Question

A number consists of two digit of which tens digit exceeds the unit by 3 the number is 7 times the sum of its digits find the number

Answer 1

Take the tens digit as x and ones digit as y
x=y+3
x-y=3. this is the first equation

second equation is the Number is 10x+y and it is equal to 7(x+y)
10x+y=7x+7y
3x=6y
3x-6y=0
Solve for the both equation
y=3
x=6
no. is 10x+y
63..

Answer 2

Answer: The the required number is 63.      

Step-by-step explanation:

Given:A number consists of two digit of which tens digit exceeds the unit by 3 the number is 7 times the sum of its digits

To find:We have to find the number.

Explanation:

Step 1:Let ones digit of the number be x.

Step 2:As tens digit of the number exceeds the unit digit by 3.

           ∴ tens digit will be $x+3$.

Step 3:As we know that,

⇒                   $Number=$ 10×tens digit+unit digit

⇒                   $Number=10(x+3)+x$

⇒                  $Number=10x+30+x$

⇒                  $Number=11x+30$.............(1)

Step 4:Sum of the digits will be (x+x+3).

Step 5:According to the question,the number is 7 times the sum of its digits

∴                $11x+30=7(x+x+3)$

⇒               $11x+30=7x+7x+21$

⇒               $11x+30=14x+21$

⇒             $11x-14x=21-30$

⇒                      $-3x=-9$

⇒                          $x=\frac{-9}{-3}$

⇒                          $x=3$

Step 6:Therefore,unit digit of the number is $x=3$

            Tens digit of the number $=$ $x+3$

                                                      $=3+3$  

                                                      $= 6$      

∴                                      $Number=11x+30$

                                       $Number=11(3)+30$    

                                       $Number=33+30$  

                                       $Number= 63$      

∴      The the required number is 63.      

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