x
According to the condition
\(\frac { x }{ 5 } +5=\frac { x }{ 4 } -5\)
=> \(\\ \frac { x }{ 5 } \) – \(\\ \frac { x }{ 4 } \)= – 5 – 5
=> \(\\ \frac { 4x-5x }{ 20 } \) = – 10
=> \(\\ \frac { -x }{ 20 } \) = – 10 = \(\\ \frac { x }{ 20 } \) = 10
=> x = 10 × 20 = 200
Hence number = 200 Ans.

Question 6.
Solution:
Ratio between three numbers = 4:5:6
Let the largest number = 6x
smallest number = 4x
and third number = 5x
According to the condition,
6x + 4x = 5x + 55
=> 10x = 5x + 55
=> 10x – 5x = 55
=> 5x = 55
=> x = \(\\ \frac { 55 }{ 5 } \) = 11
Numbers will be 4x = 4 × 11 = 44,
5x = 5 × 31 = 55 and 6x = 6 × 11 = 66
Hence numbers are 44, 55, 66 Ans.

Question 7.
Solution:
Let the number = x
According to the condition,
4x + 10 = 5x – 5
=> 4x – 5x = – 5 – 10
=> – x = – 15
x = 15
Hence number = 15 Ans.

Question 8.
Solution:
Ratio between two numbers 3: 5
Let first number = 3x
and second number = 5x
Now according to the condition.
3x + 10 : 5x + 10 = 5 : 7
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 8.1

Question 9.
Solution:
Let first odd number = 2x + 1
second number 2x + 3
and third number = 2 + 5
According to the condition.
2x + 1 + 2x + 3 + 2x + 5 = 147
=> 6x + 9 = 147
=> 6x = 147 – 9 = 138
=> x = \(\\ \frac { 138 }{ 6 } \) = 23
Hence first odd number 2x + 1
= 23 x 2 + 1 = 46 + 1 = 47
Second number 47 + 2 = 49
and third number = 49 + 2 = 51 Ans.

Question 10.
Solution:
Let first even number = 2x
second number = 2x + 2
and third number = 2x + 4
According to the condition,
2x + 2x + 2 + 2x + 4 = 234
=> 6x + 6 = 234
=> 6x = 234 – 6
=> 6x = 228
=> x = \(\\ \frac { 228 }{ 6 } \) = 38
First even number = 2x = 38 x 2 = 76
second number = 76 + 2 = 78
and third number 78 + 2 = 80 Ans.

Question 11.
Solution:
The sum of two digits = 12
Let the ones digit of the number = x
then tens digit = 12 – x
and number = x + 10 (12 – x)
= x + 120 – 10x = 120 – 9x
Reversing the digits,
ones digit of new number = 12 – x
and tens digit = x
the number = 12 – x + 10x = 12 + 9x
According to the condition,
12 + 9x = 120 – 9x + 54
=> 9x + 9x
=> 120 + 54 – 12
=> 174 – 12
=> 18x = 162
=> x = \(\\ \frac { 162 }{ 18 } \) = 9
Original number = 120 – 9x
= 120 – 9 x 9
= 120 – 81
= 39
Hence number 39 Ans.
Check :Original number= 39
Sum of digits = 3 + 9 = 12
Now reversing its digit the new number
will be = 93
and 93 – 39 = 54 which is given.

Question 12.
Solution:
Let units digit of the number = x
then tens digit = 3x
and number = x + 10
3x = x + 30x = 31x
on reversing the digits.
units digit = 3x
and tens digit = x
then number 3x + 10x = 13x
According to the condition,
31x – 36 = 13x
=> 31x – 13x = 36
=> 18 x 36
=> x = \(\\ \frac { 36 }{ 18 } \) = 2
The original number = 31x = 31 x 2 = 62
Hence number = 62 Ans.
Check : Number = 62
tens digit = 2 x 3 = 6
On reversing the digit, the new number will be = 26
62 – 26 = 36 which is given.

Question 13.
Solution:
Let numerator of a rational number = x
Then its denominator = x + 7
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 13.1

Question 14.
Solution:
Let numerator of a fraction = x
The denominator = 2x – 2
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 14.1

Question 15.
Solution:
Let breadth of the rectangle = x cm
then length = x + 7
Area = l × b = (x + 7) × x
In second case,
Length of the new rectangle = x + 7 – 4
= x + 3 cm
and breadth = x + 3
Area = (x + 3)(x + 3)
According to the condition,
(x + 3)(x + 3) = x(x + 7)
x2 + 3x + 3x + 9 = x2 + 7x
=> x2 + 6x – 7x – x2 = – 9
=> x = – 9
=> x = 9
Length of the original rectangle
= > x + 7 = 9 + 7 = 16 cm
and breadth = x = 9 cm. Ans.

