The sum of the digits of a two-digit number is 14. If the digit in the ten's place is 1 less than twice the digit in the unit's place, find the number. 15. The difference between the digits of a two-digit number is 2. If the digit in the ten's place and the digit in the unit's place are in the ratio 3 : 2, find the number.
Answers
Answer:
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Step-by-step explanation:
Let x be the digit at unit’s place and y be the digit at ten’s place.
Since y is at ten’s place, then the number formed is 10y+x.
By reversing the digits, it becomes 10x+y.
As the difference of the numbers is 18, so,
(10y+x)−(10x+y)=18
9(y−x)=18
y−x=2 .... (1)
As the sum of digits is 8, so,
x+y=8 .... (2)
On adding equations (1) and (2), we get
2y=10⇒y=5
Putting this in (2), we get x=8−5=3
x=3,y=5
Hence, number =10y+x=10×5+3=53.
Step-by-step explanation:
Solutions :-
1)
Let the digit at ones place in a two digit number be X
The digit at tens place
= 1 less than twice the digit in the unit's place
= 2X-1
Given that
The sum of the digits of a two-digit number is 14.
=> X+2X-1 = 14
=> 3X-1 = 14
=> 3X = 14+1
=> 3X = 15
=> X = 15/3
=> X = 5
The digit at ones place = 5
The digit at tens place = 2X-1
=> 2(5)-1
=> 10-1
=> 9
or
14-5 = 9
Therefore, The number = 95
The required two digit number is 95
2)
Given that
The digit in the ten's place and the digit in the unit's place are in the ratio 3 : 2
Let they be 3X and 2X
Their difference = 3X-2X = X
According to the given problem
Their difference = 2
=>X = 2
Now,
3X = 3(2) = 6
the digit at tens place = 6
2X = 2(2) = 4
The digit at ones place = 4
The required number is 64