Question 14 Marks
If $u, v, w$, and $x$ are in continued proportion, then prove that $(2 u+3 x):(3 u+4 x)::\left(2 u^3+3 v^3\right):$
$\left(3 u ^3+4 v ^3\right)$
$( pqr )^2\left(\frac{1}{ p ^4}+\frac{1}{ q ^4}+\frac{1}{ r ^4}\right)=\frac{ p ^4+ q ^4+ r ^4}{ q ^2}$
Answer$p: q:: q: r \Rightarrow q^2=p r$
$( pqr )^2\left(\frac{1}{ p ^4}+\frac{1}{ q ^4}+\frac{1}{ r ^4}\right)=\frac{ p ^4+ q ^4+ r ^4}{ q ^2}$
LHS
$(p q r)^2\left(\frac{1}{p^4}+\frac{1}{q^4}+\frac{1}{r^4}\right) $
$=\left(q \times q^2\right)^2\left(\frac{q^4 r^4+p^4 r^4+p^4 q^4}{p^4 q^4 r^4}\right) $
$=q^6\left(\frac{q^4 r^4+q^8+p^4 q^4}{q^8 q^4}\right) $
$=q^6\left(\frac{r^4+q^4+p^4}{q^8}\right) $
$=\left(\frac{r^4+q^4+p^4}{q^2}\right)=\text { RHS }$
LHS $=$ RHS , Hence Proved.
View full question & answer→Question 24 Marks
If $u, v, w$, and $x$ are in continued proportion, then prove that $(2 u+3 x):(3 u+4 x)::\left(2 u^3+3 v^3\right):\left(3 u^3+4 v^3\right)$
Answer$\frac{u}{v}=\frac{v}{w}=\frac{w}{x}=a $
$w=a x $
$v=a w=a^2 x $
$u=a v=a^3 x $
$\text { LHS } $
$\frac{2 u+3 x}{3 u+4 x} $
$=\frac{2 a^3 x+3 x}{3 a^3 x+4 x} $
$=\frac{2 a^3+3}{3 a^3+4} $
$\text { RHS } $
$\frac{2 u^3+3 v^3}{3 u^3+4 v^3} $
$=\frac{2 a^9 x^3+3 a^6 x^3}{3 a^9 x^3+4 a^6 x^3} $
$=\frac{2 a^3+3}{3 a^3+4}$
LHS $=$ RHS. Hence, proved.
View full question & answer→Question 34 Marks
If $p, q, r$ ands are In continued proportion, then prove that $(p^3+q^3+r^3) ( q^3+r^3+s^3) : : P : s$
Answer$\frac{p}{q}=\frac{q}{r}=\frac{r}{s}=k $
$r=k s $
$q=k r=k^2 s $
$p=k q=k^3 s$
LHS
$\frac{p^3+q^3+r^3}{q^3+r^3+s^3} $
$=\frac{k^9 s^3+k^6 s^3+k^3 s^3}{k^6 s^3+k^3 s^3+s^3} $
$=\frac{s^3 k^3\left(k^6+k^3+1\right)}{s^3\left(k^6+k^3 s^3+s^3\right)} $
$=k^3$
RHS
$\frac{ p }{ s }=\frac{ k ^3 s }{ s }= k ^3$
LHS $=$ RHS . Hence proved.
View full question & answer→Question 44 Marks
If $a: b=c: d$, then prove that $\frac{a^2+a b+b^2}{a^2-a b+b^2}=\frac{c^2+c d+d^2}{c^2-c d+d^2}$
Answer$\frac{ a }{ b }=\frac{ c }{ d } \Rightarrow a =\frac{ bc }{ d }$
LHS
$\frac{a^2+a b+b^2}{a^2-a b+b^2} $
$=\frac{\left(\frac{b c}{d}\right)^2+\left(\frac{b c}{d}\right) b+b^2}{\left(\frac{b c}{d}\right)^2-\left(\frac{b c}{d}\right) b+b^2} $
$=\frac{b^2 c^2+b^2 c d+d^2 b^2}{b^2 c^2-b^2 c d+d^2 b^2} $
$=\frac{b^2\left(c^2+c d+d\right)}{b^2\left(c^2-c d+d^2\right)}=\frac{c^2+c d+d^2}{c^2-c d+d^2}=\text { RHS }$
LHS $=$ RHS
Hence, proved.
