Tag: RS Aggarwal Class 7 Maths Solutions

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

• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper

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

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

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

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

Hence number of apples = 700

Question 9.
Solution:
Let total number of examinees = x
the numbers of examinees who passed = 72% of x
= \(\frac { x x 72 }{ 100 }\)
= \(\frac { 18x }{ 25 }\)

Question 10.
Solution:
Let the gross value of moped = x
Amount of commission = 5% of x

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

Question 13.
Solution:
Let total number of days on which the
school open = x
and Sonal’s attendance = 75%
x x 75% = 219

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

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%

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

Hence, original price of sweater = Rs. 1450

Question 18.
Solution:
Let total income = x

Question 19.
Solution:
Let the given number = 100
Then increase % = 20%

Decrease = 100 – 96 = 4
Decrease per cent = 4%

Question 20.
Solution:
Let original salary of the officer = Rs. 100
Increase = 20%

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%

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

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

Question 26.
Solution:
Present population = 60000
Rate of increase = 10%
Increased population after 2 years

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

Hope given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B are helpful to complete your math homework.

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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

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

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) 2y² z
(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:

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:

Question 8.
Solution:
We have:

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:

Question 12.
Solution:

Question 13.
Solution:

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,

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)

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

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

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:
Let the required number be x

Question 3.
Solution:
Let the required number be x.
(-20)^-1 ÷ x = (-10)^-1

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 :

Question 6.
Solution:
We have :

Mark (✓) against the correct answer in each of the following :
Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

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:

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

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper are helpful to complete your math homework.

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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

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(d)

Question 2.
Solution:
(d)

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)

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)

Question 7.
Solution:
(a)

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)

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

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

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

OBJECTIVE QUESTIONS
Mark (✓) tick against the correct answer in each of the following :
Question 1.
Solution:
(d)

Question 2.
Solution:
(c)

Question 3.
Solution:
(c)

Question 4.
Solution:
(b)

Question 5.
Solution:
(c)

Question 6.
Solution:
(b)

Question 7.
Solution:
(b)

Question 8.
Solution:
(a)

Question 9.
Solution:
(c)

Question 10.
Solution:
(b)

Question 11.
Solution:
(b)

Question 12.
Solution:
(b)

Question 13.
Solution:
(d)

Question 14.
Solution:
(a)

Question 15.
Solution:
(c)

Question 16.
Solution:
(a)

Question 17.
Solution:
(c)

Question 18.
Solution:
(b)

Question 19.
Solution:
(c)

Question 20.
Solution:
(c)
Required number = 10^-1 ÷ (-8)^-1

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.

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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

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

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

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

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

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 = \(\frac { 26 }{ 5 }\)

Question 4.
Solution:
Let the required number = x
and half of .the number = \(\frac { x }{ 2 }\)

Question 5.
Solution:
Let the required number = x
Two third of the number = \(\frac { 2 }{ 3 }\) x

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

Question 9.
Solution:
Let the second number = x
then first number = \(\frac { 2 }{ 5 }\) x
their sum = 70

Question 10.
Solution:
Let the required number = x

Question 11.
Solution:
Let the required number = x
Fifth part of the number = \(\frac { x }{ 5 }\)
Fourth part of the number = \(\frac { x }{ 4 }\)

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

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

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

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

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,

⇒ 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

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.

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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

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
Product of two numbers = 4

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A are helpful to complete your math homework.

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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

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

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

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

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)

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:

Question 8.
Solution:

L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:

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:

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:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
0.18 (5x – 4) = 0.5x + 0.8

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:

Question 32.
Solution:

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

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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

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:
Let the required number be x.

Question 4.
Solution:
Let the other number be x.

Question 5.
Solution:

Question 6.
Solution:
Let the required number be x

Question 7.
Solution:
Length of rope = 45 m

Question 8.
Solution:

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

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.

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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

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

OBJECTIVE QUESTIONS
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(b)

Question 2.
Solution:
(b)

Question 3.
Solution:
(a)

Question 4.
Solution:
(c)

Question 5.
Solution:
(b)

Question 6.
Solution:
(a)

Question 7.
Solution:
(a)

Question 8.
Solution:
(c)

Question 9.
Solution:
(b)

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

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

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