Question 12 Marks
If ax = by = cz, prove that
$\frac{x^2}{y z}+\frac{y^2}{z x}+\frac{z^2}{x y}=\frac{b c}{a^2}+\frac{c a}{b^2}+\frac{a b}{c^2}$.
AnswerLet $ax = by = cz = k$, then $x=\frac{k}{a}, y=\frac{k}{b}$ and $z=\frac{k}{c}$
$\text { L.H.S. }=\frac{x^2}{y z}+\frac{y^2}{z x}+\frac{z^2}{x y} $
$=\frac{k^2}{a^2 \times \frac{k}{b} \times \frac{k}{c}}+\frac{k^2}{b^2 \times \frac{k}{c} \times \frac{k}{a}}+\frac{k^2}{c^2 \times \frac{k}{a} \times \frac{k}{b}} $
$ =\frac{b c}{a^2}+\frac{c a}{b^2}+\frac{a b}{c^2}$
$= R.H.S.$
Hence proved.
View full question & answer→Question 22 Marks
Show, that a, b, c, d are in proportion if:
(6a + 7b) : (6c + 7d) : : (6a - 7b) : (6c - 7d)
AnswerWe have,
$\frac{a}{c}=\frac{c}{d}$
$\left(\right.$ Both sides are multiplied by $\left.\frac{6}{7}\right)$
$\Rightarrow \frac{6 a}{7 b}=\frac{6 c}{7 d}$
Applying componendo and dividendo
$\frac{6 a+7 b}{6 a-7 b}=\frac{6 c+7 d}{6 c-7 d}$
Applying alternendo
$\frac{6 a+7 b}{6 c+7 d}=\frac{6 a-7 b}{6 c-7 d}$
(6a + 7b ) : (6c + 7d) :: (6a - 7b) (6c - 7d).
View full question & answer→Question 32 Marks
If $x$ and $y$ be unequal and $x$ : $y$ is the duplicate ratio of $(x+z)$ and $(y+z)$ prove that $z$ is mean proportional between $x$ and $y$.
AnswerSince $x : y$ is duplicate ratio of $(x + z)$ and $(y + z)$
$\therefore x : y = (x + z)^2 : (y + z)^2$
$x(y + z)^2 = y(x + z)^2$
On simplifying we get
$xy^2+ xz^2 = yx^2 + yz^2$
$\Rightarrow x^2y - xy^2 = xz^2 - yz^2$
$\Rightarrow xy(x -y) = z^2(x - y)$
$\Rightarrow xy = z^2$
$\Rightarrow x : z : : z : y.$
Hence proved.
View full question & answer→Question 42 Marks
If $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$, show that each ratio is equal to $\frac{1}{2}$ or $-1.$
Answer$\text { Now } \frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$
$ =\frac{a+b+c}{2(a+b+c)}=\frac{1}{2} \text { if } a+b+c \neq 0 .$
But if $a + b + c = 0, ...$ (Applying prop. law)
Then $\frac{a}{b+c}=\frac{a}{-a}=-1$
and $\frac{b}{c+a}=\frac{b}{-b}=-1$
Also, $\frac{c}{a+b}=\frac{c}{-c}=-1$.
Hence each of the given ratio is either $\frac{1}{2}$ or $-1 .$
View full question & answer→Question 52 Marks
If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$, Prove that each of these ratios is equal to $\frac{a+c+e}{b+d+f}$.
AnswerLet $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ = k then a = bk, c = dk and e = fk.
Now $\frac{a+c+e}{b+d+f}=\frac{b k+d k+f k}{b+d+f}$
$=\frac{k(b+d+f)}{(b+d+f)}= k$
Hence, $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+c+e}{b+d+f}$
Hence proved.
View full question & answer→Question 62 Marks
If $a : b = 5 : 3,$ show that $(5a + 8b) : (6a – 7b) = 49 : 9.$
AnswerGiven $a : b = 5 : 3$
$\Rightarrow \frac{a}{b}=\frac{5}{3}$
$ \therefore \frac{5 a+8 b}{6 a-7 b}=\frac{5\left(\frac{a}{b}\right)+8}{6\left(\frac{a}{b}\right)-7}$
Numberator and Denominator is divided by 'b'.
