RS Aggarwal Solutions for Class 7 Maths Chapter 7 Linear Equations in One Variables

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rs aggarwal solution class 7 maths chapter 7
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rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7
rs aggarwal solution class 7 maths chapter 7

Also, access RS Aggarwal Solutions for Class 7 Chapter 7 Exercises

Exercise 7A

Exercise 7B

Exercise 7C

Exercise 7A

Solve the following equations. Check your result in each case.

  1. 3x – 5 = 0Solution:-

    The value of the variable which makes the equation a true statement is called the root of the equation.

    Now, find the value of x by considering the given equation,

    = 3x – 5 = 0

    Transposing – 5 to RHS it becomes 5

    = 3x = 5

    Multiplying both side by (1/3)

    = 3x × (1/3) = 5 × (1/3)

    = x = (5/3)

    By substituting (5/3) in the place of x in given equation, we get

    = 3 × (5/3) – 5 = 0

    = 5 – 5 = 0

    = 0= 0

    ∴ LHS = RHS

    Hence, the result is verified.

  2. 8x – 3 = 9 – 2xSolution:-

    Now, find the value of x by considering the given equation,

    =8x – 3 = 9 – 2x

    Transposing – 3 to RHS and it becomes 3 and -2x to LHS it becomes 2x

    = 8x + 2x = 9 + 3

    = 10x = 11

    Multiplying both side by (1/10)

    = 10x × (1/10) = 12 × (1/10)

    = x = (12/10)

    = x = (6/5)

    By substituting (6/5) in the place of x in given equation, we get

    LHS,

    = 8 × (6/5) – 3

    = (48/5) – 3

    = (48-15/5)

    = (33/5)

    RHS,

    = 9 – 2 × (6/5)

    = 9 – (12/5)

    = (45 – 12)/5

    = (33/5)

    By comparing LHS and RHS

    = (33/5) = (33/5)

    ∴ LHS = RHS

    Hence, the result is verified.

  3. 7 – 5x = 5 – 7xSolution:-

    Now, find the value of x by considering the given equation,

    =7 – 5x = 5 – 7x

    Transposing 7 to RHS and it becomes -7 and -7x to LHS it becomes 7x

    = 7x – 5x = 5 – 7

    = 2x = -2

    Multiplying both side by (1/3)

    = 2x × (1/2) = -2 × (1/2)

    = x = (-2/2)

    = x = -1

    By substituting -1 in the place of x in given equation, we get

    LHS,

    = 7 – 5 × (-1)

    = 7 – (-5)

    = 7 + 5

    = 12

    RHS,

    = 5 – 7 × (-1)

    = 5 – (-7)

    = 5 + 7

    = 12

    By comparing LHS and RHS

    = 12 = 12

    ∴ LHS = RHS

    Hence, the result is verified.

  4. 3 + 2x = 1 – xSolution:-

    Now, find the value of x by considering the given equation,

    =3 + 2x = 1 – x

    Transposing 3 to RHS and it becomes -3 and -x to LHS it becomes x

    = 2x + x = 1 – 3

    = 3x = -2

    Multiplying both side by (1/3)

    = 3x × (1/3) = -2 × (1/3)

    = x = (-2/3)

    By substituting -2/3 in the place of x in given equation, we get

    LHS,

    = 3 + 2 × (-2/3)

    = 3 + (-4/3)

    = 3 – (4/3)

    = (9 – 4) / 3

    = 5/3

    RHS,

    = 1 – (-2/3)

    = 1 + (2/3)

    = (3 + 2)/ 3

    = 5/3

    By comparing LHS and RHS

    = 5/3 = 5/3

    ∴ LHS = RHS

    Hence, the result is verified.

  5. 2 (x – 2) + 3 (4x – 1) = 0Solution:-

    The above question can be written as,

    = 2x – 4 + 12x – 3 = 0

    Now, find the value of x by considering the above equation,

    = 2x + 12x – 4 – 3 = 0

    = 14x – 7 = 0

    Transposing -7 to RHS and it becomes 7

    = 14x = 7

    Multiplying both side by (1/14)

    = 14x × (1/14) = 7 × (1/14)

    = x = (7/14)

    = x = 1/2

    By substituting 1/2 in the place of x in given equation, we get

    LHS,

    = (2 × (1/2)) – 4 + (12 × (1/2)) – 3

    = 1– 4 + 6 – 3

    = 7 – 7

    = 0

    By comparing LHS and RHS

    = 0 = 0

    ∴ LHS = RHS

    Hence, the result is verified.

