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

• 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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