$=\frac{5 \times \frac{5}{3}+8}{6 \times \frac{5}{3}-7} $
$ =\frac{49}{9} $
$ =49: 9$
Hence proved.
View full question & answer→Question 72 Marks
If $\frac{3 x+4 y}{3 u+4 v}=\frac{3 x-4 y}{3 u-4 v}$, then show that $\frac{x}{y}=\frac{u}{v}$.
AnswerWe have
$\frac{3 x+4 y}{3 u+4 v}=\frac{3 x-4 y}{3 u-4 v} \ldots[$ Applying alternendo]
$\frac{3 x+4 y}{3 u-4 v}=\frac{3 x+4 y}{3 u-4 v} \quad \ldots$ [By componendo and dividendo]
$\frac{3 x+4 y+3 x-4 y}{3 x+4 y-3 x+4 y}=\frac{3 u+4 v+3 u-4 v}{3 u+4 v-3 u+4 v} $
$\Rightarrow \frac{6 x}{8 y}=\frac{6 u}{8 v} $
$\Rightarrow \frac{x}{y}=\frac{u}{v}$
Hence proved.
View full question & answer→Question 82 Marks
If a : b = c : d, show that (2a - 7b) (2c + 7d) = (2c - 7d) (2a + 7b).
AnswerWe have
a: b = c : d
$\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{2 a}{7 b}=\frac{2 c}{7 d} \quad \ldots\left[\right.$ Multiplying both side $\left.\frac{2}{7}\right]$
[Using componendo and dividendo]
$\frac{2 a+7 b}{2 a-7 b}=\frac{2 c+7 d}{2 c-7 d} \quad \ldots$ [By cross multiplication $]$ $\Rightarrow(2 a-7 b)(2 c+7 d)=(2 a+7 b)(2 c-7 d)$.
Hence proved.
View full question & answer→Question 92 Marks
if $\frac{3 a+4 b}{3 c+4 d}=\frac{3 a-4 b}{3 c-4 d}$ Prove that $\frac{a}{b}=\frac{c}{d}$.
AnswerGiven $\frac{3 a+4 b}{3 c+4 d}=\frac{3 a-4 b}{3 c-4 d}$
App. alternendo $=\frac{3 a+4 b}{3 a-4 d}=\frac{3 c+4 b}{3 c-4 d}$
App. componendo and dividendo
$\frac{3 a+4 b+3 a-4 b}{3 a+4 b-3 a+4 b}=\frac{3 c+4 d+3 c-4 d}{3 c+4 d-3 c+4 d}$
$ \therefore \frac{6 a}{8 b}=\frac{6 c}{8 d}$
or
$\frac{a}{b}=\frac{c}{d}$
Hence proved.
View full question & answer→Question 102 Marks
If $a, b, c, d$ are in continued proportion, prove that:
$(a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2.$
Answer$L.H.S.$
$= (d^2k^6 + d^2k^4 + d^2k^2)(d^2k^4 + d^2k^2 + d^2)$
$= d^2k^2(k^4+ k^2 + 1)d^2 (k^4 + k^2 + 1)$
$= d^4k^2 (k^4 + k^2 + 1)^2$
$R.H.S.$
$= (ab + bc + cd)^2$
$= (dk^3·dk^2 + dk^{2·}dk + dk·d)^2$
$= d^4·k^2 (k^4 + k^2+ 1)^2$
$L.H.S. = R.H.S.$
Hence proved.
View full question & answer→Question 112 Marks
If $a: b$ with $a ≠ b$ is the duplicate ratio of $a + c: b + c$, show that $c^2 = ab.$
AnswerThe duplicate ratio of
$\frac{a}{b}=\frac{(a+c)^2}{(b+c)^2} $
$\Rightarrow \frac{a}{b}=\frac{a^2+c^2+2 a c}{b^2+c^2+2 b c} $
$\Rightarrow a b^2+a c^2+2 a b c=a^2 b+b c^2+2 a b c $
$\Rightarrow c^2-b c^2=a^2 b-a b^2 $
$\Rightarrow c^2(a-b)=a b(a-b)$
Hence, $c^2=a b$.
Hence proved.
