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

Given:

    x + (x + !) = 49 

⇒ x + x +1 = 49     ( Grouping the like terms )

⇒ (x + x) + 1 = 49

2x = 49 - 1       ( on transposing +1 becomes - 1 )

2x = 48

$\frac{\mathrm{2x}}{2}$ = $\frac{48}{2}$   ( Dividing both the sides by 2 )

x = 24.

x + 1 = 24 + 1 = 25

Hence, the consecutive numbers are 24 and 25.

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

Given:

    y + (y + !) = 53 

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

⇒ (y + y) + 1 = 53

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

2y = 52

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

y = 26.

y + 1 = 26 + 1 = 27

Hence, the consecutive numbers are 26 and 27.

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 y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 27 

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

⇒ (y + y) + 1 = 27

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

2y = 26

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

y = 13.

y + 1 = 13 + 1 = 14

Hence, the consecutive numbers are 13 and 14.

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

Given:

    y + (y + !) = 59 

⇒ y + y +1 = 59     ( Grouping the like terms )

⇒ (y + y) + 1 = 59

2y = 59 - 1       ( on transposing +1 becomes - 1 )

2y = 58

$\frac{\mathrm{2y}}{2}$ = $\frac{58}{2}$   ( Dividing both the sides by 2 )

y = 29.

y + 1 = 29 + 1 = 30

Hence, the consecutive numbers are 29 and 30.

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

Given:

    x + (x + !) = 31 

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

⇒ (x + x) + 1 = 31

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

2x = 30

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

x = 15.

x + 1 = 15 + 1 = 16

Hence, the consecutive numbers are 15 and 16.

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

Given:

    y + (y + !) = 41 

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

⇒ (y + y) + 1 = 41

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

2y = 40

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

y = 20.

y + 1 = 20 + 1 = 21

Hence, the consecutive numbers are 20 and 21.

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

Given:

    x + (x + !) = 57 

⇒ x + x +1 = 57     ( Grouping the like terms )

⇒ (x + x) + 1 = 57

2x = 57 - 1       ( on transposing +1 becomes - 1 )

2x = 56

$\frac{\mathrm{2x}}{2}$ = $\frac{56}{2}$   ( Dividing both the sides by 2 )

x = 28.

x + 1 = 28 + 1 = 29

Hence, the consecutive numbers are 28 and 29.

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

Given:

    x + (x + !) = 25 

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

⇒ (x + x) + 1 = 25

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

2x = 24

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

x = 12.

x + 1 = 12 + 1 = 13

Hence, the consecutive numbers are 12 and 13.

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

Given:

    y + (y + !) = 35 

⇒ y + y +1 = 35     ( Grouping the like terms )

⇒ (y + y) + 1 = 35

2y = 35 - 1       ( on transposing +1 becomes - 1 )

2y = 34

$\frac{\mathrm{2y}}{2}$ = $\frac{34}{2}$   ( Dividing both the sides by 2 )

y = 17.

y + 1 = 17 + 1 = 18

Hence, the consecutive numbers are 17 and 18.