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Page No 139:

Question 1:

Write each of the following statements as an equation:
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice  a number is 16.
(vi) If 12 is subtracted from a number, the result is 24.
(vii) Twice a number subtracted from 19 is 11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the number.

Answer:

(i) Let the required number be x.
  So, five times the number will be 5x.
  ∴ 5x = 40

(ii) Let the required number be x.
     So, when it is increased by 8, we get x + 8.
     ∴ x + 8 = 15

(iii) Let the required number be x.
     So, when 25 exceeds the number, we get 25 - x.
 ∴ 25 - x  = 7

(iv) Let the required number be x.
    So, when the number exceeds 5, we get x - 5.
   ∴ x - 5  = 3

(v) Let the required number be x.
     So, thrice the number will be 3x.
  ∴ 3x - 5 = 16

(vi) Let the required number be x.
     So, 12 subtracted from the number will be x - 12.
   ∴ x - 12 = 24

(vii) Let the required number be x.
     So, twice the number will be 2x.
   ∴ 19 - 2x = 11

(viii) Let the required number be x.
     So, the number when divided by 8 will be x8.
   ∴ x8 = 7

(ix) Let the required number be x.
     So, four times the number will be 4x.
 ∴ 4x - 3 = 17

(x) Let the required number be x.
     So, 6 times the number will be 6x.
    ∴ 6x = x + 5



Page No 140:

Question 2:

Write a statement for each of the equations, given below:
(i) x − 7 = 14
(ii) 2y = 18
(iii) 11 + 3x = 17
(iv) 2x − 3 = 13
(v) 12y − 30 = 6
(vi) 2z3=8

Answer:

(i) 7 less than the number x equals 14.
(ii) Twice the number y equals 18.
(iii) 11 more than thrice the number x equals 17.
(iv) 3 less than twice the number x equals 13.
(v) 30 less than 12 times the number y equals 6.
(vi) When twice the number z is divided by 3, it equals 8.

Page No 140:

Question 3:

Verify by substitution that
(i) the root of 3x − 5 = 7 is x = 4
(ii) the root of 3 + 2x = 9 is x = 3
(iii) the root of 5x − 8 = 2x − 2 is x = 2
(iv) the root of 8 − 7y = 1 is y = 1
(v) the root of z7=8 is z = 56

Answer:

(i)

   3x - 5 = 7Substituting x = 4  in the given equation:L.H.S.: 3×4 -5or, 12 - 5 = 7 = R.H.S.L.H.S. = R.H.S. Hence,  x = 4  is the root of the given equation. 

(ii)

    3 + 2x= 9Substituting x = 3 in the given equation:L.H.S.: 3 + 2×3or,  3 + 6 = 9 = R.H.S. L.H.S. = R.H.S. Hence,  x = 3  is the root of the given equation. 

(iii)

    5x - 8 = 2x -2 Substituting x = 2 in the given equation:L.H.S.:                                   R.H.S. :5×2- 8                           = 2×2-2or, 10 - 8 = 2                      = 4 -2 = 2     L.H.S. = R.H.S. Hence, x =2  is the root of the given equation. 

(iv)

     8 - 7y = 1 Substituting y = 1  in the given equation:L.H.S.: 8 -7×1or, 8 - 7 = 1 = R.H.S. L.H.S. = R.H.S. Hence, y =1  is the root of the given equation. 

(v)

     z7 = 8 Substituting z = 56 in the given equation:L.H.S.: 567 = 8 =R.H.S.L.H.S. = R.H.S. Hence, z =56  is the root of the given equation.

Page No 140:

Question 4:

Solve each of the following equations by the trial-and-error method:
(i) y + 9 = 13
(ii) x − 7 = 10
(iii) 4x = 28
(iv) 3y = 36
(v) 11 + x = 19
(vi) x3=4
(vii) 2x − 3 = 9
(viii) 12x + 7 = 11
(ix) 2y + 4 = 3y
(x) z − 3 = 2z − 5

Answer:

(i) y + 9 = 13
    We try several values of y until we get the  L.H.S. equal to the R.H.S.

   y    L.H.S.    R.H.S. Is LHS =RHS ?
 1 1 + 9 = 10 13 No
2 2 + 9 = 11 13 No
3 3 + 9 = 12 13 No
4 4 + 9 = 13 13 Yes
 ∴ y = 4

(ii) x − 7= 10
We try several values of x until we get the  L.H.S. equal to the R.H.S.
  x    L.H.S.    R.H.S. Is L.H.S. = R.H.S.?
 10 10 − 7 = 3 10 No
11 11 − 7 = 4 10 No
12 12 − 7 = 5 10 No
13 13 − 7 = 6 10 No
14 14 − 7 = 7 10 No
15 15 − 7 = 8 10 No
16 16 − 7 = 9 10 No
17 17 − 7 = 10 10 Yes
 
∴ x = 17

(iii) 4x = 28
     We try several values of x until we get the  L.H.S. equal to the R.H.S.
  x    L.H.S.    R.H.S. Is L.H.S. = R.H.S.?
 1 4 × 1 = 4 28 No
2 4 × 2 = 8 28 No
3 4 × 3 = 12 28 No
4 4 × 4 = 16 28 No
5 4 × 5 = 20 28 No
6 4 × 6 = 24 28 No
7 4 × 7 = 28 28 Yes
  ∴ x = 7

(iv) 3y = 36
    We try several values of x until we get the L.H.S. equal to the R.H.S.
  y   L.H.S.   R.H.S. Is L.H.S. = R.H.S.?
 6 3 × 6 = 18 36 No
7 3 × 7 = 21 36 No
8 3 × 8 = 24 36 No
9 3 × 9 = 27 36 No
10 3 × 10 = 30 36 No
11 3 ×11 = 33 36 No
12 3 × 12 = 36 36 Yes
 ∴ y = 12
 
