RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A

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

Question 1.
Solution:
(i) 24 to 56
= \(\\ \frac { 24 }{ 56 } \)
= \(\frac { 24\div 8 }{ 56\div 8 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q1.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q1.2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q1.3

Question 2.
Solution:
(i) 36 : 90
= \(\\ \frac { 36 }{ 90 } \)
= \(\frac { 36\div 18 }{ 90\div 18 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q2.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q2.2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q2.3
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q2.4
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q2.5

Question 3.
Solution:
(i) The given ratio = Rs. 6.30 : Rs. 16.80
= \(\\ \frac { Rs.6.30 }{ Rs.16.80 } \)
= \(\\ \frac { 630 }{ 1680 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q3.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q3.2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q3.3
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q3.4

Question 4.
Solution:
Earning of Sahai = Rs. 16800
and of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q4.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q4.2

Question 5.
Solution:
Rohit monthly earnings = Rs. 15300
and his savings = Rs. 1224
So, his expenditure = Rs. 15300 – 1224
= Rs. 14076
Now,
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q5.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q5.2

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
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q8.1

Question 9.
Solution:
Total amount = Rs. 3450
Ratio in A, B and C shares = 3 : 5 : 7
Sum of share = 3 + 5 + 7 = 15
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q9.1

Question 10.
Solution:
Let the numbers be 11x and 12x.
Then. 11x + 12x = 460
=> 23x = 460
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q10.1

Question 11.
Solution:
Length of line segment = 35 cm
Ratio = 4 : 3
Sum of ratio = 4 + 3 = 7
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q11.1

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
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q12.1

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.
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q17.1

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
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q18.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q18.2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q18.3
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q18.4

Question 20.
Solution:
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q20.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q20.2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A Q20.3

 

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A are helpful to complete your math homework.

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

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

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
RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8B Q2.1
RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8B Q2.2

Question 3.
Solution:
(i) Substituting p = – 2, q = – 1 and r = 3
in the given expression, we get :
RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8B Q3.1
RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8B Q3.2

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.

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

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

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

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

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.

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D

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

Other Exercises

Question 1.
Solution:
27.86 from 53.74
= 53.74 – 27.86
= 25.88 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q1.1

Question 2.
Solution:
64.98 from 103.87
103.87 – 64.98
= 38.89 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q2.1

Question 3.
Solution:
59.63 from 92.4
92.40 – 59.63
= 32.77 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q3.1

Question 4.
Solution:
56.8 from 204
204.0 – 56.8
= 147.2 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q4.1

Question 5.
Solution:
127.38 from 216.2
216.20 – 127.38
= 88.82 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q5.1

Question 6.
Solution:
39.875 from 70.68
70.680 – 39.875
= 30.805 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q6.1

Question 7.
Solution:
523.120 – 348.237
= 174.883 Ans
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q7.1

Question 8.
Solution:
600.000 – 458.573
= 141.427 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q8.1

Question 9.
Solution:
206.321 – 149.456
= 56.865 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q9.1

Question 10.
Solution:
3.400 – 0.612
= 2.788 Ans
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q10.1

Question 11.
Solution:
Converting them in like decimals
37.600 + 72.850 – 58.678 – 6.090
= (37.600 + 72.850) – (58.678 + 6.090)
= 110.450 – 64.768
= 45.682
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q11.1

Question 12.
Solution:
75.3 – 104.645 + 178.96 – 47.9
= 75.300 – 104.645 + 178.960 – 47.900
(Converting into like decimals)
= 75.300 + 178.960 – 104.645 – 47.900
= (75.300 + 178.960) – (104.645 + 47.900)
= 254.260 – 152.545
= 101.715 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q12.1

Question 13.
Solution:
213.4 – 56.84 – 11.87 – 16.087
= 213.400 – 56.840 – 11.870 – 16.087
(Converting into like decimals)
= 213.400 – (56.840 + 11.870 + 16.087)
= 213.400 – 84.797
= 128.603 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q13.1

