Question

The sum of a two digit number and the number obtained by reversing the order of its digits is 121, And the two digits differ by 3. Find the number, if the digit at tens place is the greater digit​

Answer 1

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The sum of a two digit number and the number obtained by reversing the digits is 121.

• The two digits differ by 3.

• The tens digits is the greater digit.

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The original number

$\bf{\underline{\underline\green{Solution:-}}}$

Let the tens digit of the number be x

And ones digit be y

The original number = 10x + y

And the number obtained by

Reversing the digits = 10y + x

According to the 1st condition:-

The sum of a two digit number and the number obtained by reversing the digits is 121.

$\rm \implies(10x + y) + (10y + x) = 121$

$\rm \implies10x + y+ 10y + x = 121$

$\rm \implies11x + 11y = 121$

Dividing the whole equation by 11

$\rm \implies \dfrac{11x}{11} + \dfrac{11y}{11} = \dfrac{121}{11}$

$\rm \implies x + y = 11......(i)$

According to the 2nd condition :-

The digits differ by 3.

$\rm \implies x - y = 3......(ii)$

Adding equation (i) and (ii)

$\rm x + \not y = 11$

$\rm x - \not y = 3$

________________

$\rm \implies 2x = 11 + 3$

$\rm \implies 2x = 14$

$\rm \implies x = \dfrac{14}{2} = 7$

Substituting x = 7 in equation (i)

$\rm \implies x + y = 11$

$\rm \implies 7 + y = 11$

$\rm \implies y = 11 - 7$

$\rm \implies y =4$

The original number = 10x + y

= 10(7) + 4

= 70 + 4

= 74

$\underline{\boxed{ \rm \therefore The \: number \: = 74}}$

Answer 2

» To Find :

The original number.

» Taken :

Let the digits of the two-digit number be a and b.

So, the two-digit number is (10a + b) and the

number obtained on reversing the digits is (10b + a).

» Given :

• Difference of the digits of the two-digit number .

$\mathtt{\underline{\boxed{a - b = 3}}}$

• Sum of the original number and the number obtained on reversing the digits :

$\mathtt{\underline{\boxed{(10a + b) + (10b + a) = 121}}}$

» Concept :

According to the question ,if we find the two linear equation of the number ,and solve it by using the elimination method ,we can find the original two-digit number.

First Equation :

$\sf{a - b = 3}$➝ Equation (i)

Second equation :

Given Equation :

$\sf{(10a + b) + (10b + a) = 121}$

By solving it ,we get :

$\sf{\Rightarrow 10a + b + 10b + a = 121}$

$\sf{\Rightarrow 11a + 11b = 121}$

Taking the common i.e 11, from the equation,

$\sf{\Rightarrow 11(a + b) = 121}$

$\sf{\Rightarrow a + b = \dfrac{121}{11}}$

$\sf{\Rightarrow a + b = \dfrac{\cancel{121}}{\cancel{11}}}$

$\sf{\Rightarrow a + b = 11}$

Hence,

$\sf{a + b = 11}$➝ Equation (ii)

» Solution :

• Equation (i)

$\sf{a - b = 3}$

• Equation (ii)

$\sf{a + b = 11}$

Putting the two Equations together and solving them linearly ,we get :

$\sf{a - b = 3}$

$\sf{a + b = 11}$

$\sf{a - \cancel{b} = 3}$

$\sf{a + \cancel{b} = 11}$

__________[By adding]

$\sf{2a = 14}$

Solving , the equation (2a + 14) ,we get :

$\sf{\Rightarrow 2a = 14}$

$\sf{\Rightarrow a = \cancel{\dfrac{14}{2}}}$

$\sf{\Rightarrow a = 7}$

Putting the value of a in the equation (i) ,we get :

$\sf{a - b = 3}$

$\sf{\Rightarrow 7 - b = 3}$

$\sf{\Rightarrow - b = 3 - 7}$

$\sf{\Rightarrow - b = - 4}$

$\sf{\Rightarrow \cancel{-} b = \cancel{-} 4}$

$\sf{\Rightarrow b = 4}$

Hence ,the value of a is 7 and the value of b is 4.

Putting the value of a and b in the equation (10a + b) ,we get :

$\mathtt{\Rightarrow 10 \times 7 + 4}$

$\mathtt{\Rightarrow 70 + 4}$

$\mathtt{\Rightarrow 74}$

Hence, the two-digit number is 74.

Additional information :

Methods of Solving two linear equations :

• Elimination by cancellation

• Elimination by Substitution

• Graphical method

• Adding and Subtracting one Equation from the other one