RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B

RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B.

Other Exercises

Question 1.
Solution:
Rupesh seemed 495 marks out of 750
Percentage of marks = \(\frac { 495 }{ 750 }\) x 100 = 66%

Question 2.
Solution:
Monthly salary = Rs. 15625
Increase = 12%
Amount of increase = \(\frac { 15625 x 12 }{ 100 }\) = Rs. 1875
New salary = Rs. 15625 + Rs. 1875 = Rs. 17500

Question 3.
Solution:
Excise duty in the beginning = Rs. 950
Reduced duty = Rs. 760
Reduction = Rs. 950 – Rs. 760 = Rs. 190
Reduction percent = \(\frac { 190 x 100 }{ 950 }\) = 20%

Question 4.
Solution:
Let x be the total cost of the T.V
96% of x = 10464
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 1
Total cost of T.V = Rs. 10900

Question 5.
Solution:
Let number of students = x
In a school boys = 70%
girls = 100 – 70 = 30%
Now 30% of x = 504
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 2
Number of boys = 1680 – 504 = 1176
and number of total students = 1680

Question 6.
Solution:
Copper required = 69 kg
copper in ore = 12%
Let quantity of ore = x kg
12% of x = 69
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 3
Quantity of ore = 575kg

Question 7.
Solution:
Pass marks = 36%
A students gets marks = 123
But failed by 39 marks
Pass marks = 123 + 39 = 162
Now, 36% of maximum marks = 162
Maximum marks = \(\frac { 162 x 100 }{ 36 }\) = 450 marks

Question 8.
Solution:
Let number of apples = x
Number of apples sold = 40% of x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 4
Hence number of apples = 700

Question 9.
Solution:
Let total number of examinees = x
the numbers of examinees who passed = 72% of x
= \(\frac { x x 72 }{ 100 }\)
= \(\frac { 18x }{ 25 }\)
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 5

Question 10.
Solution:
Let the gross value of moped = x
Amount of commission = 5% of x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 6

Question 11.
Solution:
Total gunpowder = 8 kg
Amount of nitre = 75%
amount of sulphur = 10%
Rest of powder which is charcoal = 100 – (75 + 10) = 100 – 85 = 15 = 15%
Amount of charcoal = 8 x \(\frac { 15 }{ 100 }\) = \(\frac { 120 }{ 100 }\)
= \(\frac { 6 }{ 5 }\) kg = 1 kg 200 grams = 1.2 kg

Question 12.
Solution:
Quantity of chalk = 1 kg or 1000 g
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 7

Question 13.
Solution:
Let total number of days on which the
school open = x
and Sonal’s attendance = 75%
x x 75% = 219
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 8
No. of days on which was school open = 292 days

Question 14.
Solution:
Rate of commission = 3%
Amount of commission = Rs. 42660
Let value of property = x
then 3% of x = 42660
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 9

Question 15.
Solution:
Total votes of the constituency = 60000
Votes polled = 80% of total votes
= \(\frac { 80 }{ 100 }\) x 60000 = 48000
Votes polled in favour of A = 60% of polled votes
= \(\frac { 60 }{ 100 }\) x 48000 = 28800
Votes polled in favour of B = 48000 – 28800 = 19200

Question 16.
Solution:
Let original price of shirt = Rs. x
Discount = 12%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 10
Original price of shirt = Rs. 1350

Question 17.
Solution:
Let original price of sweater = x
Rate of increase = 8%
Increased price = x + 8% of x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 11
Hence, original price of sweater = Rs. 1450

Question 18.
Solution:
Let total income = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 12
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 13

Question 19.
Solution:
Let the given number = 100
Then increase % = 20%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 14
Decrease = 100 – 96 = 4
Decrease per cent = 4%

Question 20.
Solution:
Let original salary of the officer = Rs. 100
Increase = 20%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 15
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 16

Question 21.
Solution:
Rate of commission = 2% on first Rs. 200000
1 % on next Rs. 200000 and 0.5% on remaining price
Sale price of property = Rs.200000 + 200000 +140000 = Rs. 540000
Now commission earned by the
= Rs. 200000 x 2% + Rs. 200000 x 1% + 140000 x 0.5%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 17

Question 22.
Solution:
Let Akhil’s income = Rs. 100
Then income of Nikhil’s will be = Rs. 100 – 20 = Rs. 80
Amount which is more than that of Akhil’s = 100 – 80 = Rs. 20
% age = \(\frac { 20 x 100 }{ 80 }\) = 25%

Question 23.
Solution:
Let income of Mr Thomas = Rs. 100
then income of John = Rs. 100 + 20 = Rs. 120
Income of Mr Thomas is less than John = Rs. 120 – 100 = Rs. 20
% age = \(\frac { 20 x 100 }{ 120 }\)
= \(\frac { 50 }{ 3 }\) = 16\(\frac { 2 }{ 3 }\) %

