RS Aggarwal Maths Solutions Class 6

RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8C.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8A
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8B
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8C
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D

Question 1.
Solution:
(i) The required sum
= 3x + 7x
= (3 + 7) x
= 10x

Question 2.
Solution:
(i) Adding columnwise,
we get

Question 3.
Solution:
(i) Arranging the like terms columnwise and adding, we get :

Question 4.
Solution:
(i) We have :
2x – 5x = (2 – 5)x = – 3x
(ii) We have :
6x – y – (- xy) = 6xy + xy = 7xy
(iii) We have : 5b – 3a
(iv) We have : 9y – ( – 7x) = 9y + 7x
(v) We have : – 7x² – 10x² = ( – 7 – 10)x²
= – 17x²
(vi) We have : b² – a² – (a² – b²)
= b² – a² – a² + b²
= b² + b² – a² – a²
= (1 + 1) b² + ( – 1 – 1) a²
= 2b² – 2a²

Question 5.
Solution:
(i) Arranging the like terms columnwise, we get :

Question 6.
Solution:
(i) Rearranging and collecting the like terms, we get :

Question 7.
Solution:
We have:

Question 8.
Solution:
We have :
A = 7x² + 5xy – 9y²
B = – 4x² + xy + 5y²
C = 4y² – 3x² – 6xy

= 0+0+0 = 0
Hence the result

Question 9.
Solution:
Required expression

Question 10.
Solution:
Substituting the values of P, Q, R and S, we have :
P + Q + R + S = (a² – b² + 2ab)

Question 11.
Solution:
Required expression

Question 12.
Solution:
Required expression

Question 13.
Solution:
Required expression

Question 14.
Solution:
Required expression

Question 15.
Solution:
Sum of 5x – 4y + 6z and – 8x + y – 2z
= 5x – 4y + 6z – 8x + y – 2z
= 5x – 8x – 4y + y + 6z – 2z
= – 3x – 3y + 4z
Sum of 12x – y + 3z and – 3x + 5y – 8z
= 12x – y + 3z – 3x + 5y – 8z
= 12x – 3x – y + 5y + 3z – 8z
= 9x + 4y – 5z

Question 16.
Solution:
Required expression

Question 17.
Solution:
Required expression

Hope given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions 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 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

Question 1.
Solution:
(i) 24 to 56
= \(\\ \frac { 24 }{ 56 } \)
= \(\frac { 24\div 8 }{ 56\div 8 } \)

Question 2.
Solution:
(i) 36 : 90
= \(\\ \frac { 36 }{ 90 } \)
= \(\frac { 36\div 18 }{ 90\div 18 } \)

Question 3.
Solution:
(i) The given ratio = Rs. 6.30 : Rs. 16.80
= \(\\ \frac { Rs.6.30 }{ Rs.16.80 } \)
= \(\\ \frac { 630 }{ 1680 } \)

Question 4.
Solution:
Earning of Sahai = Rs. 16800
and of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300

Question 5.
Solution:
Rohit monthly earnings = Rs. 15300
and his savings = Rs. 1224
So, his expenditure = Rs. 15300 – 1224
= Rs. 14076
Now,

Question 6.
Solution:
Let the number of male and female workers in the mill be 5x and 3x respectively. Then,
5x = 115
=> \(\\ \frac { 5x }{ 5 } \) = \(\\ \frac { 115 }{ 5 } \)
(Dividing both sides by 5)
=> x = 23
Number of female workers in the mill
= 3x
= 3 x 23 = 69.

Question 7.
Solution:
Let the number of boys and girls in the school be 9x and 5x respectively.
According to the question,
9x + 5x = 44
=> 14x = 448
=> \(\\ \frac { 14x }{ 14 } \) = \(\\ \frac { 448 }{ 14 } \)
(Dividing both sides by 14)
=> x = 32.
Number of girls =5x
= 5 x 32
= 160

Question 8.
Solution:
Total amount = Rs. 1575
Ratio in Kamal and Madhu’s share = 7 : 2
Sum of ratios = 7 + 2 = 9

Question 9.
Solution:
Total amount = Rs. 3450
Ratio in A, B and C shares = 3 : 5 : 7
Sum of share = 3 + 5 + 7 = 15