  6. 5 (2x – 3) – 3 (3x – 7) = 5Solution:-

    The above question can be written as,

    = 10x – 15 – 9x + 21 = 5

    Now, find the value of x by considering the given equation,

    = 10x – 15 – 9x + 21 = 5

    = x + 6 = 5

    Transposing 6 to RHS and it becomes -6

    = x = 5 – 6

    = x = -1

    By substituting -1 in the place of x in given equation, we get

    LHS,

    = 10 (-1) -15 -9(-1) + 21

    = -10 – 15 + 9 + 21

    = -25 + 30

    = 5

    By comparing LHS and RHS

    = 5 = 5

    ∴ LHS = RHS

    Hence, the result is verified.

  7. 2x – (1/3) = (1/5) – xSolution:-

    Now, find the value of x by considering the given equation,

    =2x – (1/3) = (1/5) – x

    Transposing – (1/3) to RHS and it becomes (1/3) and – x to LHS it becomes x

    = 2x + x = (1/5) + (1/3)

    = 3x = (3 + 5) / 15

    = 3x = 8/15

    Multiplying both side by (1/3)

    = 3x × (1/3) = (8/15) × (1/3)

    = x = (8/45)

    By substituting (8/45) in the place of x in given equation, we get

    LHS,

    = 2 × (8/45) – (1/3)

    = (16/45) – (1/3)

    = (16 – 15) / 45

    = (1/45)

    RHS

    = (1/5) – (8/45)

    = (9-8) / 45

    = (1/45)

    By comparing LHS and RHS

    = (1/45) = (1/45)

    ∴ LHS = RHS

    Hence, the result is verified.

  8. (1/2)x- 3 = 5 + (1/3)xSolution:-

    Now, find the value of x by considering the given equation,

    = (1/2)x- 3 = 5 + (1/3)x

    Transposing -3 to RHS and it becomes 3 and (1/3)x to LHS it becomes (-1/3)x

    = (1/2)x – (1/3)x = 5 + 3

    = ((3-2)/6)x = 8

    = (1/6)x = 8

    Multiplying both side by 6

    = 6 × (1/6)x = 8 × 6

    = x = 48

    By substituting 48 in the place of x in given equation, we get

    LHS,

    = (1/2) × (48) – 3

    = (48/2) – 3

    = 24 – 3

    = 21

    RHS,

    = 5 + (1/3) × (48)

    = 5 + (48/3)

    = 5 + 16

    = 21

    By comparing LHS and RHS

    = 21 = 21

    ∴ LHS = RHS

    Hence, the result is verified.

  9. (x/2) + (x/4) = (1/8)Solution:-

    Now, find the value of x by considering the given equation,

    = (x/2) + (x/4) = (1/8)

    = ((2 + 1) / 4) = (1/8)

    = (3/4)x = (1/8)

    Multiplying both side by (4/3)

    = (3/4)x × (4/3) = (1/8) × (4/3)

    = x = (1/6)

    By substituting (1/6) in the place of x in given equation, we get

    LHS,

    = (3/4)x

    = (3/4) × (1/6)

    = (1/8)

    RHS,

    = (1/8)

    By comparing LHS and RHS

    = (1/8) = (1/8)

    ∴ LHS = RHS

    Hence, the result is verified.

