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Given number = p 

By subtracting 7 from the number = p – 7 

∴ p – 7 = 21

Given 8y – 9 = 15 

⇒ 8y – 9 + 9 = 15 + 9 (Add 9 on both sides) 

⇒ 8y = 24

⇒ 8y/8 = 24/8 (Divide by 8 on both sides)

⇒ y = 3

Perimeter of rectangle = 36 cm 2(length + breadth) = 36 

⇒ 2(x + 4 + x) = 36 

⇒ 2(2x + 4) = 36

⇒ \(\frac {2(2x+4)}{2} = \frac {36}{2}\)(Divide by 2 on both sides) 

⇒ 2x + 4 = 18 

⇒ 2x + 4 – 4 = 18 – 4 (Subtract 4 on both sides)

⇒ 2x = 14 .

= 2x/2 = 14/2 (Divide by 2 on both sides) 

∴ x = 7

Let the number = x 

By adding 5 to the number = 5 + x 

∴ 5 + x = 9

Let us take the number of students get down at first stop = x 

No. of students get down at the second bus stop = 2x 

No. of students get down at the third bus stop = 8

Remaining students in the bus = 5 

⇒ x + 2x +8 + 5 = 40 

⇒ 3x + 13 = 40 

⇒ 3x = 40 – 13 

⇒ 3x = 27 

⇒ x = 27/3 

⇒ x = 9 

∴ The number of students got down in the first bus stop = 9. 

No. of students got down in second stop = 2x = 2 × 9 = 18.

Correct option is: (a) = × +

Given 7 + 2 = 2 × 3 ,

 ⇒ 7 = 2 × 2 + 3 

⇒ 7 = 4 + 3 

∴ 7 = 7

Correct option is D) p + q

Correct option is C) 6
 

Changing of term from one side of equation to other side is called Transposition.

A) 14

Let the required number be x.

\(\therefore\) According to given condition, we have

3x + 5 = 47

⇒ 3x = 47- 5 = 42

⇒ x = 42/3 = 14

x – 8 = – 8 

x – 8 + 8 = – 8 + 8 (Add 8 on both sides) 

x = 0 is not a natural number. 

So, given equation has no solution.

Correct option is C) 21

Given = 5m/3 10

5m/3 x 3 = 10 × 3 (Multiply with 3 on both sides)

⇒ 5m = 30

⇒ 5m/5 = 30/5 (Divide both sides by 5)

⇒ m = 6

Check:

Substitute m = 6 in the given equation,

LHS = 5m/3 = 5(6)/3 = 30/3 = 10 = RHS.

Hence verified.

Given 3m/5 – 7 = 1

Substitute m = 10 in the given equation

\(\frac {3(10)}{5}\) – 7 = 1

⇒ 3 × 2 – 7 = 1 

⇒ 6 – 7 = 1 

∴ – 1 = 1 LHS ≠ RHS 

So, m = 10 is not the solution of given equation.

Given 5 – 2x = 19 

When x = – 7

LHS = 5 – 2x 

= 5 – 2(- 7) = 5 + 14 = 19 

RHS = 19 

19 = 19 

Here, LHS = RHS 

So, x = – 7 is a solution of the given equation.

When k = 7 

LHS: 2k – 11 = 2(7) – 11 = 14 – 11 = 3 

RHS: 5 

Here LHS ≠ RHS, 

So k = 7 is not a solution.

Given 2 + 3(m – 1) = 5 

When m = – 2 

LHS = 2 + 3(m – 1) 

= 2 + 3(- 2 – 1) 

= 2 + 3(- 3) = 2 – 9 = – 7 

RHS = 5 

– 7 ≠ 5 

Here, LHS ≠ RHS 

So, m = – 2 is not a solution of given equation.

Given 5n – 7 = 23 

When n = 6 

L.H.S = 5n – 7 

= 5(6) – 7 

= 30 – 7 

= 23 

R.H.S =23 

Here, L.H.S = R.H.S 

So, n = 6 is a solution of the given equation.

Given 6(q – 5) = 42

⇒ \(\frac {6(q-5)}{6} = \frac {42}{6} \)(Divide by 6 on both sides) 

⇒ q – 5 = 7 

⇒ q – 5 + 5 = 7 + 5 (Add 5 on both sides) 

⇒ q = 12 

Check: 

Substitute q = 12 in 

6(q – 5) = 42 

LHS = 6(q – 5) 

= 6(12 – 5) 

= 6(7) = 42 = RHS 

Hence verified.