(v) 11 + x = 19
     We try several values of x until we get the L.H.S. equal to the R.H.S.
     
  x    L.H.S.    R.H.S. Is L.H.S. = R.H.S.?
 1 11 + 1 = 12 19 No
2 11 + 2 = 13 19 No
3 11 + 3 = 14 19 No
4 11 + 4 = 15 19 No
5 11 + 5 = 16 19 No
6 11 + 6 = 17 19 No
7 11 + 7 = 18 19 No
8 11 + 8 = 19 19 Yes
   ∴ x = 8

(vi) x3 = 4
    Since R.H.S. is an natural number so L.H.S. must also be a natural number. Thus, x has to be a multiple of 3.
   
  x   L.H.S.   R.H.S. Is L.H.S. = R.H.S.?
3 33=1 4 No
6 63=2 4 No
9 93=3 4 No
12 123=4 4 Yes
  ∴ x = 12

(vii) 2x − 3 = 9
 
  We try several values of x until we get the L.H.S. equal to the R.H.S.
  x    L.H.S.    R.H.S. Is L.H.S. = R.H.S.?
 1 2 × 1 − 3 = −1 9 No
2 2 × 2 − 3 = 1 9 No
3 2 × 3 − 3 = 3 9 No
4 2 × 4 − 3 = 5 9 No
5 2 × 5 − 3 = 7 9 No
6 2 × 6 − 3 = 9 9 Yes
  ∴  x = 6

(viii) 12x + 7 = 11
     Since, R.H.S. is a natural number so L.H.S. must be a natural number Thus, we will try values if x which are multiples of 'x'
    
  x    L.H.S.    R.H.S. Is L.H.S. = R.H.S.?
2 2/2 + 7 = 8 11 No
4 4/2 + 7 = 9 11 No
6 6/2 + 7 = 10 11 No
8 8/2 + 7 = 11 11 Yes
    ∴ x = 8

(ix) 2y + 4 = 3y
      We try several values of y until we get the L.H.S. equal to the R.H.S.
 y    L.H.S.   R.H.S. Is L.H.S. = R.H.S.?
1 2 × 1 + 4 = 6 3 × 1 = 3 No
2 2 × 2 + 4 = 8 3 × 2 = 6 No
3 2 × 3 + 4 = 10 3 × 3 = 9 No
4 2 × 4 + 4 = 12 3 × 4 = 12 Yes
  ∴ y = 4

(x) z − 3 = 2z − 5
 We try several values of z till we get the L.H.S. equal to the R.H.S.
 z L.H.S.    R.H.S. Is L.H.S. = R.H.S.?
1 1 − 3 = −2 2 × 1 − 5 = −3 No
2 2 − 3 = −1 2 × 2 − 5 = −1 Yes
∴ z = 2



Page No 143:

Question 1:

Solve each of the following equations and verify the answer in each case:
x
+ 5 = 12

Answer:

 x + 5 = 12

Subtracting 5 from both the sides:
⇒ x + 5 − 5 = 12 − 5              
⇒ x = 7
Verification:
Substituting x = 7 in the L.H.S.:
⇒ 7 + 5 = 12 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 2:

Solve each of the following equations and verify the answer in each case:
x
+ 3 = −2

Answer:

 x + 3 = −2

Subtracting 3 from both the sides:
⇒ x + 3 − 3 = −2 − 3             
⇒ x = −5

Verification:
Substituting x = −5 in the L.H.S.:
⇒ −5 + 3 =  −2 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 3:

Solve each of the following equations and verify the answer in each case:
x
 − 7 = 6

Answer:

 x − 7 = 6
Adding 7 on both the sides:
⇒ x − 7 + 7 = 6 + 7               
⇒ x = 13

Verification:
Substituting x = 13 in the L.H.S.:
⇒ 13 − 7 =  6 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 4:

Solve each of the following equations and verify the answer in each case:
x
 − 2 = −5

Answer:

 x − 2 = −5

Adding 2 on both sides:
⇒ x − 2 + 2 = −5 + 2            
⇒ x = −3
Verification:
Substituting x = −3 in the L.H.S.:
⇒  −3 − 2 = −5 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 5:

Solve each of the following equations and verify the answer in each case:
3x − 5 = 13

Answer:

3x − 5 = 13
⇒ 3x − 5 + 5 = 13 + 5             [Adding 5 on both the sides]
⇒ 3x = 18
3x3 = 183                        [Dividing both the sides by 3]
⇒ x = 6
Verification:
Substituting x = 6 in the L.H.S.:
⇒  3 × 6 − 5 = 18 − 5 = 13 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 6:

Solve each of the following equations and verify the answer in each case:
4x + 7 = 15

Answer:

4x + 7 = 15
⇒ 4x + 7 − 7 = 15 − 7              [Subtracting 7 from both the sides]
⇒ 4x = 8
4x4 = 84                        [Dividing both the sides by 4]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
⇒  4×2 + 7 = 8 + 7 = 15 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 7:

Solve each of the following equations and verify the answer in each case:
x5=12

Answer:

x5 = 12
x5×5  = 12×5                                  [Multiplying both the sides by 5]
⇒ x = 60
Verification:
Substituting x = 60 in the L.H.S.:
605 = 12 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 8:

Solve each of the following equations and verify the answer in each case:
3x5=15

Answer:

     3x5 = 15
3x5× 5  = 15 × 5                                  [Multiplying both the sides by 5]
⇒ 3x = 75
3x3 = 753
⇒ x = 25
Verification:
Substituting x = 25 in the L.H.S.:
3 × 255 = 15 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 9:

Solve each of the following equations and verify the answer in each case:
5x − 3 = x + 17

Answer:

5x − 3 = x + 17
⇒ 5x − x = 17 + 3                  [Transposing x to the L.H.S. and 3 to the R.H.S.]
⇒ 4x = 20
4x4 = 204                        [Dividing both the sides by 4]
⇒ x = 5
Verification:
Substituting x = 5 on both the sides:
L.H.S.:  5(5) − 3
⇒ 25 − 3
⇒ 22

R.H.S.:  5 + 17 = 22
⇒ L.H.S. = R.H.S.
 Hence, verified.

