RD Sharma 2017 Solutions for Class 7 Math Chapter 8 Linear Equations In One Variable are provided here with simple step-by-step explanations. These solutions for Linear Equations In One Variable are extremely popular among class 7 students for Math Linear Equations In One Variable Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 7 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 8.12:

Question 1:

Solve each of the following equations and check your answers:
x − 3 = 5

Answer:

x − 3 = 5
Adding 3 to both sides, we get
x − 3 + 3 = 5 + 3  
⇒ x = 8
Verification:
Substituting x= 8 in LHS, we get
LHS =
x − 3 and RHS = 5
LHS = 8
− 3 = 5 and RHS = 5
LHS = RHS
Hence, verified
.

Page No 8.12:

Question 2:

x − 3 = 5
Adding 3 to both sides, we get
x − 3 + 3 = 5 + 3  
⇒ x = 8
Verification:
Substituting x= 8 in LHS, we get
LHS =
x − 3 and RHS = 5
LHS = 8
− 3 = 5 and RHS = 5
LHS = RHS
Hence, verified
.

Answer:

x + 9 = 13
Subtracting 9 from both sides, we get
=> x + 9 - 9 = 13 - 9
=> x = 4
Verification:
Substituting x = 4 on LHS, we get
LHS = 4 + 9 = 13 = RHS
LHS = RHS
Hence, verified.

Page No 8.12:

Question 3:

Solve each of the following equations and check your answers:
x-35=75

Answer:

x 35 = 75
Adding 35 to both sides, we get
=> x 3535 = 75 + 35
=> x = 7+35
=> x = 105
⇒ x = 2
Verification:
Substituting x = 2 in LHS, we get
LHS =
2 35=10-35=75, and RHS = 75
LHS = RHS
Hence, verified.

Page No 8.12:

Question 4:

Solve each of the following equations and check your answers:
3x = 0

Answer:

3x = 0
Dividing both sides by 3, we get
⇒ 3x3
  = 03
⇒ x = 0
Verification:
Substituting x = 0 in LHS = 3x, we get
LHS = 3 × 0 = 0 and RHS = 0

LHS = RHS
Hence, verified
.

Page No 8.12:

Question 5:

Solve each of the following equations and check your answes:
x2=0

Answer:

 x2 = 0
Multiplying both sides by 2, we get
x2×2 = 0 × 2
⇒ x = 0

Verification:
Substituting x= 0 in LHS, we get
LHS =02= 0 and RHS = 0
LHS = 0 and RHS = 0
LHS = RHS
Hence, verified.

Page No 8.12:

Question 6:

Solve each of the following equations and check your answers:
x-13=23

Answer:

x 13 = 23
⇒ Adding 1 to both sides, we get

x 13 +1 = 23 + 1
=> x = 2+13
=> x = 33
⇒ x = 1
Verification:
Substituting x= 1 in LHS, we get
LHS =
1 1 =3-13=2,  and RHS = 2
LHS = RHS
Hence, verified.

Page No 8.12:

Question 7:

Solve each of the following equations and check your answers:
x+12=72

Answer:

x + 12 = 72
⇒ Subtracting 12 from both sides, we get
x + 12 12 = 72 12
x=7-12=62
⇒ x = 3
Verification:
Substituting x = 3 in LHS, we get
LHS = 3
+ 12=6+12=72, and RHS = 72
LHS = RHS
Hence, verified.

Page No 8.12:

Question 8:

Solve each of the following equations and check your answers:
10 − y = 6

Answer:

10 − y = 6
Subtracting 10 from both sides, we get
10 − y − 10 = 6 − 10
⇒ −y = −4.
⇒ Multiplying both sides by −1, we get
⇒ −y ×−1 = −4 ×−1
⇒ y = 4
Verification:
Substituting y = 4 in LHS, we get
LHS = 10
− y = 10-4 = 6 and RHS = 6
LHS = RHS
Hence, verified
.

Page No 8.12:

Question 9:

Solve each of the following equations and check your answers:
7 + 4y = −5

Answer:

7 + 4y = −5
Subtracting 7 from both sides, we get
⇒ 7 + 4y − 7 = −5 − 7
⇒  4y  = −12
Dividing both sides by 4, we get

⇒ y= -12  4
⇒  y  = −3

Verification :
Substituting y = −3 in LHS, we get
LHS = 7 + 4y = 7 
+ 4(−3) = 7 − 12 = −5, and RHS = −5
LHS = RHS
Hence, verified
.

Page No 8.12:

Question 10:

Solve each of the following equations and check your answers:
45-x=35

Answer:

45 − x = 35
Subtracting45 from both sides, we get
45
− x − 45= 35  +    6+45
⇒ −x = 3-45
-x=-15
Multiplying both sides by -1, we get
⇒ −x × (−1) = −15× (−1)
⇒ x = 15
Verification:
Substituting x= 15 in LHS, we get
LHS =
45 15=4-15=35, and RHS = 35
LHS = RHS
Hence, verified.

Page No 8.12:

Question 11:

Solve each of the following equations and check your answers:
2y-12=-13

Answer:

2y  12 =-13
Adding 12to both sides, we get
⇒ 2y  − 12+12=-13 + 12
⇒ 2y = -2 + 36
⇒ 2y = 16
Dividing both sides by 2, we get
2y2 = 16×2
⇒ y = 112
Verification:
Substituting y = 112 in LHS, we get
LHS = 2×112 - 12 = 16 - 12 =1-36=-26 =
-13and RHS = -13
LHS = RHS
Hence, verified.

Page No 8.12:

Question 12:

Solve each of the following equations and check your answers:
14=7x10-8

Answer:

 14=7x10 - 8
Adding 8 to both sides, we get
⇒ 14 + 8 = 7x10 − 8 + 8
⇒ 22 = 7x10
Multiplying both sides by 10, we get
⇒ 22 × 10 = 7x10× 10
⇒ 220 = 7x
Dividing both sides by 7, we get
2207 = 7x7
⇒ x = 2207
Verification:

Substituting x = 2207 in RHS, we get
LHS = 14, and RHS = 7220710 - 8 = 22010 - 8 = 22 - 8 = 14
LHS = RHS
Hence, verified.