Let the number be x. 

Twice the number = 2x 

Sum of twice a number and 4 = 80 

2x + 4 = 80 

⇒ 2x + 4 – 4 = 80 – 4(Subtract 4 on both sides) 

⇒ 2x = 76

⇒ 2x/2 = 76/2 (Divide by 2 on both sides) 

⇒ x = 38 

∴ Number = 38

Given b/7 = -2

⇒ b/7 × 7 = – 2 × 7 (Multiply by 7 on both sides)

⇒ b = – 14

Check: 

Substitute b = – 14 in b/7 = -2

LHS = b/7 = -14/7 = -2 = RHS

Hence verified

Given – 2x = – 10

⇒ -2x/-2 = -10/-2 (Divide by – 2 on both sides)

⇒ x = 5 

Check: 

Substitute x = 5 in – 2x = – 10 

LHS = – 2x 

= – 2(5) = – 10 = RHS 

Hence verified.

Given 29 – 7y = 1 

⇒ 29 – 7y – 29 = 1 – 29 (Subtract 29 on both sides)

⇒ – 7y = -28 .

⇒ -7y/-7 = -28/-7 (Divide by – 7 on both sides) 

⇒ y = 4 

Check: 

Substitute y = 4 in 29 – 7y = 1

LHS = 29 – 7y 

= 29 – 7(4) 

= 29 – 28 = 1 = RHS 

Hence verified.

Given 5x – 17 = 18 

⇒ 5x – 17 + 17 = 18 + 17 (Add 17 on both sides) 

⇒ 5x = 35

⇒ 5x/5 = 35/5 (Divide by 5 on both sides)

⇒ x = 7

Check: 

Substitute x = 7 in 5x – 17 = 18

LHS = 5x – 17

= 5(7) – 17 

= 35 – 17 = 18 = RHS 

Hence verified.

From the given diagram 

Height of the statue (BC) = x m 

Height of the Pedastal (AB) = 1.9 m

Total Height (AC) = 3.6 m

BC + AB = 3.6 m 

⇒ x + 1.9 m = 3.6 m 

⇒ x + 1.9 – 1.9 = 3.6 – 1.9 (Subtract 1.9 on both sides) 

⇒ x = 1.7 m 

∴ Height of the statue = 1.7 m

1. Subtract 9 from both sides. 

2. Add 4 to both sides. 

3. Divide both sides by 8. 

4. Multiply both sides by 6.

Given 2y – 1 = 5 

⇒ 2y – 1 + 1 = 5 + 1 (Add 1 on both sides) 

⇒ 2y = 6

⇒ 2y/2 = 6/2 (Divide by 2 on both sides)

∴ y = 3 

then 5y + 3 = 5(3) + 3 

= 15 + 3 = 18

Correct option is: (d) Dividing both sides of the equation by the same number.

Correct option is: (c) 5 – 3m = 1

(a) 2(x – 3) = 10 ⇒ 2x – 6 = 10 (Distributive property) 

⇒ 2x – 6 + 6 = 10 + 6 (Add 6 on both sides) 

⇒ 2x = 16

⇒ 2x/2 = 16/2 (Divide by 2 on both sides)

⇒ x = 8 is an integer.

(b) x/3 = 5

⇒ x/3 × 3 = 5 × 3 (Multiply by 3 on both sides) 

⇒ x = 15 is an integer.

(c) 5 – 3m = 1 

⇒ 5 – 3m – 5 = 1 – 5 (Subtract 5 on both sides) 

⇒ – 3m = – 4

⇒ -3m/-3 = -4/-3

⇒ m = 4/3 is not an integer.

(d) 2k + 1 = 1 

⇒ 2k + 1 – 1 = 1 – 1 (Subtract 1 on both sides) 

⇒ 2k = 0 

⇒2k/2 = 0/2 (Divide by 2 on both sides) 

⇒ k = 0 is an integer 

So, our answer is c. 

