Question

4. The sum of a two-digit number and the number
obtained by reversing its digits in 121. Find the
number, if its units place digit is greater than
the tens place digit by 7.
​

Answer 1

Answer:29

Step-by-step explanation:

Let the required number be 10a + b.

Also, b -a = 7                                           - [ Given in question]   -1st eqn

Also, [10a + b] + [10b+ a ] = 121               - [given]

=> 11a + 11b = 121

=> a+ b = 11                                              -2nd eqn

From 1st and 2nd eqn,

=> (11 - a) -a = 7

=> 4 = 2a

=> a = 2

Therefore, a = 2and b= 9 .

Thus, the required number is 29.

PLEASE MARK IT AS THE BRAINLIEST....

Answer 2

AnswEr :

• Original number is 29.

GivEn :

• Sum of a two-digit number and the number
• obtained by reversing its digits is 121 .

• Units place digit is greater than the tens place digit by 7 .

To find :

• Original number = ?

SoluTion :

Let us assume that the tens place digit be x and one's place digit be y.

• 1st Case :

$:\implies \sf \: \: \: 10x + y + 10y + x = 121$

$:\implies \sf \: \: \: 11x + 11y = 121$

$:\implies \sf \: \: \: 11(x + y) = 121$

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

$:\implies \sf \: \: \: x + y = 11 \: \: \: \: \: \: \: \: - eqn \: 1$

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• 2nd Case :

$:\implies \sf \: \: \: x - y = 7 \: \: \: \: \: \: \: - eqn \: 2$

Add eqn 1 and 2,

$:\implies \sf \: \: \: x + y + x - y= 11 + 7$

$:\implies \sf \: \: \: 2x = 18$

$:\implies \sf \: \: \: x = \dfrac{18}{2}$

$:\implies \sf \: \: \: { \underline{ \boxed{ \sf{ \blue{x = 9}}}}} \: \bigstar$

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Put the value of x in eqn 1,

$:\implies \sf \: \: \: 9 + y = 11$

$:\implies \sf \: \: \: y= 11 - 9$

$:\implies \sf \: \: \: { \underline{ \boxed{ \sf{ \purple{y = 2 }}}}} \: \bigstar$

✩ Original number :

• One's place = y = 2

• Ten's place = x = 9

Hence , the original number is 29.

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