Ratio and proportion tutorial problems: pre RMO or PRMO or IITJEE foundation maths

Example 1:

If $(2ma+6mb+3nc+9nd)(2ma-6mb-3nc+9nd)=(2ma-6mb+3nc-9nd)(2ma+6mb-3nc-9nd)$, prove that a, b, c, and d are proportionals.

Solution 1:

Given that $\frac{2ma+6mb+3nc+9nd}{2ma-6mb+3nc-9nd} = \frac{2mn+6mb-3nc-9nd}{2ma-6mb-3nc+9nd}$

We also know that if $\frac{x}{y} = \frac{p}{q}$, then the property of componendo and dividendo says: $\frac{x+y}{x-y} = \frac{p+q}{p-q}$. Applying this property to the above “huge” fraction, we get: $\frac{2(2ma+3nc)}{2(6mb+9nd)} = \frac{2(2ma-3nc)}{2(6mb-9nd)}$

We know that if $\frac{x}{y} = \frac{p}{q}$, then $\frac{x}{p} = \frac{y}{q}$, which is the property called alternendo. Applying this property to the above fraction, we get $\frac{2ma+3nc}{2ma-3nc} = \frac{6mb+9nd}{6mb-9nd}$,

again, applying componendo and dividendo to the above, we get $\frac{4ma}{6nc} = \frac{12mb}{18nd}$

hence, $\frac{a}{c} = \frac{b}{d}$, that is, a, b, c and d are proportionals. Hence, the proof.

Example 2:

Solve the equation: $\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \frac{4x-1}{2}$

Solution 2:

By componendo and dividendo, we get $\frac{\sqrt{x+1}}{\sqrt{x-1}} = \frac{4x+1}{4x-3}$

Now, squaring both sides of the above equation, we get $\frac{x+1}{x-1} = \frac{16x^{2}+8x+1}{16x^{2}-24x+9}$.

Again, applying componendo and dividendo, $\frac{2x}{2} = \frac{32x^{2}-16x+10}{32x-8}$ $x = \frac{16x^{2}-8x+5}{16x-4}$

so, we get $16x^{2}-4x=16x^{2}-8x+5$

so, we get finally $x=\frac{5}{4}$

Cheers,

Nalin Pithwa