Question

A two digit number is such that the product of the digits is 20. If 9 is subtracted from the number, the digits interchange their place. Find the number.

Answer 1

Answer :

Let the unit place is x

Let the tenth palce is y

So equation will be -  10y + x

Interchange will be - 10x + y

Given ,

Now according to first case :

xy = 20 ................1] Equation

Now according to seco

10y + x - 9 = 10x + y.

→ 10y - y = 10x - x + 9

→9y = 9x + 9

→9y - 9x = 9

So we also written in this form :

→ y - x =  1

→y - 1 = x

→x = y - 1 ...............................2] Equation

Now put the value of y in the First equation :

xy = 20

y[y - 1 ] = 20

$=> y^{2} - y = 20$

$=> y^{2} - y - 20 = 0$ ...................3] Equation

Now we find the value of x and y from 3 equation through factorisation method :

∴ $y^{2} - 5y + 4y -20 = 0$

→y[ y - 5] + 4[y- 5] = 0

→[y - 5] [  y + 4] = 0

Now , $\boxed{y = 5}$  and $\boxed{y = - 4}$

So Always Number should be in positive form so Correct value of y = 5

Now put the value of y in 2 equation :

→x =   y - 1

→x = 5 - 1

→x = 4

Therfore Number will be :

$\boxed{\boxed{ 10y + x => 10 x 5 + 4 = 54 Answer }}$

Answer 2

Question :A two digit number is such that the product of the digits is 20. If 9 is subtracted from the number, the digits interchange their place. Find the number.

Solution :

Given :

• Product of the digits is 20.

To Find :

• The number

Let the digit in tens place be x and the digit at unit place be y.

Product of the digits = 20

$= > x \times y = 20$

$\fbox{\orange{x= \frac {20}{y} \:\:\:\:\: ....(1)}}$

In original number we write the number in the form of 10x+y and in reverse number we write in the form of 10y+x because when digit interchange their places then, number becomes reverse.

We write digit of tens place at unit place and digit of unit place in tens place in reverse number.

Original number = 10x+y

Reverse number = 10y+x

= > 10y + x= (10x + y) - 9

$= > (10y + x) - (10x + y )= - 9 \\ = > 10y + x - 10x - y = - 9 \\ = > 9y - 9x = - 9 \\ = >9 (y - x) = - 9 \\ = > y - x = \frac{ - 9}{9}$

$\fbox{\red { y - x = - 1 \: \: \: \: \: \: .....(2)}}$

Now putting value of equation (1) in equation (2) :

$=>y-x=-1$

$=>y-\frac {20}{y}=-1$

$=>\frac {y^2-20}{y}=-1$

$=>y^2-20 = -y$

$=>y^2+y-20=0$

$=>y^2 +(5-4)y-20=0$

$=>y^2+5y-4y-20=0$

$=>y(y+5)-4(y+5)=0$

$=>(y+5)(y-4)=0$

$=>y+5=0$

$=>y =0-5$

$=>y=-5$

and

$=>y-4=0$

$=>y=0+4$

$=>y=4$

Value of y=4 because we can't take value of y in negative.

$x\times y=20$

$x\times 4=20$

$x=\frac {20}{4}$

$x=5$

Original number

=10x+y

=10×5+4

=50+4

=54

$\fbox {\green {Original \:number=54}}$

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