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

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B.

Other Exercises

By using the formula, find the amount and compound interest on :

Question 1.
Solution:
Principal (P) = Rs. 6000
Rate (R) = 9% p.a.
Period (n) = 2 years

Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 11% p.a.
Period (n) = 2 years

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

Question 3.
Solution:
Principal (P) = Rs. 31250
Rate (R) = 8% p.a.
Period (n) = 3 years

Question 4.
Solution:
Principal (P) = Rs. 10240
Rate (R) = $$12\frac { 1 }{ 2 }$$% = $$\\ \frac { 25 }{ 2 }$$% p.a.
Period (n) = 3 years

Question 5.
Solution:
Principal (P) = Rs. 62500
Rate (R) = 12% p.a.
Period (n) = 2 years 6 months

Question 6.
Solution:
Principal (P) = Rs. 9000
Rate (R) = 10% p.a.
Period (n) = 2 years 4 months

Question 7.
Solution:
Principal (P) = Rs. 8000
Period (n) = 2 years

Question 8.
Solution:
Principal (p) = Rs. 1, 25,000
Rate of interest (r) = 8% p.a.
Period (n) = 3 years

Question 9.
Solution:
Price of a buffalo (P) = Rs. 11000
Rate of interest (R) = 10% p.a.
Period (n) = 3 years
Price of buffalo at present

Question 10.
Solution:
Amount of loan taken (P)
= Rs. 18000

Question 11.
Solution:
Amount borrowed from Bank (P) = Rs. 24000
Rate (R) = 10% p.a.
Period (n) = 2 years 3 months

Question 12.
Solution:
In case of Abhay,
Principal (p) = Rs. 16000

Question 13.
Solution:
Simple interest (S.I.) = Rs. 2400
Rate (R) = 8% p.a.
Period (T) = 2 years

Question 14.
Solution:
Difference between C.I. and S.I.
= Rs. 90
Rate (R) = 6% p.a.
Period (n) = 2 years
Let principal (P) = Rs. 100

Question 15.
Solution:
Let sum (p) = Rs. 100
Rate (r) 10% p.a.
Period (t) = 3 years.

Question 16.
Solution:
Amount (A) = Rs. 10240
Rate (r) = $$6\frac { 2 }{ 3 }$$% = $$\\ \frac { 20 }{ 3 }$$% p.a.
Period (n) = 2 years
Let sum = P, then

Question 17.
Solution:
Amount (A) = Rs. 21296
Rate (r) = 10% p.a.
Period (n) = 3 years.
Let P be the sum, Then

Question 18.
Solution:
Principal (P) = 4000
Amount (A) = Rs. 4410
Period (n) = 2 years
Let r be the rate per cent per annum
We know that,

Question 19.
Solution:
Principal (P) = Rs. 640
Amount (A) = Rs. 774.40
Period (n) = 2 years
Let r be the rate per cent per annum.
We know that

Question 20.
Solution:
Principal (P) = Rs. 1800
Amount (A) = Rs. 2178
Rate (r) = 10% p.a.
Let n be the number of years,
We know that

Question 21.
Solution:
Principal (P) = Rs. 6250
Amount (A) = Rs. 7290
Rate (R) = 8% p.a.
Let n be the time, then

Question 22.
Solution:
Present population (P) = 125000
Rate of increasing (R) = 2% p.a.
Period (n) = 3 years

Question 23.
Solution:
3 years ago, the population was = 50000
Rate of increase successively (r1, r2, r3) = 4%, 5% and 3% p.a.
Period (n) = 3 years
Present Population

Question 24.
Solution:
Population of a city in 2013 = 120000
Increase in next year = 6%
and decrease in the following year = 5%
Population in 2015

Question 25.
Solution:
Initially bacteria = 500000
Increase in bacteria = 2% per hour
Period (n) = 2 hours

Question 26.
Solution:
Growth of bacteria in a culture (R1) = 10% in first hour
Decrease in next hour (R2) = 10%
Increase in the third hour (R3) = 10%
Bacteria in the beginning = 20000
Bacteria after 3 hours