Question 16.
Solution:
Let length of rectangle = x m
then width = \(\frac { 2 }{ 3 } x\)m
Perimeter = 2 (l + b) m
=> \(2\left( x+\frac { 2 }{ 3 } x \right) =180\)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 16.1

Question 17.
Solution:
Let the length of the base of the triangle = x cm
then altitude = \(\frac { 5 }{ 3 } x\)cm
Area = \(\\ \frac { 1 }{ 2 } \) base x altitude
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 17.1
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 17.2

Question 18.
Solution:
Let ∠A, ∠B and ∠C are the three angles of a triangle and
Let ∠A + ∠B = ∠C
But ∠A + ∠B + ∠C = 180°
∠C + ∠C = 180°
=> 2∠C = 180°
∠C = 90°
and ∠A + ∠B = 90°
Let ∠A = 4x and ∠B = 5x
4x + 5x = 90°
=> 9x = 90°
x = \(\frac { { 90 }^{ o } }{ 9 } \) = 10°
∠A = 4x = 4 x 10° = 40°
and ∠B = 5x = 5 x 10° = 50°
and ∠C = 90° Ans.

Question 19.
Solution:
Time taken downstream = 9 hour
and time taken upstream = 10 hour
Speed of stream = 1 km/h
Let the speed of a steamer in still water
= x km/h.
Distance downstream = 9(x + 1) km.
and upstream = 10 (x – 1) km.
According to the condition,
10(x – 1) = 9(x + 1)
l0x – 10 = 9x + 9
=> 10x – 9x = 9 + 10
=> x = 19
Hence speed of steamer in still water = 19 km/h
and distance = 9(x + 1)
= 9(19 + 1)
= 9 x 20km
= 180km. Ans.

Question 20.
Solution:
The distance between two stations = 300 km
Let the speed of first motorcyclists = x km/h
and speed of second motorcyclists = (x + 7)km/h
Distance covered by the first = 2x km
and distance covered by the second = 2 (x + 7) km
= 2x + 14 km
Distance uncovered by them = 300 – (2x + 2x + 14)kms.
According to the condition,
300 – (4x + 4) = 34
=> 300 – 4x – 14 = 34
=> 300 – 14 – 34 = 4x
=> 4x = 300 – 48
=> 4x = 252
=> x = \(\\ \frac { 252 }{ 4 } \) = 63
Speed of the first motorcyclists = 63km/h
and speed of second = 63 + 7 = 70 km/h
Check. Distance covered by both of them
= 2 x 63 + 2 x 70 = 126 + 140 = 266
Total distance = 300 km.
Distance between them = 300 – 266 = 34 km.
which is given.

Question 21.
Solution:
Sum of three numbers = 150
Let first number = x
then second number = \(\frac { 5 }{ 6 } x\)
and third number = \(\\ \frac { 4 }{ 5 } \) of second
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 21.1

Question 22.
Solution:
Sum of two pans = 4500
Let first part = x
then second part = 4500 – x
According to the condition,
5% of x = 10% of (4500 – x)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 22.1

Question 23.
Solution:
Let age of Rakhi = x years
then her mother’s age = 4x
After 5 years,
Rakhi’s age = x + 5
and her mother’s age = 4x + 5
According to the conditions,
4x + 5 = 3 (x + 5)
=> 4x + 5 = 3x + 15
=> 4x – 3x = 15 – 5
=> x = 10
Rakhi ‘s present age = 10 years
and her mother’s age = 4 x 10
= 40 years Ans.

Question 24.
Solution:
Let age of Monu = x year
His father’s age = x + 29
and his grandfather’s age = x + 29 + 26
= x + 55
and sum of their ages = 135 years
Now,
x + x + 29 + x + 55 = 135
=> 3x + 84 = 135
=> 3x = 135 – 84 = 51
=> x = \(\\ \frac { 51 }{ 3 } \) = 17 years
Monu’s age = 17 years
His father age = 17 + 29 = 46 years
and his grandfather’s age = 46 + 26 = 72 years

Question 25.
Solution:
Let age of grandson = x year
Then his age = 10x
But 10x = x + 54
=> 10x – x = 54
=> 9x = 54
=> x = \(\\ \frac { 54 }{ 9 } \) = 6
Grand’s son age = 6 years
and his age = 6 x 10 = 60 years

Question 26.
Solution:
Let age of elder cousin = x years
and age of younger = (x – 10) years.
15 years ago,
age of older, cousin = x – 15 years
and age of younger = x – 10 – 15
= (x – 25) years.
According to the condition,
x – 15 = 2(x – 25)
=> x – 15 = 2x – 50
=> 2x – x = 50 – 15
=> x = 35
Age of elder cousin = 35 years
and age of younger = 35 – 10
= 25 years Ans.