View full question & answer→Question 54 Marks
If $\frac{7 a +12 b }{7 c +12 d }$ then prove that $\frac{ a }{ b }=\frac{ c }{ d }$
Answer$\frac{7 a +12 b }{7 c +12 d }=\frac{7 a -12 b }{7 c -12 d }$
Applying alternendo,
$\frac{7 a +12 b }{7 a -12 b }=\frac{7 c +12 d }{7 c -12 d }$
Applying componendo and dividendo,
$\frac{7 a +12 b +7 a -12 b }{7 a +12 b -7 a +12 b }=\frac{7 c +12 d +7 c -12 d }{7 c +12 d -7 c +12 d } $
$\Rightarrow \frac{14 a }{24 b }=\frac{14 c }{24 d }$
Dividing both sides by $\frac{14}{24}$
$\frac{ a }{ b }=\frac{ c }{ d }$
Hence, proved.
View full question & answer→Question 64 Marks
Show that the value of $x$ is 11 , when $\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91}$
Answer$\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91} $
$\Rightarrow 91\left(x^3+3 x\right)=341\left(3 x^2+1\right) $
$\Rightarrow 91 x^3-1023 x^2+273 x-341=0$
$LHS$
$91 x^3-1023 x^2+273 x-341 $
$=91(11 \times 11 \times 11)-1023(11 \times 11)+273(11)-341 $
$=(91 \times 1331)-(1023 \times 121)+(273 \times 11)-341 $
$=121121-123783+3003-341 $
$=0=\text { RHS }$
$L H S= RHS.$
Hence $x=11$
View full question & answer→Question 74 Marks
If a : b : : c : d, then prove that
ax+ by): (cx + dy)=(ax - by) : (cx - dy)
Answer$(a x+b y):(c x+d y)=(a x-b y):(c x-d y) $
$\frac{a}{b}=\frac{c}{d}$
Multiplying both sides by $\frac{x}{y}$
$\Rightarrow \frac{a}{b} \times \frac{x}{y}=\frac{c}{d} \times \frac{x}{y} $
$\Rightarrow \frac{a x}{b y}=\frac{c x}{d y}$
Applying componendo and dividendo,
$\frac{a x+b y}{a x-b y}=\frac{c x+d y}{c x-d y} $
$\Rightarrow \frac{a x+b y}{c x+d y}=\frac{a x-b y}{c x-d y}$
Hence, $a x+b y: c x+d y=a x-b y: c x-d y$
View full question & answer→Question 84 Marks
If $\frac{ a }{ b + c }=\frac{ b }{ c + a }=\frac{ c }{ a + b }$, then prove that $a ( b - c )+ b ( c - a )+ c ( a - b )=0$
Answer$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k $
$a=k(b+c) $
$b=k(c+a) $
$c=k(a+b) $
$a(b-c)+b(c-a)+c(a-b)=0$
LHS
$a(b-c)+b(c-a)+c(a-b) $
$=k(b+c)(b-c)+k(c+a)(c-a)+k(a+b)(a-b) $
$=k\left(b^2-c^2\right)+k\left(c^2-a^2\right)+k\left(a^2-b^2\right) $
$=k b^2-k c^2+k c^2-k a^2+k a^2-k b^2 $
$=0=\text { RHS }$
LHS $=$ RHS. Hence, proved.
View full question & answer→Question 94 Marks
If $\frac{ x }{ b + c - a }=\frac{ y }{ c + a - b }=\frac{ z }{ a + b - c }$, then prove that each ratio is equal to the ratio of $\frac{ x + y + z }{ a + b + c }$
Answer$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k $
$x=k(b+c-a) $
$Y=k(c+a-b) $
$z=k(a+b-c)$
Now,
$\frac{x+y+z}{a+b+c} $
$=\frac{k(b+c-a)+k(c+a-b)+k(a+b-c)}{a+b+c} $
$=\frac{k(b+c-a+c+a-b+a+b-c)}{a+b+c} $
$=\frac{k(a+b+c)}{a+b+c}=k$
Hence,
$\frac{ x }{ b + c - a }=\frac{ y }{ c + a - b }=\frac{ z }{ a + b - c }=\frac{ x + y + z }{ a + b + c }$
Proved.