Page No 8.12:

Question 13:

Solve each of the following equations and check your answers:
3 (x + 2) = 15

Answer:

3 (x + 2) = 15
Dividing both sides by 3, we get
3 (x + 2)3= 153
⇒ (x + 2) = 5
Subtracting 2 from both sides, we get
x + 2 - 2 = 5 - 2
x = 3
Verification:
Substituting x = 3 in LHS, we get
LHS = 3 (x + 2)= 3 (3+2) = 3×5 = 15, and RHS = 15
LHS = RHS
Hence, verified.

Page No 8.12:

Question 14:

Solve each of the following equations and check your answers:
x4=78

Answer:

 x4 = 78
Multiplying both sides by 4, we get
⇒ 
x4× 4 = 78 × 4
⇒ x = 72
Verification:
Substituting x = 72 in LHS, we get
LHS = 72×478, and RHS = 78
LHS = RHS
Hence, verified.

Page No 8.12:

Question 15:

Solve each of the following equations and check your answers:
13-2x=0

Answer:

 13 - 2x = 0
Subtracting 13 from both sides, we get
⇒ 
13 - 2x - 13 = 0 -13
-2x = -13
Multiplying both sides by -1, we get
-2x × (-1) = -13 ×(-1)
⇒ 2x = 13
Dividing both sides by 2, we get
2x2 = 13×2
⇒ x = 16
Verification:
Substituting x = 16 in LHS, we get
LHS = 13 - 2 × 16 = 13- 13 = 0, and RHS = 0
LHS = RHS
Hence, verified
.

Page No 8.12:

Question 16:

Solve each of the following equations and check your answers:
3(x + 6) = 24

Answer:

3(x + 6) = 24

Dividing both sides by 3, we get
3 (x + 6)3=243
⇒ (x + 6) = 8
Subtracting 6 from both sides, we get
x + 6 - 6 = 8 - 6
x = 2
Verification:
Substituting x = 2 in LHS, we get
LHS = 3 (x + 6) = 3 (2 + 6) = 3×8 = 24, and RHS = 24
LHS = RHS
Hence, verified.

Page No 8.12:

Question 17:

Solve each of the following equations and check your answers:
3(x + 2) − 2(x − 1) = 7

Answer:

3(x + 2) − 2(x − 1) = 7
On expanding the brackets, we get 
3× x + 3 × 2 - 2 × x + 2 × 1 = 7
⇒ 3x + 6 - 2x + 2 = 7
⇒ 3x - 2x + 6 + 2 = 7
⇒ x + 8 = 7
Subtracting 8 from both sides, we get
x + 8 -8 = 7 - 8
x = -1
Verification:
Substituting x = -1 in LHS, we get
LHS = 3 (x + 2) -2(x -1), and RHS = 7
LHS = 3 (-1 + 2) -2(-1-1) = (3×1) - (2×-2) = 3 + 4  = 7, and RHS = 7
LHS = RHS
Hence, verified.

Page No 8.12:

Question 18:

Solve each of the following equations and check your answers:
8(2x  5) − 6(3x − 7) = 1

Answer:

8(2x  5) − 6(3x − 7) = 1
On expanding the brackets, we get
⇒ (8×2x) - (8 × 5) - (6 × 3x) + (-6)×(-7)  =  1
⇒ 16x - 40 - 18x + 42 = 1
⇒ 16x - 18x + 42 - 40 = 1
-2x + 2 = 1
Subtracting 2 from both sides, we get
-2x + 2 - 2 = 1 - 2
-2x  = -1
Multiplying both sides by -1, we get
-2x ×(-1) = -1×(-1)
⇒ 2x = 1
Dividing both sides by 2, we get
2x2 = 12
⇒ x = 12

Verification:
Substituting x = 12 in LHS, we get
= 8(2×12- 5) - 6(3×12 − 7)

= 8(1 − 5) − 6(32 − 7)

= 8×(−4) − (6 ×32)  + (6 ×7)
= −32 − 9 + 42 = − 41 + 42 = 1 = RHS
LHS = RHS
Hence, verified.

Page No 8.12:

Question 19:

Solve each of the following equations and check your answers:
6(1 4x) + 7(2 + 5x) = 53

Answer:

6(1 4x) + 7(2 + 5x) = 53
On expanding the brackets, we get
⇒ (6×1) - (6 × 4x) + (7 ×2) + (7×5x)  =  53
⇒ 6 - 24x + 14 + 35x = 53
⇒ 6 + 14 + 35x - 24x = 53
⇒ 20 + 11x = 53
Subtracting 20 from both sides, we get
⇒ 20 + 11x - 20  = 53 - 20
⇒ 11x  = 33
⇒ Dividing both sides by 11, we get
11x11 = 3311
⇒ x = 3


Verification:
Substituting x = 3 in LHS, we get
=
6(1 4×3) + 7(2 + 5×3)
= 6(1 − 12) + 7(2 + 15)
= 6(−11) + 7(17)
= −66 + 119 = 53 = RHS
LHS = RHS
Hence, verified.

Page No 8.12:

Question 20:

Solve each of the following equations and check your answers:
5(2 3x) − 17(2x − 5) = 16

Answer:

5(2 3x) − 17(2x − 5) = 16
On expanding the brackets, we get
⇒ (5×2) - (5 × 3x) − (17 ×2x ) + (17×5)  =  16
⇒ 10 - 15x − 34x + 85 = 16
⇒ 10 + 85 − 34x - 15x = 16
⇒ 95 - 49x = 16
Subtracting 95 from both sides, we get
⇒ - 49x + 95 - 95  = 16 - 95
⇒ - 49x  = -79
Dividing both sides by -49, we get
-49x-49 = -79-49

⇒ x = 7949

Verification:
Substituting x = 7949  in LHS, we get
=
5(2 3×7949) − 17(2×7949 − 5)
= (5×2) - (5 × 3×7949 ) − (17 ×2×7949 ) + (17×5) 
= 10 -118549268649 + 85
= 490 - 1185 - 2686 + 416549
= 78449
= 16
= RHS
So, LHS = RHS
Hence, verified.