Question 27.
Solution:
Let number of deer = x
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 27.1

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8B 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.

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

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7E

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7E

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

Other Exercises

OBJECTIVE QUESTIONS.
Tick the correct answer in each of the following :

Question 1.
Solution:
7a2 – 63b2 = 7 (a– 9b2)
= 7 {(a)2 – (3b)2}
= 7 (a – 3b) (a + 36) (d)

Question 2.
Solution:
2x – 32x3 = 2x (1 – 16x2)
= 2x {(1)2 – (4x)2}
= 2x (1 – 4x) (1 + 4x) (d)

Question 3.
Solution:
x3 – 144x
= x (x2 – 144)
= x {(x)2 – (12)2}
= x (x – 12) (x + 12) (c)

Question 4.
Solution:
2 – 50x2 = 2 (1 – 25x2)
= 2 {(1)2 – (5x)2}
= 2 (1 – 5x) (1 + 5x) (d)

Question 5.
Solution:
a2 + bc + ab + ac
= a2 + ab + ac + bc
= a (a + b) + c (a + b)
= (a + b) (a + c)   (a)

Question 6.
Solution:
pq2 + q (p – 1) – 1
= pq2 + pq – q – 1
= pq (q + 1) – 1 (q + 1)
= (q + 1) (pq – 1)
= (pq – 1) (q + 1) (d)

Question 7.
Solution:
ab – mn + an – bm
= ab + an – bm – mn
= a (b + n) – m (b + n)
= (b + n) (a – m) (b)

Question 8.
Solution:
ab – a – b + 1
= a (b – 1) – 1 (b – 1)
= b – 1) (a – 1) (a)

Question 9.
Solution:
x2 – xz + xy – yz
= x (x – z) + y (x – z)
= (x – z) (x + y) (c)

Question 10.
Solution:
12m2 – 21
= 3 (4m2 – 9)
= 3 {(2m)2 – (3)2}
= 3 (2m – 3) (2m + 3) (c)

Question 11.
Solution:
x3 – x
= x (x2 – 1)
= x (x – 1) (x + 1) (d)

Question 12.
Solution:
1 – 2ab – (a2 + b2)
= 1 – 2ab – a2 – b2
= 1 – (a2 + b2 + 2ab)
= 1 – (a + b)2
= (1 + a + b) (1 – a – b) (c)

Question 13.
Solution:
x2 + 6x + 8
= x2 + 4x + 2x + 8
{8 = 4 x 2, 6 = 4 + 2}
= x (x + 4) + 2 (x + 4)
= (x + 4) (x + 2) (c)

Question 14.
Solution:
x2 + 4x – 21
= x2 + 7x – 3x – 21
{ – 21 = + 7 x ( – 3), 4 = 7 – 3}
= x (x + 7) – 3 (x + 7)
= (x + 7) (x – 3) (b)

Question 15.
Solution:
y2 + 2y – 3
{ – 3 = 3 x ( – 1), 2 = 3 – 1}
= y2 + 3y – y – 3
= y (y + 3) – 1 (y + 3)
= (y + 3) (y – 1) (a)

Question 16.
Solution:
40 + 3x – x2
= 40 + 8x – 5x -x2
{40 = 8 x ( – 5), 3 = 8 – 5}
= 8(5 + x) – x(5 + x)
= (5 + x)(8 – x) (c)

Question 17.
Solution:
2x2 + 5x + 3
= 2x2 + 2x + 3x + 3
{2 x 3 = 6, 6 = 2 x 3, 5 = 2 + 3}
= 2x(x + 1) + 3(x + 1)
= (x + 1)(2x + 3) (b)

Question 18.
Solution:
6a2 – 13a + 6
= 6a2 – 9a – 4a + 6
{6 x 6 = 36, 36 = ( – 9)x( – 4), – 13 = – 9 – 4}
= 3a (2a – 3) – 2 (2a – 3)
= (2a – 3) (3a – 2) (c)

Question 19.
Solution:
4z2 – 8z + 3
= 4z2 – 6z – 2z + 3
{4 x 3 = 12,12 = ( – 6)x( – 2), – 8 = – 6 – 2}
= 2z (2z – 3) – 1 (2z – 3)
= (2z – 3) (2z – 1) (a)

Question 20.
Solution:
3 + 23y – 8y2
= 3 + 24y – y – 8y2
{3 x ( – 8) = – 24, – 24 = 24 x ( – 1), 23 = 24 – 1}
= 3(1 + 8y) – y(1 + 8y)
= (1 + 8y) (3 – y) (b)

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E 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.