View full question & answer→Question 104 Marks
If $a, b, c$ and dare in continued proportion, then prove that
ad $(c^2 + d^2) = c^3 (b + d)$
Answer$a d\left(c^2+d^2\right)=c^3(b+d) $
$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k $
$\Rightarrow c=k d $
$b=k c=k^2 d $
$a=k b=k^3 d $
$a c\left(c^2+d^2\right)=c^3(b+d)$
LHS
$a c\left(c^2+d^2\right)$
$=k^3 d x c\left(k^2 d^2+d^2\right)$
$=k^3 d^3\left(k^2 d+d\right)$
RHS
$c^3(b+d) $
$=k^3 d^3\left(k^2 d+d\right)$
LHS $=$ RHS. Hence, proved.
View full question & answer→Question 114 Marks
If y is the mean proportional between x and y ; show that $y ( x + z )$ is the mean p roporti ona I between $x ^2+ y ^2$ and $y ^2+ z ^2$
AnswerSince y is the mean proportion between $x$ and $z$
Therefore,$ y^2=xz$
Now, we have to prove that xy+yz is the mean proportional between $x^2+y^2$ and $y^2+z^2.$
$(xy + yz)^2 = (x^2 + y^2)(y^2 + z^2)$
$LHS = (xy + yz)^2$
$= [y (x + z)]^2$
$= y^2(x + z)^2$
$= xz (x + z)^2$
$RHS = (x^2 + y^2)(y^2 + z^2)$
$= (x^2 + xz) (xz+ z^2)$
$= x (x + z) z (x + z)$
$= xz (x + z)^2$
$LHS = RHS$
View full question & answer→Question 124 Marks
If y is the mean proportional between x and z, show that :
$xyz (x+y+z)^3 =(xy+yz+xz)^3$
AnswerSince y is the mean proportion between x and z
$y^2 = xz$
$LHS$
$xyz (x + y + z)^3$
$= yy^2 (x + y + z)^3$
$= y^3 (x + y +z)^3$
$= [y (x + y + z)]^3$
$= (xy + y^2 + yz)^3$
$= (xy + y^2 + yz)^3 = RHS$
$LHS = RHS$
View full question & answer→Question 134 Marks
Find two numbers whose mean proportional is $18$ and the third proportional is $486.$
AnswerLet $a$ and $b$ be the two numbers, whose mean proportional is $18 .$
'therefore $"ab" =18^2\Rightarrow \sim " a b "=324 \Rightarrow \sim " b "=324 / " a^{\prime 2}$......(i)
Now, third proportional is 486
$\therefore a: b:: b: 486$
$\Rightarrow b^2=486 a $
$\Rightarrow\left(\frac{324}{a}\right)^2=486 a $
$\Rightarrow \frac{(324)^2}{a^2}=486 a $
$\Rightarrow a^3=\frac{324 \times 324}{486} $
$\Rightarrow a^3=216 $
$\Rightarrow a=6 $
$b=\frac{324}{a}=\frac{324}{6}=54$
Therefore, numbers are $6$ and $54 .$
View full question & answer→Question 144 Marks
Find two nurnbers whose mean proportional is 12 and the third proportional is 324.
AnswerLet $a$ and $b$ be the two numbers, whose mean proportional is 12 .
$
\therefore a b=12^2 \Rightarrow a b=144 \Rightarrow b=\frac{144}{a}\ldots(i)
$
Now, third proportional is 324
$\therefore a: b:: b: 324$
$
\Rightarrow\left(\frac{144}{ a }\right)^2=324 a
$
$
\Rightarrow \frac{(144)^2}{ a ^2}=324 a
$
$
\Rightarrow a ^3=\frac{144 \times 144}{324}
$
$
\Rightarrow a ^3=64
$
$
\Rightarrow a =4
$
$
b=\frac{144}{a}=\frac{144}{4}=36
$
Therefore, numbers are 4 and 36
View full question & answer→Question 154 Marks
Find the smallest number that must be subtracted from each of the numbers $20, 29, 84$ and $129$ so that they are in proportion.
AnswerLet $x$ be subtracted from each number so that $20-x, 29-x, 84-x$ and $129-x$ are in proportion.