Question 24.
Solution:
Present value of machine = Rs. 387000
Rate of depreciation = 10%
Let 1 year ago the value of machine was = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 18
1 year ago, value of machine = Rs. 430000

Question 25.
Solution:
Present value of car = Rs. 450000
Rate of decreasing of value = 20%
Value after 2 years
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 19

Question 26.
Solution:
Present population = 60000
Rate of increase = 10%
Increased population after 2 years
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 20

Question 27.
Solution:
Let the price of sugar = Rs. 100
and consumption = 100 kg.
Increase price of 100 kg = Rs. 100 + 25 = Rs. 125
Now increased amount on 100 kg = Rs. 125
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 21

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RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper.

Other Exercises

Question 1.
Solution:
We have:
x3 + y3 + z3 – 3xyz
= (-2)3 + (-1)3 + (3)3 – 3 x (-2) x (-1) x 3
= – 8 – 1 + 27 – 18 = -27 + 27 = 0

Question 2.
Solution:
Co-efficient of x in the given numbers are
(i) -5y
(ii) 2yz
(iii) \(\frac { -3 }{ 2 }\) ab

Question 3.
Solution:
We have:
(4xy – 5x2 – y2 + 6) – (x2 – 2xy + 5y2 – 4)
= 4xy – 5x2 – y2 + 6 – x2 + 2xy – 5y2 + 4
= -6x2 – 6y2 + 6xy +10
= -2 (3x2 + 3y – 3xy – 5)

Question 4.
Solution:
We have:
(2x2 – 3y2 + xy) – (x2 – 2xy + 3y2)
= 2x2 – 3y2 +xy – x2 + 2xy – 3y2
= 2x2 – x2 – 3y2 – 3y2 + xy + 2xy
= x2 – 6y2 + 3xy
x2 – 2xy + 3y2 is less than 2x2 – 3y2 + xy by x2 – 6y + 3xy.

Question 5.
Solution:
We have:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 1

Question 6.
Solution:
We have:
(3a + 4) (2a – 3) + (5a – 4) (a + 2)
= {3a (2a – 3) + 4 (2a – 3)} + {5a (a + 2) – 4 (a + 2)}
= (6a2 – 9a + 8a – 12) + (5a2 + 10a – 4a – 8)
= (6a2 – a – 12) + (5a2 + 6a – 8)
= (11a2 + 5a – 20)

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 2
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 3

Question 8.
Solution:
We have:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 4

Question 9.
Solution:
Let the consecutive odd number be
x and (x + 2)
x + (x + 2) = 68
2x + 2 = 68
2x = 68 – 2 = 66
x = 33
The required numbers are 33 and (33 + 2), i.e., 35.

Question 10.
Solution:
Let Reenu’s present age be x
Then, her father’s present age will be 3x
Reenu’s age after 12 years = (x + 12)
3x + 12 = 2x + 24
x = 12
Reenu’s present age = x = 12 yrs
And her father’s age = 3x = (3 x 12) = 36 yrs

Mark (✓) against the correct answer in each of the following :
Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 5
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 6

Question 12.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 7

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 8

Question 14.
Solution:
(c) 18
Let the number be x.
According to the equation, we have:
4x = x + 54
⇒ 3x = 54
⇒ x = 18

Question 15.
Solution:
(b) 52°
Let the two complementary angles be x° and (90 – x)°.
According to the equation, we have:
x – (90 – x) = 14
⇒ 2x = 104
⇒ x = 52
(90° – x)° = 90° – 52° = 38°
The larger angle is 52°.

Question 16.
Solution:
(c) 32 m
Let the length and breadth of the rectangle be l m and b m, respectively.
According to the questions, we have:
l = 2b ……(i)
2 (l + b) = 96 …..(ii)
Now, 2 (2b+ b) = 96
⇒ 6b = 96
⇒ b = 16
Length = 16 x 2 m = 32 m

Question 17.
Solution:
(b) 12 years
Let the ages of A and B be x and y years respectively,
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 9

Question 18.
Solution:
(i) -2a2 b is a monomial.
(ii) (a2 – 2b2) is a binomial.
(iii) (a + 2b – 3c) is a trinomial.
(iv) In -5ab, the coefficient of a is -5.
(v) In x2 + 2x – 5, the constant term is -5.