Correct option is: (a) positive number

If a and b are positive integers and b < a, then ax ± b also positive integer.

 i. y – 3 = 11 

∴ y – 3 + 3 = 11 + 3 

…. (Adding 3 to both sides) 

∴ y + 0 = 14 ∴ y = 14

ii. 17 = n + 7 

∴ 17 – 7 = n + 7 – 7 

…. (Subtracting 7 from both sides) 

∴ 17 + (-7) = n + 7 – 7 

∴ 10 = n 

∴  n = 10

iii. 30 = 5x

\(\therefore\) \(\cfrac{30}{5}\) = \(\cfrac{5x}{5}\)

…. (Dividing both sides by 5) 

∴  6 = 1x 

∴ 6 = x 

∴  x = 6

iv. \(\cfrac{m}{2}\)= 14

∴ \(\cfrac{m}{2}\)× 2 = 14 × 2

…. (Multiplying both sides by 2)

\(\cfrac{m\times2}{2\times1}\) = 28

∴ m = 28

Correct option is: (b) 7 + 3x = 1

(a) x + 2 = 5 

⇒ x + 2 – 2 = 5 – 2 (Subtract 2 on both sides) 

⇒ x = 3 it is not – 2

(b) 7 + 3x = 1 

⇒ 7 + 3x – 7 = 1 – 7 (Subtract 7. on both sides)

⇒ 3x = – 6

⇒ 3x/3 = -6/3 (Divide by 3 on both sides)

∴ x = – 2 

so, solution is – 2 

Similarly, check (c) and (d) also.

Correct option is: (c) – 4

– 6 + m = -10 

⇒ – 6 + m + 6 = – 10 + 6 (Add 6 on both sides) 

⇒ m = – 4

Correct option is: (a) + – =

Given 5 – 6 + 8 = 3 

⇒ 5 + 6 – 8 = 3

⇒ 11 – 8 = 3 

∴ 3 = 3

Correct option is:  (c) × + =

Given 7 + 2 × 6 = 20 

⇒ 7 × 2 + 6 = 20 

⇒ 14 + 6 = 20 

∴ 20 = 20

Correct option is: (b) + ×

Given 8 ÷ 2 = 2 × 8 

⇒ 8 ÷ 2 × 2 = 8 

⇒ 4 × 2 – 8 

∴8 = 8

Correct option is: (c) – × =

Given 3 = 3 – 7 + 0 

⇒ 3 – 3 × 7 = 0 

⇒ 0 × 7 = 0 

∴ 0 = 0

Correct option is: (c) + × =

Given 3 + 1 ÷ 4 = 16 

⇒3 + 1 × 4 = 16 

⇒ 4 × 4 = 16 

∴ 16 = 16

Correct option is: (a) × ÷ =

Given 2 × 2 + 2 = 2 

⇒ 2 × 2 ÷ 2 = 2 

⇒ 2 × 1 = 2 

∴ 2 = 2

Correct option is: (d) ÷ × =

Given 6 = 3 – 6 ÷ 12 

⇒ 6 ÷ 3 × 6 =12 

⇒ 2 × 6 = 12

∴ 12 = 12

Correct option is: (b) ÷ = ×

Given 8 ÷ 4 = 2 + 1 

⇒ 8 ÷ 4 = 2 × 1 

∴ 2 = 2

Correct option is: (d) ÷ = +

Given 15 ÷ 5 = 2 × 1 

⇒ 15 ÷ 5 = 2 + 1 

∴ 3 = 3

1. 16 ÷ 2 = 8 

2. 5 × 2 = 10 

3. 9 + 4 = 13 

4. 72 ÷ 3 = 24 

5. 4 + 5 = 9

6. 8 × 3 = 24 

7. 19 – 10 = 9 

8. 10 – 2 = 8 

9. 37 – 27 = 10 

10. 6 + 7 = 13

∴ The equations are

1. 16 ÷ 2 = 10 – 2 

2. 5 × 2 = 37 – 27 

3. 9 + 4 = 6 + 7 

4. 72 ÷ 3 = 8 x 3 

5. 4 + 5 = 19 – 10

i. y – 5 = 1 

∴ y – 5 + 5 = 1 + 5 

…. (Adding 5 to both sides) 

∴ y + 0 = 6 

∴ y = 6

ii. 8 = t + 5 

∴ 8 – 5 = t + 5 – 5 

……(Subtracting 5 from both sides) 

∴ 8 + (-5) = t + 0 

∴ 3 = t 

∴ t = 3

iii. 4x = 52

∴ \(\cfrac{4x}{4} = \cfrac{52}{4}\)

…. (Dividing both sides by 4) 

∴ 1x = 13 

∴ x = 13

iv. 19 = m -4 

∴ 19 + 4 = m – 4 + 4 

…. (Adding 4 to both sides) 

∴ 23 = m + 0 

∴ m = 23

i. Let the number of sheep before selling be x. 

