The sum of a two-digit number and the number obtained the digits is 66. If the digits of the number differ by 2, find the number. How numbers are there? The sum of a two-digit number and the number obtained the digits is 66. If the digits of the number differ by 2, find the number. How numbers are there? The sum of a two-digit number and the number obtained
the digits is 66. If the digits of the number differ by 2, find the number. How
numbers are there? The sum of a two-digit number and the number obtained
the digits is 66. If the digits of the number differ by 2, find the number. How
numbers are there? The sum of a two-digit number and the number obtained
the digits is 66. If the digits of the number differ by 2, find the number. How
numbers are there?
Answers
Step-by-step explanation:
Answer
Let the two digit be x and y
∴ Number (2−digit) =10×x+y
Sum of 2−digit and reverse of it
⟹10x+y+10y+x=66 (given)
∴x+y=6 (equation 1)
Digits differ by 2
∴x−y=2 (equation 2)
On adding,
2x=8
⟹x=4
∴y=2
or, x+y=6 From 1 equation
and y−x=2 From 2 equation
On adding:
2y=8
y=4
x=2
∴ Two digit numbers are: 42 and 24
There are two such numbers.
Answer:
From the question it’s told as, the two digits of the number are differing by 2. Thus, we can write
x – y = ±2………….. (i)
Now after reversing the order of the digits, the number becomes 10x + y.
Again from the question it’s given that, the sum of the numbers obtained by reversing the digits and the original number is 66. Thus, this can be written as;
(10x+ y) + (10y+x) = 66
⇒ 10x + y + 10y + x = 66
⇒ 11x +11y = 66
⇒ 11(x + y) = 66
⇒ x + y = 66/11
⇒ x + y = 6………….. (ii)
Now, we have two sets of systems of simultaneous equations
x – y = 2 and x + y = 6
x – y = -2 and x + y = 6
Let’s first solve the first set of system of equations;
x – y = 2 …………. (iii)
x + y = 6 ………….. (iv)
On adding the equations (iii) and (iv), we get;
(x – y) + (x + y) = 2+6
⇒ x – y + x + y = 8
⇒ 2x =8
⇒ x = 8/2
⇒ x = 4
Putting the value of x in equation (iii), we get
4 – y = 2
⇒ y = 4 – 2
⇒ y = 2
Hence, the required number is 10 × 2 +4 = 24
Now, let’s solve the second set of system of equations,
x – y = -2 …………. (v)
x + y = 6 ………….. (vi)
On adding the equations (v) and (vi), we get
(x – y)+(x + y )= -2 + 6
⇒ x – y + x + y = 4
⇒ 2x = 4
⇒ x = 4/2
⇒ x = 2
Putting the value of x in equation 5, we get;
2 – y = -2
⇒ y = 2+2
⇒ y = 4
Hence, the required number is 10×4+ 2 = 42
Therefore, there are two such possible numbers i.e, 24 and 42.
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