Page No 143:

Question 10:

Solve each of the following equations and verify the answer in each case:
2x-12=3

Answer:

2x-12 = 3
⇒ 2x -12 + 12 = 3 + 12                              [Adding 12 on both the sides]
⇒ 2x  = 6 + 12
⇒ 2x = 72
2x2 = 72 × 2                                            [Dividing both the sides by 3]
⇒ x = 74
Verification:
Substituting  x = 74 in the  L.H.S.:
274 - 12= 72 - 12 = 62 = 3 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 11:

Solve each of the following equations and verify the answer in each case:
3(x + 6) = 24

Answer:

3(x + 6) = 24
3×x + 3×6 = 24                     [On expanding the brackets]
⇒  3x + 18 = 24
⇒ 3x + 18 - 18 = 24 - 18            [Subtracting 18 from both the sides]
⇒ 3x = 6
3x3 = 63                                [Dividing both the sides by 3]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
 3(2 + 6) = 3 ×8 = 24  = R.H.S.
 L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 12:

Solve each of the following equations and verify the answer in each case:
6x + 5 = 2x + 17

Answer:

6x + 5 = 2x + 17
6x  - 2x = 17 - 5                          [Transposing 2x to the L.H.S. and 5 to the R.H.S.]
4x = 12
4x4= 124                                      [Dividing both the sides by 4]
x = 3
Verification:
Substituting x = 3 on both the sides:
L.H.S.: 6(3) + 5
=18 + 5
=23
R.H.S.:   2(3) + 17
= 6 + 17
= 23
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 13:

Solve each of the following equations and verify the answer in each case:
x4-8=1

Answer:

x4- 8 = 1
x4- 8 + 8 = 1 + 8                  [Adding 8 on both the sides]
x4 = 9
x4 × 4 = 9 × 4                        [Multiplying both the sides by 4]
or, x = 36
Verification:
Substituting x = 36 in the L.H.S.:
or, 364 - 8 = 9 − 8 = 1 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 14:

Solve each of the following equations and verify the answer in each case:
x2=x3+1

Answer:

x2 = x3 + 1
x2 - x3 = 1                                      [Transposing x3 to the L.H.S.]
3x - 2x6 = 1
x6 = 1
x6 × 6 = 1 × 6                                    [Multiplying both the sides by 6]
or, x = 6
Verification:
Substituting x = 6 on both the sides:
L.H.S.: 62 = 3
R.H.S.: 63 + 1 =  2 + 1 =  3
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 15:

Solve each of the following equations and verify the answer in each case:
3(x + 2) − 2(x − 1) = 7

Answer:

3(x + 2) − 2(x − 1) = 7
3×x + 3×2 - 2×x -2×(-1) = 7              [On expanding the brackets]
or, 3x + 6 -2x + 2 = 7
or, x + 8 = 7
or, x + 8 - 8 = 7 - 8                                        [Subtracting 8 from both the sides]
or, x = -1
Verification:
Substituting x = -1 in the L.H.S.:
3(-1+2) -2(-1-1)or, 3(1) -2(-2)or, 3 + 4 = 7 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 16:

Solve each of the following equations and verify the answer in each case:

Answer:

5(x-1) +2(x+3) + 6 = 0
5x -5 +2x +6 +6 = 0        (Expanding within the brackets)
7x +7 = 0
x +1 = 0       (Dividing by 7)
x = -1

Verification:
Putting x = -1 in the L.H.S.:
L.H.S.: 5(-1 -1) + 2(-1 + 3) + 6
          = 5(-2) + 2(2) + 6
          = -10 + 4 + 6  = 0 = R.H.S.

Hence, verified.

Page No 143:

Question 17:

Solve each of the following equations and verify the answer in each case:
6(1 − 4x) + 7(2 + 5x) = 53

Answer:

6(1 − 4x) + 7(2 + 5x) = 53
or, 6 × 1 - 6 × 4x + 7 × 2 + 7 × 5x = 53              [On expanding the brackets]
or, 6 - 24x + 14 + 35x = 53
or, 11x + 20 = 53
or, 11x + 20 - 20 = 53 - 20                                        [Subtracting 20 from both the sides]
or, 11x = 33
or, 11x11= 3311                                                      [Dividing both the sides by 11]
or, x = 3
Verification:
Substituting x = 3 in the L.H.S.:
6(1 - 4 × 3) + 7(2 + 5 × 3)6(1 - 12) + 7(2 + 15)6(-11) + 7(17)-66 + 119  = 53 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 18:

Solve each of the following equations and verify the answer in each case:
16(3x − 5) − 10(4x − 8) = 40

Answer:

16(3x − 5) − 10(4x − 8) = 40
or, 16 × 3x - 16 × 5 -10 × 4x - 10 × (-8) = 40              [On expanding the brackets]
or, 48x - 80 - 40x + 80 = 40
or, 8x  = 40
or, 8x8=408                                                      [Dividing both the sides by 8]
or, x = 5
Verification:
Substituting x = 5 in the L.H.S.:

16(3 × 5 - 5) - 10( 4 × 5 - 8)16(15 - 5) - 10(20 - 8)16(10) -10(12)160 - 120  = 40 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 19:

Solve each of the following equations and verify the answer in each case:
3(x + 6) + 2(x + 3) = 64

Answer:

3(x + 6) + 2(x + 3) = 64
3 × x   +   3 × 6 + 2 × x  + 2 × 3   = 64            [On expanding the brackets]
3x + 18 +  2x + 6 = 64
⇒5x + 24 = 64
⇒5x + 24 - 24 = 64 - 24                                       [Subtracting 24 from both the sides]
⇒5x = 40
5x5 = 405                                                           [Dividing both the sides by 5]
⇒x = 8
Verification:
Substituting x = 8 in the L.H.S.:
3(8 + 6) + 2(8 + 3)3(14) + 2(11)42 + 22 = 64 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 20:

Solve each of the following equations and verify the answer in each case:
3(2 − 5x) − 2(1 − 6x) = 1

Answer:

3(2 − 5x) − 2(1 − 6x) = 1
or, 3 × 2  + 3 × (−5x) − 2 × 1 − 2 × (−6x) = 1           [On expanding the brackets]
or, 6 − 15x −  2 + 12x = 1
or, 4 - 3x = 1
or,  3  =3x                                                        
or, x = 1

Verification:
Substituting x = 1 in the L.H.S.:
3(2 - 5 × 1) - 2(1 - 6 × 1)3(2 - 5) - 2(1- 6)3(-3) -2(-5)-9 + 10 = 1 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 21:

Solve each of the following equations and verify the answer in each case:
n4-5=n6+12

Answer:

n4-5 = n6 + 12
or, n4 - n6 = 12 + 5                                     [Transposing n/6 to the L.H.S. and 5 to the R.H.S.]
or, 3n-2n12 = 1+102
or, n12 = 112
or, n12×12 = 112×12                                    [Dividing both the sides by 12]
or, n = 66
Verification:
Substituting n = 66 on both the sides:

L.H.S.:
664-5 =332 - 5  =33 - 102 =232 = 232  R.H.S.:  666+12= 11 + 12 = 22+12= 232 
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 22:

Solve each of the following equations and verify the answer in each case:
2m3+ 8=m2-1

Answer:

2m3 + 8 = m2- 1
or, 2m3 - m2 = -1 -8                      [Transposing m/2 to the L.H.S. and 8 to the R.H.S.]
or, 4m-3m6 = -9or, m6 = -9
or, m6×6 = -9×6                            [Multiplying both the sides by 6]
or, m = -54
Verification:
Substituting x = −54 on both the sides:

L.H.S.: 2(-54)3 +  8 = -542-1= -1083 + 8 = -36+ 8 = -28  R.H.S.:-542-1 = -27 - 1= -28
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 23:

Solve each of the following equations and verify the answer in each case:
2x5-32=x2+1

Answer:

2x5 -32 = x2 + 1
or, 2x5- x2 = 1+ 32                        [Transposing x/2 to the L.H.S. and 3/2 to R.H.S.]
or, 4x-5x10= 2+32or, -x10 = 52
or, -x10(-10) =52 ×(-10)               [Multiplying both the sides by −10]
or, x = −25
Verification:
Substituting x = −25 on both the sides:
L.H.S.: 2(-25)5 - 32 = -505 - 32  = -10 - 32 = -232R.H.S.: -252+ 1= -25 + 22 = -232
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 24:

Solve each of the following equations and verify the answer in each case:
x-35-2=2x5

Answer:

x-35 - 2 = 2x5 
or, x5- 35 -2 = 2x5
or, - 35- 2 = 2x5-x5                        [Transposing x/5 to the R.H.S.]
or, -3- 105 = x5
or, -135 = x5
or, -135(5) =x5 ×(5)               [Multiplying both the sides by 5]
or, x = −13
Verification:
Substituting x = −13 on both the sides:
L.H.S.: -13 - 35 - 2 =-165 - 2= -16 - 105 = -265R.H.S.: 2×(-13)5  = -265  

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 25:

Solve each of the following equations and verify the answer in each case:
3x10-4=14

Answer:

3x10 - 4 = 14 
or, 3x10- 4  + 4= 14 + 4                        [Adding 4 on both the sides]
or, 3x10 = 18
or, 3x10×10 = 18×10                              [Multiplying both the sides by 10]
or, 3x = 180              
or, 3x3 = 1803                  [Dividing both the sides by 3]
or, x = 60
Verification:
Substituting x = 60 on both the sides:
3×6010 - 4 =18010 - 4 = 18 - 4 = 14 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 26:

Solve each of the following equations and verify the answer in each case:
34 (x - 1) = x - 3

Answer:

34x-1 = x - 3
 34×x  - 34 × 1= x - 3                   [On expanding the brackets]
3x4- 34  = x - 3                      
3x4- x = -3 + 34                           [Transposing x to the L.H.S. and -34 to the R.H.S.]
3x-4x4=  -12+34
-x4 =  -94
-x4×-4 =  -94×-4                              [Multiplying both the sides by -4]
or, x = 9             

Verification:
Substituting x = 9 on both the sides:
L.H.S. : 349-1 = 34(8) = 6  R.H.S.: 9 - 3 = 6

L.H.S. = R.H.S.
Hence, verified.



Page No 144:

Question 1:

If 9 is added to a certain number, the result is 36. Find the number.

Answer:

Let the required number be x.
According to the question:
9 + x = 36
or, x + 9 - 9 = 36 - 9                        [Subtracting 9 from both the sides]
or, x = 27
Thus, the required number is 27.