Page No 8.12:

Question 21:

Solve each of the following equations and check your answers:
x-35-2=-1

Answer:

 x - 35- 2 = -1
Adding 2 to both sides, we get
x - 35 - 2 + 2 = -1 + 2
x - 35 = 1
Multiplying both sides by 5, we get
x - 3 5 × 5 = 1 × 5
⇒ x - 3 = 5
Adding 3 to both sides, we get
⇒ x - 3 + 3  = 5 + 3
⇒ x = 8

Verification:
Substituting x = 8  in LHS, we get
 = 8 -3 5-2
 = 55 - 2
= 1 - 2
= -1
= RHS
LHS = RHS
Hence, verified.

Page No 8.12:

Question 22:

Solve each of the following equations and check your answers:
5(x 2) + 3(x + 1) = 25

Answer:

5(x 2) + 3(x + 1) = 25
On expanding the brackets, we get
⇒ (5 × x) - (5 × 2) + (3 × x) + (3×1)  =  25
⇒ 5x - 10 + 3x + 3 = 25
⇒ 5x  + 3x - 10 + 3 = 25
⇒ 8x - 7  = 25
Adding 7 to both sides, we get
⇒ 8x - 7 + 7 = 25 +7
⇒ 8x  = 32
Dividing both sides by 8, we get
8x8 = 328
⇒ x =
4

Verification:
Substituting x = 4  in LHS, we get

=
5(4 2) + 3(4 + 1)
= 5(2) + 3(5)
= 10 + 15
= 25
= RHS
LHS = RHS
Hence, verified.



Page No 8.19:

Question 1:

Solve each of the following equations. Also, verify the result in each case.
6x + 5 = 2x + 17

Answer:

We have
⇒ 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
⇒ 6x - 2x = 17 - 5
⇒ 4x = 12   
Dividing both sides by 4, we get  
4x4=124
⇒ x = 3
Verification:
Substituting x =3 in the given equation, we get
6×3 + 5 = 2×3 + 17
18 + 5 = 6 + 17
23 = 23
LHS = RHS
Hence, verified.

Page No 8.19:

Question 2:

Solve each of the following equations. Also, verify the result in each case.
2(5x − 3) − 3(2x − 1) = 9

Answer:

We have
⇒2(5x − 3) − 3(2x − 1) = 9
Expanding the brackets, we get
2×5x - 2×3 -3×2x  + 3×1 = 9
⇒ 10x − 6 − 6x + 3 = 9
⇒ 10x − 6x − 6 + 3 = 9
⇒ 4x  − 3 = 9
Adding 3 to both sides, we get    
⇒ 4x − 3 + 3= 9 + 3
⇒ 4x = 12
Dividing both sides by 4, we get
4x4 = 124
⇒ Thus, x = 3.
Verification:
Substituting x =3 in LHS, we get
=2(5×3 − 3) − 3(2×3 − 1)
=2×12 − 3 × 5
=24 − 15
= 9

LHS = RHS

Hence, verified.

Page No 8.19:

Question 3:

Solve each of the following equations. Also, verify the result in each case.
x2=x3+1

Answer:

x2 = x3 + 1
Transposing x3 to LHS, we get
x2-x3  = 13x-2x6 = 1 
x6 = 1      
Multiplying both sides by 6, we get                 
x6×6 = 1×6                            
 x = 6
Verification:
Substituting x = 6 in the given equation, we get
62= 63+ 1
3 = 2 + 1
3 = 3
LHS = RHS
Hence, verified.

Page No 8.19:

Question 4:

Solve each of the following equations. Also, verify the result in each case.
x2+32=2x5-1

Answer:

 x2+32=2x5-1   
Transposing 2x5 to LHS and 32 to RHS, we get
=> x2-2x5 = -1 -32
                
=> 5x-4x10= -2-32
=> x10= -5  2 
Multiplying both sides by 10, we get                 
=> x10×10 = -52×10                                    
=>  x = -25
Verification:
Substituting x = -25 in the given equation, we get
-252+32= 2×-255-1
-222= -10 -1
-11= -11
LHS = RHS
Hence, verified.

Page No 8.19:

Question 5:

Solve each of the following equations. Also, verify the result in each case.
34(x-1)=x-3

Answer:

 34x-1 = x-3
On expanding the brackets on both sides, we get
=>34x- 34= x-3
Transposing 34x to RHS and 3 to LHS, we get
=> 3-34= x-34x                       
=> 12-34=4x-3x4
=> 94=x4
Multiplying both sides by 4, we get
=> x = 9                                       

Verification:
Substituting x = 9 on both sides, we get
349-1=9-3
34×8=6
6 = 6
LHS = RHS
Hence, verified.

Page No 8.19:

Question 6:

Solve each of the following equations. Also, verify the result in each case.
3(x − 3) = 5(2x + 1)

Answer:

6. 3(x − 3) = 5(2x + 1)
On expanding the brackets on both sides, we get
=> 3×x - 3×3 = 5×2x + 5×1           
=> 3x - 9 = 10x + 5
Transposing 10x to LHS and 9 to RHS, we get
=> 3x - 10x = 9 + 5                            
=> -7x = 14
Dividing both sides by 7, we get
=> -7x7  =147                                   
=> x = -2
Verification:
Substituting x = -2 on both sides, we get
3-2-3 = 52-2 +1
3-5 = 5-3
-15 = -15
LHS = RHS
Hence, verified.

Page No 8.19:

Question 7:

Solve each of the following equations. Also, verify the result in each case.
3x − 2 (2x − 5) = 2(x + 3) − 8

Answer:

3x − 2 (2x − 5) = 2(x + 3) − 8
On expanding the brackets on both sides, we get
=> 3x-2×2x+2×5 = 2×x + 2×3 -8                 
=> 3x -4x + 10 = 2x + 6 -8
=> -x + 10 = 2x - 2
Transposing x to RHS and 2 to LHS, we get
=> 10 + 2 = 2x + x                                                  
=> 3x = 12
Dividing both sides by 3, we get
=> 3x 3 = 123                                                       
 => x = 4
Verification:
Substituting x = 4 on both sides, we get
34 - 224-5 = 24+3-8
12-2 (8 - 5) = 14-8
12 - 6 = 6
6 = 6
LHS = RHS
Hence, verified.

