Math, asked by SupriyaPandit, 7 months ago


4. The sum of digits of a two digit number is 7. The number obtained, on reversing the order of digits
is greater than original number by 9. Find the number.

Answers

Answered by rjayaditya
6
Answer:

ANSWER
Let the original number be
yx

; i.e., 10y+x. We know x+y=7. The number obtained by reversing the digits in
xy

, i.e., 10x+y. The second condition gives 10x+y−2=2(10y+x). Thus we have two equations:
x+y=7....(1)
8x−19y=2....(2)
Multiply the equation (1) by 19 and get
19x+19y=133.
Adding this to (2), we obtain 27x=135. This gives x=5. Hence y=7−x=7−5=2.
The required number is 25.
Answered by samrajansari33
1

Step-by-step explanation:

The sum of digits of two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the numbers?

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Let x and y be the digits at tens and ones place respectively.

Let A be the number formed by these digits.

Therefore, A=10x+y……….(1)

According to the question.

x+y=7…………(2)

Let B be the new number formed by reversing the digits.

Therefore, B=10y+x……….(3)

According to the question.

B=27+A

=>(10y+x)=27+(10x+y)

=>(10y+x)-(10x+y)=27

=>9y-9x=27

=>9(y-x)=27

=>y-x=27/9=3…………..(4)

Adding equations (2) and (4).

(x+y)+(y-x)=7+3=10

=>2y=10

=>y=5 and x=7–5=2……….from equation(2)

Hence,the number is A=10x+y=10*2+5=25

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