Question 14.
Solution: 76.3 . 7.666 . 6.77
= 76.300 – 7.666 – 6.770
(Converting into like decimals)
= 76.300 – 14.436
= 61.864 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q14.1

Question 15.
Solution:
In order to get the required number, we have to subtract 74.5 from 91.
Required number = 91 – 74.5
= 91.0 – 74.5
= 16.5 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q15.1

Question 16.
Solution:
In order to get the required numbers, we have to subtract 0.862 from 7.3.
Required number = 7.3 – 0.862
= 7.300 – 0.862
= 6.438 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q16.1

Question 17.
Solution:
In order to get the required number, we have to subtract 23.754 from 50
Required number = 50 – 23.754
= 50.000 – 23.754
= 26.246 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q17.1

Question 18.
Solution:
In order to get the required number, we should subtract 27.84 from 84.5
Required number = 84.5 – 27.84
= 84.50 – 27.84
= 56.66 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q18.1

Question 19.
Solution:
Weight of Neelam’s bag = 6 kg 80 g
Weight of Garima bag = 5 kg 265 g
Difference in their weights = 6 kg 80 g – 5 kg 265 g
= 6.080 kg – 5.265 kg
= 0.815 kg
= 815 g
Neelam’s bag is heavier by 815 g Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q19.1

Question 20.
Solution:
Cost of a notebook = Rs. 19.75
Cost of a pencil = Rs. 3 .85
Cost of a pen = Rs. 8.35
Total cost = Rs. 19.75 + Rs. 3.85 + Rs. 8.35
= Rs. 31.95
Amount given to the bookshop = Rs. 50
Balance amount to get back = Rs. 50.00 – Rs. 31.95
= Rs. 18.05 Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q20.1

Question 21.
Solution:
Weight of fruits = 5 kg 75 g .
Weight of vegetables = 3 kg 465 kg
Total weight of both = 5 kg 75 g + 3 kg 465 g
= 5.075 kg + 3.465 kg
= 8.540 kg
Gross weight of bag with these things = 9 kg
Net weight of bag = 9.000 – 8.540
= 0.460 kg
= 460 g Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q21.1

Question 22.
Solution:
Total distance = 14 km
Distance covered by scooter = 10 km 65 m
Distance covered by bus = 3 km 75 m
Total distance covered by scooter and by bus = 10 km 65 m + 3 km 75 m
= 10.065 km + 3 075 m
= 13.140 km
Remaining distance covered by walking
= (14.000 – 13.140) km
= 0.860 km
= 860 m Ans.
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7D Q22.1

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals 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 6 Solutions Chapter 7 Decimals Ex 7C

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C

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

Other Exercises

Add the following decimals :

Question 1.
Solution:
9.6, 14.8, 37 and 5.9
Converting these decimals into like decimals and then adding 9.6 + 14.8 + 37.0 + 5.9
= 67.3 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q1.1

Question 2.
Solution:
23.7, 106.94, 68.9 and 29.5
Converting them into like decimals and then adding
23.70 + 106.94 + 68.90 + 29.50
= 229.04 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q2.1

Question 3.
Solution:
72.8, 7.68, 16.23 and 0.7
Converting them into like decimals and then adding
72.80 + 7.68 + 16.23 + 0.70
= 97.41 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q3.1

Question 4.
Solution:
18.6, 84.75, 8.345 and 9.7
Converting them into like decimals and then adding
18.600 + 84.750 + 8.345 + 9.700
= 121.395 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q4.1

Question 5.
Solution:
8.236, 16.064, 63.8 and 27.53
Converting them into like decimals and then adding
8.236 + 16.064 + 63.800 + 27.530
= 115.630 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q5.1

Question 6.
Solution:
28.9, 19.64, 123.697 and 0.354
Converting them into like decimals and then adding
28.900 + 19.640 + 123.697 + 0.354
= 172.591 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q6.1