$\therefore \frac{20-x}{29-x}=\frac{84-x}{129-x} $
$\Rightarrow(20-x)(129-x)=(29-x)(84-x) $
$\Rightarrow 2580-129 x-20 x+x^2=2436-84 x-29 x+x^2 $
$\Rightarrow 2580-149 x=2436-113 x $
$\Rightarrow 36 x=144 $
$\Rightarrow x=4$
Hence, $4$ is to be subtracted from $20,29, 84$ and $129$ for them to be in proportion.
View full question & answer→Question 164 Marks
If $( a+c) : b = 5 : 1$ and $(bc + cd) : bd = 5 : 1,$ then prove that $a : b = c : d$
Answer$\frac{a+c}{b}=\frac{5}{1} $
$\Rightarrow a+c=5 b $
$\frac{b c+c d}{b d}=\frac{5}{1} $
$\Rightarrow b c+c d=5 b d $
$\Rightarrow b c+c d=(a+c) d $
$\Rightarrow b c+c d=a d+c d $
$\Rightarrow b c=a d $
$\Rightarrow \frac{a}{b}=\frac{c}{d}$
Hence $a: b=c: d$
View full question & answer→Question 174 Marks
Find the compounded ratio of the following:
$(m-n):(m+n), (m+n)^2 : (m^2+n^2) and (m^4 - n^4): (m^2-n^2) ^2$
AnswerCompounded ratio of $(m-n):(m+n),(m+n)^2:\left(m^2+n^2\right)$ and $\left(m^4-n^4\right):\left(m^2-n^2\right)^2$
$=\frac{ m - n }{ m + n } \times \frac{( m + n )^2}{\left( m ^2+ n ^2\right)} \text { and } \frac{ m ^4- n ^4}{\left( m ^2- n ^2\right)^2} $
$=\frac{ m - n }{1} \times \frac{ m + n }{ m ^2+ n ^2} \times \frac{\left( m ^2+ n ^2\right)\left( m ^2- n ^2\right)}{\left( m ^2- n ^2\right)^2} $
$=\frac{ m ^2- n ^2}{ m ^2+ n ^2} \times \frac{\left( m ^2+ n ^2\right)\left( m ^2- n ^2\right)}{\left( m ^2- n ^2\right)^2} $
$=\frac{1}{1}$
Compounded ratio $=1: 1$
View full question & answer→Question 184 Marks
Find the compounded ratio of the following:
$(a^2 - b^2 ): (a^2 + b^2)$ and $(a^4 - b^4 ): (a+ b)^4$
AnswerCompounded ratio of $\left(a^2-b^2\right):\left(a^2+b^2\right)$ and $\left(a^4-b^4\right):(a+b)^4$
$=\frac{a^2-b^2}{a^2+b^2} \times \frac{a^4-b^4}{(a+b)^4} $
$=\frac{(a+b)(a-b)}{\left(a^2+b^2\right)} \times \frac{\left(a^2+b^2\right)\left(a^2-b^2\right)}{(a+b)^2(a+b)^2} $
$=\frac{(a-b)(a+b)(a-b)(a+b)}{(a+b)^2(a+b)^2} $
$=\frac{(a-b)^2}{(a+b)^2}$
Compounded ratio $=(a-b)^2:(a+b)^2$
View full question & answer→Question 194 Marks
In an examination the ratio of the number of suo:essful candidates to unsuccessful candidates is 7:5. Had 30 more appeared and 10 more passed, the ratio of successful candidates to unsuccessful candidates would have been 4:3. Find the number of candidates who appeared in the examination originally.
AnswerLet successful candidates be $7 x$ and unsuccessful candidates be $5 x$.
Now,$\frac{7 x +10}{5 x +20}=\frac{4}{3}$
$
\Rightarrow 21 x+30=20 x+80
$
$
\Rightarrow x=50
$
$
7 x=7 \times 50=350
$
$
5 x=5 \times 50=250
$
$
\therefore 7 x+5 x=350+250=600
$
Therefore, there were 600 candidates originally.
View full question & answer→Question 204 Marks
A cistern of milk caitains a mixture of milk and water in the ratio of $11 :4.$ If the cistern contains $60$ litres of milk, how much more water must be added to make the
ratio of milk to water as $11:6 ?$
AnswerQuantity of milk $=60 \times \frac{11}{15}=44$ litres
Quantity of water in it $=60-44=16$ litres
New ratio $=11: 6$
Let quantity of water to be added further be $x$ litres.
