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Let the required number be y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 29 

⇒ y + y +1 = 29     ( Grouping the like terms )

⇒ (y + y) + 1 = 29

2y = 29 - 1       ( on transposing +1 becomes - 1 )

2y = 28

$\frac{\mathrm{2y}}{2}$ = $\frac{28}{2}$   ( Dividing both the sides by 2 )

y = 14.

y + 1 = 14 + 1 = 15

Hence, the consecutive numbers are 14 and 15.

Let the required number be x. ⇒ 2 times = 2x

Given:

    x + (x + !) = 53 

⇒ x + x +1 = 53     ( Grouping the like terms )

⇒ (x + x) + 1 = 53

2x = 53 - 1       ( on transposing +1 becomes - 1 )

2x = 52

$\frac{\mathrm{2x}}{2}$ = $\frac{52}{2}$   ( Dividing both the sides by 2 )

x = 26.

x + 1 = 26 + 1 = 27

Hence, the consecutive numbers are 26 and 27.

Let the required number be y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 45 

⇒ y + y +1 = 45     ( Grouping the like terms )

⇒ (y + y) + 1 = 45

2y = 45 - 1       ( on transposing +1 becomes - 1 )

2y = 44

$\frac{\mathrm{2y}}{2}$ = $\frac{44}{2}$   ( Dividing both the sides by 2 )

y = 22.

y + 1 = 22 + 1 = 23

Hence, the consecutive numbers are 22 and 23.

Let the required number be x. ⇒ 2 times = 2x

Given:

    x + (x + !) = 41 

⇒ x + x +1 = 41     ( Grouping the like terms )

⇒ (x + x) + 1 = 41

2x = 41 - 1       ( on transposing +1 becomes - 1 )

2x = 40

$\frac{\mathrm{2x}}{2}$ = $\frac{40}{2}$   ( Dividing both the sides by 2 )

x = 20.

x + 1 = 20 + 1 = 21

Hence, the consecutive numbers are 20 and 21.

Let the required number be y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 25 

⇒ y + y +1 = 25     ( Grouping the like terms )

⇒ (y + y) + 1 = 25

2y = 25 - 1       ( on transposing +1 becomes - 1 )

2y = 24

$\frac{\mathrm{2y}}{2}$ = $\frac{24}{2}$   ( Dividing both the sides by 2 )

y = 12.

y + 1 = 12 + 1 = 13

Hence, the consecutive numbers are 12 and 13.

Let the required number be x. ⇒ 2 times = 2x

Given:

    x + (x + !) = 21 

⇒ x + x +1 = 21     ( Grouping the like terms )

⇒ (x + x) + 1 = 21

2x = 21 - 1       ( on transposing +1 becomes - 1 )

2x = 20

$\frac{\mathrm{2x}}{2}$ = $\frac{20}{2}$   ( Dividing both the sides by 2 )

x = 10.

x + 1 = 10 + 1 = 11

Hence, the consecutive numbers are 10 and 11.

Let the required number be x. ⇒ 2 times = 2x

Given:

    x + (x + !) = 27 

⇒ x + x +1 = 27     ( Grouping the like terms )

⇒ (x + x) + 1 = 27

2x = 27 - 1       ( on transposing +1 becomes - 1 )

2x = 26

$\frac{\mathrm{2x}}{2}$ = $\frac{26}{2}$   ( Dividing both the sides by 2 )

x = 13.

x + 1 = 13 + 1 = 14

Hence, the consecutive numbers are 13 and 14.

Let the required number be y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 51 

⇒ y + y +1 = 51     ( Grouping the like terms )

⇒ (y + y) + 1 = 51

2y = 51 - 1       ( on transposing +1 becomes - 1 )

2y = 50

$\frac{\mathrm{2y}}{2}$ = $\frac{50}{2}$   ( Dividing both the sides by 2 )

y = 25.

y + 1 = 25 + 1 = 26

Hence, the consecutive numbers are 25 and 26.

Let the required number be y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 31 

⇒ y + y +1 = 31     ( Grouping the like terms )

⇒ (y + y) + 1 = 31

2y = 31 - 1       ( on transposing +1 becomes - 1 )

2y = 30

$\frac{\mathrm{2y}}{2}$ = $\frac{30}{2}$   ( Dividing both the sides by 2 )

y = 15.

y + 1 = 15 + 1 = 16

Hence, the consecutive numbers are 15 and 16.

Let the required number be x. ⇒ 2 times = 2x

Given:

    x + (x + !) = 23 

⇒ x + x +1 = 23     ( Grouping the like terms )

⇒ (x + x) + 1 = 23

2x = 23 - 1       ( on transposing +1 becomes - 1 )

2x = 22

$\frac{\mathrm{2x}}{2}$ = $\frac{22}{2}$   ( Dividing both the sides by 2 )

x = 11.

x + 1 = 11 + 1 = 12

Hence, the consecutive numbers are 11 and 12.