Question 10.
Solution:
Let the numbers be 11x and 12x.
Then. 11x + 12x = 460
=> 23x = 460

Question 11.
Solution:
Length of line segment = 35 cm
Ratio = 4 : 3
Sum of ratio = 4 + 3 = 7

Question 12.
Solution:
Total bulbs produced per day = 630
Out of every 10 bulbs, defective bulb = 1
Out of every 10 bulbs, lighting bulbs = 10 – 1 = 9

Question 13.
Solution:
Price of 20 pencils = Rs. 96
(1 score = 20 pencils)
Price of 1 pencil = Rs. (96 ÷ 20)
= Rs. 4.80
Price of 12 ball pens = Rs. 50.40
(1 dozen = 12)
Price of 1 ball pen = Rs. (50.40 ÷ 12)
= Rs. 4.20.
Ratio of the price of a pencil to that of a ball pen = Rs. 4.80 : Rs. 4.20
= 480 paise : 420 paise
= 480 : 420
= 48 : 42
= 8 : 7.
Required ratio = 8 : 7.

Question 14.
Solution:
It is given that the ratio of the length of a field to its width is 5 : 3.
If the width of the field is 3 metres then length = 5 metres.
If the width of the field is 1 metres than length = \(\\ \frac { 5 }{ 3 } \) metres.
If the width of the field is 42 metres then length
= \(\\ \frac { 5 }{ 3 } \) x 42 metres
= 5 x 14 metres
= 70 metres.

Question 15.
Solution:
Ratio in income and savings of a family = 11 : 2
But Total savings = Rs. 1520
Let income = x
11 : 2 = x : 1520
=> x = \(\\ \frac { 11\times 1520 }{ 2 } \) = 11 x 760
= Rs 8360
Expenditure = total income – savings
= Rs 8360 – 1520
= Rs 6840

Question 16.
Solution:
Ratio in income and expenditure = 7 : 6
Total income = Rs. 14000
Let expenditure = x, then
7 : 6 :: 14000 : x
=>x = \(\\ \frac { 6\times 14000 }{ 7 } \) = Rs. 12000
Savings = Total income – Expenditure
= Rs. 14000 – 12000
= Rs. 2000

Question 17.
Solution:
It is given that the ratio of zinc and copper in an alloy is 7 : 9.
If the weight of zinc in the alloy is 7 kg then the weight of copper in the alloy is 9 kg.

Question 18.
Solution:
A bus covers in 2 hours = 128 km
128 It will cover in 1 hour = \(\\ \frac { 128 }{ 2 } \) = 64 km
A train cover in 3 hours = 240 km
It will cover in 1 hour = \(\\ \frac { 240 }{ 3 } \)
= 80 km
Ratio in their speeds = 64: 80
= 4 : 5
{Dividing by 16, the LCM of 64, 80}

Question 19.
Solution:
(i) (3 : 4) or (9 : 16)
LCM of 4, 16 = 16

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method 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 6 Solutions Chapter 8 Algebraic Expressions Ex 8B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8B.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8A
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8B
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8C
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D

Question 1.
Solution:
(i) Substituting a = 2 and b = 3 in the , given expression, we get :
a + b = 2 + 3 = 5
(ii) Substituting a = 2 and b = 3 in the given expression, we get :
a² + ab = (2)² + 2 x 3
= 4 + 6 = 10
(iii) Substituting a = 2 and b = 3 in the given expression, we get :
ab – a² = 2 x 3 – (2)²
= 6 – 4 = 2
(iv) Substituting a = 2 and b = 3 in the given expression, we get :
2a – 3b = 2 x 2 – 3 x 3
= 4 – 9 = – 5
(v) Substituting a = 2 and b = 3 in the given expression, we get :
5a² – 2ab = 5 x (2)² – 2 x 2 x 3
= 5 x 4 – 4 x 3
= 20 – 12 = 8
(vi) Substituting a = 2 and b = 3 in the given expression, we get :
a³ – b³ = (2)³ – (3)³ = 2 x 2 x 2 – 3 x 3 x 3
= 8 – 27 = – 19

Question 2.
Solution:
(i) Substituting x = 1, y = 2 and z = 5 in the given expression, we get :
3x – 2y + 4z = 3 x 1 – 2 x 2 + 4 x 5
= 3 – 4 + 20 = 23 – 4 = 19

