Question

What is the sum of two numbers?
I. The bigger of these numbers is 6 more than the smaller number.
II. $40%$of the smaller number is equal to 30% of the bigger number.
III. The ratio between half of the bigger number and one-third of the smaller number is $2:1$
A. Only II and III are sufficient
B. Only I and II are sufficient
C. I and either II or III is sufficient
D. All, II and III together are sufficient

Answer 1

Hint: In this type of question we have not to solve the question, rather we have to find the information so that the question can be solved. So, look at the given information and write it in a mathematical equation. In this question we have to find the value of two numbers so we have to form two independent equations which give unique value after the solution. We have three statements in the question. Write their equations in the form of $\alpha x+\beta y=c,\text{ }{{\alpha }_{1}}x+{{\beta }_{1}}={{c}_{2}}\text{ }and\text{ }{{\alpha }_{2}}x+{{\beta }_{2}}={{c}_{2}}$ are straight line so they have unique solution only if $\dfrac{\alpha }{{{\alpha }_{1}}}\ne \dfrac{\beta }{{{\beta }_{1}}}$ if taken two equation at a time.
Write the I equation as$x=y+6$, similarly write equation for second and third statement and then proceed.

Complete step-by-step answer:
Let us assume that the two numbers are $x$ and $y$ $x>y$.
The first statement says that the bigger number is 6 more than the smaller number.
So, we can write
$\begin{align}
  & x=y+6 \\
 & x-y=6------(a) \\
\end{align}$
Second statement says that 40% of smaller numbers is equal to 30% of bigger numbers.
As we know that $x%\text{ of }y=\left( \dfrac{x}{100} \right)(y)$
So, we can write
$\begin{align}
  & 40\%\text{ of y}=\left( \dfrac{40}{100} \right)(y) \\
 & =.4y---------(1) \\
\end{align}$
Also, we have from second statement
$\begin{align}
  & 30\%\text{ of }x=\left( \dfrac{30}{100} \right)(x) \\
 & =.3x-------(2) \\
\end{align}$
Hence from equation (1) and (2) we can write
$\begin{align}
  & .4x=.3y \\
 & \Rightarrow 4x=3y \\
\end{align}$
We multiply both sides by 10.
Hence, we can write
$4x-3y=0-----(b)$
Now, statement III says that the ratio between half of the bigger number and one third of the smaller number is 2:1, hence we can write
\[\begin{align}
  & \left( \dfrac{{}^{x}/{}_{2}}{{}^{y}/{}_{3}} \right)=\dfrac{2}{1} \\
 & \Rightarrow \dfrac{x}{2}=(2)(\dfrac{y}{3}) \\
\end{align}\]
On cross multiplication we can write,
$\begin{align}
  & 3x=4y \\
 & \Rightarrow 3x-4y=0-----(c) \\
\end{align}$
Now we have three equation (a), (b) and (c)
As we now the any line in the form of
$\alpha x+\beta y=0$ is a straight line which always passes through origin, so if we solve the equation (b) and (c) we get solution $x=0$,$y=0$as both line (b) and (c) passes through origin, hence they cut each other at origin.
So, both are equal which is not possible as per question.
Now taking equation (a) and (b) simultaneously, we have
$x-y=6$ and $4x-3y=0$
Here we see that
$\dfrac{1}{4}\ne \dfrac{-1}{-3}$
Hence both line cuts at a unique point hence unique solution.
Now taking equation (a) and (c) simultaneously, we have
$x-y=6$ and $3x-4y=0$
Here we see that
$\dfrac{1}{3}\ne \dfrac{-1}{-4}$
Hence both line cuts at a unique point hence unique solution.
In the above both cases there is a unique solution.
Hence option C is correct.


Note: This type of question falls in the category of data sufficiency, here we have to look which data is redundant data and which data is sufficient to achieve the solution of the given question.
 It should be noted that the equation in the form of $ax+by+c=0$is a straight line whose slope is defined as $m=\dfrac{-a}{b}$. Two lines cut each other only if their slope is not equal.