# Pre RMO Practice Sheet 2

Question 1:

A man walks a certain distance and rides back in 3.75 hours, he could ride both ways in 2.5 hours. How many hours would it take him to walk both ways?

Let the man walk distance x km in time t hours. His walking speed is $\frac{x}{t}$ kmph.

Let him ride distance x km in time T hours. His riding speed is $\frac{x}{T}$ kmph.

First journey:

$\frac{x}{t} + \frac{x}{T} = 3.75$

Second journey:

$\frac{2x}{T} = 2.5$

Hence, $\frac{x}{T} = 1.25$

Using above in first equation: $\frac{x}{t} + 1.25 = 3.75$

Hence, $\frac{x}{t} = 2.50$ = his speed of walking. Hence, it would take him 5 hours to walk both ways.

Question 2:

Positive integers a and b are such that $a+b=\frac{a}{b} + \frac{b}{a}$. What is the value of $a^{2}+b^{2}$?

Given that a and b are positive integers.

Given also $\frac{a}{b} + \frac{b}{a} = a+b$.

Hence, $a^{2}(b-1)+b^{2}(a-1)=0$

As a and b are both positive integers, so $a-1$ and $b-1$ both are non-negative.

So, both the terms are non-negative and hence, sum is zero if both are zero or $a=1$ and $b=1$.

Hence, $a^{2}+b^{2}=2$

Question 3:

The equations $x^{2}-4x+k=0$ and $x^{2}+kx-4=0$ where k is a real number, have exactly one common root. What is the value of k?

$x^{2}-4x+k=0$…equation I

$x^{2}+kx-4=0$…equation II

Let $k \in \Re$ and let $\alpha$ is a common root.

Hence, $\alpha$ satisfies both the equations. So, by plugging in the value of $\alpha$ we get the following:

$\alpha^{2} -4\alpha + k =0$ and $x^{2}+kx-4=0$ and so using these two equations, we get the following:

$(k+4)(1-\alpha)=0$.

Case 1: $\alpha \neq 1$ then $k=-4$. But hold on, we havent’t checked thoroughly if this is the real answer. We got to check now if both equations with these values of alpha and k have only one common root.

Equation I now goes as : $x^{2}-4x-4=0$ so this equation has irrational roots. On further examination, we see that if $\alpha=1$, then k can be any value. So, what are the conditions on k? We get that from equation I: plug in the value of alpha:

$1-4+k=0$ so $k-3=0$ and $k=3$.

So, we now recheck if both equations have only one common root when alpha is 1 and k is 3:

$x^{2}-4x+3=0$ so the roots of first equation are 3 and 1.

$x^{2}+3x-4=0$ so the roots of second equation are -4 and 1.

Clearly so $k=3$ is the final answer. 🙂

Cheers,

Nalin Pithwa

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