Question 19.
Solution:
(i) False.
The coefficient of x is -1.
(ii) False.
The coefficient of x in – 3.
(iii) False.
LHS = (5x – 7) – (3x – 5) = 5x – 7 – 3x + 5 = 2x – 2.
(iv) True.
LHS = (3x + 5y) (3x – 5y)
= 3x (3x – 5y) + 5y (3x – 5y)
= 9x2 – 15xy + 15xy – 25y2
= 9x2 – 25y2
(v) True
(a2 + b2)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper 10

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 1

Question 2.
Solution:
Let the required number be x
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 2

Question 3.
Solution:
Let the required number be x.
(-20)-1 ÷ x = (-10)-1
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 3

Question 4.
Solution:
(i) 2000000 = 2.000000 x 106 [Since the decimal point is moved 6 to the left]
= 2 x 106
(ii) 6.4 x 105 = 6.4 x 100000 = 640000

Question 5.
Solution:
We have :
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 4

Question 6.
Solution:
We have :
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 5

Mark (✓) against the correct answer in each of the following :
Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 6

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 7

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 8
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 9

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 10

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 11
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 12

Question 12.
Solution:
(c) 3.263 x 105
A given number is said to be in standard form if it can be expressed as
k x 10n, where k is a real number such that 1 < k < 10 and n is a positive integer.
For example : 3.263 x 105

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 13
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 14

Question 14.
Solution:
(i) True
645 = 6.45 x 102
[Since the decimal point is moved 2 places to the left]
(ii) False
27000 = 2.7 x 104
[Since the decimal point is moved 4 places to the left]
(iii) False
(3° + 4° + 5°) = 1
(iv) False
Reciprocal of 56 = Reciprocal of
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 15

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RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C

RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C.

Other Exercises

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 1

Question 2.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 2
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 3

Question 3.
Solution:
(a)
2n + 5 = 3 (3n – 10)
⇒ 2n + 5 = 9n – 30
⇒ 9n – 2n = 5 + 30
⇒ 7n = 35
⇒ n = 5

Question 4.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 4

Question 5.
Solution:
(c)
8 (2x – 5) – 6 (3x – 7) = 1
⇒ 16x – 40 – 18x + 42 = 1
⇒ -2x + 2 = 1
⇒ -2x = 1 – 2 = -1
x = \(\frac { 1 }{ 2 }\)

Question 6.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 5

Question 7.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 6

Question 8.
Solution:
(b)
Let first whole number=x
Then second number = x + 1
and sum = 53
x + x + 1 = 53
⇒ 2x = 53 – 1
⇒ 2x = 52
⇒ x = 26
Smaller number = 26

Question 9.
Solution:
Let first even number = 2x
Then second number = 2x + 2
and sum = 86
2x + 2x + 2 – 86
⇒ 4x = 86 – 2 = 84
⇒ x = 21
Larger even number = 2x + 2 = 2 x 21 + 2 = 42 + 2 = 44

Question 10.
Solution:
(b)
Let first odd number = 2x + 1
Second number = 2x + 3
2x + 1 + 2x + 3 = 36
⇒ 4x + 4 = 36
⇒ 4x = 36 – 4 = 32
⇒ x = 8
Smaller number = 2x + 1 = 2 x 8 + 1 = 16 + 1 = 17

Question 11.
Solution:
(d)
Let number = x
2x + 9 = 31
⇒ 2x = 31 – 9 = 22
⇒ x = 11

Question 12.
Solution:
(a)
Let number = x then
3x + 6 = 24
⇒ 3x = 24 – 6 = 18
⇒ x = 6
Number = 6

Question 13.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 7

Question 14.
Solution:
(b)
Let first angle = x
Then second = 90° – x
x – (90° – x) = 10
⇒ x – 90° + x = 10°
⇒ 2x = 10° + 90° = 100°
x = 50°
Second angle = 90° – 50° = 40°
Larger angle = 50°

Question 15.
Solution:
(b)
Let first angle = x
Then second = 180° – x
x – (180° – x) = 20°
⇒ x – 180° + x = 20°
⇒ 2x = 20° + 180° = 200°
x = 100°
Second angle = 180° – 100° = 80°
Smaller angle = 80°

Question 16.
Solution:
(c)
Let age of A = 5x
Then age of B = 3x
After 6 years,
A’s age = 5x + 6
and B’s age = 3x + 6
\(\frac { 5x + 6 }{ 3x + 6 }\) = \(\frac { 7 }{ 5 }\)
⇒ 25x + 30 = 21x + 42
⇒ 25x – 21x = 42 – 30
⇒ 4x = 12
⇒ x = 3
A’s age = 5x = 5 x 3 = 15 years

Question 17.
Solution:
(b)
Let the number = x
According to the condition,
5x = 80 + x
⇒ 5x – x = 80
⇒ 4x = 80
⇒ x = 20
Number = 20

Question 18.
Solution:
(c)
Let width of rectangle = x m
Then length = 3x m
Perimeter = 96 m
2 (x + 3x) = 96
⇒ x + 3x = \(\frac { 96 }{ 2 }\) = 48
⇒ 4x = 48
⇒ x = 12
Length = 3x = 12 x 3 = 36 m

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable 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 7 Solutions Chapter 5 Exponents Ex 5C

RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C.

