PUC
Question : If $a^2+b^2+c^2=16$, $x^2+y^2+z^2=25$ and $ax+by+cz=20$, then the value of $\frac{a+b+c}{x+y+z}$ is:
Option 1: $\frac{3}{5}$
Option 2: $\frac{5}{3}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{5}{4}$
Correct Answer: $\frac{4}{5}$
Solution : Given: $a^2+b^2+c^2=16$----(equation 1) $x^2+y^2+z^2=25$-----(equation 2) $ax+by+cz=20$-----(equation 3) Substitute the values of $b=0$ and $c=0$ in the given equation 1, we get, $a^2+0^2+0^2=16$ $⇒a^2=16$ $\therefore a=4$ Substitute the values of $y=0$ and $z=0$ in the given equation 2, we get, $x^2+0^2+0^2=25$ $⇒x^2=25$ $\therefore x=5$ Substitute the
Question : The highest population density in India, according to the 2011 Census, is in ______________.
Option 1: Bihar
Option 2: Union Territory of Chandigarh
Option 3: National Capital Territory of Delhi
Option 4: Madhya Pradesh
Correct Answer: National Capital Territory of Delhi
Solution : The correct option is the National Capital Territory of Delhi.
As of the 2011 Census of India, the National Capital Territory of Delhi had the highest population density among all states and union territories in India. Delhi is a densely
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