Page No 8.19:

Question 8:

Solve each of the following equations. Also, verify the result in each case.
x-x4-12=3+x4

Answer:

x - x4-12= 3 + x4
Transposing x4 to LHS and -12 to RHS, we get
=> x - x4 -x4 = 3 + 12               
=> 4x-x - x4 = 6 + 12
=> 2x4 = 72
Multiplying both sides by 4, we get
=> 2x4×4 = 72×4                         
 => 2x = 14
Dividing both sides by 2, we get
=> 2x2 = 142                                 
=> x = 7
Verification:
Substituting x = 7 on both sides, we get

7 -74- 12 = 3 + 74
28 - 7 - 24 = 12 + 74
194 = 194
LHS = RHS
Hence, verified.

Page No 8.19:

Question 9:

Solve each of the following equations. Also, verify the result in each case.
6x-29+3x+518=13

Answer:

6x - 29 + 3x + 518 = 13
=> 6x×2 - 2×2 + 3x + 518 = 13
=> 12x - 4 + 3x + 518 = 13
=> 15x + 118 = 13
Multiplying both sides by 18, we get
=> 15x + 118×18 = 13×18                 
=> 15x + 1 = 6
Transposing 1 to RHS, we get
=> 15x = 6-1                                           
=> 15x = 5
Dividing both sides by 15, we get
=> 15x15 = 515                                         
=> x = 13
Verification:
Substituting x = 13 on both sides, we get
613-29 + 313 +518 = 13
2 - 29 + 1 +5 18 = 13
0 + 618 = 13
13 = 13
LHS = RHS
Hence, verified.

Page No 8.19:

Question 10:

Solve each of the following equations. Also, verify the result in each case.
m-m-12=1-m-23

Answer:

m-m-12 = 1 - m-23
=> 2m-m -(- 1) 2 = 3 -m -(-2)3
=> m + 12 = 3 - m + 23
=> m+ 1 2 = 5-m3
=> m2 + 12 = 53 -m3
Transposing m/3 to LHS and 1/2 to RHS, we get
=> m2 + m3 = 53 -12                       
=> 3m+2m6 = 10-36
Multiplying both sides by 6, we get
=> 5m6×6 = 76×6                               
 => 5m = 7
Dividing both sides by 5, we get
=> 5m5 = 75                                         
=> m = 75
Verification:
Substituting m =75 on both sides, we get
75-75-12 = 1 -75-23
75-7-552 = 1 - 7-1053
75-25×2 = 1 - -35×3
75- 15 = 1 + 15
65 = 65
LHS = RHS
Hence, verified.

Page No 8.19:

Question 11:

Solve each of the following equations. Also, verify the result in each case.
(5x-1)3-(2x-2)3=1

Answer:

5x -13- 2x - 23 = 1
 5x - 1 - 2x - (-2)3= 15x - 1 - 2x + 23 = 15x - 2x + 2 -13  = 13x + 13 = 1
Multiplying both sides by 3, we get
3x + 1 3×3 = 1× 3                        
=> 3x + 1 = 3
Subtracting 1 from both sides, we get
=> 3x + 1 - 1 = 3 - 1                               
=> 3x = 2
Dividing both sides by 3, we  get
=> 3x 3= 23                                             
=> x = 23
Verification:
Substituting x =23 in LHS, we get
=523 - 13-223 - 23=103-13 - 43- 23=10-333-4-633=73×3--23×3=79 + 29=  99=  1 = RHS
LHS = RHS
Hence, verified.

Page No 8.19:

Question 12:

Solve each of the following equations. Also, verify the result in each case.
0.6x+45=0.28x+1.16

Answer:

0.6x + 45 = 0.28x + 1.16
Transposing 0.28x to LHS and 4/5 to RHS, we get
=> 0.6x - 0.28x = 1.16 -45                   
=> 0.32x = 1.16 - 0.8
 => 0.32x = 0.36
Dividing both sides by 0.32, we get
  => 0.32x0.32 = 0.360.32                               
 => x = 98
Verification:
Substituting x = 98 on both sides, we get
0.698 + 45 = 0.2898 + 1.165.48 + 45 = 2.528 + 1.160.675 + 0.8 = 0.315 + 1.161.475 = 1.475

    LHS = RHS
Hence, verified.

Page No 8.19:

Question 13:

Solve ech of the following question. Also, verify the result in each case.
0.5x+x3=0.25x+7

Answer:

0.5x + x3 = 0.25x + 70510x+x3=25x100+7x2 + x3 =  x4 + 7
Transposing x/4 to LHS, we get

x2 + x3 -x4 = 7                     
6x + 4x - 3x12 = 77x12 = 7
Multiplying both sides by 12, we get
=> 7x12×12 = 7 × 12                      
=> 7x = 84
Dividing both sides by 7, we get
=> 7x7 = 847                                   
=> x = 12
Verification:
Substituting x = 12 on both sides, we get
0.512 + 123 = 0.2512 + 7
6 + 4 = 3 + 7
10 = 10
LHS =RHS
Hence, verified.



Page No 8.26:

Question 1:

If 5 is subtracted from three times a number, the result is 16. Find the number.

Answer:

Let the required number be 'x'. Then, 5 subtracted from 3 times x = 3x - 5.
⇒ 3x - 5 = 16
Adding 5 to both sides, we get
⇒ 3x - 5 + 5 = 16 + 5                        
⇒ 3x= 21
Dividing both sides by 3, we get
3x3 = 213                                     
⇒ x =7
Thus, the required number is 7.

Page No 8.26:

Question 2:

Find the number which when multiplied by 7 is increased by 78.

Answer:

Let the required number be 'x'. Thus, when multiplied by 7, it gives 7x, and x increases by 78.
⇒ 7x = x + 78
Transposing x to LHS, we get
⇒ 7x - x = 78                        
⇒ 6x = 78
Dividing both sides by 6, we get
6x6 = 786                                     
⇒ x =13
Thus, the required number is 13.

Page No 8.26:

Question 3:

Find three consecutive natural numbers such that the sum of the first and second is 15 more than the third.

