Question

The sum of the digits of a three-digit number is 11.If the digits are reversed,the new number is 46 more than five times the former number.If the hundreds digit plus twice the tens digit is equal to the units digit,then find the three equations using which we can find the original three-digit number.

• A. 4x+7y+9z=135
• B. x +y +z=11
  
• C. 499x+40y-95z = -46
• D. x+2y-z=0

Answer 1

Answer: (B), (C), (D)

The correct options are
B x +y +z=11

C 499x+40y-95z = -46
D x+2y-z=0

Let the digits in hundreds,tens and unit's place be x ,y and z respectively.

As given in the question,

x +y+z =11…..........(1)

Original number=100x+10y+z

If the digits are reversed,the new number is given by:

New number=100z+10y+x

Given:" If the digits are reversed,the new number is 46 more than five times the former number "

​​​​​​$\Rightarrow$ 100z+10y+x=46+5 (100x+10y+z )

$\Rightarrow$ 100z+10y+x=46+500x+50y+5z

$\Rightarrow$ 500x+50y+5z-100z-10y-x =-46

$\Rightarrow$ 499x +40y -95z = -46.............(2)

Given:"Hundreds digit plus twice the tens digit is equal to the units digit "

$\Rightarrow$ x +2y =z

$\Rightarrow$ x +2y - z =0 ................(3)

By solving equations (1), (2) and (3), we can find the original three-digit number by putting x,y and z values in the expression for original number.