Then, milk $:$ water $=\frac{44}{16+ x }$
Now ,
$\frac{44}{16+x}=\frac{11}{6} $
$\Rightarrow 264=176+11 x $
$\Rightarrow 11 x=88 $
$\Rightarrow x=8$
Therefore, $8$ litres of water need to be added further.
View full question & answer→Question 214 Marks
The present age of two persons are in the ratio $4: 3.$ Nine years hence their ages will be in the ratio $23: 18.$ Find their present ages.
AnswerLet present ages be $4 x$ and $3 x$.
$\frac{4 x+9}{3 x+9}=\frac{23}{18} $
$\Rightarrow 72 x+162=69 x+207 $
$\Rightarrow 3 x=45 $
$\Rightarrow x=15 $
$\therefore 3 x=45 $
$4 x=60$
Hence, their present ages are $60$ years and $45$ years.
View full question & answer→Question 224 Marks
A bag contains Rs 1800 in the form of Rs 1 , Rs 2 and Rs 5 ooins. The ratio of the number of the respective coins is 3: 7: 11. Find the total number of ooi ns in the bag.
AnswerLet the number of Re 1, Rs 2 and Rs 5 coins be 3x, 7x and 11x
The sum of their value =
3x+ (2 × 7x) + (5 × 11x) = Rs 1800
3x + 14x + 55x = 1800
72x = 1800
x = 25
3x =3 × 25 = 75
7x = 7 × 25 = 175
11 x = 11 x 25 = 275
total = 75 + 175 + 275 = 525
Total number of coins = 525
View full question & answer→Question 234 Marks
Two numbers are in the ratio $5 : 7$ and the difference of their squares is $600$. Find the numbers.
AnswerLet the two numbers be 5x and 7x, since the ratio between them is $5:7$
Now,
$(7x)^2 - (5x)^2 = 600$
$\Rightarrow 49x^2 - 25x^2 - 600$
$\Rightarrow 24x^2 - 600$
$\Rightarrow x^2 = 25$
$\Rightarrow x = 5$
$\therefore 5x = 25$
$7x = 35$
Therefore, the two numbers are $25$ and $35.$
View full question & answer→Question 244 Marks
Two positive numbers are in the ratio $3: 4$ and the sum of their squares is $1225$. Find the numbers.
AnswerLet the two numbers be ::$lx$ and $4x$, since the ratio between them is $3:4$
Now,
$(3x)^2 + (4x)^2 = 1225$
$\Rightarrow 9x^2 + 16x^2 = 1225$
$\Rightarrow 25x^2 = 1225$
$\Rightarrow x^2 = 49$
$\Rightarrow x = 7$
$\therefore 3x = 21$
$4x = 28$
Therefore, the two numbers are $21$ and $28$
View full question & answer→Question 254 Marks
Two numbers are in the ratio $7 : 10.$ If $8$ is added to each number, the ratio becomes $3 : 4.$ Find the nLimbers.
AnswerLet the two numbers be $7 x$ and $10 x$; since the ratio between numbers is $7: 10$
Now,
$\frac{7 x+8}{10 y+8}=\frac{3}{4} $
$\Rightarrow 28 x+32=30 x+24 $
$\Rightarrow 2 x=8 $
$\Rightarrow x=4 $
$\therefore 7 x=28 $
$10 x=40$
Therefore, two numbers are $28$ and $40.$
View full question & answer→Question 264 Marks
Which of the following ratio is greater?
