[3 marks sum] - Mathematics STD 10 Questions - Vidyadip

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[3 marks sum]

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Question 13 Marks

If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²), prove that a : b :: c : d.

Answer

$
\begin{aligned}
& \left(11 a^2+13 b^2\right)\left(11 c^2-13 d^2\right)=\left(11 a^2-13 b^2\right)\left(11 c^2+13 d^2\right) \\
& \Rightarrow \frac{11 a+13 b^2}{11 a^2-13 b^2}=\frac{11 c^2+13 d^2}{11 c^2-13 d^2}
\end{aligned}$
Applying componendo and dividendo
$
\begin{aligned}
& \frac{11 a^2+13 b^2+11 a^2-13 b^2}{11 a^2+13 b^2-11 a^2+13 b^2}=\frac{11 c^2+13 d^2+11 c^2-13 d^2}{11 c^2+13 d^2-11 c^2+13 d^2} \\
& \Rightarrow \frac{22 a^2}{26 b^2}=\frac{22 c^2}{26 d^2} \\
& \Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2} \ldots\left(\text { Dividing by } \frac{22}{26}\right) \\
& \Rightarrow \frac{a}{b}=\frac{c}{d}
\end{aligned}
$
Hence $a: b:: c: d$.

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Question 23 Marks

If (pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd) prove that a : b : : c : d

Answer

$
\begin{aligned}
& (p a+q b):(p c+q d)::(p a-q b):(p c-q d) \\
& \Rightarrow \frac{p a+q b}{p c+q d}=\frac{p q-q b}{p c-q d} \\
& \Rightarrow \frac{p a+q b}{p c-q d}=\frac{p q+q b}{p c-q d}
\end{aligned}$
Applying componendo and dividendo
$
\begin{aligned}
& \Rightarrow \frac{ pa + qb + pa - qb }{ pa + qb - pa + qb }=\frac{ pc + qs + pc - qd }{ pc - qd - pc + qd } \\
& \Rightarrow \frac{2 p a}{2 q b}=\frac{2 p c}{2 q d} \\
& \Rightarrow \frac{a}{b}=\frac{c}{d} \quad \ldots\left(\text { Dividing by } \frac{2 p}{2 q}\right)
\end{aligned}
$
Hence $a : b :: c = d$.

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Question 33 Marks

If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proporton.

Answer

$
\begin{aligned}
& (4 a+5 b)(4 c-5 d)=(4 a-5 d)(4 c+5 d) \\
& \Rightarrow \frac{4 a+5 b}{4 a-5 b}=\frac{4 c+5 d}{4 c-5 d}
\end{aligned}$
Applying componendo and dividendo
$
\begin{aligned}
& \frac{4 a+5 b+4 a-5 b}{4 a+5 b-4 a+5 b}=\frac{4 c+5 d+4 c-5 d}{4 c+5 d-4 c+5 d} \\
& \Rightarrow \frac{8 a}{10 b}=\frac{8 c}{10 d} \\
& \Rightarrow \frac{a}{b}=\frac{c}{d}
\end{aligned}
$
Hence, $a, b, c, d$ are in proportion.

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Question 43 Marks

If $\frac{8 a-5 b}{8 c-5 a}=\frac{8 a+5 b}{8 c+5 d}$, prove that $\frac{a}{b}=\frac{c}{d}$

Answer

$
\begin{aligned}
& \frac{8 a-5 b}{8 c-5 a}=\frac{8 a+5 b}{8 c+5 d} \\
& \Rightarrow \frac{8 a+5 b}{8 a-5 b}=\frac{8 c+5 b}{8 c-5 d} \ldots \text { (using alternendo) }
\end{aligned}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{8 a+5 b+8 a-5 b}{8 a+5 b-8 a+5 b}=\frac{8 c+5 d+8 c-5 c}{8 c+5 d-8 c+5 d} \\
& \therefore \frac{16 a}{10 b}=\frac{16 c}{10 d} \\
& \Rightarrow \frac{a}{b}=\frac{c}{d} \quad \ldots\left(\text { Dividing by } \frac{16}{10}\right)
\end{aligned}
$
Hence proved.