View full question & answer→Question 122 Marks
If $\frac{p}{q}=\frac{r}{s}$, prove that $\frac{2 p+3 q}{2 p-3 q}=\frac{2 r+3 s}{2 r-3 s}$.
AnswerWe have
$\frac{p}{q}=\frac{r}{s}$..[Multiplying both side by 2 / 3]
$\frac{2 p}{3 q}=\frac{2 r}{3 s} \quad$... By componendo and dividendo]
$
\therefore \frac{2 p+3 q}{2 p-3 q}=\frac{2 r+3 s}{2 r-3 s} \text {. }
$
Hence proved.
View full question & answer→Question 132 Marks
If $\frac{a}{b}=\frac{c}{d}$, show that (9a + 13b) (9c - 13d) = (9c + 13b) (9a - 13d).
AnswerWe have $\frac{a}{b}=\frac{c}{d}$
$\left[\right.$ Multiplying both sides by $\left.\frac{9}{13}\right]$
$
\frac{9 a}{13 b}=\frac{9 c}{13 d}
$
[By componendo and dividendo]
$
\frac{9 a+13 b}{9 a-13 b}=\frac{9 c+13 d}{9 c-13 d}
$
(By cross multiplication)
$
(9 a+13 b)(9 c-13 d)=(9 a-13 b)(9 c+13 d) \text {. }
$
Hence proved.
View full question & answer→Question 142 Marks
If p + r = 2q and $\frac{1}{q}+\frac{1}{s}=\frac{2}{r}$ then prove that p : q = r : s.
Answer$\frac{1}{q}+\frac{1}{s}=\frac{2}{r} $
$\Rightarrow \frac{s+q}{q s}=\frac{2}{r}$
$\Rightarrow 2 qs = r ( s + q ) $
$ \Rightarrow( p + r ) s = r ( s + q ) $
$ \Rightarrow ps + rs = rs + rq $
$ \Rightarrow ps = rq $
$ \Rightarrow \frac{p}{q}=\frac{r}{s} .$
Hence proved.
View full question & answer→Question 152 Marks
If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}$ then show that (b - c)x + (c - a)y + (a - b) z = 0.
Answer$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}= k , \ldots[\text { By k method }] $
$x =( b + xc - a ) K, $
$y =( c + a - b ) k , $
$z=( a + b - c ) k $
$\Rightarrow b ^2+ bc - ab - bc - c ^2+ ca + c ^2+ ca - bc - ab - bc - c ^2+ ca + c ^2+ ab - ac - ab - b ^2+ bc =0 .$
Hence proved.
View full question & answer→Question 162 Marks
If (a – x) : (b – x) be the duplicate ratio of a: b show that:
$\frac{1}{x}=\frac{1}{a}+\frac{1}{b}$
AnswerHere (a - x) : (b - x) is duplicate ratio of a : b
$\therefore \frac{a^2}{b^2}=\frac{a-x}{b-x} $
$ \Rightarrow a^2 b-a^2 x=b^2 a-b^2 x $
$ \Rightarrow a^2 b-b^2 a=a^2 x-b^2 x $
$ \Rightarrow ab ( a - b )= x ( a - b )( a + b ) $
$ \Rightarrow ab = x ( a + b ) $
$ \Rightarrow \frac{1}{x}=\frac{a+b}{a b}$
$ \Rightarrow \frac{1}{x}=\frac{1}{a}+\frac{1}{b}$
Hence proved.
View full question & answer→Question 172 Marks
If $(x - 9) : (3x + 6)$ is the duplicate ratio of $4 : 9,$ find the value of x.
AnswerAs given $(x - 9) : (3x + 6)$ is duplicate ratio of $4 : 9.$
i.e.,$\frac{x-9}{3 x+6}=\left(\frac{4}{9}\right)^2$
$\frac{x-9}{3 x+6}=\frac{16}{81} $
$ \Rightarrow 81(x-9)=16(3 x+6) $
$ 81 x-729=48 x+96$
$ 81 x-48 x=96+729$
$33 x=825 $
$ x=\frac{825}{33}$
$ =25$
Thus required value of x is $25.$
View full question & answer→Question 182 Marks
The ratio between two numbers is 3: 4. If their L.C.M., is 180. Find the numbers.