Question 27.
Solution:
Value of machine (P) = Rs. 625000
Rate of depreciation (R) = 8% p.a.
Period (n) = 2 years

Question 28.
Solution:
Value of scooter (P) = Rs. 56000
Rate of depreciation (R) = 10% p.a.
Period = 3 years
Value of scooter after 3 years

Question 29.
Solution:
Cost of car = Rs. 34800
Rate of depreciation (R1) = 10% p.a. for first year

Question 30.
Solution:
Rate of depreciation (R) = 10% p.a.
Period (n) = 3 years
Present value (A) = Rs. 291600
Value of machine 3 years ago

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

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## 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:
Answer = (c)
C.P. of toy Rs. = 75
S.P. = Rs. 100

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

Question 3.
Solution:
Answer = (b)
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 7 Factorisation Ex 7C

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

Other Exercises

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C are helpful to complete your math homework.

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

## RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B

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.

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

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

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 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

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 7 Factorisation Ex 7D

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

Other Exercises

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

Question 2.
Solution:
y2 + 10y + 24
= y2 + 6y + 4y + 24
{24 = 6 x 4, 10 = 6 + 4}
= y (y + 6) + 4 (y + 6)
= (y + 6) (y + 4)

Question 3.
Solution:
z2 + 12x + 27
{27 = 9 x 3, 12 = 9 + 3}
= z2 + 9z + 3z + 27
= z (z + 9) + 3 (z + 9)
= (z + 9) (z + 3)

Question 4.
Solution:
p2 + 6p + 8
= p2 + 4p + 2p + 8
{ 8 = 4 x 2, 6 = 4 + 2}
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

Question 5.
Solution:
x2 + 15x + 56
= x2 + 8x + 7x + 56
{56 = 8 x 7, 15 = 8 x 7}
= x(x + 8) + 7(x + 8)
= (x + 8) (x + 7) Ans.

Question 6.
Solution:
y2 + 19y + 60
= y2 + 15y + 4y + 60
{60 = 15 x 4, 19 = 15 + 4}
= y (y + 15) + 4 (y + 15)
= (y + 15) (y + 4)

Question 7.
Solution:
x2 + 13x + 40
= x2 + 5x + 8x + 40
{40 = 5 x 8, 13 = 5 + 8}
= x(x + 5) + 8(x + 5)
= (x + 5) (x + 8) Ans.

Question 8.
Solution:
q2 – 10q + 21
= q2 – 7q – 3q + 21
{21 = ( – 7)x( – 3), – 10 = – 7 – 3}
= q {q – 7) – 3 (q – 7)
= (q – 7)(q – 3)

Question 9.
Solution:
p+ 6p – 16
= p2 + 8p – 2p – 16
{ – 16 = 8x( – 2),6 = 8 – 2}
= p (p + 8) – 2 (p + 8)
= (p + 8) (p – 2)

Question 10.
Solution:
x2 – 10x + 24
= x2 – 6x – 4x + 24
{24 = ( – 6) x ( – 4), – 10 = – 6 – 4 }
= x(x – 6) – 4(x – 6)
= (x – 6) (x – 4) Ans.

Question 11.
Solution:
x2 – 23x + 42
= x2 – 2x – 21x + 42
{42 = ( – 2) x ( – 21), – 23 = – 2 – 21}
= x(x – 2) – 21(x – 2)
= (x – 2) (x – 21) Ans.

Question 12.
Solution:
x2 – 17x + 16
= x2 – x – 16x + 16
{ 16 = ( – 1) x ( – 16), 17 = – 1 – 16}
= x (x – 1) – 16(x – 1)
= (x – 1) (x – 16) Ans.

Question 13.
Solution:
y2 – 21y + 90
= y2 – 15y – 6y + 90
{90 = ( – 15)x ( – 6), – 21 = – 15 – 6}
= y(y – 15) – 6 (y – 15)
= (y – 15) (y – 6)

Question 14.
Solution:
x2 – 22x + 117
= x2 – 13x – 9x + 117
{117 = ( – 13) x ( – 9), – 22 = – 13 – 9}
= x(x – 13) – 9(x – 13)
= (x – 9) (x – 13) Ans.