  10. 3x + 2(x + 2) = 20 – (2x – 5)Solution:-

    The above question can be written as,

    = 3x + 2x + 4 = 20 – 2x + 5

    Now, find the value of x by considering the above equation,

    = 5x + 4 = 20 – 2x + 5

    Transposing 4 to RHS and it becomes -4 and -2x to LHS it becomes 2x

    = 5x + 2x = 20 – 4 + 5

    = 7x = 25 – 4

    = 7x = 21

    Multiplying both side by (1/7)

    = 7x × (1/7) = 21 × (1/7)

    = x = 21/7

    = x = 3

    By substituting 3 in the place of x in given equation, we get

    LHS,

    = 5x + 4

    = 5 × (3) + 4

    = 15 + 4

    = 19

    RHS,

    = 20 – 2x + 5

    = 20 – 2 × (3) + 5

    = 20 – 6 + 5

    = 25 – 6

    = 19

    By comparing LHS and RHS

    = 19 = 19

    ∴ LHS = RHS

    Hence, the result is verified.

  11. 13( y – 4) – 3(y-9) – 5(y + 4) = 0Solution:-

    The above question can be written as,

    = 13y – 52 – 3y + 27 – 5y – 20 = 0

    Now, find the value of y by considering the above equation,

    = 13y – 3y – 5y – 52 + 27 – 20 = 0

    = 13y – 8y – 72 + 27 = 0

    = 5y – 45 = 0

    Transposing -45 to RHS and it becomes 45

    = 5y = 45

    Multiplying both side by (1/5)

    = 5y × (1/5) = 45 × (1/5)

    = y = 9

    By substituting 9 in the place of y in given equation, we get

    LHS,

    = 5y – 45

    = 5 × (9) – 45

    = 45 – 45

    = 0

    RHS,

    = 0

    By comparing LHS and RHS

    = 0 = 0

    ∴ LHS = RHS

    Hence, the result is verified.

  12. (2m + 5) / 3 = (3m – 10)Solution:-

    The above question can be written as,

    = 3 × [(2m + 5)/ 3] = 3 × (3m -10) … [Multiplying both side by 3]

    = (2m + 5) = 9m – 30

    Transposing 2m to RHS and it becomes -2m and -30 to LHS it becomes 30

    = 30 + 5 = 9m -2m

    = 35 = 7m

    Multiplying both side by (1/7),

    = 7m × (1/7) = 35 × (1/7)

    = m = (35/7)

    = m = 5

    By substituting 5 in the place of m in given equation, we get

    LHS,

    = 2m + 5

    = 2 × (5) + 5

    = 10 + 5

    = 15

    RHS,

    = 9m – 30

    = 9 × (5) – 30

    = 45 – 30

    = 15

    By comparing LHS and RHS

    = 15 = 15

    ∴ LHS = RHS

    Hence, the result is verified.

  13. 6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9xSolution:-

    The above question can be written as,

    = 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x

    Now, find the value of x by considering the above equation,

    Transposing 12, 5 to RHS and it becomes -12, -5 and -35x, 9x, 3x to LHS it becomes 35x, -9x, -3x

    = 18x – 30x + 35x – 9x – 3x = -24 + 30 – 12 – 5

    = 53x – 42x = 30 – 41

    = 11x = -11

    Multiplying both side by (1/11),

    = 11x × (1/11) = -11 × (1/11)

    = x = (-11/11)

    = x = -1

    By substituting 5 in the place of x in given equation, we get

    LHS,

    = 18x + 12 – 30x + 5

    = 18 (-1) + 12 – 30 (-1) + 5

    = -18 + 12 + 30 + 5

    = -18 + 47

    = 29

    RHS,

    = 3x – 24 – 35x + 30 + 9x

    = 3 (-1) – 24 – 35 (-1) + 30 + 9 (-1)

    = -3 – 24 + 35 + 30 – 9

    = -36 + 65

    = 29

    By comparing LHS and RHS

    = 15 = 15

    ∴ LHS = RHS

    Hence, the result is verified.

