## 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

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

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

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

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

= Rs 79507

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

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

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

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## 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:
C.P. of toy Rs. = 75
S.P. = Rs. 100

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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## 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%

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.

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## 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

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

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.

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## 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

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

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.

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## 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

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.

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## 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:

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.

## 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 – $$\\ \frac { 7 }{ 2 }$$

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

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.

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## 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.

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

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

Question 14.
Solution:
Let numerator of a fraction = x
The denominator = 2x – 2

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
=> $$2\left( x+\frac { 2 }{ 3 } x \right) =180$$

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

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

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)

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

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

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## 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

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

Question 10.
Solution:
$$\\ \frac { 3t-2 }{ 4 }$$ – $$\\ \frac { 2t+3 }{ 3 }$$ = $$\\ \frac { 2 }{ 3 }$$ – t

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.

Question 12.
Solution:
$$\frac { 5x-4 }{ 6 } =4x+1-\frac { 3x+10 }{ 2 }$$
Multiplying by 6, the L.C.M. of 6 and 2

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

Question 14.
Solution:
$$4-\frac { 2\left( z-4 \right) }{ 3 } =\frac { 1 }{ 2 } \left( 2z+5 \right)$$

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

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:
$$\\ \frac { 2x-(7-5x) }{ 9x-(3x+4x) }$$ = $$\\ \frac { 7 }{ 6 }$$

Question 25.
Solution:
$$m-\frac { \left( m-1 \right) }{ 2 } =1-\frac { \left( m-2 \right) }{ 3 }$$

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

Question 27.
Solution:
$$\\ \frac { 9x-7 }{ 3x+5 }$$ = $$\\ \frac { 3x-4 }{ x+6 }$$

Question 28.
Solution:
$$\\ \frac { 2-7x }{ 1-5x }$$ = $$\\ \frac { 3+7x }{ 4+5x }$$

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

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.

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