Question

sum of the digits of three digit number is 20.the middle digit is equal to one fourth of the other two. if the order of the digits is to be reversed the number increases the original number by 198.find the original number

Answer 1

ur required answer is here

Answer 2

Answer:

749

Step-by-step explanation:

Let the number be xyz

We are given that sum of digits of number is 20

⇒x+y+z=20  ---1

Now we are given that the middle digit is equal to one fourth of the sum of other two.

⇒$y= \frac{1}{4}(x+z)$    --2

Substitute the value of y from 2 in 1

$x+ \frac{1}{4}(x+z)+z=20$

$5x+5z=80$

$x+z=16$  ---3

Now we are given that if the order of the digits is to be reversed the number increases the original number by 198

⇒$100z+10y+x=100x+10y+z+198$

⇒$100z+x=100x+z+198$

⇒$99z=99x+198$  ---4

Now solve 3 and 4

Substitute the value of x from 3 in 4

$99z=99(16-z)+198$

$99z=1584-99z+198$

$198z=1782$

$z=\frac{1782}{198}$

$z=9$

Substitute the value of z in 3 to get value of x

$x+9=16$

$x=7$

Now substitute the value of z and x in 1 to get value of y

7+y+9=20

16+y=20

y=4

Thus the number is 749