  14. t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t -4)Solution:-

    The above question can be written as,

    = t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12

    Now, find the value of t by considering the above equation,

    Transposing -5, -5 to RHS and it becomes 5, 5 and 8t, -3t to LHS it becomes -8t, 3t

    = t – 2t + 10t – 8t + 3t = 6 + 12 + 5 + 5

    = 14t – 10t = 28

    = 4t = 28

    Multiplying both side by (1/4),

    = 4t × (1/4) = 28 × (1/4)

    = t = 7

    By substituting 7 in the place of t in given equation, we get

    LHS,

    = t – 2t – 5 – 5 + 10t

    = 7 – 2× (7) – 5 – 5 + 10(7)

    = 7 – 14 – 5 – 5 + 70

    = 77 – 24

    = 53

    RHS,

    = 6 + 8t – 3t + 12

    = 6 + 8 × (7) – 3 × (7) + 12

    = 6 + 56 – 21 + 12

    = 74 – 21

    = 53

    By comparing LHS and RHS

    = 15 = 15

    ∴ LHS = RHS

    Hence, the result is verified.

  15. (2/3)x = (3/8)x + (7/12)Solution:-

    Find the value of x by considering the above equation,

    Transposing (3/8)x to RHS and it becomes -(3/8)x.

    = (2/3)x – (3/8)x = (7/12)

    = [(16 – 9) / 24]x = (7/12)

    = (7/24) x = (7/12)

    Multiplying both side by (24/7),

    = (7/24)x × (24/7) = (7/12) × (24/7)

    = x = 2

    By substituting 2 in the place of x in given equation, we get

    LHS,

    = (2/3)x

    = (2/3) × (2)

    = (4/3)

    RHS,

    = (3/8)x + (7/12)

    = (3/8) × (2) + (7/12)

    = (3/4) + (7/12)

    = (9 + 7) / (12)

    = (16/12)

    = (4/3)

    By comparing LHS and RHS

    = (4/3) = (4/3)

    ∴ LHS = RHS

    Hence, the result is verified.

  16. [(3x – 1)/ 5] – (x/7) = 3Solution:-

    The above question can be written as,

    = [(7 × (3x -1)) – (5 × (x))/35] = 3

    = ((21x – 7) – (5x))/35 = 3

    = (21x – 7 – 5x)/35 = 3

    = (16x – 7)/35 = 3 … [Multiplying both side by 35]

    = [(16x – 7)/35] × 35 = 3 × 35

    = 16x – 7 = 105

    = 16x = 105 + 7

    = 16x = 112 … [Multiplying both side by 1/16]

    = 16x × (1/16) = 112 × (1/16)

    = x = 7

    By substituting 7 in the place of x in given equation, we get

    LHS,

    = (16x – 7)/35

    = ((16 × 7) – 7)/35

    = (112 – 7)/ 35

    = (105/35)

    = 3

    RHS,

    = 3

    By comparing LHS and RHS

    = (4/3) = (4/3)

    ∴ LHS = RHS

    Hence, the result is verified.


Exercise 7B

  1. Twice a number when decreased by 7 gives 45. Find the number.Solution:-

    Let the required number be x.

    Then,

    = 2x – 7 = 45

    Transposing -7 to RHS and it becomes 7.

    = 2x = 45 + 7

    = 2x = 52

    Multiplying both side by (1/2)

    = 2x × (1/2) = 52 × (1/2)

    = x = 26

    ∴ the required number is 26

  2. Thrice a number when increased by 5 gives 44. Find the number.Solution:-

    Let the required number be x.

    Then,

    = 3x + 5 = 44

    Transposing 5 to RHS and it becomes -5.

    = 3x = 44 – 5

    = 3x = 39

    Multiplying both side by (1/3)

    = 3x × (1/3) = 39 × (1/3)

    = x = 13

    ∴ the required number is 13

  3. Four added to twice a number yields (26/5). Find the fraction.Solution:-

    Let the required number be x.

    Then,

    = 2x + 4 = (26/5)

    Transposing 4 to RHS and it becomes -4.

    = 2x = (26/5) – 4

    = 2x = (26 – 20)/5

    = 2x = 6/5

    Multiplying both side by (1/2)

    = 2x × (1/2) = (6/5) × (1/2)

    = x = 3/5

    ∴ the required number is 3/5

  4. A number when added to its half gives 72. Find the number.Solution:-

    Let the required number be x.

    Then,

    = x + (x/2) = 72

    = (2x + x)/ 2 = 72

    = 3x/2 = 72

    Multiplying both side by (2)

    = (3x/2) × (2) = 72 × 2

    = 3x = 144

    Again, multiplying both side by (1/3)

    = 3x × (1/3) = 144 × (1/3)

    = x = 48

    ∴ the required number is 48

  5. A number added to its two-thirds is equal to 55. Find the number.Solution:-

    Let the required number be x.