Question 3.
Solution:
(i) Substituting p = – 2, q = – 1 and r = 3
in the given expression, we get :

Question 4.
Solution:
(i) The coefficient of x in 13x is 13
(ii) The coefficient of y in – 5y is – 5
(iii) The coefficient of a in 6ab is 6b
(iv) The coefficient of z in – 7xz is – 7x
(v) The coefficient of p in – 2pqr is – 2qr
(vi) The coefficient of y² in 8xy²z is 8xz
(vii) The coefficient of x³ in x³ is 1
(viii) The coefficient of x² in – x² is -1

Question 5.
Solution:
(i) The numerical coefficient of ab is 1
(ii) The numerical coefficient of – 6bc is – 6
(iii) The numerical coefficient of 7xyz is 7
(iv) The numerical coefficient of – 2x³y²z is – 2.

Question 6.
Solution:
(i) The constant term is 8
(ii) The constant term is – 9
(iii) The constant term is \(\\ \frac { 3 }{ 5 } \)
(iv) The constant term is \(– \frac { 8 }{ 3 } \)

Question 7.
Solution:
(i) The given expression contains only one term, so it is monimial.
(ii) The given expression contains only two terms, so it is binomial.
(iii) The given expression contains only one term, so it is monomial.
(iv) The given expression contains three terms, so it is trinomial.
(v) The given expression contains three terms, so it is trinomial.
(vi) The given expression contains only one term, so it is monomial.
(vii) The given expression contains four terms, so it is none of monomial, binomial and trinomial.
(viii) The given expression contains only one term so it is monomial.
(ix) The given expression contains two terms, so it is binomial.

Question 8.
Solution:
(i) The terms of the given expression 4x⁵ – 6y⁴ + 7x²y – 9 are :
4x⁵, – 6y⁴, 7x²y, – 9
(ii) The terms of the given expression 9x³ – 5z⁴ + 7x³y – xyz are :
9x³, – 5z⁴, 7x³y, – xyz.

Question 9.
Solution:
(i) We have : a², b², – 2a², c², 4a
Here like terms are a², – 2a²
(ii) We have : 3x, 4xy, – yz, \(\\ \frac { 1 }{ 2 } \) zy
Here like terms are – yz, \(\\ \frac { 1 }{ 2 } \) zy
(iii) We have : – 2xy², x²y, 5y²x, x²z
Here like terms are – 2xy², 5y²x
(iv) We have :
abc, ab²c, acb², c²ab, b²ac, a²bc, cab²
Here like terms are ab²c, acb², b²ac, cab².

Hope given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions 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 6 Solutions Chapter 8 Algebraic Expressions Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8A.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8A
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8B
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8C
• RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D

Question 1.
Solution:
(i) x + 12
(ii) y – 7
(iii) a – b
(iv) (x + y) + xy
(v) \(\\ \frac { 1 }{ 3 } \)x (a + b)
(vi) 7y + 5x
(vii) \(x+ \frac { y }{ 5 } \)
(viii) 4 – x
(ix) \(\\ \frac { x }{ y } -2\)
(x) x²
(xi) 2x + y
(xii) y² + 3x 
(xiii) x – 2y
(xiv) y³ – x³
(xv) \(\\ \frac { x }{ 8 } \times y\)

Question 2.
Solution:
Marks scored in English = 80
Marks scored in Hindi = x
∴ Total score in the two subjects = 80 + x

Question 3.
Solution:
We can write :
(i) b × b × b ×….15 times = 6¹⁵
(ii) y × y × y ×…..20 times = y²⁰
(iii) 14 × a × a × a × a × b × b × b= 14a⁴ b³
(iv) 6 × x × x × y × y = 6x²y²
(v) 3 × z × z × z × y × y × x= 3z³y²x

Question 4.
Solution:
We can write :
(i) x²y⁴ = x × x × y × y × y × y
(ii) 6y⁵ = 6 × y × y × y × y × y
(iii) 9xy²z = 9 × x × y × y × z
(iv) 10a³b³c³ = 10 × a × a × a × b × b × b × c × c × c

Hope given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions 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 6 Solutions Chapter 7 Decimals Ex 7E

