Question

Difference between two numbers is 9 and the difference between their squares is 981. Lowest of the two numbers is
A. 40
B. 50
C. 55
D. 59

Answer 1

Hint: In this question assume two numbers as x and y. Write the equation as per question. We have two condition in the question so we have to write two equation in variable x and y as
 $ x-y=9 $ and $ {{x}^{2}}-{{y}^{2}}=981 $
 As we have two unknown and we have two equations so we can get the result by solving the equation. Form two equations in the form of \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\], solve the equation so obtained to get the values of unknowns.

Complete step-by-step answer:
We start our solution by assuming that the two numbers are $ x $ and $ y $ . Now from question it is given that difference of two number is 9 so we here assume that \[x>y\]
hence, we can write
 $ x-y=9------(a) $
Also, it is also given from question that difference between their square is 981, Hence we can write this in equation form as
 $ {{x}^{2}}-{{y}^{2}}=981----(b) $
As we know that $ {{a}^{2}}-{{b}^{2}}=(a+b)(a-b) $
So, we can write equation (b) as
 $ (x+y)(x-y)=981---(c) $
Now here we can substitute the value of $ x-y=9 $ from equation (a) in equation (c), so we can write
 $ (x+y)(9)=981 $
Dividing both side by 9 we have
 $ \begin{align}
  & \dfrac{(x+y)(9)}{9}=\dfrac{981}{9} \\
 & \Rightarrow x+y=109----(d) \\
\end{align} $
Now we see here that we get two linear equations that are equation $ (a) $ and equation $ (d) $ . Adding equation $ (a) $ and $ (d) $ we can write
 $ \begin{align}
  & (x+y)+(x-y)=109+9 \\
 & \Rightarrow 2x=118 \\
\end{align} $
Dividing both side by 2 we get
 $ x=59 $
Now in order to find out the value of $ y $ we substitute the value of $ x=59 $ in equation $ (a) $ hence we can write
 $ 59-y=9 $
So, we can write by transposing the terms so that variables terms are on one side and constant terms on another side we can write
 $ y=50 $
Hence, we get the values
 $ \begin{align}
  & x=59 \\
 & y=50 \\
\end{align} $
As we have to find lowest number so $ y=50 $
Hence option B is correct.

Note: We can solve the above equation by substitution method also. Whether two linear equation have unique solution or not we check the ratio of coefficient of variables as,
If \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]are two linear equation then they have unique solution if $ \dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}} $ . Also we can draw the graph of equation $ (a) $ and ( $ (d) $ as the equations satisfy the above condition, both lines cut each other. The point of intersection is the solution of the equation on the $ x $ and $ y $ axis respectively.