∴ x – 34 = 176 

∴ x – 34 + 34 = 176 + 34 

….(Adding 34 to both sides) 

∴ x + 0 = 210 

∴ x = 210

The number of sheep with Haraba before selling is 210.

ii. Let the total number of bottles be x. 

∴ x – 7 = 12

∴ x – 7 + 7 = 12 + 7 

….(Adding 7 to both sides)

∴ x + 0 = 19 

∴ x = 19 

Weight of jam in each bottle = 250g 

∴ Total weight of jam = 19 × 250g = 4750 g 

= \(\cfrac{4750}{1000}\) kg = 4.75 kg

∴ The total number of bottles of jam made by Sakshi is 19, and the total weight of jam made is 4.75 kg.

iii. Let the total wheat bought by Archana be x kg. Wheat used in 1 month = 12 kg

∴ Wheat used in 3 months = 3 × 12 = 36 kg 

∴ x – 36 = 14 

∴ x – 36 + 36 = 14 + 36 

….(Adding 36 to both sides) 

∴ x + 0 = 50 

∴ x = 50 

∴ The total amount of wheat bought by Archana was 50 kg.

8 times of number ‘m’ is 24.

i. \(\cfrac{p}{4}\) = 9

∴ \(\cfrac{p}{4}\)× 4 = 9 × 4 

…. (Multiplying both sides by 4)

∴ \(\cfrac{p\times4}{4\times1}\)= 36

∴ 1p = 36

∴ p = 36

ii. x + 10 = 5 

∴ x + 10 – 10 = 5 – 10 

…. (Subtracting 10 from both sides) 

∴ x + 0 = 5 + (-10) 

∴ x = -5

iii. m – 5 = -12 

∴m – 5 + 5 = – 12 + 5 

…. (Adding 5 to both sides)

∴ m + 0 = -7 

∴ m = -7

iv. p + 4 = – 1 

∴ p + 4 – 4 = -1 – 4 

…. (Subtracting 4 from both sides) 

∴ p + 0 = (-1) + (-4) 

∴ P = -5

Given 4(3y + 4) = 7.6.

\(\frac {4(3y+4)}{4} = \frac {7.6}{4}\) (Divide by 4 on both sides) 

⇒ 3y + 4 = 1.9 

⇒ 3y + 4 – 4= 1.9 – 4 (Subtract 4 on both sides) 

⇒ 3y = – 2.1

⇒ 3y/3 = - 2.1/3 (Divide by 3 on both sides) 

⇒ y = – 0.7

Check: 

Substitute y = – 0.7 in 

4(3y + 4) = 7.6 

LHS = 4(3y + 4) 

= 4[3 (- 0.7) + 4] 

= 4[- 2.1 + 4] 

= 4 × 1.9 = 7.6 = RHS 

Hence verified.

Given 3(2k + 1) – 2(k T 5) – 5(5 – 2k) = 16 

⇒ 6k + 3-2k + 10-25 + 10k = 16 (Distributive property) 

⇒ 14k – 12 = 16 

⇒ 14k – 12 + 12 – 16 + 12 (Add 12 on both sides) 

⇒ 14k = 28

⇒ 14k/14 = 28/14 (Divide by 14 on both sides) 

⇒ k = 2 

Check:

Substitute k = 2 in 

3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16 

LHS = 3(2k + 1) – 2(k – 5) – 5(5 – 2k) 

= 3[2 × 2 + 1] – 2[2 – 5] – 5[5 – 2 × 2] 

= 3[4 + 1] – 2(- 3) – 5(5 – 4) 

= 15 + 6 – 5 – 16 = RHS

Hence verified.

Given 10 + 6a = 40 

⇒ 10 + 6a – 10 = 40 – 10 (Subtract 10 on both sides) 

⇒ 6a = 30

⇒ 6a/6 = 30/6 (Divide by 6 on both sides)

⇒ a = 5 

Check: 

Substitute a = 5 in 10 + 6a = 40 

LHS = 10 + 6a 

= 10 + 6(5) 

= 10 + 30 

= 40 = RHS 

Hence verified.