Question

A combination of locks requires 3 numbers to open. The second number is $2d+5$ greater than the first number. The third number is $3d-20$ less than the second number. The sum of the three numbers is $10d+9$. The first number is

• A. $5d-11$
• B. $3d-7$
• C. $2d+19$
• D. $3d-11$

Answer 1

The correct option is B $3d-7$

Let the first number be $x$
Then, second number = $x+2d+5$
Third number = $x+2d+5-(3d-20)$ = $x-d+25$
Sum of the three numbers = $x+x+2d+5+x-d+25$ = $3x+d+30$
Thus, $3x+d+30$ = $10d+9$
$3x=10d+9-d-30$ = $9d-21$
$x=3d-7$