Question

A combination of locks requires three 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

Hint: According to the question we know that the lock requires three numbers to open, therefore let us suppose that the three numbers are ‘x’ , ‘y’ and ‘z’, use the given 3 conditions to get 3 linear equations and solve to get value of x, y, z.

Complete step-by-step answer:
Then from the question we know that the first number is ‘x’ , second number is ‘y’ and third number is ‘z’. We have been given that the second number is 2d+5 greater than the first number therefore it can be written as
 $ \Rightarrow y = 2d + 5 + x $ (let this be the equation number 1)
The third number is 3d-20 less than the second number that is it can be written as:
 $
   \Rightarrow z = y - (3d - 20) \\
   \Rightarrow z = y - 3d + 20 \;
 $
(let this be the equation number 2 )
Also we have been given that sum of all the three numbers is 10d+9 that is
 $ x + y + z = 10d + 9 $ (let this be the equation number 3)
Now we can get the value for the first number by substituting the value of the second number ‘y’ from the equation number 1 in the equation number 2 to get the third number ‘z’ in the terms of x. therefore we have,
 $
   \Rightarrow y = 2d + 5 + x \\
   \Rightarrow z = y - 3d + 20 \\
   \Rightarrow z = 2d + 5 + x - 3d + 20 \\
   \Rightarrow z = x - d + 25 \;
 $
 Now we the number y and the number z in the terms of x so on substituting the values of all the numbers in the equation number three we will get all the terms of x and hence eliminating we can get the value of x which is the first number,
 $
   \Rightarrow x + y + z = 10d + 9 \\
   \Rightarrow x + 2d + 5 + x + x - d + 25 = 10d + 9 \\
   \Rightarrow 3x + d + 30 = 10d + 9 \;
 $
Keeping x on one side and moving all the other terms on R.H.S, we have
 $
   \Rightarrow 3x = 10d - d + 9 - 30 \\
   \Rightarrow 3x = 9d - 21 \\
   \Rightarrow x = \dfrac{{9d - 21}}{3} = 3d - 7 \;
 $
So, the correct answer is “Option B”.

Note: Whenever we have a linear equation to solve in more than one or two variables like the case above we need to have three equations. So that we can change the three variables to have similar terms in one variable. Then it can be solved easily by transposition and other methods.