Question 15.
Solution:
x2 – 9x + 20
= x2 – 5x – 4x + 20
{ 20 = ( – 5) x ( – 4), – 9 = – 5 – 4}
= x(x – 5) – 4(x – 5)
= (x – 5) (x – 4) Ans.

Question 16.
Solution:
x2 + x – 132
= x2 + 12x – 11x – 132
{ – 132 = 12 x ( – 11), 1 = 12 – 11}
= x(x + 12) – 11(x + 12)
= (x + 12) (x – 11) Ans.

Question 17.
Solution:
x2 + 5x – 104
= x2 + 13x – 8x – 104
{ – 104 = 13 x ( – 8), 5 = 13 – 8}
= x(x + 13) – 8(x + 13)
= (x + 13) (x – 8) Ans.

Question 18.
Solution:
y2 + 7y – 144
= y2 + 16y – 9y – 144
{144 = – 16 x 9, 7 = 16 – 9}
= y(y +16) – 9(y + 16)
= (y + 16) (y – 9) Ans.

Question 19.
Solution:
z2 + 19z – 150
= z2 + 25z – 6z – 150
{ – 150 = 25 x ( – 6), 19 = 25 – 6}
= z(z + 25) – 6(z + 25)
= (z + 25) (z – 6) Ans.

Question 20.
Solution:
y2 + y – 72
= y2 + 9y – 8y – 72
{ – 72 = 9x( – 8), 1 = 9 – 8}
= y(y + 9) – 8(y + 9)
= (y + 9) (y – 8) Ans

Question 21.
Solution:
a2 + 6a – 91
= a2 + 13a – 7a – 91
{ – 91 = 13x( – 7), 6 = 13 – 7}
= a (a + 13) – 7 (a + 13)
= (a + 13) (a – 7)

Question 22.
Solution:
p2 – 4p – 11
= p2 – 11p + 7p – 77
{ – 77 = – 11 x 7, – 4 = – 11 + 7}
= p(p – 11) + 7 (p – 11)
= (p – 11)(p + 7)

Question 23.
Solution:
x2 – 7x – 30
= x2 – 10x + 3x – 30
{ – 30 = – 10 x 3, – 7 = – 10 + 3}
= x(x – 10) + 3(x – 10)
= (x – 10) (x + 3) Ans.

Question 24.
Solution:
x2 – 14 x + 3 x – 42
{ – 11 = – 14 + 3], – 42 = – 14 x 3}
= x (x – 14) + 3 (x – 14)
= (x – 14) (x + 3) Ans.

Question 25.
Solution:
x2 – 5x – 24
= x2 – 8x + 3x – 24
{ – 24 = – 8 x 3, – 5 = – 8 + 3}
= x(x – 8) + 3(x – 8)
= (x – 8) (x + 3) Ans.

Question 26.
Solution:
y2 – 6y – 135
= y2 – 15y + 9y – 135
{ – 135 = – 15 x 9, – 6 = – 15 + 9}
= y (y – 15) + 9 (y – 15)
= (y – 15) (y + 9)

Question 27.
Solution:
z2 – 12z – 45
= z2 – 15z + 3z – 45
= z (z – 15) + 3 (z – 15)
{ – 45 = – 15 x 3, – 12 = – 15 + 3}
= (z – 15)(z + 3)

Question 28.
Solution:
x2 – 4x – 12
= x2 – 6x + 2x – 12
{ – 12 = – 6 x 2, – 4 = – 6 + 2}
= x (x – 6) + 2 (x – 6)
= (x – 6) (x + 2)

Question 29.
Solution:
3x2 + 10x + 8
= 3x2 + 6x + 4x + 8
{ 3 x 8 = 24, 24 = 6 x 4,10 = 6 + 4}
= 3x(x + 2) + 4(x + 2)
= (x + 2) (3x + 4) Ans.

