Question

the digit at unit place of a two-digit number is increased by 100% and the digit at ten places of the same number is incresed by 50% the new number thus formed is 19 more then the original number. what is the original number?​

Answer 1

Answer:

Step-by-step explanation:

let the 2 digit no be 10x+y where x be the digit at tens place and y be the digit at ones place.

according to the question;

(10x+50% of 10x)+(y+100% of y)-(10x+y)=19

10x+5x+y+y-10x-y=19

5x+y=19

Answer 2

29 or 34

Step-by-step explanation:

Let two digit number be xy and

unit digit is y ad tens digit be x

then the number is 10x+y

unit place is increased by 100% = 2y

tens place is increased by 50% = 1.5 x

new number = 10 (1.5x) +2y = 15x+2y

now , this new number is 19 more than the original number

i,e

15x+2y -19 = 10x+y

5x +y = 19

5x +y -19 =

possible values are x = 2, y = 9  and x= 3 , y = 4

then the original number is 29 or 34

verification for 29

10 (2(1.5) +2(9) - 19

= 30 +18 -19 = 29 = original number

verification for 34

10 (3(1.5) + 2(4)) - 19

45 +8 -19 = 34 = original number

#Learn more :

A number has 2-digits. the units digit of the number is 3 times the ten's digit. if the digits are reversed, the new number thus formed, is 36 more than the original number. find the original number.

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