Question 7.
Solution:
4.37, 9.638, 17.007 and 6.8
Converting them into like decimals and then adding
4. 370 + 9.638 + 17.007 + 6.800
= 37.815 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q7.1

Question 8.
Solution:
14.5, 0.038, 118.573 and 6.84
Converting them into like decimals and then adding
14.500 + 0.038 + 118.573 + 6.840
= 139.951 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q8.1

Question 9.
Solution:
Earning for the first day = 32.60 rupees
Earning for the second day = 56.80 rupees
Earning for the third day = 72 rupees
Total earning = Rs. 32.60 + Rs. 56.80 + Rs. 72
= Rs. 161.40 Ans.
Working
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q9.1

Question 10.
Solution:
Cost of almirah = Rs. 11025
Cartage = Rs. 172.50
Cost on repair = Rs. 64.80
Total cost = Rs. 11025 + Rs. 172.50 + Rs. 64.80
= Rs. 11262.30 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q10.1

Question 11.
Solution:
Distance covered by taxi = 36 km 235 m
= 36.235 km
Distance covered by Rickshaw = 4 km 85 m
= 4.085 km
and distance covered on foot
= 1 km 80 m
= 1.080 m
Total distance covered = 36.235 km + 4.085 km + 1.080 km
= 41.400 km
= 41 km 400 m Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q11.1

Question 12.
Solution:
Weight of sugar in a bag = 45 kg 80 g
= 45.080 kg
Mass (weight) of empty bag = 950 g
= 0.950 kg
Total weight of the bag with sugar = 45 kg 80 g + 950 g
= 45.080 kg + 0.950 kg
= 46.030 kg
= 46 kg 30 g Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q12.1

Question 13.
Solution:
Length of cloth for shirt = 2 m 70 cm
= 2.70 m
Length of cloth for pyjamas = 2 m 60 cm
= 2.60 m
Total length of cloth = 2.70 m + 2.60 m
= 5.30 m
= 5 m 30 cm Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q13.1

Question 14.
Solution:
Cloth of salwar = 2 m 5 cm = 2.05 m
Cloth for shirt = 3 m 35 cm = 3.35 m
Total length of cloth = 2.05 m + 3.35 m
= 5.4.0 m
= 5 m 40 cm Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q14.1

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7C are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B

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

Other Exercises

Convert each of the following into a fraction in its simplest form :

Question 1.
Solution:
.9 = \(\\ \frac { 9 }{ 10 } \)

Question 2.
Solution:
0.6
= \(\\ \frac { 6 }{ 10 } \)
= \(\frac { 6\div 2 }{ 10\div 2 }\)
= \(\\ \frac { 3 }{ 5 } \)
(Dividing by 2, the HCF of 6, 10)

Question 3.
Solution:
.08
= \(\\ \frac { 8 }{ 100 } \)
= \(\frac { 8\div 4 }{ 100\div 4 }\)
= \(\\ \frac { 2 }{ 25 } \)
(Dividing by 4, the HCF of 7, 100)

Question 4.
Solution:
0.15
= \(\\ \frac { 15 }{ 100 } \)
= \(\frac { 15\div 5 }{ 100\div 5 }\)
= \(\\ \frac { 3 }{ 20 } \)
(Dividing by 5, the HCF of 15, 100)

Question 5.
Solution:
0.48
= \(\\ \frac { 48 }{ 100 } \)
= \(\frac { 48\div 4 }{ 100\div 4 }\)
= \(\\ \frac { 12 }{ 25 } \)
(Dividing by 4, the HCF of 48, 100)

Question 6.
Solution:
0.53
= \(\\ \frac { 53 }{ 1000 } \)

Question 7.
Solution:
= \(\\ \frac { 125 }{ 1000 } \)
= \(\frac { 125\div 125 }{ 1000\div 125 }\)
= \(\\ \frac { 1 }{ 8 } \)
(Dividing by 125, the HCF of 125, 1000)