Page No 144:

Question 2:

If 11 is subtracted from 4 times a number, the result is 89. Find the number.

Answer:

Let the required number be x.
According to the question:
4x -11 = 89
or, 4x - 11 +11 = 89 + 11                        [Adding 11 on both the sides]
or, 4x = 100
or, 4x4 = 1004                                       [Dividing both the sides by 4]
or, x = 25
Thus, the required number is 25.

Page No 144:

Question 3:

Find a number which when multiplied by 5 is increased by 80.

Answer:

Let the required number be x.
According to the question:
or, 5x = x + 80
or, 5x - x = 80                       [Transposing x to the L.H.S.]
or, 4x = 80
or, 4x4 = 804                                        [Dividing both the sides by 4]
or, x = 20
Thus, the required number is 20.

Page No 144:

Question 4:

The sum of three consecutive natural numbers is 114. Find the numbers.

Answer:

Let the three consecutive natural numbers be x, (x+1), (x+2).
According to the question:
x + (x + 1) + (x + 2) = 114
or, x + x + 1 + x + 2 = 114
or, 3x + 3 = 114
or, 3x + 3 - 3 = 114 - 3                     [Subtracting 3 from both the sides]
or, 3x = 111
or, 3x3 = 1113                                   [Dividing both the sides by 3]
or, x = 37
Required numbers are:
x = 37
or, x + 1 = 37 + 1 = 38
or ,x + 2 = 37 + 2 = 39
Thus, the required numbers are 37, 38 and 39.

Page No 144:

Question 5:

When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find that number.

Answer:

Let the required number be x.
When Raju multiplies it with 17, the number becomes 17x.
According to the question :
17x + 4 = 225
or, 17x + 4 - 4 = 225 - 4                          [Subtracting 4 from both the sides]
or, 17x = 221
or, 17x17 = 22117                                        [Dividing both the sides by 17]
or, x = 13
Thus, the required number is 13.

Page No 144:

Question 6:

If a number is tripled and the result is increased by 5, we get 50. Find the number.

Answer:

Let the required number be x.
According to the question, the number is tripled and 5 is added to it
∴ 3x + 5
or, 3x  + 5 = 50
or, 3x + 5 - 5 = 50 - 5                         [Subtracting 5 from both the sides]
or, 3x = 45
or, 3x3 =453                              [Dividing both the sides by 3]
or, x = 15
Thus, the required number is 15.

Page No 144:

Question 7:

Find two numbers such that one of them exceeds the other by 18 and their sum is 92.

Answer:

Let one of the number be x.
∴ The other number = (x + 18)
According to the question:
x + (x + 18) = 92
or, 2x + 18 - 18 = 92 - 18                         [Subtracting 18 from both the sides]
or, 2x =74
or, 2x2 = 742                                           [Dividing both the sides by 2]
or, x = 37
Required numbers are:
x = 37
or, x + 18 = 37 + 18 = 55

Page No 144:

Question 8:

One out of two numbers is thrice the other. If their sum is 124, find the numbers.

Answer:

Let one of the number be 'x'
∴ Second number = 3x
According to the question:
x + 3x = 124
or, 4x  = 124                         
or, 4x4 = 1244                      [Dividing both the sides by 4]
or, x = 31
Thus, the required number is x = 31 and 3x = 3×31 = 93.

Page No 144:

Question 9:

Find two numbers such that one of them is five times the other and their difference is 132.

Answer:

Let one of the number be x.
∴ Second number = 5x
According to the question:
5x - x = 132
or, 4x = 132
or, 4x4 = 1324                        [Dividing both the sides by 4]
or, x = 33
Thus, the required numbers are x = 33 and 5x = 5×33 = 165.

Page No 144:

Question 10:

The sum of two consecutive even numbers is 74. Find the numbers.

Answer:

Let one of the even number be x.
Then, the other consecutive even number is (x + 2).
According to the question:
x + (x + 2)  = 74
or, 2x + 2  = 74
or, 2x + 2 - 2 = 74 - 2           [Subtracting 2 from both the sides]
or, 2x = 72
or, 2x2 = 722                         [Dividing both the sides by 2]
or, x = 36
Thus, the required numbers are x = 36 and x+ 2 = 38.

Page No 144:

Question 11:

The sum of three consecutive odd numbers is 21. Find the numbers.

Answer:

Let the first odd number be x.
Then, the next consecutive odd numbers will be (x + 2) and (x + 4).
According to the question:
x + (x + 2) + (x + 4)  = 21
or, 3x + 6  = 21
or, 3x + 6 - 6 = 21 - 6           [Subtracting 6 from both the sides]
or, 3x = 15
or, 3x3 = 153                         [Dividing both the sides by 3]
or, x = 5
∴ Required numbers are:
x = 5
x + 2 = 5 + 2 = 7
x + 4 = 5 + 4 = 9

Page No 144:

Question 12:

Reena is 6 years older than her brother Ajay. If the sum of their ages is 28 years, what are their present ages?

Answer:

Let the present age of Ajay be x years.
Since Reena is 6 years older than Ajay, the present age of Reena will be (x+ 6) years.
According to the question:
x + (x + 6) = 28
or, 2x + 6 = 28
or, 2x + 6 - 6 = 28 - 6                 [Subtracting 6 from both the sides]
or, 2x = 22
or, 2x2 = 222                              [Dividing both the sides by 2]
or, x = 11
∴ Present age of Ajay = 11 years
Present age of Reena  = x +6 = 11 + 6
                                 = 17 years

Page No 144:

Question 13:

Deepak is twice as old as his brother Vikas. If the difference of their ages be 11 years, find their present ages.