$\frac{5}{2}: \frac{15}{4}$ and $\frac{5}{3}: \frac{11}{6}$
Answer$\frac{5}{2}: \frac{15}{4} \text { and } \frac{5}{3}: \frac{11}{6} $
$ \frac{5}{2}: \frac{15}{4}=\frac{5}{2} \times \frac{4}{15}=\frac{2}{3}$
$ \frac{5}{3}: \frac{11}{6}=\frac{5}{3} \times \frac{6}{11}=\frac{10}{11}$
$ \frac{2}{3}: \frac{10}{11} $
$ \Rightarrow 2 \times 11<3 \times 10 $
$ \Rightarrow 22<30 $
$ \Rightarrow 2: 3<10: 11 $
$ \Rightarrow \frac{5}{2}: \frac{15}{4}<\frac{5}{3}: \frac{11}{6} $
$ \frac{5}{3}: \frac{11}{6} \text { is greater. }$
View full question & answer→Question 274 Marks
Find the sub-triplicate ratio of the following:
$\frac{64 m ^3}{729 n ^3}: \frac{216 m ^3}{27 n ^3}$
Answer$\frac{64 m ^3}{729 n ^3}: \frac{216 m ^3}{27 n ^3} $
$=\left(\frac{64 m ^3}{729 n ^3}\right)^{\frac{1}{3}}:\left(\frac{216 m ^3}{27 n ^3}\right)^{\frac{1}{3}}$
$=\frac{4 m }{9 n }: \frac{6 m }{3 n }$
$=\frac{4 m }{9 n } \times \frac{6 m }{3 n } $
$=\frac{2}{9}$
Sub-triplicate raoo $=2: 9$
View full question & answer→Question 284 Marks
Find the sub-triplicate ratio of the following :
$512: 216$
Answer$512: 216$
$=(512)^{\frac{1}{3}}:(216)^{\frac{1}{3}} $
$=\left(8^3\right)^{\frac{1}{3}}:\left(6^3\right)^{\frac{1}{3}}$
$=8: 6 $
$=4: 3$
Sub-triplicate rabo $=4: 3$
View full question & answer→Question 294 Marks
Find the sub-duplicate ratio of the following:
$\frac{1}{16}: \frac{1}{36}$
Answer$\frac{1}{16}: \frac{1}{36} $
$=\sqrt{\frac{1}{16}}: \sqrt{\frac{1}{36}} $
$=\frac{1}{4}: \frac{1}{6} $
$=\frac{1}{4} \times 6 $
$=\frac{3}{2}$
Sub-duplicate ratio $=3: 2$
View full question & answer→Question 304 Marks
If $a: b=c: d$, then prove that $\frac{a^2+c^2}{b^2+d^2}=\frac{a c}{b c}$
Answer$\frac{ a }{ b }=\frac{ c }{ d } $
$\Rightarrow a =\frac{ bc }{ d }$
To prove,
$\frac{ a ^2+ c ^2}{ b ^2+ d ^2}=\frac{ ac }{ bd }$
$\ce{LHS}$
$\frac{a^2+c^2}{b^2+d^2}$
$=\frac{\left(\frac{b c}{d}\right)^2+c^2}{b^2+d^2}$
$=\frac{\frac{b^2 c^2}{d^2}+c^2}{b^2+d^2}$
$=\frac{c^2\left(b^2+d^2\right)}{d^2\left(b^2+d^2\right)}$
$=\frac{c^2}{d^2}$
$\text { RHS }$
$\frac{ ac }{ bd }$
$=\frac{\frac{ bc }{ d c}}{ bd }$
$=\frac{ bc ^2}{ bd ^2}$
$=\frac{ c ^2}{ d ^2}$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 314 Marks
If $x =\frac{ pab }{ a + b }$, then prove that $\frac{ x + pa }{ x - pa }+\frac{ x + pb }{ x - pb }=\frac{2\left( a ^2- b ^2\right)}{ ab }$
Answer$x=\frac{p a b}{a+b} $
$\Rightarrow \frac{x}{p a}=\frac{b}{a+b}$
Applying componendo and dividendo
$\frac{ x + pa }{ x - pa }=\frac{ b + a + b }{ b - a - b }=\frac{2 b + a }{- a }\ldots(i)$
Again , $x=\frac{ pab }{ a + b } $
$\Rightarrow \frac{ x }{ pb }=\frac{ a }{ a + b }$
Applying componendo and dividendo
$\frac{ x + pb }{ x - pb }=\frac{ a + a + b }{ a - a - b }=\frac{2 a + b }{- b }\ldots(ii)$
Adding $(i)$ and $(ii)$
$\frac{x+p a}{x-p a}+\frac{x+p b}{x-p b}$
$=\frac{2 b+a}{-a}+\frac{2 a+b}{-b}$
$=\frac{a-2 b}{a}+\frac{b-2 a}{b}$
$=\frac{-2 b^2+a b-2 a^2+a b}{a b}$
$=\frac{-2 b^2+2 a b-2 a^2}{a b}$
$=\frac{2\left(a^2-b^2\right)}{a b}=2=\text { RHS }$
$\text { LHS = RHS }$
Hence proved.