Answer:

Let the first number be 'x'. Hence, the second number = x + 1 and the third number = x + 2.
⇒ Sum of first and second numbers = (x) + (x + 1).
ATQ:
⇒ (x) + (x + 1)  = 15 + (x + 2)
⇒ 2x + 1 = 17 + x
Transposing x to LHS and 1 to RHS, we get
⇒ 2x - x = 17 - 1         
⇒ x = 16
So, first number = x = 16
   Second number = x + 1 = 16 + 1 = 17
   Third number = x + 2 = 16 + 2 = 18
Thus, the required consecutive natural numbers are 16, 17 and 18.

Page No 8.26:

Question 4:

The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.

Answer:

Let the smaller number be 'x'. So, the larger number = x + 7.
ATQ:
⇒ 6x + (x + 7) = 77
⇒ 6x + x + 7 = 77
⇒ 7x + 7 = 77
Subtracting 7 from both sides, we get
⇒ 7x + 7 - 7 = 77 - 7           
⇒ 7x = 70
Dividing both sides by 7, we get
7x7 = 707                         
 x = 10
Thus, the smaller number = x = 10, and the larger number = x + 7 = 10 + 7 = 17.
The two required numbers are 10 and 17.

Page No 8.26:

Question 5:

A man says, "I am thinking of a number. When I divide it by 3 and then add 5, my answer is twice the number I thought of". Find the number.

Answer:

Let the number thought of by the man be 'x'.
So, ATQ:
x3 + 5 = 2x
Transposing x/3 to RHS, we get
⇒ 5 = 2x - x3                           
⇒ 5 = 6x - x3
⇒ 5 = 5x3
Multiplying both sides by 3, we get
5×3 = 5x3×3                       
⇒ 15 = 5x
Dividing both sides by 5, we get
155= 5x5                             
⇒ x = 3
Thus, the number thought of by the man is 3.

Page No 8.26:

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'.
So, ATQ:
⇒ 3x + 5 = 50
Subtracting 5 from both sides, we get
⇒ 3x + 5 - 5 = 50 - 5                 
⇒ 3x = 45
Dividing both sides by 3, we get
3x3 = 453                               
⇒ x = 15
Thus, the required number is 15.

Page No 8.26:

Question 7:

Shikha is 3 years younger to her brother Ravish. If the sum of their ages is 37 years, what are  their present ages?

Answer:

Let the present age of Shikha = 'x' years.
So, the present age of Shikha's brother Ravish = (x + 3) years.
So, sum of their ages  = x + (x+ 3)
⇒ x  + ( x + 3 ) = 37
⇒ 2x + 3 = 37
Subtracting 3 from both sides, we get
⇒ 2x + 3 - 3 = 37 - 3         
⇒ 2x = 34
Dividing both sides by 2, we get
2x2 = 342                       
⇒ x = 17
So, the present age of Shikha = 17 years, and the present age of Ravish = x+ 3 = 17 + 3 = 20 years.

Page No 8.26:

Question 8:

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

Answer:

 Let the present age of Nilu  = 'x' years.
Therefore, the present age of Nilu's mother, Mrs. Jain = (x + 27) years.
So, after 8 years,
Nilu's age = (x + 8), and Mrs. Jain's age  = (x + 27 + 8) = (x + 35) years
⇒ x + 35 = 2(x + 8)
Expanding the brackets, we get
⇒ x + 35 = 2x + 16
Transposing x to RHS and 16 to LHS, we get
⇒ 35 - 16 = 2x - x                    
⇒ x = 19
So, the present age of Nilu  = x =  19 years, and the present age of Nilu's mother  = x+ 27 = 19 + 27 = 46 years.

Page No 8.26:

Question 9:

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

Answer:

Let the present age of the son  = 'x' years.
Therefore, the present age of his father  = '4x'  years.
So, after 16 years,
Son's age = (x + 16) and father's age  = (4x + 16) years
ATQ:
⇒ 4x + 16 = 2(x + 16)
⇒ 4x + 16 = 2x + 32
Transposing 2x to LHS and 16 to RHS, we get
⇒ 4x - 2x = 32 - 16                    
⇒ 2x = 16
Dividing both sides by 2, we get
2x2 = 162                             
⇒ x = 8
So, the present age of the son  = x =  8 years, and the present age of the father = 4x = 4(8) = 32 years.

Page No 8.26:

Question 10:

The difference in age between a girl and her younger sister is 4 years. The younger sister in turn is 4 years older than her brother. The sum of the ages of the younger sister and her brother is 16. How old are the three children?

Answer:

Let the age of the girl = 'x' years.
So, the age of her younger sister = (x - 4) years.
Thus, the age of the brother = (x - 4 - 4) years = (x - 8) years.
ATQ:
⇒ (x - 4) + (x - 8) = 16
⇒  x + x - 4 - 8 = 16
⇒ 2x - 12 = 16
Adding 12 to both sides, we get
⇒ 2x - 12 + 12 = 16 + 12           
⇒ 2x = 28
Dividing both sides by 2, we get
2x2 = 282                             
⇒ x = 14
Thus, the age of the girl = x = 14 years, the age of the younger sister  = x - 4 = 14 - 4 = 10 years,
and the age of the younger brother = x - 8 = 14 -  8 = 6 years.

Page No 8.26:

Question 11:

One day, during their vacation at a beach resort, Shella found twice as many sea shells as Anita and Anita found 5 shells more than sandy. Together sandy and Shella found 16 sea shells. How many did each of them find?

Answer:

 Let the number of sea shells found by Sandy  = 'x'.
So, the number of sea shells found by Anita = (x + 5).
The number of sea shells found by Shella = 2 (x + 5 ).
According to the question,
⇒ x + 2 (x + 5)  = 16
⇒ x + 2x  + 10  = 16
⇒ 3x + 10 = 16
Subtracting 10 from both sides, we get
⇒ 3x + 10 - 10 = 16 - 10             
⇒ 3x = 6
Dividing both sides by 3, we get
3x3 = 63                                 
⇒ x = 2
Thus, the number of sea shells found by Sandy = x = 2, the number of sea shells found by Anita = x + 5 = 2 + 5 = 7,
and the number of sea shells found by Shella  = 2(x + 5) = 2(2 + 5) = 2(7 ) = 14.

Page No 8.26:

Question 12:

Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. Andy has 110 marbles which is 115 marbles less than Sandy. How many does each of them have?