Other Exercises

OBJECTIVE QUESTIONS
Mark (✓) tick against the correct answer in each of the following :
Question 1.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 1

Question 2.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 2

Question 3.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 3

Question 4.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 4

Question 5.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 5

Question 6.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 6

Question 7.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 7

Question 8.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 8

Question 9.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 9

Question 10.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 10

Question 11.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 11

Question 12.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 12

Question 13.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 13

Question 14.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 14

Question 15.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 15

Question 16.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 16

Question 17.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 17

Question 18.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 18

Question 19.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 19

Question 20.
Solution:
(c)
Required number = 10-1 ÷ (-8)-1
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 20

Question 21.
Solution:
(c)
The number which is in standard form is 2.156 x 106

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C 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 7 Solutions Chapter 5 Exponents Ex 5B

RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B.

Other Exercises

Question 1.
Solution:
We can write in standard form
(i) 538 = 5.38 x 102
(ii) 6428000 = 6.428000 x 106 = 6.428 x 106
(iii) 82934000000 = 8.2934000000 x 1010 = 8.2934 x 1010
(iv) 940000000000 = 9.40000000000 x 1011 = 9.4 x 1011
(v) 23000000 = 2.3000000 x 107 = 2.3 x 107

Question 2.
Solution:
(i) Diameter of Earth = 12756000 m = 1.2756000 x 107m = 1.2756 x 107 m
(ii) Distance between Earth and Moon = 384000000 m = 3.84000000 x 108 = 3.84 x 108
(iii) Population of India in March 2001 = 1027000000 = 1.027000000 x 109 = 1.027 x 109
(iv) Number of stars in a galaxy = 100000000000 = 1.00000000000 x 1011 = 1 x 1011
(v) The present age of universe = 12000000000 years = 1.2000000000 x 1010 years = 1.2 x 1010 years

Question 3.
Solution:
The expanded form will be
(i) 684502 = 6 x 105 + 8 x 104 + 4 x 103 + 5 x 102 + 2
(ii) 4007185 = 4 x 106 + 0 x 105 + 0 x 104 + 7 x 103 + 1 x 102 + 8 x 101 + 5 x 100
(iii) 5807294 = 5 x 106 + 8 x 105 + 0 x 104 + 7 x 103 + 2 x 102 + 9 x 101 + 4 x 100
(iv) 50074 = 5 x 104 + 0 x 103 + 0 x 102 + 7 x 101 + 4 x 100

Question 4.
Solution:
(i) 6 x 104 + 3 x 103 + 0 x 102 + 7 x 101 + 8 x 100
= 60000 + 3000 + 0 + 70 + 8
= 63078
(ii) 9 x 106 + 7 x 105 + 0 x 104 + 3 x 103 + 4 x 102 + 6 x 101 + 2 x 100
= 9000000 + 700000 + 0 + 3000 + 400 + 60 + 2
= 9703462
(iii) 8 x 105 + 6 x 104 + 4 x 103 + 2 x 102 + 9 x 101 + 6 x 100
= 800000 + 60000 + 4000 + 200 + 90 + 6
= 864296

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B 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 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B

RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B.

Other Exercises

Question 1.
Solution:
Let the required number = x
Then 2x – 7 = 45
2x = 45 + 7 = 52
x = 26
Required number = 26

Question 2.
Solution:
Let the required number = x Then
3x + 5 = 44
⇒ 3x = 44 – 5 = 39
x = 13
Required number = 13

Question 3.
Solution:
Let the required fraction = x
then 2x + 4 = \(\frac { 26 }{ 5 }\)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 1

Question 4.
Solution:
Let the required number = x
and half of .the number = \(\frac { x }{ 2 }\)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 2
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 3

Question 5.
Solution:
Let the required number = x
Two third of the number = \(\frac { 2 }{ 3 }\) x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 4

Question 6.
Solution:
Let the required number = x
Then, 4x = x + 45
⇒ 4x – x = 45
⇒ 3x = 45
⇒ x = 15
Required number = 15

Question 7.
Solution:
Let the required number = x
Then x – 21 = 71 – x
⇒ x + x = 71 + 21
⇒ 2x = 92
⇒ x = 46

Question 8.
Solution:
Let the original number = x
Then \(\frac { 2 }{ 3 }\) of the number = \(\frac { 2 }{ 3 }\) x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 5

Question 9.
Solution:
Let the second number = x
then first number = \(\frac { 2 }{ 5 }\) x
their sum = 70
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 6

Question 10.
Solution:
Let the required number = x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 7
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 8

Question 11.
Solution:
Let the required number = x
Fifth part of the number = \(\frac { x }{ 5 }\)
Fourth part of the number = \(\frac { x }{ 4 }\)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 9