    Then,

    = x + (2x/3) = 55

    = (3x + 2x)/3 = 55

    = 5x/3 = 55

    Multiplying both side by (3/5)

    = (5x/3) × (3/5) = 55 × (3/5)

    = x= 11 × 3

    = x = 33

    ∴ the required number is 33

  6. A number when multiplied by 4, exceeds itself by 45. Find the number.Solution:-

    Let the required number be x.

    Then,

    = 4x – x = 45

    = 3x = 45

    Multiplying both side by (1/3)

    = 3x × (1/3) = 45 × (1/3)

    = x = 15

    ∴ the required number is 15

  7. A number is as much greater than 21 as it is less than 71. Find the number.Solution:-

    Let the required number be x.

    Then,

    = (x – 21) = (71 –x)

    Transposing -21 to RHS and it becomes 21 and –x to LHS it becomes x.

    = x + x = 71 + 21

    = 2x = 92

    Multiplying both side by (1/2)

    = 2x × (1/2) = 92 × (1/2)

    = x = 46

    ∴ the required number is 46

  8. (2/3) of a number is less than the original number by 20. Find the number.Solution:-

    Let the required number be x.

    Then,

    = (2x/3) = x – 20

    Transposing x to LHS and it becomes –x.

    = (2x/3) – x = -20

    = (2x – 3x)/3 = -20

    = -x/3 = -20

    Multiplying both side by – 3

    = (-x/3) × -3 = -20 × -3

    = x = 60

    ∴ the required number is 60

  9. A number is (2/5) times another number. If their sum is 70. Find the numbers.Solution:-

    Let the required number be x.

    Then,

    = x + (2x/5) = 70

    = (5x + 2x)/5 = 70

    = (7x/5) = 70

    Multiplying both side by (5/7)

    = (7x/5) × (5/7) = 70 × (5/7)

    = x = 50

    The other number is (2x/5)

    = (2 × 50)/5

    = 100/5

    = 20

    ∴the required numbers are 50 and 20

  10. Two- thirds of a number is greater than one-third of the number by 3. Find the number.Solution:-

    Let the required number be x.

    Then,

    = (2x/3) = (1x/3) + 3

    Transposing (1x/3) to LHS and it becomes – (1x/3)

    = (2x/3) – (1x/3) = 3

    = (2x – 1x)/ 3 = 3

    = (1x/3) = 3

    Multiplying both side by (3)

    = (1x/3) × 3 = 3 × 3

    = x = 9

    ∴the required numbers are 9

  11. The fifth part of a number when increased by 5 equals its fourth part decreased by 5. Find the number.Solution:-

    Let the required number be x.

    Then,

    = (x/5) + 5 = (x/4) – 5

    Transposing (x/4) to LHS and it becomes – (x/4) and 5 to RHS it becomes -5

    = (x/5) – (x/4) = -5 – 5

    = (4x – 5x)/20 = -10

    = -x/20 = -10

    Multiplying both side by (-20)

    = (-x/20) × (-20) = (-10) × (-20)

    = x = 200

    ∴the required numbers are 200

  12. Find two consecutive natural numbers whose sum is 63.Solution:-

    Let the required two consecutive natural numbers be x and (x + 1)

    Then,

    = x + (x+1) = 63

    = x + x + 1 = 63

    = 2x + 1 = 63

    Transposing 1 to LHS and it becomes – 1

    = 2x = 63 -1

    = 2x = 62

    Multiplying both side by (1/2)

    = 2x × (1/2) = 62 × (1/2)

    = x = 31

    The other number is (x + 1) = 31 + 1 = 32

    ∴the required two consecutive natural numbers are 31 and 32

  13. Find two consecutive positive odd integers whose sum is 76Solution:-

    Let the required two consecutive positive odd integers be x and (x + 2)

