a number consist of two dight whose sum is 5.when the dight are reversed the numbers become greater bye 9.find the number
Answers
Question:
A number consist of two digit whose sum is 5.when the digits are reversed the numbers become greater by 9.find the number.
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Answer:
original number = 23
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Step-by-step explanation:
given that,
a number consist of two digits
sum of the digits = 5
let the digits be x and y
so,
number be 10y + x
Then,
ACCORDING TO THE QUESTION
x + y = 5 .....(1)
now,
also given that,
when digits are reversed the numbers become greater by 9.
here,
original number = 10y + x
reversed number = 10x + y
ACCORDING TO THE QUESTION,
10x + y = 10y + x + 9 [greater by 9]
10 - x + y - 10y = 9
9x - 9y = 9
9(x - y) = 9
x - y = 9/9
x - y = 1 .....(2)
now,
by adding (1) and (2),
x + y + x - y = 5 + 1
2x = 6
x = 6/2
x = 3
now,
putting the value of x on (1)
x + y = 5
3 + y = 5
y = 5 - 3
y = 2
now,
value of x = 3
value of y = 2
so,
original number = 10y + x
by putting the value of x and y ,
10y + x
➡ 10(2 )+ 3
➡ 20 + 3
✔ 23
so,
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Original number = 23
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Question: A number consists of two digits whose sum is 5. When the digits are reversed, the numbers become greater by 9. Find the numbers.
Answer:
The original digit is 23.
Step-by-step explanation:
Let the digit at the tens place be x, and the digit at the unit place be y.
So, the number will become (10x+y) [10x is used instead of just x because x is at the tens place]
Now, its given that, sum of the digits at the tens and units place is 5.
Therefore, (x+y)=5 ---(1)
Now, if we reverse the digits of this number, the original number increases by 9.
The equation for this can be given as
(10y+x)=(10x+y)+9
10y-y+x-10x=9
9y-9x=9 9(y-x)=9
Therefore y-x=1 ---(2)
Adding eq. (1) and (2)
x+y+y-x=1+5
2y=6
Therefore y=3
Putting y=3 in eq. (1)
x+3=5
Therefore x=2
So, the original number is (10×2+3)=20+3=23