Question 8.
Solution:
.224
= \(\\ \frac { 224 }{ 1000 } \)
= \(\frac { 224\div 8 }{ 1000\div 8 }\)
= \(\\ \frac { 28 }{ 125 } \)
(Dividing by 8, the HCF of 224, 1000)

Convert each of the following as a mixed fraction

Question 9.
Solution:
6.4
= \(\\ \frac { 64 }{ 10 } \)
= \(\frac { 64\div 2 }{ 10\div 2 }\)
= \(\\ \frac { 32 }{ 6 } \)
= \(6 \frac { 2 }{ 5 } \)
(Dividing by 2, the HCF of 64, 10)

Question 10.
Solution:
16.5
= \(\\ \frac { 165 }{ 10 } \)
= \(\frac { 165\div 5 }{ 10\div 5 }\)
= \(\\ \frac { 33 }{ 2 } \)
= \(16 \frac { 1 }{ 2 } \)
(Dividing by 5, the HCF of 165, 10)

Question 11.
Solution:
8.36
= \(\\ \frac { 836 }{ 100 } \)
= \(\frac { 836\div 4 }{ 100\div 4 }\)
= \(\\ \frac { 209 }{ 25 } \)
= \(8 \frac { 9 }{ 25 } \)
(Dividing by 4, the HCF of 836, 100)

Question 12.
Solution:
4.275
= \(\\ \frac { 4275 }{ 1000 } \)
= \(\frac { 4275\div 25 }{ 1000\div 25 }\)
= \(\\ \frac { 171 }{ 40 } \)
= \(4 \frac { 11 }{ 40 } \)
(Dividing by 25 )

Question 13.
Solution:
25.06
= \(\\ \frac { 2506 }{ 100 } \)
= \(\frac { 2506\div 2 }{ 100\div 2 }\)
= \(\\ \frac { 1253 }{ 50 } \)
= \(25 \frac { 3 }{ 50 } \)
(Dividing by 2 )

Question 14.
Solution:
7.004
= \(\\ \frac { 7004 }{ 1000 } \)
= \(\frac { 7004\div 4 }{ 1000\div 4 }\)
= \(\\ \frac { 1751 }{ 250 } \)
= \(7 \frac { 1 }{ 250 } \)
(Dividing by 4)

Question 15.
Solution:
2.052
= \(\\ \frac { 2052 }{ 1000 } \)
= \(\frac { 2052\div 4 }{ 1000\div 4 }\)
= \(\\ \frac { 513 }{ 250 } \)
= \(2 \frac { 13 }{ 250 } \)
(Dividing by 4)

Question 16.
Solution:
3.108
= \(\\ \frac { 3108 }{ 1000 } \)
= \(\frac { 3108\div 4 }{ 1000\div 4 }\)
= \(\\ \frac { 777 }{ 250 } \)
= \(3 \frac { 27 }{ 250 } \)
(Dividing by 4)

Question 17.
Solution:
\(\\ \frac { 23 }{ 10 } \)
= 2.3
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q17.1

Question 18.
Solution:
\(\\ \frac { 167 }{ 100 } \)
= 1.67
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q18.1

Question 19.
Solution:
\(\\ \frac { 1589 }{ 100 } \)
= 15.89
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q19.1

Question 20.
Solution:
\(\\ \frac { 5413 }{ 1000 } \)
= 5.413
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q20.1

Question 21.
Solution:
\(\\ \frac { 21415 }{ 1000 } \)
= 21.415
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q21.1

Question 22.
Solution:
\(\\ \frac { 25 }{ 4 } \)
= 6.25
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q22.1

Question 23.
Solution:
\(3 \frac { 3 }{ 5 } \)
= \(\\ \frac { 3\times 5+3 }{ 5 } \)
= \(\\ \frac { 15+3 }{ 5 } \)
= \(\\ \frac { 18 }{ 5 } \)
= 3.6
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q23.1

