Question

a two digit number is 5 times the sum of its digits and is also equal to 5 more than twice the product of its digits.find the number.

Answer 1

let the tens digit be x and ones digit be y
number=10x+y
sum of digits =x+y
product of digits=xy
axcording to question: 10x+y=5 (x+y)
10x+y=5x+5y
5x=4y
x=4/5y
then 2nd equation is: 10x+y=2xy+5
putting the value of x in the above equation:
10×4/5y+y=2×4/5y×y +5
8y+y=8/5y^2+5
45y=8y^2+25
8y^2-45y+25=0
solving quadratic equation:
8y^2-40y-5y+25=0
8y(y-5)-5 (y-5)=0
(8y-5)(y-5)=0
y=5 as digit cannot be 5/8, i.e in fraction
x=4/5y=4/5*5=4
therefore number=45

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the tens place digit of required number be x.

And the unit place digit be y.

According to the Question,

⇒ 10x + y = 5(x + y)

⇒ 4y = 5x

⇒ y = 5x/4  .... (i)

And, 10x + y = 2xy + 5

⇒ 10x + 5x/4 = 2x × 5x/4 + 5  [From (i)]

⇒ 45x/4 = 10x²/4 + 5

⇒ 10x² - 45x + 20 = 0

⇒ 2x² - 9x + 4 = 0

⇒ 2x² - 8x - x + 4 = 0

⇒ 2x(x - 4) - 1(x - 4) = 0

⇒ (x - 4) (2x - 1) = 0

⇒ x - 4 = 0 or 2x - 1 = 0

⇒ x = 4, 1/2 (As x can't be a fraction)

⇒ x = 4

Putting x's value in Eq (i), we get

⇒ y = 5x/4

⇒ y = 5(4)/4

⇒ y = 20/4

⇒ y = 5

Number = 45

Hence, the required number be 45.

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