    Then,

    = x + (x+2) = 76

    = x + x + 2 = 76

    = 2x + 2 = 76

    Transposing 2 to LHS and it becomes – 2

    = 2x = 76 -2

    = 2x = 74

    Multiplying both side by (1/2)

    = 2x × (1/2) = 74 × (1/2)

    = x = 37

    The other number is (x + 2) = 37 + 2 = 39

    ∴the required two consecutive natural numbers are 37 and 39

  14. Find three consecutive positive even integers whose sum is 90Solution:-

    Let the required three consecutive positive even integers be x, (x + 2) and (x + 4)

    Then,

    = x + (x+2) + (x+4) = 90

    = x + x + 2 + x + 4 = 90

    = 3x + 6 = 90

    Transposing 6 to LHS and it becomes – 6

    = 3x = 90 -6

    = 3x = 84

    Multiplying both side by (1/3)

    = 3x × (1/3) = 90 × (1/3)

    = x = 30

    The other numbers are (x + 2) = 30 + 2 = 32 and (x + 4) = 30 + 4 = 34

    ∴the required three consecutive positive even integers are 30, 32 and 34

  15. Divide 184 into two parts such that one-third of one part may exceed one- seventh of the other part by 8.Solution:-

    Let the two parts be x and (184 – x)

    Then,

    = (1/3)x – (1/7) (184 – x) = 8

    = (1/3)x – (184/7) + (x/7) = 8

    Transposing – (184/7) to LHS and it becomes (184/7)

    = (1/3) x + (x/7) = 8 + (184/7)

    = [(7 + 3)/21]x = (56 +184)/7

    = (10/21) x = 240/7

    Multiplying both side by (21/10)

    = (10/21)x × (21/10) = (240/7) × (21/10)

    = x = (240 × 21)/ (7 × 10)

    = x = (24 × 3)/ (1 × 1)

    = x = 72

  16. A sum of ₹ 500 is in the form of denominations of ₹ 5 and ₹ 10.if the total number of notes is 90. Find the number of notes of each type.Solution:-

    The total number of notes is 90.

    Let the number of ₹ 5 note be x

    And the number of ₹ 10 note be (90 –x)

    Then,

    = 5x + 10(90 – x) = 500

    = 5x + 900 -10x = 500

    = – 5x + 900 = 500

    Transposing – 5x to RHS and it becomes 5x and 500 to LHS it becomes -500

    = 900 – 500 = 5x

    = 400 = 5x

    Multiplying both side by (1/5)

    = 400 × (1/5) = 5x × (1/5)

    = 80 = x

    ∴the number of ₹ 10 note be (90 –x) = (90 – 80) = 10 notes

  17. Sumitra has ₹ 34 in 50-paise and 25-paise coins. If the number of 25-paise coins is twice the number of 50-paise coins, how many coins of each kind does she have?Solution:-

    From the question,

    Sumitra has ₹34 = 3400 paise

    Let 50 paise coins be x

    And 25 paise coins be 2x

    Then,

    = 50x + 25 (2x) = 3400

    = 50x + 50x = 3400

    = 100x = 3400

    Multiplying both side by (1/100).

    = 100x × (1/100) = (3400) × (1/100)

    = x = 34

    ∴The number of 50 paise coins is 34

    The number of 25 paise coins is 2x = 2 × 34 = 68


Exercise 7c

Mark against the correct answer in each of the following:

  1. If 5x – (3/4) = 2x – (2/3), then x =?(a)(1/12) (b)(1/4) (C) 36 (d) (1/36)

    Solution:-

    (d) (1/36)

    Because,

    Transposing – (3/4) to RHS and it becomes (3/4) and 2x to LHS it becomes -2x

    = 5x – 2x = – (2/3) + (3/4)

    = 3x = (-8+9)/12

    = 3x = 1/12

    Multiplying both side by (1/3).

    = 3x × (1/3) = (1/12) × (1/3)

    = x = 1/36

  2. If 2z + (8/3) = (1/4) z + 5, then z =?(a)3 (b)3 (c)(2/5) (d)(4/3)

    Solution:-

    (d)(4/3)

    Because,

    Transposing (8/3) to RHS and it becomes – (8/3) and (1/4)z to LHS it becomes –(1/4)z

    = 2z – (1/4)z = 5 – (8/3)

    = [(8-1)/4]z = (15-8)/3

    = (7/4)z = 7/3

    Multiplying both side by (4/7).