View full question & answer→Question 324 Marks
If $x=\frac{\sqrt[3]{m+1}+\sqrt[3]{m-1}}{\sqrt[3]{m+1}+\sqrt[3]{m-1}}$ then prove that $x^3-3 m x^2+3 x=m$
Answer$\frac{x}{1}=\frac{\sqrt[3]{m+1}+\sqrt[3]{m-1}}{\sqrt[3]{m+1}+\sqrt[3]{m-1}}$
Applying componendo and dividendo
$\frac{x+1}{x-1}=\frac{\sqrt[3]{m+1}+\sqrt[3]{m-1}+\sqrt[3]{m+1}-\sqrt[3]{m-1}}{\sqrt[3]{m+1}+\sqrt[3]{m-1}-\sqrt[3]{m+1}+\sqrt[3]{m-1}}$
$\Rightarrow \frac{x+1}{x-1}=\frac{2 \sqrt[3]{m+1}}{2 \sqrt[3]{m-1}}$
Cubing both sides
$\Rightarrow \frac{( x +1)^3}{( x -1)^3}=\frac{8( m +1)}{8( m -1)}$
$\Rightarrow \frac{ x ^3+3 x ^2+3 x +1}{ x ^3-3 x ^2+3 x -1}=\frac{ m +1}{ m -1}$
$\Rightarrow(m-1)\left(x^3+3 x^2+3 x+1\right)=(m+1)\left(x^3-3 x^2+3 x-1\right)$
$\Rightarrow m x^3+3 m x^2+3 m x+m-x^3-3 x^2-3 x-1-m x^3-3 m x^2+3 m x-m+x^3-3 x^2+3 x-1$
$\Rightarrow 6 m x^2+2 m-2 x^3-6 x=0$
$\Rightarrow 3 m x^2+m-x^3-3 x=0$
$\Rightarrow x^3-3 m x^2+3 x=m$
Hence Proved.
View full question & answer→Question 334 Marks
If $a, b, c$ and dare in continued proportion, then prove that
$(a+ d)(b+ c)-(a+ c)(b+ d)$
$= (b-c)^2$
Answer$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$
$\Rightarrow c=k d$
$b=k c=k^2 d$
$a=k b=k^3 d$
$(a+d)(b+c)-(a+c)(b+d)=(b-c)^2$
$\text{LHS}$
$(a+d)(b+c)-(a+c)(b+d)$
$=a b+b d+a c+c d-a b-b c-a d-c d$
$=b d+c a-b c-a d$
$=k^2 d^2+k^4 d^4-k^3 d^2-k^3 d^2$
$=k^2 d^2+k^4 d^4-2 k^3 d^2$
$=k^2 d^2\left(1+k^2-2 k\right)$
$\text{RHS}$
$(b-c)^2=(b-c)(b-c)$
$=b^2-2 b c+c^2$
$=k^4 d^4-2 k^3 d^2+k^2 d^2$
$=k^2 d^2\left(k^2-2 k+1\right)$
$\text{LHS} = \text{RHS}.$ Hence, proved.