Question 24.
Solution:
\(1 \frac { 4 }{ 25 } \)
= \(\\ \frac { 1\times 25+4 }{ 25 } \)
= \(\\ \frac { 25+4 }{ 25 } \)
= \(\\ \frac { 29 }{ 25 } \)
= 1.16
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q24.1

Question 25.
Solution:
\(5 \frac { 17 }{ 50 } \)
= \(\\ \frac { 5\times 50+17 }{ 50 } \)
= \(\\ \frac { 250+17 }{ 50 } \)
= \(\\ \frac { 267 }{ 50 } \)
= 5.34
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q25.1

Question 26.
Solution:
\(12 \frac { 3 }{ 8 } \)
= \(\\ \frac { 12\times 8+3 }{ 8 } \)
= \(\\ \frac { 96+3 }{ 8 } \)
= \(\\ \frac { 99 }{ 8 } \)
= 12.375
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q26.1

Question 27.
Solution:
\(2 \frac { 19 }{ 40 } \)
= \(\\ \frac { 2\times 40+19 }{ 40 } \)
= \(\\ \frac { 80+19 }{ 40 } \)
= \(\\ \frac { 99 }{ 40 } \)
= 2.475
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q27.1

Question 28.
Solution:
\(\\ \frac { 19 }{ 20 } \)
= 0.95
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q28.1

Question 29.
Solution:
\(\\ \frac { 37 }{ 50 } \)
= 0.74
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q29.1

Question 30.
Solution:
\(\\ \frac { 107 }{ 250 } \)
= 0.428
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q30.1

Question 31.
Solution:
\(\\ \frac { 3 }{ 40 } \)
= 0.075
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q31.1

Question 32.
Solution:
\(\\ \frac { 7 }{ 8 } \)
= 0.875
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q32.1

Question 33.
Solution:
(i) 8 kg 640 g in kilograms
= \(8 \frac { 640 }{ 1000 } \) kg
= 8.640kg
(ii) 9 kg 37 g in kilograms
= \(9 \frac { 37 }{ 1000 } \) kg
= 9.037 kg.
(iii) 6 kg 8 g in kilograms
= \(6 \frac { 8 }{ 1000 } \) kg
= 6.008 kg Ans.

Question 34.
Solution:
(i) 4 km 365 m in kilometres
= \(4 \frac { 365 }{ 1000 } \) km
= 4.365 km
(ii) 5 km 87 m in kilometres
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q34.1

Question 35.
Solution:
(i) 15 kg 850 g in kilograms
= \(15 \frac { 850 }{ 1000 } \) kg
= 15.850 kg
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q35.1

Question 36.
Solution:
(i) Rs. 18 and 25 paise in rupees
= \(18 \frac { 25 }{ 100 } \)
= 18.25 rupees
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q36.1

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7B are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A

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

Other Exercises

Question 1.
Solution:
(i)Fifty eight point six three = 58.63
(ii)One hundred twenty four point four two five = 124.425
(iii)Seven point seven six = 7.76
(iv)Nineteen point eight = 19.8
(v)Four hundred four point zero four four = 404.044
(vi)Point one seven three = 173
(v)Point zero one five = .015 Ans.

Question 2.
Solution:
(i) 14.83
Place value of 1 = 10,
Place value of 4 = 4,
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.1
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.2
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.3

Question 3.
Solution:
(i) 67.83 = (6 x 10) + (7 x 1)
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q3.1

Question 4.
Solution:
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q4.1

Question 5.
Solution:
(i) 7.5, 64.23, 0.074 = 7.500, 64.230, 0.074
(Here, at the most 0.074 has 3 places)
(ii) 0.6, 5.937, 2.36, 4.2 = 0.600, 5.937, 2.360, 4.200
(Here, 5.937 has at most 3 places)
(iii) 1.6, 0.07, 3.58, 2.9 = 1.60, 0.07, 3.58, 2. 90
(Here, at the most are two places)
(iv) 2.5. 0.63, 14.08, 1.637 = 2.500, 0.630. 14.080, 1.637 Ans.
(Here, at the most are three places)

