ISI2015-MMA-50

Question

Let

$$\begin{array}{} V_1 & = & \frac{7^2+8^2+15^2+23^2}{4} – \left( \frac{7+8+15+23}{4} \right) ^2,  \\ V_2 & = & \frac{6^2+8^2+15^2+24^2}{4} – \left( \frac{6+8+15+24}{4} \right) ^2 , \\ V_3 & = & \frac{5^2+8^2+15^2+25^2}{4} – \left( \frac{5+8+15+25}{4} \right) ^2 . \end{array}$$ Then

• A. $V_3<V_2<V_1$
• B. $V_3<V_1<V_2$
• C. $V_1<V_2<V_3$
• D. $V_2<V_3<V_1$

Answer 1

In all the $V_1$ ,$V_2$ ,$V_3$

$\bigg( \frac{7+8+15+23}{4} \bigg) ^2$=$\bigg( \frac{6+8+15+24}{4} \bigg) ^2$=$\bigg( \frac{5+8+15+25}{4} \bigg) ^2$=$53^{2}/4^{2}$

so it won't decide the relation.Hence no need to calculate it.

In all the $V_1$ ,$V_2$ ,$V_3$ ,first term is divided by 4 so it won't have impact on relation.Hence no need to calculate it.

also there is $8^{2}+15^{2}$ in numerator of first term of $V_1$ ,$V_2$ ,$V_3$. So it won't decide the relation.

Hence we need to calculate only $7^{2}+23^{2}$ in $V_1$, $6^{2}+24^{2}$ in $V_2$,$5^{2}+25^{2}$ in $V_3$

$7^{2}+23^{2}$=578,

$6^{2}+24^{2}$=612

$5^{2}+25^{2}$ =650

Therefore $V_3$ > $V_2$ >$V_1$

Answer 2

Here

$\small\begin{align} V_1-V_2 &=\frac{7^2+8^2+15^2+23^2}{4} -\frac{6^2+8^2+15^2+24^2}{4}– \left[ \left( \frac{7+8+15+23}{4} \right) ^2 -\left( \frac{6+8+15+24}{4} \right) ^2\right] \\&=\frac{7^2-6^2+23^2-24^2}{4}-\left[ \left( \frac{7+8+15+23}{4} \right) ^2 -\left( \frac{7-1+8+15+23+1}{4} \right) ^2\right]\\&=\frac{(7+6)(7-6)+(23+24)(23-24)}{4}- \left[ \left( \frac{7+8+15+23}{4} \right) ^2 -\left( \frac{7+8+15+23}{4} \right) ^2\right]\\&=\frac{13-47}{4}-0\\&=-\frac{34}{4}<0 \end{align}$

$\therefore V_1<V_2 \tag{i}$

 

Likewise
$$\small\begin{align} V_2-V_3 &=\frac{6^2+8^2+15^2+24^2}{4} -\frac{5^2+8^2+15^2+25^2}{4}– \left[ \left( \frac{6+8+15+24}{4} \right) ^2 -\left( \frac{5+8+15+25}{4} \right) ^2\right] \\&=\frac{(6+5)(6-5)+(24+25)(24-25)}{4}- \left[ \left( \frac{6+8+15+24}{4} \right) ^2 -\left( \frac{6+8+15+24}{4} \right) ^2\right]\\&=\frac{14-49}{4}-0\\&=-\frac{35}{4}<0\end{align}$$

$\therefore V_2<V_3 \tag{ii}$

 

Combining no$\mathrm{(i)}$ and no$\mathrm{(ii)}$, we get

$$V_1<V_2<V_3$$

 

So the correct answer is C.

 

Comments

Nice answer.

I was stuck on this earlier.