Question 12.
Solution:
Let first natural number = x then
next number = x + 1
x + x + 1 = 63
⇒ 2x = 63 – 1 = 62
x = 31
first number = 31
and second number = 31 + 1 = 32
Numbers are 31, 32

Question 13.
Solution:
Let first odd number = 2x + 1
second odd number = 2x + 3
2x + 1 + 2x + 3 = 76
⇒ 4x + 4 = 76
⇒ 4x = 76 – 4 = 72
x = 18
First number = 2x + 1 = 2 x 18 + 1 = 36 + 1 = 37
Second number = 2x + 3 = 2 x 18 + 3 = 36 + 3 = 39
Numbers are 37, 39

Question 14.
Solution:
Let first positive even number = 2x
Second number = 2x + 2
Third number = 2x + 4
2x + 2x +2 + 2x + 4 = 90
⇒ 6x + 6 = 90
⇒ 6x = 90 – 6 = 84
x = 14
First even number = 2x = 2 x 14 = 28
Second number = 2x + 2 = 2 x 14 + 2 = 28 + 2 = 30
Third number = 30 + 2 = 32
Required numbers are 28, 30, 32

Question 15.
Solution:
Sum of two numbers = 184
Let first number = x
Then second number = 184 – x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 10
First part = 72
Second part = 184 – 72 = 112
Hence parts are 72, 112

Question 16.
Solution:
Total number of notes = 90
Let number of notes of Rs. 5 = x
Then number of notes of Rs.10 = 90 – x
Then x x 5 + (90 – x) x 10 = 500
⇒ 5x + 900 – 10x = 500
⇒ -5x = 500 – 900 = -400
x = 8
Number of 5 rupees notes = 80
and ten rupees notes = 90 – 80 = 10

Question 17.
Solution:
Amount of coins = Rs. 34
Let 50 paisa coins = x
then 25 paisa coins = 2x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 11
Number of 50 paisa coins = 34
and number of 25 paisa coins = 2x = 2 x 34 = 68

Question 18.
Solution:
Let present age of Raju’s cousin = x years
then age of Raju = (x – 19) years
After 5 years,
Raju’s age = x – 19 + 5 = (x – 14) years
and his cousin age = x + 5
(x – 14) : (x + 5) = 2 : 3
⇒ \(\frac { x – 14 }{ x + 5 }\) = \(\frac { 2 }{ 3 }\)
⇒ 3(x – 14) = 2 (x + 5) (By cross multiplication)
⇒ 3x – 42 = 2x + 10
⇒ 3x – 2x = 10 + 42
⇒ x = 52
Raju’s age = x – 19 = 52 – 19 = 33 years
and his cousin age = 52 years.

Question 19.
Solution:
Let present age of son = x years
Age of father = (x + 30) years
12 years after,
Father’s age = x + 30 + 12 = (x + 42) years
and son’s age = (x + 12) years
(x + 42) = 3(x + 12)
⇒ x + 42 = 3x + 36
⇒ 3x + 36 = x + 42
⇒ 3x – x = 42 – 36
⇒ 2x = 6
⇒ x = 3
Son’s age = 3 years
Father’s age = 3 + 30 = 33 years

Question 20.
Solution:
Ratio in present ages of Sonal and Manoj = 7 : 5
Let Sonal’s age = 7x
then Manoj’s age = 5x
10 years hence,
Sonal’s age will be = 7x + 10
and Manoj’s age = 5x + 10
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 12
Sonal’s present age = 7x = 7 x 5 = 35 years
and Manoj’s age = 5x = 5 x 5 = 25 years

Question 21.
Solution:
Five years ago,
Let Son’s age = x years
and father’s age = 7x years
Present age of son = (x + 5) years
and age of father = (7x + 5) years
5 years hence,
father’s age = 7x + 5 + 5 = 7x + 10
and Son’s age = x + 5 + 5 = x + 10
(7x + 10) = 3(x + 10)
⇒ 7x + 10 = 3x + 30
⇒ 7x – 3x = 30 – 10
⇒ 4x = 20
⇒ x = 5
Father present age = 7x + 5 = 7 x 5 + 5 = 35 + 5 = 40 years
and son’s age = x + 5 = 5 + 5 = 10 years

Question 22.
Solution:
Let age of Manoj 4 years ago = x
then his present age = x + 4
After 12 years his age will be = x + 4 + 12 = x + 16
x + 16 = 3(x)
x + 16 = 3x
⇒ 16 = 3x – x
⇒ 2x = 16
x = 8
His present age = 8 + 4 = 12 years

