RS Aggarwal Maths Solutions Class 6

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

• 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:
27.86 from 53.74
= 53.74 – 27.86
= 25.88 Ans.

Question 2.
Solution:
64.98 from 103.87
103.87 – 64.98
= 38.89 Ans.

Question 3.
Solution:
59.63 from 92.4
92.40 – 59.63
= 32.77 Ans.

Question 4.
Solution:
56.8 from 204
204.0 – 56.8
= 147.2 Ans.

Question 5.
Solution:
127.38 from 216.2
216.20 – 127.38
= 88.82 Ans.

Question 6.
Solution:
39.875 from 70.68
70.680 – 39.875
= 30.805 Ans.

Question 7.
Solution:
523.120 – 348.237
= 174.883 Ans

Question 8.
Solution:
600.000 – 458.573
= 141.427 Ans.

Question 9.
Solution:
206.321 – 149.456
= 56.865 Ans.

Question 10.
Solution:
3.400 – 0.612
= 2.788 Ans

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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

• 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

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 :

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 :

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 :

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 :

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 :

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 :

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 :

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 :

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

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 :

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 :

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 :

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 :

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 :

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals 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 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

• 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

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

Question 18.
Solution:
\(\\ \frac { 167 }{ 100 } \)
= 1.67

Question 19.
Solution:
\(\\ \frac { 1589 }{ 100 } \)
= 15.89

Question 20.
Solution:
\(\\ \frac { 5413 }{ 1000 } \)
= 5.413

Question 21.
Solution:
\(\\ \frac { 21415 }{ 1000 } \)
= 21.415

Question 22.
Solution:
\(\\ \frac { 25 }{ 4 } \)
= 6.25

Question 23.
Solution:
\(3 \frac { 3 }{ 5 } \)
= \(\\ \frac { 3\times 5+3 }{ 5 } \)
= \(\\ \frac { 15+3 }{ 5 } \)
= \(\\ \frac { 18 }{ 5 } \)
= 3.6

Question 24.
Solution:
\(1 \frac { 4 }{ 25 } \)
= \(\\ \frac { 1\times 25+4 }{ 25 } \)
= \(\\ \frac { 25+4 }{ 25 } \)
= \(\\ \frac { 29 }{ 25 } \)
= 1.16

Question 25.
Solution:
\(5 \frac { 17 }{ 50 } \)
= \(\\ \frac { 5\times 50+17 }{ 50 } \)
= \(\\ \frac { 250+17 }{ 50 } \)
= \(\\ \frac { 267 }{ 50 } \)
= 5.34

Question 26.
Solution:
\(12 \frac { 3 }{ 8 } \)
= \(\\ \frac { 12\times 8+3 }{ 8 } \)
= \(\\ \frac { 96+3 }{ 8 } \)
= \(\\ \frac { 99 }{ 8 } \)
= 12.375

Question 27.
Solution:
\(2 \frac { 19 }{ 40 } \)
= \(\\ \frac { 2\times 40+19 }{ 40 } \)
= \(\\ \frac { 80+19 }{ 40 } \)
= \(\\ \frac { 99 }{ 40 } \)
= 2.475

Question 28.
Solution:
\(\\ \frac { 19 }{ 20 } \)
= 0.95

Question 29.
Solution:
\(\\ \frac { 37 }{ 50 } \)
= 0.74

Question 30.
Solution:
\(\\ \frac { 107 }{ 250 } \)
= 0.428

Question 31.
Solution:
\(\\ \frac { 3 }{ 40 } \)
= 0.075

Question 32.
Solution:
\(\\ \frac { 7 }{ 8 } \)
= 0.875

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

Question 35.
Solution:
(i) 15 kg 850 g in kilograms
= \(15 \frac { 850 }{ 1000 } \) kg
= 15.850 kg

Question 36.
Solution:
(i) Rs. 18 and 25 paise in rupees
= \(18 \frac { 25 }{ 100 } \)
= 18.25 rupees

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

• 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:
(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,

Question 3.
Solution:
(i) 67.83 = (6 x 10) + (7 x 1)

Question 4.
Solution:

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,

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.

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

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

• RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A
• RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6B

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.

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