Answer:

Let the present age of Vikas be x years.
Since Deepak is twice as old as Vikas, the present age of Deepak will be 2x years.
According to the question:
2x - x = 11
x = 11
∴ Present age of Vikas = 11 years
Present age of Deepak  = 2x = 2×11
                                   = 22 years

Page No 144:

Question 14:

Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find their present ages.

Answer:

Let the present age of Rekha be x years.
As Mrs. Goel is 27 years older than Rekha, the present age of Mrs. Goel will be (x + 27) years.
After 8 years:
Rekha's age = (x + 8) years
Mrs. Goel's age = (x + 27 + 8)
                      = (x + 35) years

According to the question:
(x + 35) = 2(x + 8)
or, x + 35 = 2×x + 2×8                [On expanding the brackets]
or, x + 35 = 2x + 16
or, 35 - 16 = 2x - x                [Transposing 16 to the L.H.S. and x to the R.H.S.]
or, x = 19
∴ Present age of Rekha = 19 years
Present age of Mrs. Goel = x + 27
                                     = 19 + 27
                                     = 46 years



Page No 145:

Question 15:

A man is 4 times as old as his son. After 16 years he will be only twice as old as his son. Find their present ages.

Answer:

Let the present age of the son be x years.
As the man is 4 times as old as his son, the present age of the man will be (4x) years.
After 16 years:
Son's age  = (x + 16) years
Man's age = (4x + 16) years

According to the question:
(4x + 16) = 2(x + 16)
or, 4x + 16 = 2×x + 2×16                [On expanding the brackets]
or, 4x + 16 = 2x + 32
or, 4x - 2x = 32 - 16                [Transposing 16 to the R.H.S. and 2x to the L.H.S.]
or, 2x = 16
or, 2x2 = 162                           [Dividing both the sides by 2]
or, x = 8
∴ Present age of the son = 8 years
Present age of the man  = 4x = 4×8
                                    = 32 years

Page No 145:

Question 16:

A man is thrice as old as his son. Five years ago the man was four times as old as his son. Find their present ages.

Answer:

Let the present age of the son be x years.
As the man is 3 times as old as his son, the present age of the man will be (3x) years.

5 years ago:
Son's age = (x - 5) years
Man's age = (3x - 5) years

According to the question:
(3x - 5) = 4(x - 5)
or, 3x - 5 = 4×x - 4×5                [On expanding the brackets]
or, 3x - 5 = 4x - 20
or, 20 - 5 = 4x - 3x                [Transposing 3x to the R.H.S. and 20 to the L.H.S.]
or, x = 15
∴ Present age of the son = 15 years
Present age of the man  = 3x = 3×15
                                    = 45 years

Page No 145:

Question 17:

After 16 years, Fatima will be three times as old as she is now. Find her present age.

Answer:

Let the present age of Fatima be x years.

After 16 years:
Fatima's age = (x + 16) years

According to the question:
x + 16 = 3(x)
or, 16 = 3x - x               [Transposing x to the R.H.S.]
or, 16 = 2x
or, 2x2 = 162                [Dividing both the sides by 2]
or, x = 8
∴ Present age of Fatima = 8 years

Page No 145:

Question 18:

After 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today?

Answer:

Let the present age of Rahim be x years.
After 32 years:
Rahim's age = (x + 32) years
8 years ago:
Rahim's age = (x - 8) years
According to the question:
x + 32 = 5(x - 8)
or, x + 32  = 5x - 5×8              
or, x + 32 = 5x - 40
or, 40 + 32 = 5x - x                      [Transposing 'x' to the R.H.S. and 40 to the L.H.S.]
or, 72 = 4x
or, 4x4 = 724                               [Dividing both the sides by 4]
or, x = 18
Thus, the present age of Rahim is 18 years.

Page No 145:

Question 19:

A bag contains 25-paisa and 50-paisa coins whose total value is Rs 30. If the number of 25-paisa coins is four times that of 50-paisa coins, find the number of each type of coins.

Answer:

Let the number of 50 paisa coins be x.
Then, the number of 25 paisa coins will be 4x.
According to the question:
0.50(x) + 0.25(4x) = 30
or, 0.5x + x = 30
or, 1.5x = 30
or, 1.5x1.5 = 301.5          [Dividing both the sides by 1.5]
or, x = 20
Thus, the number of 50 paisa coins is 20.
Number of 25 paisa coins = 4x = 4×20 = 80

Page No 145:

Question 20:

Five times the price of a pen is Rs 17 more than three times its price. Find the price of the pen.

Answer:

Let the price of one pen be Rs x.
According to the question:
5x = 3x + 17
or, 5x - 3x = 17                    [Transposing 3x to the L.H.S.]
or, 2x = 17
or, 2x2 = 172                        [Dividing both the sides by 2]
or, x = 8.50
∴ Price of one pen = Rs 8.50

Page No 145:

Question 21:

The number of boys in a school is 334 more than the number of girls. If the total strength of the school is 572, find the number of girls in the school.

Answer:

Let the number of girls in the school be x.
Then, the number of boys in the school will be (x + 334).
Total strength of the school = 572

∴ x + (x + 334) = 572
or, 2x + 334 = 572
or, 2x + 334 - 334 = 572 - 334                {Subtracting 334 from both the sides]
or, 2x = 238
or, 2x2 = 2382                                          [Dividing both the sides by 2]
or, x = 119
∴ Number of girls in the school = 119

Page No 145:

Question 22:

The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168 metres, fund its dimensions.