Question 23.
Solution:
Let total marks = x
Pass marks = 40% of x = \(\frac { 40x }{ 100 }\) = \(\frac { 2 }{ 5 }\) x
No. of marks got by Rupa = 185
No. of marks by which she failed = 15
Pass marks = 185 + 15 = 200
\(\frac { 2 }{ 5 }\) x = 200
⇒ x = \(\frac { 200 x 5 }{ 2 }\) x
⇒ x = 500
Hence total marks = 500

Question 24.
Solution:
Sum of digits = 8
Let units digit = x
Then tens digit = 8 – x
and number will be x + 10 (8 – x) ….(i)
By adding 18, the digits are reversed then
units digit = 8 – x
and tens digit = x
Number = (8 – x) = 10x
According to the condition,
(8 – x) + 10x = 18 + x + 10 (8 – x)
⇒ 8 – x + 10x = 18 + x + 80 – 10x
⇒ 10x – x – x + 10x = 18 + 80 – 8
⇒ 18x = 90
⇒ x = 5
Number is
x + 10(8 – x) = 5 + 10(8 – 5) = 5 + 10 x 3 = 35

Question 25.
Solution:
Cost of 3 tables and 2 chairs = 1850
Cost of table = Rs. 75 + cost of a chair
Let cost of chair = Rs. x,
then Cost of table = Rs. 75 + x
According to the condition,
3 (75 + x) + 2x = 1850
⇒ 225 + 3x + 2x = 1850
⇒ 225 + 5x = 1850
⇒ 5x = 1850 – 225 = 1625
x = 325
Cost of chair = Rs. 325
and cost of table = Rs. 325 + 75 = Rs. 400

Question 26.
Solution:
S.P of article = Rs. 495
gain = 10%
Let cost price = Rs. x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 13

Question 27.
Solution:
Perimeter of field = 150 m
Length + Breadth = \(\frac { 150 }{ 2 }\) = 75 m
[Perimeter = 2(l + b)]
Let length = x Then breadth = 75 – x
Then x = 2(75 – x)
⇒ x = 150 – 2x
⇒ x + 2x = 150
⇒ 3x = 150
⇒ x = \(\frac { 150 }{ 3 }\) = 50
Length = 50 m
and breadth = 75 – 50 = 25 m

Question 28.
Solution:
Perimeter of an isosceles triangle = 55 m
Let the third side of an isosceles triangle = x
Then each equal side = (2x – 5) m
According to the condition,
x + 2 (2x – 5) = 55
⇒ x + 4x – 10 = 55
⇒ 5x = 55 + 10
⇒ 5x = 65
⇒ x = 13
and 2x – 5 = 2 x 13 – 5 = 21 m
Sides will be 13m, 21m, 21m

Question 29.
Solution:
Sum of two complementary angles = 90°
Let first angle = x
then second = 90° – x
x – (90 – x) = 8
⇒ x – 90 + x = 8
⇒ 2x = 8 + 90
⇒ 2x = 98
⇒ x = 49
first angle = 49°
and second angle = 90° – 49° = 41°
Hence angles are 41°, 49°

Question 30.
Solution:
Sum of two supplementary angles = 180°
Let first angle = x
Then second angle = 180° – x
x – (180° – x) = 44°
⇒ x – 180° + x = 44°
⇒ 2x = 44° + 180° = 224°
⇒ 2x = 224°
⇒ x = 112°
First angle = 112°
and second angle = 180° – 112° = 68°
Hence angles are 68°, 112°

Question 31.
Solution:
In an isosceles triangle
Let each equal base angles = x
Then vertex angle = 2x
According to the condition,
x + x + 2x = 180° (sum of angles of a triangle)
⇒ 4x = 180°
⇒ x = 45°
Then vertex angle = 2x = 2 x 45° = 90°
Angles of the triangle are 45°, 45° and 90°

Question 32.
Solution:
Let length of total journey = x km
According to the condition,
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 14
⇒ 39x + 80 = 40x
⇒ 40x – 39x = 80
⇒ x = 80
Total journey = 80km

Question 33.
Solution:
No. of days = 20 Let no. of days he worked = x
Then he will receive amount = x x Rs. 120 = Rs. 120x
No. of days he did not work = 20 – x
Fine paid = (20 – x) x Rs. 10 = Rs. 10(20 -x)
120x – 10 (20 – x) = 1880
⇒ 120x – 200 + 10x = 1880
⇒ 130x = 1880 + 200 = 2080
x = 16
No. of days he remained absent = 20 – x = 20 – 16 = 4 days

Question 34.
Solution:
Let value of property = x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 15

