Question

1
exceeds the sum of its digits by 18. Find the number.
3. A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the
number, its digits are reversed. Find the number.
The sum of the digite of
1:​

Answer 1

Correct Question :–

A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Given :–

• A two digit numbers is 3 more than 4 times the sum of its digits.
• 18 is added to the number, the digits are reversed.

To Find :–

• The number.

Solution :–

Let,

The tens digit be y.

The units digit be x.

So, the equation will be,

• 10y + x

After reversed we get,

• 10x + y

According to the given condition,

A two digit numbers is 3 more than 4 times the sum of its digits,

So,

↣ 10y + x = 4(y + x) + 3

Now, open the bracket.

↣ 10y + x = 4y + 4y + 3

↣ 10y – 4y + y – 4y = 3

↣ 6y – 3x = 3

↣ $\dfrac{ \cancel6 \times y - \cancel3 \times x}{ \cancel3}$

↣ $\dfrac{2 \times y - 1 \times x}{1}$

↣ $\dfrac{2y - 1x}{1}$

↣ 2y – x = 1 –––––(1)

Now, after adding 18 to the number the digits get interchanged,

Now, we get,

↣ 10x + y + 18 = 10y + x

↣ 10x – x + y – 10y = 18

↣ 9x – 9y = 18

↣ $\dfrac{ \cancel9 \times x - \cancel9 \times y}{ \cancel{18}}$

↣ $\dfrac{1 \times x - 1 \times y}{2}$

↣ $\dfrac{x - y}{2}$

↣ x – y = 2 –––––(2)

Now, add equation (1) and equation (2),

We get,

↣ 2y – x + y – x = 1 + 2

↣ 2y – y – x + x = 3

↣ 1y – 0 = 3

↣ y = 3

Now, substitute the value of y in equation (2),

↣ x – y = 2

↣ x – 3 = 2

↣ x = 2 + 3

↣ x = 5

Now, we have to find the number.

So,

The number is (10y + x),

• y = 3.
• x = 5.

↣ 10(3) + 5

↣ 30 + 5

↣ 35

Hence,

The number is 35.

Answer 2

Solution :

Let the ten's place digit be r & one's place digit be m respectively;

$\boxed{\bf{Original\:number=10r+m}}\\\boxed{\bf{Reversed\:number=10m+r}}$

A/q

$\underbrace{\bf{1^{st}\:Case\::}}$

$\longrightarrow\sf{10r+m = 4(r+m)+3}\\\\\longrightarrow\sf{10r + m = 4r + 4m + 3}\\\\\longrightarrow\sf{10r-4r+m-4m=3}\\\\\longrightarrow\sf{6r - 3m = 3.......................(1)}$

$\underbrace{\bf{2^{nd}\:Case\::}}$

$\longrightarrow\sf{10r+m+18=10m+r}\\\\\longrightarrow\sf{10r -r+ m-10m =-18}\\\\\longrightarrow\sf{9r - 9m = -18}\\\\\longrightarrow\sf{9(r-m) = -18}\\\\\longrightarrow\sf{r-m=\cancel{-18/9}}\\\\\longrightarrow\sf{r-m=-2}\\\\\longrightarrow\sf{r=-2 + m......................(2)}$

Putting the value of r in equation (1),we get;

$\longrightarrow\sf{6(-2+m)-3m=3}\\\\\longrightarrow\sf{-12 + 6m -3m = 3}\\\\\longrightarrow\sf{-12 + 3m = 3}\\\\\longrightarrow\sf{3m = 3 + 12}\\\\\longrightarrow\sf{3m=15}\\\\\longrightarrow\sf{m=\cancel{15/3}}\\\\\longrightarrow\bf{m=5}$

Putting the value of m in equation (2),we get;

$\longrightarrow\sf{r=-2+5}\\\\\longrightarrow\bf{r=3}$

Thus;

The number = 10r + m = 10(3) + 5 = 30 + 5 = 35 .