Answer:

Let the number of marbles with Pandy = 'x'.
So, the number of marbles with Andy = '2x'.
Thus, the number of marbles with Sandy = 12x +2x = 3x2.
According to the question,
3x2- 115  = 110
Adding 115 to both sides, we get
3x2- 115  + 115 = 110 + 115                   
3x2 = 225
Multiplying both sides by 2, we get
3x2×2 = 225 ×2                                     
 3x = 450
Dividing both sides by 3, we get
3x3 = 4503                                             
  x = 150
So, Pandy has 150 marbles, Andy has 2x = 2(150) =  300 marbles, and Sandy has 3x2 = 3×1502 = 225 marbles.

Page No 8.26:

Question 13:

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

Answer:

Let the number of 50 paise coins  = 'x'.
So, the money value contribution of 50 paise coins = 0.5x.
The number of 25 paise coins = '4x'.
The money value contribution of 25 paise coins = 0.25(4x) = x.
According to the question,
⇒ 0.5x + x = 30
⇒ 1.5x = 30
Dividing both sides by 1.5, we get
1.5x1.5= 301.5              
⇒ x = 20
Thus, the number of 50 paise coins  = 'x' = 20, and the number of 25 paise coins  = '4x' = 4 (20) = 80.

Page No 8.26:

Question 14:

The length of a rectangular field is twice its breadth. If the perimeter of the field is 228 metres, find the dimensions of the field.

Answer:

Let the breadth of the rectangle = 'x' metres.
According to the question,
Length of the rectangle = '2x' metres
Perimeter of a rectangle = 2 (length + breadth)
So,       2 (2x + x) = 228
           =>  2 (3x) = 228
            => 6x = 228
Dividing both sides by 6, we get
           => 6x6 = 2286                   
            => x = 38
So, the breadth of the rectangle = x = 38 metres, and the length of the rectangle  = 2x = 2(38) = 76 metres.

Page No 8.26:

Question 15:

There are only 25 paise coins in a purse. The value of money in the purse is Rs 17.50. Find the number of coins in the purse.

Answer:

Let the number of 25-paise coins in the purse be 'x'.
So, the value of money in the purse = 0.25x.
But 0.25x = 17.5.
Dividing both sides by 0.25, we get
      => 0.25x0.25 = 17.50.25       
      => x = 70
Thus, the number of 25-paise coins in the purse = 70.

Page No 8.26:

Question 16:

In a hostel mess, 50 kg rice are consumed everyday. If each student gets 400 gm of rice per day, find the number of students who take meals in the hostel mess.

Answer:

Let the number of students in the hostel be 'x'.
Quantity of rice consumed by each student = 400 gm.
So, daily rice consumption in the hostel mess = 400(x).
But, daily rice consumption = 50 kg = 50 × 1000 = 50000 gm  [since 1 kg = 1000 gm].
According to the question,
           400x = 50000
Dividing both sides by 400, we get
         =>  400x400 = 50000400                                           
         =>  x = 125
Thus, 125 students have their meals in the hostel mess.



Page No 8.27:

Question 1:

Mark the correct alternative in the following question:

The zero of 3x + 2 is

a 23                                     b 32                                     c -23                                     d -32

Answer:

If 3x+2=0, then3x=-2          Transposing +2 to RHSx=-23

So, the zero of 3x + 2 is -23.

Note: A zero is that number, when put in place of the variable, makes the expression equal to zero.

Hence, the correct alternative is option (c).

Page No 8.27:

Question 2:

Mark the correct alternative in the following question:

If 2x-32=5x+34, then x=a 34                                     b -34                                     c 43                                     d -43

Answer:

As, 2x-32=5x+342x-5x=32+34           By transposing -32 to RHS and 5x to LHS-3x=64+34-3x=6+34x=94×-3                 By transposing -3 to RHSx=34×-1x=3-4 x=-34

Hence, the correct alternative is option (b).

Page No 8.27:

Question 3:

Mark the correct alternative in the following question:

If x2-4=x3-1, then x=a 3                                     b 6                                     c 18                                     d 2

Answer:

As, x2-4=x3-1x2-x3=4-1             By transposing x3 to LHS and -4 to RHS3x6-2x6=33x-2x6=3x6=3x=3×6                    By transposing 6 to RHS x=18

Hence, the correct alternative is option (c).

Page No 8.27:

Question 4:

Mark the correct alternative in the following question:

If x+2x-2=23, then x=a -10                                     b 10                                     c 43                                     d -43

Answer:

As, x+2x-2=233x+2=2x-2              By cross multiplication3x+6=2x-43x-2x=-6+4               By transposing 2x to LHS and 6 to RHS x=-10

Hence, the correct alternative is option (a).

Page No 8.27:

Question 5:

Mark the correct alternative in the following question:

If x6+x4=x2+34, then x=a 9                                     b 6                                     c -9                                     d 4

Answer:

As, x6+x4=x2+34x6+x4-x2=34           By transposing x2 to LHS2x12+3x12-6x12=342x+3x-6x12=34-x12=34-x×4=3×12              By cross multiplication-4x=36x=36-4 x=-9

Hence, the correct alternative is option (c).

Page No 8.27:

Question 6:

Mark the correct alternative in the following question:

If 2x+53=14x+4, then x=a 3                                     b 4                                     c 34                                     d 43

Answer:

As, 2x+53=14x+42x-14x=4-53                By transposing 53 to RHS and 14x to LHS2x1-x4=41-538x4-x4=123-538x-x4=12-537x4=737x×3=4×7                          By cross multiplication21x=28x=2821 x=43

Hence, the correct alternative is option (d).

Page No 8.27:

Question 7:

Mark the correct alternative in the following question:

If x2-x3=5, then x=a 8                                     b 16                                     c 24                                     d 30

Answer:

As, x2-x3=53x6-2x6=53x-2x6=5x6=5x=5×6         By transposing 6 to RHS x=30

Hence, the correct alternative is option (d).