AnswerLet the required numbers be 3x and 4x.
The L.C.M., of 3x and 4x = 12x
Then 12x = 180
⇒ x = 15
HEnce, the required numbers are
3x = 3 x 15 = 45
4x = 4 x 15 = 60.
View full question & answer→Question 192 Marks
If x : y = 2 : 3, find the value of (3x + 2y) : (2x + 5y).
AnswerWe have
$\frac{x}{y}=\frac{2}{3}$
Now $\frac{3 x+2 y}{2 x+5 y}=\frac{y\left(\frac{3 x}{y}+2\right)}{y\left(\frac{2 x}{y}+5\right)}$
(Both numerator and denominator are divided by y)
$\frac{3 \times \frac{2}{3}+2}{2 \times \frac{2}{3}+5}=\frac{4}{\frac{19}{3}} $
$ =\frac{4 \times 3}{19} \\ =12: 19 .$
View full question & answer→Question 202 Marks
Two numbers are in the ratio of 3: 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.
AnswerLet the number be 3x and 5x
3x + 8 : 5x + 8 = 2 : 3
$\frac{3 x+8}{5 x+8}=\frac{2}{3}$
⇒ 2(5x + 8) = 3(3x + 8)
⇒ 10x + 16 = 9x + 24
⇒ 10x - 9x = 24 - 16
⇒ x = 8
∴ The number are 24 and 40.
View full question & answer→Question 212 Marks
If 3x – 2y – 7z = 0 and 2x + 3y – 5z = 0 find x : y : z.
AnswerHere, 3x - 2y - 7z = 0
2x + 3y - 5x = 0
Using method of cross multiplication
$\therefore$ $\frac{x}{(-2)(-5)-3(-7)}=\frac{y}{(-7)(2)-(3)(-5)}=\frac{z}{(3)(3)-(2)(-2)}$
$\Rightarrow \frac{x}{10+21}=\frac{y}{-14+15}=\frac{z}{9+4}$
$\Rightarrow \frac{x}{31}=\frac{y}{1}=\frac{z}{13}$
⇒ x : y : z = 31 : 1 : 13.
View full question & answer→Question 222 Marks
Find:
The sub-duplicate ratio of the sub-triplicate ratio of $4096x^6: 729y^{12}.$
AnswerThe sub-triplicate ratio of $4096x^6:729y^{12}$
= $=\sqrt[3]{4096 x^6}: \sqrt[3]{729 y^{12}}$
$= 16x^2 : 9y^4$
$\therefore$ The sub-duplicate ratio of $16x^2 : 9y^4$
$=\sqrt{16 x^2}: \sqrt{9 y^4}$
$= 4x : 3y^2.$
View full question & answer→Question 232 Marks
Find:
The sub-triplicate ratio of the sub-duplicate ratio of $729x^{18}: 64y^6.$
AnswerThe sub-duplicate ratio of $729x^{18} : 64y^6$
$=\sqrt{729 x^{18}}: \sqrt{64 y^6}$
$= 27x^9 : 8y^3$
$\therefore $ The sub-triplicate ratio of $27x^9 : 8y^3$
=$\sqrt[3]{27 x^9}: \sqrt[3]{8 y^3}$
$= 3x^3 : 2y.$
View full question & answer→Question 242 Marks
If $\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}$, find x : y.
Answer$\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}$
Applying componendo and dividendo
$\frac{3 x+5 y+3 x-5 y}{3 x+5 y-3 x+5 y}=\frac{7+3}{7-3} $
$ \frac{6 x}{10 y}=\frac{10}{4} $
$ \frac{x}{y}=\frac{10 \times 10}{4 \times 6} $
$ \frac{x}{y}=\frac{25}{6} $
$ \therefore x: y=25: 6$
View full question & answer→Question 252 Marks
Find the fourth proportional to:
$2xy, x^2, y^2$
AnswerLet A be the fourth proportional then
$2 x y: x^2=y^2: A $
$ \Rightarrow \frac{2 x y}{x^2}=\frac{y^2}{ A } $
$ \Rightarrow A =\frac{x^2 y^2}{2 x y} $
$ \Rightarrow A =\frac{x y}{2} .$
View full question & answer→Question 262 Marks
Find the third proportional to:
$\frac{a}{b}+\frac{b}{c}, \sqrt{a^2+b^2}$.
