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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 + !) = 47 

⇒ x + x +1 = 47     ( Grouping the like terms )

⇒ (x + x) + 1 = 47

2x = 47 - 1       ( on transposing +1 becomes - 1 )

2x = 46

$\frac{\mathrm{2x}}{2}$ = $\frac{46}{2}$   ( Dividing both the sides by 2 )

x = 23.

x + 1 = 23 + 1 = 24

Hence, the consecutive numbers are 23 and 24.

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

Given:

    x + (x + !) = 43 

⇒ x + x +1 = 43     ( Grouping the like terms )

⇒ (x + x) + 1 = 43

2x = 43 - 1       ( on transposing +1 becomes - 1 )

2x = 42

$\frac{\mathrm{2x}}{2}$ = $\frac{42}{2}$   ( Dividing both the sides by 2 )

x = 21.

x + 1 = 21 + 1 = 22

Hence, the consecutive numbers are 21 and 22.

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.

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 + !) = 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 + !) = 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 + !) = 39 

⇒ y + y +1 = 39     ( Grouping the like terms )

⇒ (y + y) + 1 = 39

2y = 39 - 1       ( on transposing +1 becomes - 1 )

2y = 38

$\frac{\mathrm{2y}}{2}$ = $\frac{38}{2}$   ( Dividing both the sides by 2 )

y = 19.

y + 1 = 19 + 1 = 20

Hence, the consecutive numbers are 19 and 20.

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 + !) = 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.