Question 30.
Solution:
3y2 + 14y + 8
= 3y2 + 12y + 2y + 8
{3 x 8 = 24, 24 = 12 x 2, 14 = 12 + 2}
= 3y (y + 4) + 2 (y + 4)
= (y + 4) (3y + 2)

Question 31.
Solution:
3z2 – 10z + 8
= 3z2 – 6z – 4z + 8
{ 3 x 8 = 24, 24 = ( – 6)x( – 4), – 10 = – 6 – 4}
= 3z (z – 2) – 4 (z – 2)
= (z – 2) (3z – 4)

Question 32.
Solution:
2x2 + x – 45
= 2x2 + 10x – 9x – 45
{2 x ( – 45) = – 90,- 90 = (10)x( – 9), 1 = 10 – 9}
= 2x (x – 5) + 9 (x – 5)
= (x – 5) (2x + 9)

Question 33.
Solution:
6p2 + 11p – 10
= 6x2 + 15x – 4x – 10
{6x ( – 10) = 60, – 60 = 15 x ( – 4),11 = 15 – 4}
= 3x (2x + 5) – 2 (2x + 5)
= (3x – 2) (2x + 5) Ans.

Question 34.
Solution:
2x2 – 20x + 3x – 30
{2 x ( – 30) = – 60, – 17 = – 20 + 3, – 60 = – 20 x 3}
= 2x (x – 10) + 3 (x – 10)
= (x – 10) (2x + 3) Ans.

Question 35.
Solution:
7y2 -19y – 6
= 7y2 – 21 y + 2y – 6
{7 x ( – 6) = – 42, – 42 = – 21 x 2, – 19 = – 21 + 2}
= 7y(y – 3) + 2 (y – 3)
= (y – 3) (7y + 2)

Question 36.
Solution:
28 – 31x – 5x2
= 28 – 35x + 4x – 5x2
{28 x ( – 5)= – 140, -140 = – 35 x 4, – 31 = – 35 + 4}
= 7 (4 – 5x) + x (4 – 5x)
= (4 – 5x) (7 + x)

Question 37.
Solution:
3 + 23z – 8z2
= 3 + 24z – z – 8z2
{3 x ( – 8) = – 24, – 24 = 24 x ( – 1), 23 = 24 – 1}
= 3 (1 + 8z) – z (1 + 8z)
= (1 + 8z) (3 – z)

Question 38.
Solution:
6x2 – 5x – 6
= 6x2 – 9x + 4x – 6
{ 6 x ( – 6) = – 36,- 36 = – 9 x 4, – 5 = – 9 + 4}
= 3x (2x – 3) + 2 (2x – 3)
= (2x – 3) (3x + 2)

Question 39.
Solution:
3m2 + 24m + 36
= 3 {m2 + 8m + 12}
= 3 {m2 + 6m + 2m + 12)
{12 = 6 x 2, 8 = 6 + 2}
= 3 {m (m + 6) + 2 (m + 6)}
= 3 (m + 6) (m + 2)

Question 40.
Solution:
4n2 – 8n + 3
= 4n2 – 6n – 2n + 3
{4 x 3 = 12, 12 = ( – 6)x( – 2), – 8 = – 6 – 2}
= 2n (2n – 3) – 1 (2n- 3)
= (2n – 3) (2n – 1)

Question 41.
Solution:
6x2 – 17x – 3
= 6x2 – 18x + x – 3
{6 x ( – 3)= – 18, – 18 = – 18 x 1, – 17 = – 18 + 1}
= 6x (x – 3) + 1 (x – 3)
= (x – 3) (6x + 1)

Question 42.
Solution:
7x2 – 19x – 6
7x2 – 19x – 6
= 7x2 – 21x + 2x – 6
{7 x ( – 6) = – 42, – 42 = – 21 x 2, – 19 = – 21 + 2}
= 7x (x – 3) + 2 (x – 3)
= (x – 3) (7x + 2)

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

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.

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 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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