Question 6.
Solution:
Making like decimals where ever it is necessary,
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q6.1

Question 7.
Solution:
First of all making them in like decimals,
(i) 5.8, 7.2, 5.69, 7.14, 5.06
=> 5.80, 7.20, 5.69, 7.14, 5.06
Arranging in ascending order,
5:06 <5.69 <5.80 <7.14 <7.20
=> 5.06 < 5.69 < 5.8 < 7.14 < 7.2 Ans.
(ii) 0.6, 6.6, 6.06, 66.6, 0.06
=>0.60, 6.60, 6.06, 66.60, 0.06
Arranging in ascending order,
0.06 < 0.60 < 6.06 < 6.60 < 66.60
=> 0.06 < 0.6 < 6.06 < 6.6 < 66.6 Ans.
(iii) 6.54, 6.45, 6.4, 6.5, 6.05
=> 6.54, 6.45, 6.4, 6.5, 6.05
Arranging in ascending order,
6. 05 < 6.40 < 6.45 < 6.50 < 6.54
=> 6.05 < 6.4 < 6.45 < 6.5 < 6.54 Ans.
(iv) 3.3,3.303, 3.033, 0.33, 3.003
=> 3.300, 3.303, 3.033, 0.330, 3.003
Arranging in descending order,
0.330 < 3.003 < 3.033 < 3.300 < 3.303
=> 0.33 < 3.003 < 3.033 < 3.3 < 3.303 Ans.

Question 8.
Solution:
Making them in like decimals and them comparing
(i) 7.3, 8.73, 73.03, 7.33, 8.073
=> 7.300, 8.730, 73.030, 7.330, 8.073
Arranging in descending order
73.030 > 8.730 > 8.073 > 7.330 > 7.300
=> 73.03 > 8.73 > 8.073 > 7.33 > 7.3 Ans.
(ii) 3.3, 3.03, 30.3, 30.03, 3.003
=> 3.300, 3.030, 30.300, 30.030, 3.003
Arranging in descending order
30.300> 30.030 >3.300 >3.030 > 3.003
=> 30.3 > 30.03 > 3.3 > 3.03 > 3.003 Ans.
(iii) 2.7, 7.2, 2.27, 2.72, 2.02, 2.007
=> 2.700, 7.200, 2.270, 2.720, 2.020, 2.007
Arranging in descending order
7. 200 > 2.720 > 2.700 > 2.270 > 2.020 > 2.007
=> 7.2 > 2.72 > 2.7 > 2.27 > 2.02 > 2.007 Ans.
(iv) 8.88, 8.088, 88.8, 88.08, ,8.008
=> 8.880, 8.088, 88.800, 88.080, 8.008
Arranging in descending order,
88.800 > 88.080 > 8.880 > 8.088 > 8.008
=> 88.8 > 88.08 > 8.88 > 8.088 > 8.008

 

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7A are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6B

RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6B

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

Other Exercises

OBJECTIVE QUESTIONS
Mark against the correct answer in each of the following :

Question 1.
Solution:
(c) 8 + 4 ÷ 2 x 5
= 8 + 4 x \(\\ \frac { 1 }{ 2 } \) x 5
= 8 + 10
= 18.

Question 2.
Solution:
(b) 54 ÷ 3 of 6 + 9 = 54 ÷ 18 + 9
= 54 x \(\\ \frac { 1 }{ 18 } \) + 9
= 3 + 9
= 12.

Question 3.
Solution:
(b) 13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – 10
= 3

Question 4.
Solution:
(a) 1001 ÷ 11 of 13
= 1001 ÷ 143
= 1001 x \(\\ \frac { 1 }{ 143 } \)
= 7.

Question 5.
Solution:
(b) 133 + 28 ÷ 7 – 8 x 2
= 133 + 4 – 16
= 137 – 16
= 121.

Question 6.
Solution:
(a) 3640 – 14 ÷ 7 x 2
= 3640 – 2 x 2
= 3640 – 4
= 3636.

