Question

two digits number contains the smaller of two digits in the units place . the product pf the digits is 24 and difference is 5. find the number Question

Answer 1

Answer: 83

Step by step explanation:

Let the digit in unit's place be x __. Smallest number )

Let the digit in ten's place be y __. ( Largest number )

Original number :- 10y + x

Now,.

According to question,

xy = 24 __.( i )

y - x = 5

» y = 5 + x __. ii )

Putting the value of y in equation ( i )

xy = 24

x ( 5 + x ) = 24

5x + x² = 24

x² + 5x - 24 = 0

x² + 8x - 3x - 24 = 0

x ( x +8 ) - 3 (x +8) = 0

( x - 3 ) ( x + 8) = 0

( x - 3 ) = 0

» x = 3

( x +8 ) = 0

x = -8

x = -8 can't be accepted, so x = 3.

y = x + 5

y = 3 + 5

y = 8

Thus, required number is :-

10y + x

» 10 × 8 + 3

» 80 + 3

» 83

Thus, the answer is 83.

Answer 2

Given :---

• in a two digit number the smaller digit is a unit place .
• Product of digits = 24.
• Difference b/w , digits = 5

To Find :---

• Original Number ?

solution :---

Let the Original number be :- 10x+y,

Where , digit at unit place = y, ( that is smaller one )

→ Digit at tenth place = x ( Larger one) .

Now, According to Question , Difference b/w both is 5 .

So,

→ x( Larger one ) - y(smaller one) = 5

→ x = (5 + y) ------------------ Equation (1)

Now, it is also given that, Product of both digits is 24.

so,

→ x * y = 24 ---------------- Equation (2)

____________________________

Putting value of Equation (1) in Equation (2) now we get,

→ (5+y) * y = 24

→ y² + 5y - 24 = 0

Now, Solving the Equation by splitting the middle term we get,

→ y² + 5y - 24 = 0

→ y² + 8y - 3y -24 = 0

→ y(y+8) -3(y+8) = 0

→ (y+8)(y-3) = 0

Putting both Equal to 0 now , we get,

→ y + 8 = 0, or, y - 3 = 0

→ y = (-8) or,. y = 3

Negative value Not Possible Now,

So,

y = 3

Putting this in Equation (1) we get ,,

→ x = 5+y

→ x = 5+3 = 8

_____________________________

Hence, our Original Number = 10x + y = 10*8 + 3 = 83 (Ans) ..

So, original number is 83...