Question 35.
Solution:
Solution = 400 mL
Quantity of alcohol = 15% of 400 mL
= \(\frac { 400 x 15 }{ 100 }\) = 60 mL
Let pure alcohol added = x mL
Total solution = 400 + x
and total alcohol = (x + 60)
Now (400 + x) x 32% = x + 60
⇒ (400 + x) x \(\frac { 32 }{ 100 }\) = x + 60
⇒ 32 (400 + x) = 100 (x + 60)
⇒ 12800 + 32x = 100x + 6000
⇒ 12800 – 6000 = 100x – 32x
⇒ 6800 = 68x
⇒ x = 6800
Pure alcohol added = 100 mL

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable 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 7 Solutions Chapter 5 Exponents Ex 5A

RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 1

Question 2.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 2
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 3

Question 3.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 4
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 5

Question 4.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 6

Question 5.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 7

Question 6.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 8

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 9
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 10
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 11

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 12
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 13

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 14
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 15
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 16
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 17

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 18
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 19

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 20

Question 12.
Solution:
Product of two numbers = 4
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 21

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 22

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 23

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 24
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 25

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 26

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 27
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 28

Question 18.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 29

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A 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 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A

RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A.

Other Exercises

Solve the following equations. Check your result in each case.
Question 1.
Solution:
3x – 5 = 0
Adding 5 to both sides
3x – 5 + 5 = 0 + 5
⇒ 3x = 5
⇒ x = \(\frac { 5 }{ 3 }\)
Check:
L.H.S. = 3x – 5
= 3 x \(\frac { 5 }{ 3 }\) – 5
= 5 – 5
= 0
= R.H.S.
Hence x = \(\frac { 5 }{ 3 }\)

Question 2.
Solution:
8x – 3 = 9 – 2x
⇒ 8x + 2x = 9 + 3 (By transposing)
⇒ 10x = 12
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 1

Question 3.
Solution:
7 – 5x = 5 – 7x
⇒ – 5x + 7x = 5 – 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 – 5x = 7 – 5(-1) = 7 + 5 = 12
R.H.S. = 5 – 7x = 5 – 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1

Question 4.
Solution:
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 2

Question 5.
Solution:
2(x – 2) + 3(4x – 1) = 0
⇒ 2x – 4 + 12x – 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
⇒ x = \(\frac { 7 }{ 14 }\) = \(\frac { 1 }{ 2 }\)
Check : L.H.S. = 2(x – 2) + 3 (4x -1)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 3

Question 6.
Solution:
5 (2x – 3) – 3(3x – 7) = 5
⇒ 10x – 15 – 9x + 21 = 5
⇒ 10x – 9x – 15 + 21 = 5
⇒ 10x – 9x = 5 + 15 – 21 (By transposing)
⇒ x = 20 – 21 = -1
⇒ x = -1
Check:
L.H.S. = 5 (2x – 3) – 3(3x – 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 – 3] – 3[-3 – 7]
= 5 x (-5) -3 x (-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 4
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 5

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 6
L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 7

Question 10.
Solution:
3x + 2(x + 2) = 20 – (2x – 5)
⇒ 3x + 2x + 4 = 20 – 2x + 5
⇒ 5x + 4 = 25 – 2x
⇒ 5x + 2x = 25 – 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)] = 3 x 3 + 2(3 + 2) = 9 + 2 x 5 = 9 + 10 = 19
R.H.S. = 20 – (2x – 5) = 20 – (2 x 3 – 5) = 20 – (6 – 5) = 20 – 1 = 19
L.H.S. = R.H.S.
Hence x = 3

Question 11.
Solution:
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
⇒ 13y – 52 – 3y + 27 – 5y – 20 = 0
⇒ 13y – 3y – 5y – 52 + 27 – 20 = 0
⇒ 13y – 8y – 72 + 27 = 0
⇒ 5y – 45 = 0
⇒ 5y = 45 (By transposing)
⇒ y = 9
Check:
L.H.S. = 13(y – 4) – 3(y – 9) – 5(y + 4)
= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)
= 13 x 5 – 3 x 0 – 5 x 13
= 65 – 0 – 65 = 0 = R.H.S.
Hence y = 9

Question 12.
Solution:
\(\frac { 2m + 5 }{ 3 }\) = 3m – 10
⇒ 2m + 5 = 3 (3m – 10) (By cross multiplication)
⇒ 2m + 5 = 9m – 30
⇒ 2m – 9m = -30 – 5
⇒ -7m = -35
⇒ m = 5
m = 5
Check:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 8
R.H.S. = 3m – 10 = 3 x 5 – 10 = 15 – 10 = 5
L.H.S. = R.H.S.
Hence m = 5

