Question

2. The sum of the digits of a 2-digit number is 7. If the digits are reversed, the new number is equal to 3 less than 4 times the original no . find the original no
3 less than 4 times the original number. Find the original number.​

Answer 1

Solution :-

Let the digits of the two digit number be x and y

Sum of the digits = 7

⇒ x + y = 7

⇒ x = 7 - y

Original number = 10x + y

New number when digits are reversed = 10y + x

Given

New number when digits are reversed = 3 less than the 4 times the Original number

⇒ 10y + x = 4(10x + y) - 3

⇒ 10y + x = 40x + 4y - 3

⇒ 10y + x - 40x - 4y = - 3

⇒ 6y - 39x = - 3

⇒ 3(2y - 13x) = - 3

⇒ 2y - 13x = - 3/3

⇒ 2y - 13x = - 1

Substitute x = 7 - y in the above equation

⇒ 2y - 13(7 - y) = - 1

⇒ 2y - 91 + 13y = - 1

⇒ 15y = - 1 + 91

⇒ 15y = 90

⇒ y = 90/15

⇒ y = 6

Subtitute y = 6 in x = 7 - y

⇒ x = 7 - y

⇒ x = 7 - 6

⇒ x = 1

Original number = 10x + y

= 10(1) + 6

= 10 + 6

= 16

Therefore the original number is 16.

Answer 2

Answer:

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let the digit at unit's place be 'x' and digit at tens place be = Y.

Then original no. = 10x +Y

Sum of digit = 7

=> x + Y =7 ...... 1

Also , 10 y +× = 4 ( 10x+Y)-3

=> 10 y + x = 40x + 4 y -3 ..

=> 39 x -6y = 3

=> 13 x -2 y = 1.......1

Multiplying equation 1 by 2 , we get 2x +2y = 14.....3

Adding equation 2 and 3

15 x = 15

=> x= 1

Y=6

Hence , = 10x+y =10(1) +6 = 16

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