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

Given:

    x + (x + !) = 37 

⇒ x + x +1 = 37     ( Grouping the like terms )

⇒ (x + x) + 1 = 37

2x = 37 - 1       ( on transposing +1 becomes - 1 )

2x = 36

$\frac{\mathrm{2x}}{2}$ = $\frac{36}{2}$   ( Dividing both the sides by 2 )

x = 18.

x + 1 = 18 + 1 = 19

Hence, the consecutive numbers are 18 and 19.

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 + !) = 51 

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

⇒ (x + x) + 1 = 51

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

2x = 50

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

x = 25.

x + 1 = 25 + 1 = 26

Hence, the consecutive numbers are 25 and 26.

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 + !) = 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.

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 y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 21 

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

⇒ (y + y) + 1 = 21

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

2y = 20

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

y = 10.

y + 1 = 10 + 1 = 11

Hence, the consecutive numbers are 10 and 11.

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 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.