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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 + !) = 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 y. ⇒ 2 times = 2y

Given:

    y + (y + !) = 47 

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

⇒ (y + y) + 1 = 47

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

2y = 46

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

y = 23.

y + 1 = 23 + 1 = 24

Hence, the consecutive numbers are 23 and 24.

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 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 + !) = 37 

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

⇒ (y + y) + 1 = 37

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

2y = 36

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

y = 18.

y + 1 = 18 + 1 = 19

Hence, the consecutive numbers are 18 and 19.

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

Given:

    y + (y + !) = 43 

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

⇒ (y + y) + 1 = 43

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

2y = 42

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

y = 21.

y + 1 = 21 + 1 = 22

Hence, the consecutive numbers are 21 and 22.

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.