Question 7.
Solution:
(b) 100 x 10 – 100 + 2000 ÷ 100
= 1000 – 100 + 20
= 920.

Question 8.
Solution:
(b) \(27-\left[ 18-\left\{ 16-\left( 5-\overline { 4-1 } \right) \right\} \right] \)
= 27 – [18 – {16 – (5 – 4 + 1)}]
= 27 – [18 – {16 – 5 + 4 – 1}]
= 27 – [18 – 16 + 5 – 4 + 1]
= 27 – 18 + 16 – 5 + 4 – 1
= 23

Question 9.
Solution:
\(32-\left[ 48\div \left\{ 36-\left( 27-\overline { 16-9 } \right) \right\} \right] \)
= 32 – [48 ÷ {36 – (27 – 16 + 9)}]
= 32 – [48 ÷ {36 – 27 + 16 – 9}]
= 32 – [48 ÷ 16]
= 32 – 3
= 29

Question 10.
Solution:
(a) 8 – [28 ÷ {34 – (36 – 18 ÷ 9 x 8)}]
\(8-\left[ 28\div \left\{ 34-\left( 36-18\times \frac { 1 }{ 9 } \times 8 \right) \right\} \right] \)
= 8 – [28 ÷ {34 – (36 – 16)}]
= 8 – [28 ÷ {34 – 20}]
= 8 – {28 ÷ 14}
= 8 – 2
= 6.

Hope given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6B are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5G.

Other Exercises

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

Question 1.
Solution:
(c) ∴ canceling the common factor 2, we get \(\\ \frac { 3 }{ 5 } \)

Question 2.
Solution:
(c) ∴ multiplying numerator and denominator by 4, we get \(\\ \frac { 8 }{ 12 } \)

Question 3.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 3.1

Question 4.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 4.1

Question 5.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 5.1

Question 6.
Solution:
(c) each of the fractions has the same denominator.

Question 7.
Solution:
(d) none of these has greater denominator than its numerator.

Question 8.
Solution:
(a) its denominator is greater than its numerator.

Question 9.
Solution:
(b) their numerators are same and 4 < 5 , \(\frac { 3 }{ 4 } >\frac { 3 }{ 5 } \)

Question 10.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 10.1

Question 11.
Solution:
(b) In \(\frac { 4 }{ 5 } ,\frac { 2 }{ 7 } ,\frac { 4 }{ 9 } ,\frac { 4 }{ 11 } \) numerator is same then the smallest denominator’s fraction is greater.

Question 12.
Solution:
(a) Denominators are same, then fraction of smallest numerator will be smallest.

Question 13.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 13.1

Question 14.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 14.1

Question 15.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 15.1

Question 16.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 16.1

Question 17.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 17.1

Question 18.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 18.1

Question 19.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 19.1

Question 20.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 20.1

Hope given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5G are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 24 Bar Graph Ex 24

RS Aggarwal Class 6 Solutions Chapter 24 Bar Graph Ex 24

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 24 Bar Graph Ex 24

Question 1.
Solution:
(i) The given bar graph shows the marks obtained by a student in each of the four subjects in an examination.
(ii) The student is poorest in Science.
(iii) The student is best in Mathematics.
(iv) He got more than 40 marks in Hindi and Mathematics.

Question 2.
Solution:
(i) The given bar graph shows the number of members in each of the 60 families of a colony.
(ii) 10 families have 3 members each.
(iii) 5 couples have no child.
(iv) A family of 4 members is most common.

Question 3.
Solution:
(i) The production was maximum in the 2nd week.
(ii) The production was minimum in the 4th week.
(iii) The average production is 720 per week.
(iv) 2400 cycles were produced in the first three weeks.

Question 4.
Solution:
(i) The given bar graph shows the different modes of transport to a school used by 51 students of a locality.
(ii) Maximum number of students use bicycle for going to school.
(iii) 14 students use bus for going to school.
(iv) 37 students do not use bus for going to school.

 

 

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