Question 13.
Solution:
6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9x
⇒ 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x
⇒ 18x – 30x + 12 + 5 = 3x – 35x + 9x – 24 + 30
⇒ -12x + 17 = -23x + 6
⇒ – 12x + 23x = 6 – 17
⇒ 11x = -11
x = – 1
Check:
L.H.S. = 6(3x + 2) – 5(6x – 1)
= 6[3x (-1) + 2] – 5[6 x (-1) x -1]
= 6[-3 + 2] – 5[-6 – 1]
= 6 x (-1) – 5 x (-7)
= -6 + 35 = 29
R.H.S. = 3(x – 8) – 5 (7x – 6) + 9x
= 3[-1 – 8] -5 [7 x (-1) – 6] + 9 (-1)
= 3 x (-9) – 5 [-7 – 6] – 9
= -27 – 5(-13) – 9
= -27 + 65 – 9
= 65 – 36 = 29 .
L.H.S. = R.H.S.
Hence x = -1

Question 14.
Solution:
t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t – 4)
⇒ t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12t
⇒ t – 2t + 10t – 10 = 8t – 3t + 18
⇒ 9t – 10 = 5t + 18
⇒ 9t – 5t = 18 + 10 (By transposing)
⇒ 4t = 28
⇒ t = 7
Check:
L.H.S. = t – [2t + 5] -5[1 – 2t]
= 7 – [2 x 7 + 5] – 5[1 – 2 x 7]
= 7 – [14 + 5] – 5 [1 – 14]
= 7 – 19 – 5(-13)
= 7 – 19 + 65
= 72 – 19 = 53
R.H.S. = 2[3 + 4t) – 3(t – 4)
= 2 (3 + 4 x 7) – 3(7 – 4)
= 2(3 + 28) – 3(3)
= 2(31) – 9 = 62 – 9 = 53
L.H.S. = R.H.S.
Hence t = 7 Ans.

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 9
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 10

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 11

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 12
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 13

Question 18.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 14
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 15

Question 19.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 16
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 17

Question 20.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 18

Question 21.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 19
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 20

Question 22.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 21

Question 23.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 22
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 23
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 24

Question 24.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 25
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 26

Question 25.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 27
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 28

Question 26.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 29
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 30

Question 27.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 31

Question 28.
Solution:
0.18 (5x – 4) = 0.5x + 0.8
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 32
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 33

Question 29.
Solution:
2.4 (3 – x) – 0.6 (2x – 3) = 0
⇒ 7.2 – 2.4x – 1.2x + 1.8 = 0
⇒ -2.4x – 1.2x = – (7.2 + 1.8).
L.H.S. = 2.4 (3 – x) – 0.6 (2x – 3)
⇒ 2.4 (3 – 2.5) – 0.6 (2 x 2.5 – 3)
⇒ 2.4 (0.5) – 0.6 (5 – 3)
⇒ 1.2 – 0.6 x 2 = 1.2 – 1.2 = 0 = R.H.S.
Hence x = 2.5

Question 30.
Solution:
0.5x – (0.8 – 0.2x) = 0.2 – 0.3x
⇒ 0.5x – 0.8 + 0.2x = 0.2 – 0.3x
⇒ 0.5x + 0.2x + 0.3x = 0.2 + 0.8
⇒ 1.0x = 1.0
⇒ x = 1
Check :
L.H.S. = 0.5x – (0.8 – 0.2x)
= 0.5 x 1 – (0.8 – 0.2 x 1)
= 0.5 – (0.8 – 0.2) = 0.5 – 0.6 = -0.1
R.H.S. = 0.2 – 0.3x = 0.2 – 0.3 x 1 = 0.2 – 0.3 = -0.1
L.H.S. = R.H.S.
Hence x = 1

Question 31.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 34

Question 32.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 35

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable 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 7 Solutions Chapter 4 Rational Numbers CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 1
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 2

Question 2.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 3

Question 3.
Solution:
Let the required number be x.
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 4

Question 4.
Solution:
Let the other number be x.
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 5

Question 5.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 6

Question 6.
Solution:
Let the required number be x
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 7

Question 7.
Solution:
Length of rope = 45 m
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 8

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 9

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 10
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 11

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 12

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 13

Question 12.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 14

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 15

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 16

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 17

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 18
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 19

Question 17.
Solution:
(i) False.
This is because \(\frac { -15 }{ -11 }\) = \(\frac { 15 }{ 11 }\), which lies to the right of 0.
(ii) True
(iii) True
(iv) False
L.C.M. of 5 and 3 is 15.
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers CCE Test Paper 20

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper  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 7 Solutions Chapter 4 Rational Numbers Ex 4G

RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G.

Other Exercises

OBJECTIVE QUESTIONS
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 1

Question 2.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 2

Question 3.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 3

Question 4.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 4

Question 5.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 5
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 6

Question 6.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 7

Question 7.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 8
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 9

Question 8.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 10

Question 9.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 11

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 12
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 13

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 14

Question 12.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 15

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 16

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 17

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 18

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 19

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 20

Question 18.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 21

Question 19.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 22

Question 20.
Solution:
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4G 23

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G are helpful to complete your math homework.

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