Page No 8.27:

Question 8:

Mark the correct alternative in the following question:

If x-23=2x-13-1, then x=a 2                                     b 4                                     c 6                                     d 8

Answer:

As, x-23=2x-13-1x-23-2x-13=-1          By transposing 2x-13 to LHSx-2-2x-13=-1x-2-2x+13=-1-x-13=-1-x-1=-1×3                 By transposing 3 to RHS-x-1=-3-x=-3+1                       By transposing -1 to RHS-x=-2 x=2

Hence, the correct alternative is option (a).

Page No 8.27:

Question 9:

Mark the correct alternative in the following question:

The sum of two consecutive whole numbers is 43. The smaller number is

(a) 21                                 (b) 22                                 (c) 23                                 (d) 24

Answer:

Let the two consecutive whole numbers be x and x+1.As, the sum of the two consecutive whole numbers is 43.x+x+1=432x+1=432x=43-1              By transposing 1 to RHS2x=42x=422                    By transposing 2 to RHS x=21

So, the smaller number is 21.

Hence, the correct alternative is option (a).

Page No 8.27:

Question 10:

Mark the correct alternative in the following question:

The sum of two consecutive odd numbers is 36. The larger number is

(a) 17                                  (b) 15                                  (c) 19                                  (d) 21

Answer:

Let the two consecutive odd numbers be x and x+2.As, the sum of the two consecutive odd numbers is 36.x+x+2=362x+2=362x=36-22x=34x=342x=17 x+2=17+2=19

So, the larger number is 19.

Hence, the correct alternative is option (c).

Page No 8.27:

Question 11:

Mark the correct alternative in the following question:

Twice a number when increased by 7 gives 25. The number is

(a) 7                                   (b) 9                                   (c) 10                                   (d) 8

Answer:

Let the number be x.As, twice the number when increased by 7 gives 25.2x+7=252x=25-7          By transposing 7 to RHS2x=18x=182               By transposing 2 to RHS x=9

So, the number is 9.

Hence, the correct alternative is option (b).

Page No 8.27:

Question 12:

Mark the correct alternative in the following question:

The length of a rectangle is three times its width and its perimeter 56 m. The length is

(a) 7 m                               (b) 14 m                               (c) 21 m                               (d) 28 m

Answer:

Let the width of the rectangle be x. Then,the length of the rectangle=3xAs, perimeter of the rectangle=56 m2×Length+Breadth=562×3x+x=562×4x=568x=56x=568 x=7So, the length of the rectangle=3x=3×7=21 m

Hence, the correct alternative is option (c).

Page No 8.27:

Question 13:

Mark the correct alternative in the following question:

Two-third of a number is greater than one-third of the number by 5. The number is

(a) 10                                (b) 5                                (c) 15                                (d) 12

Answer:

Let the number be x.As, two-third of a number is greater than one-third of the number by 5.23x-13x=52x-x3=5x3=5x=5×3 x=15

So, the number is 15.

Hence, the correct alternative is option (c).



Page No 8.28:

Question 14:

Mark the correct alternative in the following question:

If the sum of a number and its two-fifth is 70. The number is

(a) 70                                 (b) 50                                 (c) 60                                 (d) 90

Answer:

Let the number be x.As, the sum of a number and its two-fifth is 70.x+25x=70x1+2x5=705x5+2x5=705x+2x5=707x5=707x=70×5            By transposing 5 to RHS7x=350x=3507               By transposing 7 to RHS x=50

So, the number is 50.

Hence, the correct alternative is option (b).

Page No 8.28:

Question 15:

Mark the correct alternative in the following question:

23 of a number is less than the original number by 20. The number is

(a) 30                                  (b) 40                                  (c) 50                                  (d) 60

Answer:

Let the number be x.As, 23 of the number is less than the original number by 20.x-23x=20x1-2x3=203x3-2x3=203x-2x3=20x3=20x=20×3            By transposing 3 to RHS x=60

So, the number is 60.

Hence, the correct alternative is option (d).

Page No 8.28:

Question 16:

Mark the correct alternative in the following question:

A number is as much greater than 31 as it is less than 81. The number is

(a) 46                                   (b) 56                                   (c) 66                                   (d) 76

Answer:

Let the number be x.As, the number is as much greater than 31 as it is less than 81.x-31=81-xx+x=81+31            By transposing -x to LHS and -31 to RHS2x=112x=1122                     By transposing 2 to RHS x=56

So, the number is 56.

Hence, the correct alternative is option (b).

Page No 8.28:

Question 17:

Mark the correct alternative in the following question:

Two complementary angles differ by 20°. The smaller angle is

(a) 55°                                    (b) 25°                                    (c) 65°                                    (d) 35°

Answer:

Let the smaller angle be x. Then,The larger angle=x+20°As, the sum of the two complementary angles is always 90°.x+x+20°=90°2x+20°=90°2x=90°-20°2x=70°x=70°2                     By transposing 2 to RHS x=35°

So, the smaller angle is 35°.

Hence, the correct alternative is option (d).

Page No 8.28:

Question 18:

Mark the correct alternative in the following question:

Two supplementary angles differ by 40°. The measure of the larger angle is

(a) 70°                                     (b) 80°                                     (c) 110°                                     (d) 100°

Answer:

Let the larger angle be x. Then,The smaller angle=x-40°As, the sum of the two supplementary angles is always 180°.x+x-40°=180°2x-40°=180°2x=180°+40°2x=220°x=220°2                     By transposing 2 to RHS x=110°

So, the measure of the larger angle is 110°.

Hence, the correct alternative is option (c).

Page No 8.28:

Question 19:

Mark the correct alternative in the following question:

The sum of three consecutive odd numbers is 81. The middle number is

(a) 25                                     (b) 27                                     (c) 31                                     (d) 29

Answer:

Let the three consecutive odd numbers be x, x+2 and x+4.As, the sum of the three consecutive numbers is 81.x+x+2+x+4=813x+6=813x=81-6               By transposing 6 to RHS3x=75x=753                     By transposing 3 to RHSx=25 x+2=25+2=27

So, the middle number is 27.

Hence, the correct alternative is option (b).

Page No 8.28:

Question 20:

If 2(2n + 5) = 3(3n - 10), then n =

(a) 5                                       (b) 3                                      (c) 7                                      (d) 8

Answer:

As, 22n+5=33n-104n+10=9n-304n-9n=-10-30             By transposing 10 to RHS and 9n to LHS-5n=-40n=-40-5                             By transposing -5 to RHS n=8

Hence, the correct alternative is option (d).



