An algebra problem by Manish Mayank

$P:Q:R=6:5:4$ means that $P:6=Q:5=R:4$ which can also be written as $\frac{P}{6}=\frac{Q}{5}=\frac{R}{4}$ .

Multiplying by 6 we have that $P=\frac{6}{5}Q$

Multiplying by 4 we have that $R=\frac{4}{5}Q$

Substituting these values on the equation we have $P^2 + Q^2 + R^2 = 192500 \\ \frac{36}{25} Q^2 + Q^2 + \frac{16}{25} Q^2 = 192500 \\ \frac{77}{25} Q^2 = 192500 \\ Q^2 = 62500 \\ Q = 250$

Now we can find that $P=300$ and $R=200$ and therefore $(P + Q - R)/2 = (300 + 250 - 200)/2 = 175$

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var p=6,q=5,r=4;
for(var c=1;!(p*p+q*q+r*r==192500);p=c*6;q=c*5,r=c*4) 
{

}
console.log("P="+p+"Q="+q+"R="+r+" (P+Q-R)/2="+(p+q-r)/2);

Java Script.

$300^2+250^2+200^2=192,500 =>$ $[(300+250)-200]/2=175$

Let P = 6n Q = 5n R = 4n Now, (6n)^2 + (5n)^2 + (4n)^2 = 192500

Find "n" and hence P,Q and R

nice problem

P : Q : R = 6 : 5 : 4

Let P = 6x ; Q = 5x ; R = 4x

P^2 + Q^2 + R^2 = (6x)^2 + (5x)^2 + (4x)^2 = 36x^2 + 25x^2 + 16x^2 = 77x^2

Now, 77x^2 = 192500 x^2 = 2500 x = 50

Now, (P + Q - R) / 2 = (6x + 5x - 4x)/2 = 7x/2

Substituting x=50

7x/2 = 175