Answer:

Let the breadth of the park be x metres.
Then, the length of the park will be 3x metres.
Perimeter of the park = 2 (Length + Breadth) = 2 ( 3x + x ) m
Given perimeter = 168 m

∴ 2(3x + x) = 168                           
or, 2 ( 4x ) = 168
or, 8x = 168                                      [On expanding the brackets]
or, 8x8 = 1688                                [Dividing both the sides by 8]
or, x = 21 m
∴ Breadth of the park = x = 21 m
Length of the park = 3x = 3×21 = 63 m

Page No 145:

Question 23:

The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

Answer:

Let the breadth of the hall be x metres.
Then, the length of the hall will be (x + 5) metres.
Perimeter of the hall = 2(Length + Breadth) = 2( x + 5 + x) metres
Given perimeter of the rectangular hall = 74 metres

∴ 2( x + 5 + x) = 74                             
or, 2 ( 2x + 5 ) = 74
or, 2 ×2x + 2 ×5 = 74                              [On expanding the brackets]
or, 4x + 10 = 74
or, 4x + 10 - 10 = 74 - 10                     [Subtracting 10 from both the sides]
or, 4x = 64
or, 4x4 = 644                                        [Dividing both the sides by 4]
or, x = 16 metres
∴ Breadth of the park = x
                                  = 16 metres
Length of the park = x + 5 = 16 + 5
                            = 21 metres

Page No 145:

Question 24:

A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.

Answer:

Let the breadth of the rectangle be x cm.
Then, the length of the rectangle will be (x + 7) cm.
Perimeter of the rectangle = 2(Length + Breadth) = 2( x + 7 + x) cm
Given perimeter of the rectangle = Length of the wire = 86 cm

∴ 2( x + 7 + x) = 86                            
or, 2 ( 2x + 7 ) = 86
or, 2 ×2x + 2 × 7 = 86                              [On expanding the brackets]
or, 4x + 14 = 86
or, 4x + 14 - 14 = 86 - 14                   [Subtracting 14 from both the sides]
or, 4x = 72
or, 4x4 = 724                                   [Dividing by 4 on both the sides]
or, x = 18 metres
Breadth of the hall = x
                             = 18 metres
Length of the hall = x + 7
                           = 18 + 7
                           = 25 metres



Page No 146:

Question 1:

A man earns Rs 25 per hour. How much does he earn in x hours?

Answer:

Earning of the man per hour = Rs 25

Earning of the man in x hours = Rs (25×x)
                                             = Rs 25x

Page No 146:

Question 2:

The cost of 1 pen is Rs 16 and the cost of 1 pencil is Rs 5. What is the total cost of x pens and y pencils.

Answer:

Cost of 1 pen = Rs 16
∴ Cost of 'x' pens = Rs 16 × x
                           = Rs 16x
Similarly, cost of 1 pencil = Rs 5
∴ Cost of 'y' pencils = Rs 5×y
                               = Rs 5y
∴ Total cost of x pens and y pencils = Rs (16x + 5y)

Page No 146:

Question 3:

Lalit earns Rs x per day and spends Rs y per day. How much does he save in 30 days?

Answer:

Lalit's earning per day = Rs x
∴ Lalit's earning in 30 days = Rs 30 ×x
                                          = Rs 30x

Similarly, Lalit's expenditure per day = Rs y
∴ Lalit's expenditure in 30 days = Rs 30 × y
                                                = Rs 30y
∴ In 30 days, Lalit saves = (Total earnings - Total expenditure)
                                      = Rs (30x - 30y)
                                      = Rs 30(x - y)

Page No 146:

Question 4:

Three times a number added to 8 gives 20. Find the number.

Answer:

Let the required number be x.
Three times this number is 3x.
On adding 8, the number becomes 3x + 8.
3x + 8 = 20
or, 3x + 8 - 8 = 20 - 8                [Subtracting 8 from both the sides]
or, 3x = 12
or, 3x3 = 123                             [Dividing both the sides by 3]
or, x = 4
∴ Required number = 4

Page No 146:

Question 5:

If x = 1, y = 2 and z = 3, find the value of x2 + y2 + 2xyz.

Answer:

Given:
x =1
y = 2
z = 3

Substituting x = 1, y = 2 and z = 3 in the given equation (x2 + y2 + 2xyz):

12 + 22  + 21231 + 4 + 12 = 17

Page No 146:

Question 6:

Solve: 4x + 9 = 17.

Answer:

4x + 9 = 17
or, 4x + 9 - 9 = 17 - 9                          [Subtracting 9 from both the sides]
or, 4x = 8
or, 4x4 = 84                                          [Dividing both the sides with 4]
or, x = 2

Page No 146:

Question 7:

Solve: 3(x + 2) − 2(x − 1) = 7.

Answer:

3(x + 2) − 2(x − 1) = 7.
or, 3 × x + 3 × 2 - 2 × x - 2 × (-1) = 7                    [On expanding the brackets]
or, 3x + 6 − 2x + 2 = 7
or, x + 8 = 7
or, x + 8 − 8 = 7 − 8                                            [Subtracting 8 from both the sides]
or, x = −1

Page No 146:

Question 8:

Solve: 2x5-x2=52.

Answer:

2x5 -x2 = 52
or, 4x - 5x10 = 52                               [Taking the L.C.M. as 10]
or, -x10 = 52
or, -x10×-10 = 52×-10                          [Multiplying both the sides by (−10)]
or, x = −25

Page No 146:

Question 9:

The sum of three consecutive natural numbers is 51. Find the numbers.

Answer:

Let the three consecutive natural numbers be x, (x + 1) and (x + 2).

∴ x + (x + 1) + (x + 2) = 51
3x + 3 = 51
3x + 3 - 3 = 51 - 3                   [Subtracting 3 from both the sides]
3x = 48
3x3 = 483                                [Dividing both the sides by 3]
x = 16
Thus, the three natural numbers are x = 16, x+1 = 17 and x+2 = 18.