Page No 8.6:

Question 1:

Verify by substitution that:
(i) x = 4 is the root of 3x − 5 = 7
(ii) x = 3 is the root of 5 + 3x = 14
(iii) x = 2 is the root of 3x − 2 = 8x − 12
(iv) x = 4 is the root of 3x2=6
(v) y = 2 is the root of y − 3 = 2y − 5
(vi) x = 8 is the root of 12x+7=11

Answer:

(i) x = 4 is the root of 3x − 5 = 7.
  Now, substituting x = 4 in place of 'x' in the given equation 3x − 5 = 7,
  3(4) − 5 = 7

  12 − 5 = 7
     7 = 7
  LHS = RHS
  Hence, x = 4 is the root of 3x − 5 = 7.

(ii)
x = 3 is the root of 5 + 3x = 14.
 
Now, substituting x = 3 in place of 'x' in the given equation 5 + 3x = 14,
  5 + 3(3) = 14
  5 + 9 = 14
      14 = 14
  LHS = RHS
  Hence, x = 3 is the root of 5 + 3x = 14.

(iii)
x = 2 is the root of 3x − 2 = 8x − 12.
 
Now, substituting x = 2 in place of 'x' in the given equation 3x − 2 = 8x − 12,
   3(2) − 2 = 8(2) − 12
   6 − 2 = 16 − 12
         4 = 4

   LHS = RHS
   Hence, x = 2 is the root of 3x − 2 = 8x − 12.

(iv) x = 4 is the root of 3x2 = 6.
 
Now, substituting x = 4 in place of 'x' in the given equation 3x2 = 6,
    3 × 42= 6  122 = 6    6 = 6  
   LHS = RHS
  Hence, x = 4 is the root of 3x2 = 6.
(v)
y = 2 is the root of y − 3 = 2y − 5.
Now, substituting y = 2 in place of 'y' in the given equation y − 3 = 2y − 5,
2 − 3 = 2(2) − 5
    −1 = 4 − 5
    −1 = −1

LHS = RHS
Hence, y = 2 is the root of y − 3 = 2y − 5.

(vi) x = 8 is the root of 12x + 7 = 11.
 
Now, substituting x = 8 in place of 'x' in the given equation 12x + 7 = 11,
  12× 8 + 7 = 11
   4 + 7 = 11  
    11 = 11
 
LHS = RHS
Hence, x = 8 is the root of 12x + 7 = 11.

Page No 8.6:

Question 2:

Solve each of the following equations by trial-and-error method:
(i) x + 3 = 12
(ii) x − 7 = 10
(iii) 4x = 28
(iv) x2+7=11
(v) 2x + 4 = 3x
(vi) x4=12
(vii) 15x=3
(viii) x18=20

Answer:

(i) x + 3 = 12
  Here, LHS = x + 3 and RHS = 12.
 

x LHS RHS Is LHS = RHS?
1 1+3=4 12 No
2 2+3=5 12 No
3 3+3=6 12 No
4 4+3=7 12 No
5 5+3=8 12 No
6 6+3=9 12 No
7 7+3=10 12 No
8 8+3=11 12 No
9 9+3=12 12 Yes
Therefore, if x = 9, LHS = RHS.
Hence, x = 9 is the solution to this equation.

(ii) x − 7 = 10
   Here, LHS = x −7 and RHS =10.
x LHS RHS Is LHS = RHS?
9 9−7 = 2 10 No
10 10−7 = 3 10 No
11 11−7=4 10 No
12 12−7=5 10 No
13 137=6 10 No
14 147=7 10 No
15 157=8 10 No
16 167=9 10 No
17 177=10 10 Yes
Therefore, if x = 17, LHS = RHS.
Hence, x = 17 is the solution to this equation.

(iii) 4x = 28
     Here, LHS = 4x and RHS = 28.
    
x LHS RHS Is LHS = RHS?
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
  Therefore, if x = 7, LHS = RHS.
Hence, x = 7 is the solution to this equation.

(iv) x2+ 7 = 11
  Here, LHS = x2 + 7 and RHS = 11.
Since RHS is a natural number, x2 must also be a natural number, so we must substitute values of x that are multiples of 2.
x LHS RHS Is LHS = RHS?
2 22+7=8 11 No
4 42+7=9 11 No
6 62+7=10 11 No
8 82+7=11 11 Yes
  Therefore, if x = 8, LHS = RHS.
Hence, x = 8 is the solution to this equation.

(v) 2x + 4 = 3x
  Here,
LHS = 2x + 4 and RHS = 3x.
x LHS RHS Is LHS = RHS?
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
     Therefore, if x = 4, LHS = RHS.
Hence, x = 4 is the solution to this equation.

(vi) x4 = 12
   Here, LHS =x4  and RHS = 12.
  Since RHS is a natural number, x4 must also be a natural number, so we must substitute values of x that are multiples of 4.
 
x LHS RHS Is LHS = RHS?
16 164=4 12 No
20 204=5 12 No
24 244=6 12 No
28 284=7 12 No
32 324=8 12 No
36 364=9 12 No
40 404=10 12 No
44 444=11 12 No
48 484=12 12 Yes
   Therefore, if x = 48, LHS = RHS.
Hence, x = 48 is the solution to this equation.

(vii) 15x = 3
   Here, LHS =15x  and RHS = 3.
  Since RHS is a natural number, 15x must also be a natural number, so we must substitute values of x that are factors of 15.
  
x LHS RHS Is LHS = RHS?
1 151=15 3 No
3 153=5 3 No
5 155=3 3 Yes
   Therefore, if x = 5, LHS = RHS.
Hence, x = 5 is the solution to this equation.

(viii) x18= 20
   Here, LHS =x18  and RHS = 20.
  Since RHS is a natural number, x18 must also be a natural number, so we must substitute values of x that are multiples of 18.
 
x LHS RHS Is LHS = RHS?
324 32418=18 20 No
342 34218=19 20 No
360 36018=20 20 Yes

Therefore, if x = 360, LHS = RHS.
Hence, x = 360 is the solution to this equation.



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