Method 1 to solve the given question.
Lets assume the unit's digit is y and the ten's digit is x. then, the number is 10 x + y.
According to question, after interchanging the digits, the new number is 10y + x.
Then,
New Number - 18 = Original number
10y + x = 10x + y + 18
? 9y - 9x = 18
? y - x = 2 .....................(1)
Again according to question,
Sum of digits of Original Number = 8
x + y = 8 ..............................(2)
Add the equation (1) and (2) , we will get
y - x + x + y = 2 + 8
? 2y = 10
? y = 5
Put the value of Y in equation (2) , we will get
x + 5 = 8
? x = 8 - 5 = 3
Then , the original number = 10x + y
Put the value of x and y and get the original number.
The original number = 10x + y = 10 x 3 + 5 = 30 + 5 = 35
Method 2 to solve the given question.
Given that sum of digits of original number is 8.
Let us assume that unit digit is x , then ten digit will be 8 - x. (since sum of digits is 8.)
Now Original number = 10( 8 - x) + x
After interchanging the digits , the number = 10x + (8 - x) = 9x + 8
According to question.
New number = original number + 18
? 9x + 8 = 10(8 - x) + x + 18
? 9x + 8 = 80 - 10x + x + 18
? 9x + 8 = 98 - 9x
? 9x + 9x = 98 - 8
? 18x = 90
? x = 90/18
? x = 5
The original number = 10(8 - x) + x = 10(8 - 5) + 5 = 10(3) + 5 = 30 + 5 = 35
Let there be x one rupee coin, y 50 paise coin, z 25 paise coinx + y + z = 220......(i)
Total amount = 160
Now, x + (y/2) + z = 160 ......(ii)
?4x + 2y + z = 640
Thrice as many 1 re coin as 25 paise
?x = 3z ......(iii)
On solving equation (i), (ii) and (iii), we get
z = 40x= 3* 40 = 120y = 220 ?(120 + 40 ) = 60
So, 50 paise coin =60
69 ?65 = 4
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