AnswerLet x be the third proportional then
$\frac{a}{b}+\frac{b}{c}, \sqrt{a^2+b^2}=\sqrt{a^2+b^2}: x $
$\Rightarrow \frac{a^2+b^2}{a b}: \sqrt{a^2+b^2}=\sqrt{a^2+b^2}: x$
$ \Rightarrow \frac{a^2+b^2}{a b \sqrt{a^2+b^2}}=\frac{\sqrt{a^2+b^2}}{x} $
$ \Rightarrow x =\frac{a b\left(a^2+b^2\right)}{\left(a^2+b^2\right)}$
$ \Rightarrow x = ab.$
View full question & answer→Question 272 Marks
Find the third proportional to:
$x - y, x^2- y^2$
AnswerLet A be the third proportional then
$( x - y ):\left( x ^2- y ^2\right)=\left( x ^2- y ^2\right): A $
$ \Rightarrow \frac{x-y}{x^2-y^2}=\frac{x^2-y^2}{ A }$
$ \Rightarrow A =\frac{\left(x^2-y^2\right)}{x-y} $
$ \Rightarrow A =( x + y )\left( x ^2- y ^2\right) .$
View full question & answer→Question 282 Marks
What least number must be added to each of the numbers $5, 11, 19$ and $37,$ so that they are in proportion?
AnswerLet x be the number added to $5, 11, 19, 37.$
$\frac{5+x}{11+x}=\frac{19+x}{37+x}$
App. comp. and divi.
$\frac{5+x+11+x}{5+x-11-x}=\frac{19+x+37+x}{19+x-37-x} $
$ \frac{16+2 x}{-6}=\frac{56+2 x}{-18} $
$ 3(16+2 x)=56+2 x $
$ 6 x-2 x=56-48 $
$ 4 x=8 $
$ x=2 .$
View full question & answer→Question 292 Marks
What number must be added to each of the numbers $6, 15, 20$ and $43$ to make them proportional?
AnswerLet no. be 'x'
$\therefore \frac{6+x}{15+x}=\frac{20+x}{43+x}$
$ \Rightarrow(6+ x )(43+ x )=(20+ x )(15+ x )$
$ \Rightarrow 258+49 x + x ^2=300+35 x + x ^2 $
$\Rightarrow 14 x =42 $
$x =3$
$\therefore$ The required no. is $3.$
View full question & answer→Question 302 Marks
Find the compound ratio of the following:
If $P : Q = 6 : 7, Q : R = 8 : 9$ find $P : Q : R.$
Answer$P : Q = 6 : 7, Q : R = 8 : 9$
$\frac{ P }{ Q }=\frac{6}{7} \times \frac{8}{8}, \frac{ Q }{ R }=\frac{8}{9} \times \frac{7}{7}$
$ \frac{ P }{ Q }=\frac{48}{56}, \frac{ Q }{ R }=\frac{56}{63} $
$ \therefore P : Q : R =48: 56: 63$
View full question & answer→Question 312 Marks
Find the compound ratio of the following:
If $A : B = 4 : 5, B : C = 6 : 7$ and $C : D = 14 : 15.$ Find $A : D.$
AnswerGiven $A : B = 4 : 5, B : C = 6 : 7$ and $C : D = 14 : 15$
or
$\frac{ A }{ B }=\frac{4}{5}, \frac{ B }{ C }=\frac{6}{7}, \frac{ C }{ D }=\frac{14}{15}$
Multiply all $\frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}=\frac{4}{5} \times \frac{6}{7} \times \frac{14}{15}$
$\frac{A}{D}=\frac{16}{25} $
$\therefore A: D=16: 25 .$
View full question & answer→Question 322 Marks
Which is greater 4: 5 or 19: 25.
AnswerLet a = 4, b = 5, c = 19 and d = 25
then ad = 4 x 25 = 100
bc = 5 x 19 = 95
Here 100 > 95 ⇒ ad > bc so $\frac{a}{b}>\frac{c}{d}$
⇒ 4 : 5 is greater.
Hence,$\frac{4}{5}$ is greater.
View full question & answer→