Are two digit number is 4 times the sum of its digit If 18 is added to the number the digits are reversed find the number?

Question 3 Playing with Numbers Exercise 5A

Answer:

let us consider the unit place digit as x and tens place digit as y.

The equations become 10y + x……..equation (1)

From the question, a two-digit number is 3 more than 4 times the sum of its digits

∴from the above condition, 4(y + x) + 3……… equation (2)

Combining equation 1 and 2

4(y + x) + 3 = 10y + x

4y + 4x + 3 = 10y + x

4x – x + 4y – 10y = -3

3x – 6y = -3

3(x – 2y) = -3

X -2y = -1 …………..equation (3)

From the second condition, If 18 is added to the number, its digits are reversed

∴the reversed number is 10x + y……..equation (4)

∴by the given condition

(10y + x) + 18 = 10x + y

10y – y =10x –x -18

9y – 9x = -18

9(y - x) = -18

Y – X = -2 ……….equation (5)

Solving equation 3 and 5 simultaneously we get,

Y=3 and x = 5

∴the required number is (10y + x) = (10(3) + 5) = 30 + 5 = 35

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Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is  ` 10 y + x`.

The number is 4 times the sum of the two digits. Thus, we have

` 10 y + x = 4 ( x + y)`

` ⇒ 10 y + x = 4x + 4 y `

`⇒ 4 x + 4 y - 10 y - x = 0 `

`⇒ 3 x - 6 y = 0 `

`⇒ 3( x - 2y)= 0`

` ⇒ x - 2y =0`

After interchanging the digits, the number becomes `10x + y .`.

If 18 is added to the number, the digits are reversed. Thus, we have

`(10 y + x )+ 18 =10x + y`

`⇒ 10 x + y -10y -x =18`

` ⇒ 9x -9y =18`

` ⇒ 9(x -y) = 18`

` ⇒ x - y= 18/9`

`⇒ x - y = 2`

So, we have the systems of equations

`x - 2y = 0`

`x - y = 2 `

Here x and y are unknowns. We have to solve the above systems of equations for xand y.

Subtracting the first equation from the second, we have

`( x - y)-(x - 2y)=2-0`

`⇒ x - y-x+2y=2 `

` ⇒ y = 2`

Substituting the value of y in the first equation, we have

` x - 2 xx2=0`

` ⇒ x - 4 =0`

` ⇒ x = 4`

Hence, the number is ` 10 xx2 + 4 = 24`.

Page 2

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is ` 10y + x`.

The number is 3 more than 4 times the sum of the two digits. Thus, we have

` 10 y + x = 4(x +y)+ 3`

` ⇒ 10 y + x = 4x + 4y + 3`

` ⇒ 4x + 4y -10y -x =-3`

` ⇒ 3x - 6y = -3`

` ⇒ 3 ( x - 2 y)= -3`

` ⇒ x - 2y = -3/3`

` ⇒ x - 2y = -1`

After interchanging the digits, the number becomes `10 x + y.`.

If 18 is added to the number, the digits are reversed. Thus, we have

` ( 10 y + x )+ 18 = 10x + y`

` ⇒ 10x + y -10y -x =18`

` ⇒ 9x -9y = 18`

` ⇒ 9( x - y)=18`

` ⇒ x -y = 18 /9`

` ⇒ x - y =2`

So, we have the systems of equations

` x - 2y =-1`

` x - y =2`

Here x and y are unknowns. We have to solve the above systems of equations for xand y.

Subtracting the first equation from the second, we have

` ( x - y)-(x - 2y )=2 -(-1)`

` ⇒ x - y -x + 2y =3`

` ⇒ y = 3`

Substituting the value of y in the first equation, we have

` x - 2xx3 =-1`

`⇒ x - 6 = -1 `

` ⇒ x = -1+6`

` ⇒ x = 5`

Hence, the number is ` 10 xx3 + 5 = 35`

Page 3

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is `10y+x.`.

The number is 4 more than 6 times the sum of the two digits. Thus, we have

` 10 y + x = 6 (x+y)+4`

` ⇒ 10y +x =6x + 6y + 4`

`⇒ 6x + 6y -10y -x=-4 `

` ⇒ 5x -5y =-4`

After interchanging the digits, the number becomes `10x + y.`.

If 18 is subtracted from the number, the digits are reversed. Thus, we have

` ( 10y + x )- 18 =10x + y`

`⇒ 10x + y -10y -x = -18 `

` ⇒ 9x -9y =-18`

` ⇒ x -y =-18/9`

` ⇒ x - y = -2`

So, we have the systems of equations

` 5x - 4y = -4 `

` x - y =-2`

Here x and y are unknowns. We have to solve the above systems of equations for xand y.

Multiplying the second equation by 5 and then subtracting from the first, we have

`(5x-4y)-(5x-5y)=-4-(-2xx5)`

` ⇒ 5 x -4y -5x +5y =-4+10`

` ⇒ y = 6`

Substituting the value of y in the second equation, we have

` x - 6=-2`

`⇒ x = 6-2 `

` ⇒ x =4`

Hence, the number is `10 xx6+4=64.`

Page 4

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is `10 y + x`.

The number is 4 times the sum of the two digits. Thus, we have

` 10 y +x =4( x + y)`

` ⇒ 10y + x = 4x + 4y`

`⇒ 4x + 4y -10y -x =0 `

` ⇒ 3x -6y =0`

`⇒ 3(x - 2y)=0`

` ⇒ x- 2y =0`

` ⇒ x = 2y`

After interchanging the digits, the number becomes `10x + y`.

The number is twice the product of the digits. Thus, we have  `10y+x=2xy`

So, we have the systems of equations

` x = 2y,`

` 10y +x =2xy`

Here x and y are unknowns. We have to solve the above systems of equations for xand y.

Substituting  `x = 2y` in the second equation, we get

` 10y + 2y = 2xx2yxxy`

` ⇒ 12y = 4y^2`

` ⇒ 4y^2-12y =0`

` ⇒ y ( y -3)=0`

` ⇒ y =0` OR `y = 3`

Substituting the value of y in the first equation, we have

Hence, the number is `10 xx 3+6= 36.`

Note that the first pair of solution does not give a two digit number.