Page No 146:

Question 10:

After 16 years, Seema will be three times as old as she is now. Find her present age.

Answer:

Let the present age of Seema be x years.
After 16 years:
Seema's age = x + 16

After 16 years, her age becomes thrice of her age now
∴ x + 16 = 3x
or, 16 = 3x - x                     [Transposing x to the R.H.S.]
or, 2x = 16
or, 2x2 = 162                      [Dividing both the sides by 2]
or, x = 8 years

Page No 146:

Question 11:

By how much does I exceed 2x − 3y − 4?
(a) 2x − 3y − 5
(b) 2x − 3y − 3
(c) 5 − 2x + 3y
(d) none of these

Answer:

(c) 5 − 2x + 3y

1 exceeds 2x − 3y − 4.

∴1 - (2x − 3y − 4) = 1 -2x + 3y + 4
                              = 5 - 2x + 3y

∴ 1 exceeds 2x − 3y − 4 by 5 - 2x + 3y.

Page No 146:

Question 12:

What must be added to 5x3 − 2x2 + 6x + 7 to make the sum x3 + 3x2x + 1?
(a) 4x3 − 5x2 + 7x + 6
(b) −4x3 + 5x2 − 7x − 6
(c) 4x3 + 5x2 − 7x + 6
(d) none of these

Answer:

  (b) −4x3 + 5x2 − 7x − 6
In order to find what must be added, we subtract (5x3 − 2x2 + 6x + 7) from (x3 + 3x2x + 1).

    (x3 + 3x2x + 1)  − ( 5x3 − 2x2 + 6x + 7)
or,  x3 + 3x2x + 1 − 5x3  + 2x2 − 6x − 7
or,  x3 − 5x3+ 3x2+ 2x2x − 6x+ 1 − 7
or, −4x3 + 5x2 − 7x − 6

Page No 146:

Question 13:

2x − [3y − {2x − (yx)}] = ?
(a) 5x − 4y
(b) 4y − 5x
(c) 5y − 4x
(d) 4x − 5y

Answer:

(a) 5x − 4y

2x − [3y − {2x − (yx)}]
= 2x − [3y − {2x −  y + x}]
= 2x − [3y − {3x −  y}]
= 2x −  [3y −  3x + y]
= 2x −  [4y −  3x]
= 2x −  4y + 3x
= 5x −  4y

Page No 146:

Question 14:

The coefficient of x in −5xyz is
(a) −5
(b) 5yz
(c) −5yz
(d) yz

Answer:

(c) −5yz

All the terms in the expression −5xyz barring x will be the coefficient of x, i.e. −5yz.

Page No 146:

Question 15:

13(x + 7 + z) is a
(a) monomial
(b) binomial
(c) trinomial
(d) quadrinomial

Answer:

(b) trinomial
Since it contains three variables, i.e. 'x', 'y' and 'z', it is a trinomial.

Page No 146:

Question 16:

If x5=1, then
(a) x=15
(b) x = 5
(c) x = (5 + 1)
(d) none of these

Answer:

(b) x = 5

x5 = 1or, x5 × 5= 1 × 5   [Multiplying both the sides by 5]
or, x = 5

Page No 146:

Question 17:

If x = 1, y = 2 and z = 3 then (x2 + y2 + z2) = ?
(a) 6
(b) 12
(c) 14
(d) 15

Answer:

(c) 14
Substituting x = 1, y = 2 and z = 3 in (x2 + y2 + z2):
12  + 22 + 32
or, 1 + 4 + 9 = 14

Page No 146:

Question 18:

If 13 x + 5 = 8, then x = ?
(a) 3
(b) 6
(c) 9
(d) 12

Answer:

(c) 9

13x + 5 = 8
or, 13x + 5 - 5 = 8-5                                  [Subtracting 5 from both the sides]
or, 13x = 3
or, 13x × 3 = 3×3                                        [Multiplying both the sides by 3]
or, x = 9

Page No 146:

Question 19:

Fill in the blanks.
(i) An expression having one term is called a ...... .
(ii) An expression having two term is called a ...... .
(i) An expression having three term is called a ...... .
(iv) 3x - 5 = 7 - x  x = ...... .
(v) (b2 - a2) - (a2 - b2) = ...... .

Answer:

(i) monomial
(ii) binomial
(iii) trinomial
(iv) x = 3
3x - 5 = 7 - x
or, 3x + x = 7 + 5                   [Transposing x to the L.H.S. and 5 to the R.H.S.]
or, 4x = 12
or, 4x4 = 124
or, x = 3
(v) 2b2 - 2a2
or, b2 - a2 - a2 + b2
or, 2b2 - 2a2
( b2  a2)  (a2  b2)

(b2  a2)  (a2  b2
(b2  a2)  (a2  b2)
(b2  a2)  (a2  b2)



Page No 147:

Question 20:

Write 'T' for true and 'F' for false for each of the statements given below:
(i) −3xy2z is a monomial.
(ii) x=23 is solution of 2x + 5 = 8.
(iii) 2x + 3 = 5 is a linear equation.
(iv) The coefficient of x in 5xy is 5.
(v) 8 - x = 5  x = 3.

Answer:

(i) True
Since it has one term, it is a monomial.

(ii) False
2x + 5 = 8
or, 2x + 5 - 5 = 8 - 5               [Subtracting 5 from both the sides]
or, 2x = 3
or, x = 3/2 and not x = 2/3

(iii) True
This is because the maximum power of the variable x is 1.

(iv) False
The coefficient of x in 5xy would be 5y.

(v) True
    8 − x = 5
  or, 8 − 5 = x
    or, 3 = x



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