The unit digit of the two digit number exceeds its tens digit by 6 and the product of two digit is less by 12 from the number.Find the number.
Answers
Answer: 28 or 36
Step-by-step explanation:
•••••••••••••••••••••••••••••••••••••••••••••••••••••••
Let unit digit be y and tens digit be x
≈ Then the two-digit no. will be = 10x + y
≈ y-x =6(unit digit exceeds the tens digit by 6)
...........(i)
≈ product = xy
≈ (10x + y)-xy = 12 ( the product is less by 12 from the no.).
≈ 10x + y = 12 + xy ............(ii)
Solving (i) and (ii) simultaneously:
From (i) we get : y = 6 + x. ............(iii)
≈ putting the value of (iii) in (ii)
≈ {10x + (6 + x)} = 12 + x(6 + x)
≈ (10x + x + 6) = 12 + 6x + x^2
≈ 11x + 6 = x^2 + 6x + 12
≈ 5x = x^2 + 6
≈ Solving the equation, we get x= 2 or 3 Putting the value of x one by one for both values in (i)
If x = 2:
y - 2 = 6
≈ y = 8 ; then the no. would be 10*2 + 8
≈ 20 + 8
≈ 28....ANS
Or if x = 3:
y - 3 = 6
≈ y = 9 ; then the no. would be 10*3 + 6
≈ 30 + 6
≈ 36....ANS
•••••••••••••••••••••••••••••••••••••••••••••••••••••••
Hope it would help you. Plz give stars, it took a lot of tym to type this.
Answer:
8 OR 9
Step-by-step explanation:
Let digit in tens place = x
therefore, the digit in unit place be x+ 6
therefore, the number= ( x + 6 )+ 10x
= x+6+10x
= 11x + 6
product of two digit number= x(x+6)
= x2+6x
According to the question
→ x2 + 6x= ( 11x + 6 ) - 12
→ x2+6x=11x-6
→x2+6x-11x+6=0
→x2-5x+6=0
→x2-(3+2)x+6=0
→x2-3x-2x+6=0
→x(x-3) -2(x-3)=0
→(x-3) (x-2)=0
Therefore, Either, (x-3)=0
x=3
OR, (x-2)=0
x=2
Possible value for the digit unit place be
(i) Either, x+6=3+6=9
(ii)or, x+6=2+6=8
This is a correct answer