View full question & answer→Question 344 Marks
If $a, b, c$ and dare in continued proportion, then prove that $\sqrt{(a+b+c)(b+c+d)}=\sqrt{a b}+\sqrt{b c}+\sqrt{c d}$
Answer$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$
$\Rightarrow c=k d$
$b=k c=k^2 d$
$a=k b=k^3 d$
$\sqrt{(a+b+c)(b+c+d)}=\sqrt{a b}+\sqrt{b c}+\sqrt{c d}$
$\text { LHS }$
$\sqrt{(a+b+c)(b+c+d)}$
$=\sqrt{\left(k^3 d+k^2 d+k d\right)\left(k^2 d+k d+d\right)}$
$=\sqrt{k d\left(k^2+k+1\right) \times d\left(k^2+k+1\right)}$
$=\sqrt{k^2\left(k^2+k+1\right)^2}$
$=d \sqrt{k}\left(k^2+k+1\right)$
$\text { RHS }$
$=\sqrt{a b}+\sqrt{b c}+\sqrt{c d}$
$=\sqrt{k^3 d \times k^2 d}+\sqrt{k^2 d \times k d}+\sqrt{k d \times d}$
$=\sqrt{k^5 d^2}+\sqrt{k^3 d^2}+\sqrt{k} d^2$
$=k^2 d \sqrt{k}+k d \sqrt{k}+d \sqrt{k}$
$=d \sqrt{k}\left(k^2+k+1\right)$
View full question & answer→Question 354 Marks
Find the fourth proportion to the following:
$(p^2q - qr^2 ), (pqr - pr^2 )$ and $(pq^2 - pr^2)$
AnswerLet $x$ be the fourth proportion
$\left( p ^2 q - qr ^2\right):\left( pqr - pr ^2\right)::\left( pq ^2- pr ^2\right): x$
$\Rightarrow\left( p ^2 q - qr ^2\right) \times x =\left( pqr - pr ^2\right) \times\left( pq ^2- pr ^2\right)$
$\Rightarrow x =\frac{\left( pqr - pr ^2\right) \times\left( pq ^2- pr ^2\right)}{\left( p ^2 q - qr ^2\right)}$
$\Rightarrow x =\frac{\operatorname{pr}( q - r ) \times p \left( q ^2- r ^2\right)}{ q \left( p ^2- r ^2\right)}$
$\Rightarrow x =\frac{\operatorname{pr}( q - r ) \times p ( q - r )( q + r )}{ q \left( p ^2- r ^2\right)}$
$\Rightarrow x =\frac{ p ^2 r ( q - r )^2( q + r )}{ q \left( p ^2- r ^2\right)}$
The fourth proportion is $\frac{ p ^2 r ( q - r )^2( q + r )}{ q \left( p ^2- r ^2\right)}$
View full question & answer→Question 364 Marks
Given four quantities $p, q, r$ and $s$ are in proportion, show that
$q^2(p - r) : rs (q - s)$
$=(p^2- q^2- pq): ( r^2-s^2-rs). $
Answer$p, q, r$ and $s$ are $1 n$ proportion
then, $p: q:: r: s$
Let $\frac{ p }{ q }=\frac{ r }{ s }= k$
Then $p = kq$ and $r = ks$
Now, we have to prove that
$\frac{(p-r) q^2}{(q-s) r s}=\frac{p^2-q^2-p q}{r^2-s^2-r s}$
$\text { LHS }$
$=\frac{(p-r) q^2}{(q-s) r s}$
$=\frac{(k q-k s) q^2}{(q-s) k s \times s}$
$=\frac{k(q-s) q^2}{k s^2(q-s)}$
$=\frac{q^2}{s^2}$
$\text{RHS}$
$=\frac{p^2-q^2-p q}{r^2-s^2-r s}$
$=\frac{k^2 q^2-q^2-k q \times q}{k^2 s^2-s^2-k s \times s}$
$=\frac{q^2\left(k^2-1-k\right)}{s^2\left(k^2-1-k\right)}$
$=\frac{q^2}{s^2}$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 374 Marks
What quantity must be added to each term of the ratio $(p + q) : (p - q)$ to make it equal to $(p + q)^2 : (p - q)^{2}$ ?
AnswerLet $x$ be added from each term such that
$\frac{(p+q)+x}{(p-q)+x}=\frac{(p+q)^2}{(p-q)^2}$
$\Rightarrow(p+q+x)\left(p^2-2 p q+q^2\right)=\left(p^2+2 p q+q^2\right)(p-q+x)$
$\Rightarrow p^3+p^2 q+p^2 x+q^2 p+q^3+q^2 x-2 p^2 q-2 p q^2-2 p q x$
$\Rightarrow p^3-p^2 q+p^2 x+q^2 p-q^3+q^2 x+2 p^2 q-2 p q^2+2 p q x$
$\Rightarrow p^2 q+q^3-2 p^2 q-2 p q x=-p^2 q-q^3+2 p^2 q+2 p q x$
$\Rightarrow 2 q^3-p^2 q=4 p q x$
$\Rightarrow 2 q\left(q^2-p^2\right)=4 p q x$
$\Rightarrow x=\frac{q^2-p^2}{2 p}$
$\frac{q^2-p^2}{2 p}$ should be added to each term.
View full question & answer→