Ratio and Proportion — Important Questions

Answer: (a) $9:16$.

The duplicate ratio of $a:b$ is $a^2:b^2=3^2:4^2=9:16$.

Answer: (a) $12$.

Mean proportional $=\sqrt{9\times16}=\sqrt{144}=12$.

Answer: (a) $8:15$.

Make $b$ common: $a:b=8:12$ and $b:c=12:15$, so $a:b:c=8:12:15$ and $a:c=8:15$.

Answer: (a) $24$.

If $6,12,x$ are in continued proportion then $12^2=6x\Rightarrow x=\dfrac{144}{6}=24$.

Answer: (a) By R, the triplicate ratio of $2:3$ is $2^3:3^3=8:27$, which is exactly the assertion, so R correctly explains A.

If $3:12=5:x$, then $3x=12\times5=60$.

$x=\dfrac{60}{3}=\mathbf{20}$.

Compound ratio $=\dfrac{2\times6\times11}{3\times11\times2}=\dfrac{132}{66}=\dfrac{2}{1}$.

So the compound ratio is $\mathbf{2:1}$.

Since $a,b,c$ are in continued proportion, $b^2=ac$.

Take the right-hand side and substitute $b^2=ac$:

$\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a(a+c)}{c(a+c)}=\frac{a}{c}.$

Hence $\dfrac{a}{c}=\dfrac{a^2+b^2}{b^2+c^2}$, as required.

Since $x$ is the mean proportional between $(x-2)$ and $(x+6)$:

$x^2=(x-2)(x+6).$

$x^2=x^2+4x-12\Rightarrow 0=4x-12\Rightarrow x=\mathbf{3}.$

(Check: mean proportional between $1$ and $9$ is $\sqrt{9}=3$.)

Make $B$ common: $A:B=8:12$ and $B:C=12:15$, so $A:B:C=8:12:15$.

Total parts $=8+12+15=35$; value of one part $=\dfrac{1400}{35}=40$.

A $=8\times40=$ ₹$\mathbf{320}$, B $=12\times40=$ ₹$\mathbf{480}$, C $=15\times40=$ ₹$\mathbf{600}$.

The ratio equals $\dfrac{9}{1}$. Apply componendo and dividendo:

$\frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})}=\frac{9+1}{9-1}.$

$\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\frac{10}{8}\Rightarrow\frac{\sqrt{3x+4}}{\sqrt{3x-5}}=\frac{5}{4}.$

Squaring: $\dfrac{3x+4}{3x-5}=\dfrac{25}{16}$.

$16(3x+4)=25(3x-5)\Rightarrow 48x+64=75x-125\Rightarrow 189=27x$.

So $x=\mathbf{7}$.

Let $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$. Then $x=ak$, $y=bk$, $z=ck$.

Left-hand side:
$\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{a^3k^3}{a^3}+\frac{b^3k^3}{b^3}+\frac{c^3k^3}{c^3}=k^3+k^3+k^3=3k^3.$

Right-hand side:
$\frac{3xyz}{abc}=\frac{3(ak)(bk)(ck)}{abc}=\frac{3abc\,k^3}{abc}=3k^3.$

Since LHS $=$ RHS $=3k^3$, the identity is proved.

(i) $\dfrac{\text{boys}}{\text{girls}}=\dfrac{5}{4}\Rightarrow\dfrac{500}{\text{girls}}=\dfrac{5}{4}\Rightarrow\text{girls}=\dfrac{4\times500}{5}=\mathbf{400}$.

(ii) Total $=500+400=\mathbf{900}$ students.

(iii) New girls $=400+50=450$. New ratio $=500:450=\mathbf{10:9}$.