    = (7/4)z × (4/7) = ((7/3) × (4/7)

    = x = 4/3

  3. If (2n + 5) = 3(3n – 10), then z =?(a)5 (b)3 (c)(2/5) (d)(2/5)

    Solution:-

    (a)5

    Because,

    = 2n + 5 = 9n – 30

    Transposing 2n to RHS and it becomes – 2n and -30 to LHS it becomes 30

    = 30 + 5 = 9n -2n

    = 35 = 7n

    Multiplying both side by (1/7).

    = 35 × (1/7) = 7n × (1/7)

    = 5 = n

  4. If (x-1) / (x +1) = (7/9), then x =?(a)6 (b)7 (c)8 (d)10

    Solution:-

    (c)8

    Because,

    = (x-1) / (x +1) = (7/9)

    By cross multiplication,

    = 9 × (x-1) = 7 × (x +1)

    = 9x -9 = 7x + 7

    = 9x -7x = 9 + 7

    = 2x = 16

    = x = (16/2)

    = x = 8

  5. If 8(2x – 5) – 6 (3x – 7) = 1, then x =?(a)2 (b)3 (c)1/2 (d)(1/3)

    Solution:-

    (c)1/2

    Because,

    = 8(2x – 5) – 6 (3x – 7) = 1

    = 16x -40 – 18x + 42 = 1

    = -2x +2 = 1

    = 2-1 = 2x

    = 1 = 2x

    = x = (1/2)

  6. If (x/2)-1 = (x/3) +4, then x=?(a)8 (b)16 (c)24 (d)30

    Solution:-

    (d)30

    Because,

    = (x/2)-1 = (x/3) +4

    = (x/2) – (x/3) = 4 +1

    = (3x-2x)/6 = 5

    = x/6 =5

    = x = 30

  7. If (2x-1)/3 = ((x-2)/3) + 1, then x =?(a)2 (b)4 (c)6 (d)8

    Solution:-

    (a)2

    Because,

    = (2x-1)/3 = ((x-2)/3) + 1

    = [(2x-1)/3)] – [(x-2)/3)] = 1

    = [(2x -1) – (x -2)] / 3 = 1

    = (2x – 1 – x +2)/ 3 = 1

    = x +1 = 3

    = x = 3-1

    = x = 2

  8. The sum of two consecutive whole numbers is 53. The smaller number is (a)25 (b)26 (c)29 (d)23

    Solution:-

    (b)26

    Because,

    Let the two consecutive whole numbers be x and (x +1)

    = x + (x+1) = 53

    = x + x + 1 = 53

    = 2x + 1 = 53

    = 2x = 53 – 1

    = 2x = 52

    = x = 52/2

    = x = 26

  9. The sum of two consecutive even number is 86. The larger of two is(a)46 (b)36 (c)38 (d)44

    Solution:-

    (d)44

    Because,

    Let the two consecutive even numbers be x and (x +2)

    = x + (x+2) = 86

    = x + x + 2 = 86

    = 2x + 2 = 86

    = 2x = 86 – 2

    = 2x = 84

    = x = 84/2

    = x = 42

    The larger number is (x +1) = 42 +2 = 44


 

RS Aggarwal Solutions for Class 7 Maths Chapter 7 – Linear Equations in One Variables

Chapter 7 – Linear Equations in One Variables contains 3 exercises and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the essential concepts discussed in this chapter.

  • Linear Equations
  • Problems Based on Linear Equations

Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 7 – Linear Equations in One Variables

RS Aggarwal Solutions for Class 7 Maths Chapter 7 – Linear Equations in One Variable. In this chapter, we study the meaning of an equation and also that of a linear equation. Meanwhile, also discuss the purpose of the solutions of a linear equation and two methods of obtaining it, One by trial-and-error and another by a systematic method involving addition, subtraction, multiplication or division of some non-zero number on both sides of the equation.At the end of the chapter, we study the formulation and solution of equations for some real-life problems.