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7E.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7A
• RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7B
• RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7C
• RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7D
• RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7E

Question 1.
Solution:
(c) \(\\ \frac { 7 }{ 10 } \) = 0.7

Question 2.
Solution:
(d) \(\\ \frac { 5 }{ 100 } \) = .05

Question 3.
Solution:
(b) \(\\ \frac { 9 }{ 1000 } \) = 0.009

Question 4.
Solution:
(a) \(\\ \frac { 16 }{ 1000 } \) = 0.016

Question 5.
Solution:
(c) \(\\ \frac { 134 }{ 1000 } \) = 0.134

Question 6.
Solution:
(a) \(2 \frac { 17 }{ 100 } \) = 2.17

Question 7.
Solution:
(b) \(4 \frac { 3 }{ 1000 } \) = 4.03

Question 8.
Solution:
(b) 6.25 = \(6 \frac { 25 }{ 100 } \) = \(6 \frac { 1 }{ 4 } \)

Question 9.
Solution:
(b) \(\\ \frac { 6 }{ 25 } \)
= \(\\ \frac { 6\times 4 }{ 25\times 4 } \)
= \(\\ \frac { 24 }{ 100 } \)
= 0.24

Question 10.
Solution:
(c) \(4 \frac { 7 }{ 8 } \) = \(\\ \frac { 39 }{ 8 } \) = 4.875

Question 11.
Solution:
(a) 24.8 = \(24 \frac { 8 }{ 10 } \)
= \(24 \frac { 4 }{ 5 } \)

Question 12.
Solution:
(b) \(2 \frac { 1 }{ 25 } \)
= 2 + \(\\ \frac { 1 }{ 25 } \) x \(\\ \frac { 4 }{ 4 } \)
= 2 + \(\\ \frac { 4 }{ 100 } \)
= 2.04

Question 13.
Solution:
(c) 2 + \(\\ \frac { 3 }{ 10 } \) + \(\\ \frac { 4 }{ 100 } \)
= 2 + \(\\ \frac { 30 }{ 100 } \) + \(\\ \frac { 4 }{ 100 } \)
= 2.34

Question 14.
Solution:
(b) \(2 \frac { 6 }{ 100 } \)
= 2 + 0.06
= 2.06

Question 15.
Solution:
(c) \(\\ \frac { 4 }{ 100 } \) + \(\\ \frac { 7 }{ 10000 } \)
= 0.04 + 0.0007
= 0.0407

Question 16.
Solution:
(c) 2.06
= \(\left( 2\times 1 \right) +\left( 6\times \frac { 1 }{ 100 } \right) \)
= \(2+\frac { 6 }{ 100 } \)

Question 17.
Solution:
(d) Among 2.600, 2.006, 2.660,2.080, 2.660 is the largest.

Question 18.
Solution:
(b) 2.002 < 2.020 < 2.200 < 2.222 is the correct.

Question 19.
Solution:
(a) 2.1 = 2.100 and 2.005
2.100 > 2.055
=> 2.1 > 2.055

Question 20.
Solution:
(b) 1cm = \(\\ \frac { 1 }{ 100 } \) m
= 0.01

Question 21.
Solution:
(b) 2 m 5 cm = 2.05 m

Question 22.
Solution:
(c) 2 kg 8 g = 2 + 0.008 = 2.008

Question 23.
Solution:
(b) 2 kg 56 g = 2.056 kg
(∵ 1000 g = 1 kg)

Question 24.
Solution:
(c) 2 km 35 m = 2.035 km
(∵ 1000 m = 1 km)

Question 25.
Solution:
(c) ∵ 0.4 + 0.004 + 4.4
= 4.804

Question 26.
Solution:
(a) ∵ 3.5 + 4.05 – 6.005
= 3.500 + 4.050 – 6.005
= 7.550 – 6.005
= 1.545

Question 27.
Solution:
(b) ∵6.3 – 2.8 = 3.5

Question 28.
Solution:
(c) ∵ 5.01 – 3.6 = 5.01 – 3.60
= 1.41

Question 29.
Solution:
(a) ∵ 2 – 0.7 = 2.0 – 0.7 = 1.3

Question 30.
Solution:
(a) ∵ 1.1 – 0.3
= 0.8

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals 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.