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How to solve a hard number system question for CAT confidently and quickly 3

how to solve a hard number system question for CAT confidently ad quickly 3

Solving needs inequality analysis and inequality algebra

The problem: 3rd Hard number system question for CAT

Denominator of a fraction is less than the square of the numerator by 1. If 2 is added to both, the fraction will be more than $\frac{1}{3}$. And when 3 is subtracted from both, the fraction remains positive and is smaller than $\frac{1}{10}$. Find the fraction.

Hint: For quick solution of this 3rd hard number system question for CAT, explore deeply inequality analysis and inequality algebra.

Solution to the 3rd hard number system question for CAT: Inequality algebra with strategic focus on variable elimination

First step: Problem definition in mathematical terms

Let the fraction be $\displaystyle\frac{n}{d}$, where integers $n$ and $d$ are the positive numerator and the denominator.

By the first statement,

$d=n^2-1$.................................(1)

By the second statement,

$\displaystyle\frac{n+2}{d+2} > \displaystyle\frac{1}{3}$......................(2)

And by the third statement,

$0 < \displaystyle\frac{n-3}{d-3} < \displaystyle\frac{1}{10}.$..............(3)

How can we solve this awkward problem in the shortest possible way? Which of the three relations individually or combined with another would give us the first breakthrough?

The first equality relation just states the equality and we get no clue on the values of $n$ and $d$ except,

$n < d$.

The second relation is not responsive, but the third provides the first breakthrough.

As the fraction is positive, both numerator and denominator are positive.

Conclusion:

$n > 3$

$d > 3$.

Both $n$ and $d$ are greater than 3.

This is the time to eliminate $d$ using mainly the first equation and from the inequalities involving only $n$ that is greater than 3 we should get more clarity into the problem.

As planned, substitute $n^2-1$ for $d$ in inequality (2),

$\displaystyle\frac{n+2}{n^2+1} > \displaystyle\frac{1}{3}$,

Or, $3n+6 > n^2+1$..........(4), by cross-multiplying that is permissible for inequalities.

Substitute $n^2-1$ for $d$ in inequality (3),

$\displaystyle\frac{n-3}{n^2-4} < \displaystyle\frac{1}{10}$,

Or, $10n-30 < n^2-4$.......(5).

It is clear that $n^2$ can be eliminated from inequalities (4) and (5) that would give us for the first time an inequality on $n$ only and knowing $n$ to be greater than 3, we should have a clearer grip on the value of $n$.

But how to eliminate $n^2$? These two are inequalities.

Well, we can. Easily. 

First convert the inequality (5) to a greater than inequality reversing its original nature by multiplying both sides with $-1$,

$30-10n > 4-n^2$.........(6)

Now inequalities (4) and (6) can be added together as both are of same nature. Two LHSs lesser than two larger RHSs when added together would always remain less than the sum of the RHSs.

Adding (4) and (6),

$36 -7n > 5$.

Divide through by 7,

$n < \displaystyle\frac{36}{7} -\displaystyle\frac{5}{7}$,

Or, $n \leq 4$.

Knowing $n > 3$ we get the big breakthrough,

$n=4$, and ,

$d=n^2-1=15$.

The fraction is, $\displaystyle\frac{4}{15}$.

Answer: $\displaystyle\frac{4}{15}$.

Let's verify.

Verification of the values against given conditions

For equation 1: $15 = 4^2-1=16-1$,

For inequality 2: $\displaystyle\frac{4+2}{15+2}=\frac{6}{17} > \displaystyle\frac{1}{3}$

For inequality 3: $\displaystyle\frac{4-3}{15-3}=\frac{1}{12} < \displaystyle\frac{1}{10}$.

Inequality rules used

Cross-multiplication of numerator and denominator between LHS and RHS

If $\displaystyle\frac{a}{b} < \displaystyle\frac{c}{d}$, then,

$ad < bc$.

Reversing nature of inequality

Multiply both sides of an inequality by $-1$ to reverse the nature of the inequality.

If $a > b$, then,

$-a < -b$.

Addition of two inequalities

If $a > b$ and $c > d$,

$a+c > b+d$.

Adding two smaller quantities remains less than the sum of two larger quantities obviously.

So the inequality rules are nothing that we didn't know. All these are based on very basic arithmetic concepts only.

Though inequality algebra looks so normal and obvious, usually we don't require to use these but when used suitably, a difficult problem can easily be broken open.

The main motive that provided the push though had been the necessity to eliminate $d$ which played the role of the core element in the systematic and directed problem solving.

This is systematic, confident and quick problem solving using suitable concepts and techniques at each step, in short, systematic problem solving.


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