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Question 53 Marks

If $\frac{5 x+7 y}{5 u+7 v}=\frac{5 x-7 y}{5 u-7 v}$, show that $\frac{x}{y}=\frac{u}{v}$

Answer

$
\frac{5 x+7 y}{5 u+7 v}=\frac{5 x-7 y}{5 u-7 v}$
Applying alternendo $\frac{5 x+7 y}{5 u+7 v}=\frac{5 x-7 y}{5 u-7 v}$
Applying componendo and dividendo
$
\begin{aligned}
& \frac{5 x+7 y+5 x-7 y}{5 x+7 y-5 x+7 y}=\frac{5 u+7 v+5 u-7 v}{5 u+7 u-5 u+7 v} \\
& \Rightarrow \frac{10 x}{14 y}=\frac{10 u}{14 v} \\
& \Rightarrow \frac{x}{y}=\frac{u}{v}
\end{aligned}
$
Hence proved. ... ( Dividing by $\left.=\frac{10}{14}\right)$

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Question 63 Marks

If $\frac{x+y}{a x+b y}=\frac{y+z}{a y+b z}=\frac{z+x}{a z+b x}$, prove that each of these ratio is equal to $\frac{2}{a+b}$ unless $x + y + z =0$

Answer

$
\begin{aligned}
& \frac{x+y}{a x+b y}=\frac{y+z}{a y+b z}=\frac{z+x}{a z+b x} \\
& =\frac{x+y+y+z+z+x}{a x+b y+a y+b x+a z+b x} \\
& =\frac{2(x y+z)}{x(a+b)+y(a+b)+z(a+b)} \\
& =\frac{2(x+y+z)}{(a+b)(x+y+z)} \\
& =\frac{2}{a+b}
\end{aligned}
$
$
\text { if } x+y+z \neq 0 \text {. }
$Hence proved.

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Question 73 Marks

Using the properties of proportion, solve the following equation for $x$; given $\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91}$

Answer

$
\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91}
$Applying componendo and dividendo
$
\begin{aligned}
& \frac{x^3+3 x+3 x^2++1}{x^3+3 x-3 x^2-1}=\frac{341+91}{341-91} \\
& \Rightarrow \frac{x^3+3 x^2+3 x+1}{x^3-3 x^2+3 x-1}=\frac{432}{250}=\frac{216}{125} \\
& \Rightarrow \frac{(x+1)^3}{(x-1)^3}=\frac{216}{125}=\left(\frac{6}{5}\right)^3 \\
& \therefore \frac{x+1}{x-1}=\frac{6}{5} \\
& \Rightarrow 6 x-6=5 x+5 \\
& \Rightarrow 6 x-5 x=5+6 \\
& \Rightarrow x=11 .
\end{aligned}
$

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Question 83 Marks

If a : b : : c : d, prove that (la + mb) : (lc + mb) :: (la – mb) : (lc – mb)

Answer

$
\begin{aligned}
& \because a : b :: c : d \\
& \therefore \frac{a}{b}=\frac{c}{d} \\
& \Rightarrow \frac{ la }{ mb }=\frac{ lc }{ md } \ldots\left(\text { Multiply by } \frac{ l }{ m }\right)
\end{aligned}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{ la + mb }{ la - mb }=\frac{ lc + md }{ lc - md } \\
& \Rightarrow \frac{ la + mb }{ lc + md }=\frac{ la - mb }{ lc - md } \ldots(\text { (By alternendo) } \\
& \Rightarrow( la + mb ):( lc + md )::( la - mb ):( lc - md ) .
\end{aligned}
$

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Question 93 Marks

If a : b : : c : d, prove that (2a + 3b)(2c – 3d) = (2a – 3b)(2c + 3d)

Answer

$
\begin{aligned}
& \because a : b :: c : d \\
& \therefore \frac{a}{b}=\frac{c}{d} \\
& \Rightarrow \frac{2 a}{3 b}=\frac{2 c}{3 d} \ldots\left(\text { Multiply by } \frac{2}{3}\right)
\end{aligned}$
Applying componendo and dividendo,
$
\frac{2 a+3 b}{2 a-3 b}=\frac{2 c+3 d}{2 c-3 d}
$
$
\Rightarrow(2 a +3 b )(2 c -3 d )
$
$=(2 a-3 b)(2 c+3 d) \ldots($ By corss multiplication $)$.

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Question 103 Marks

If $a : b :: c : d$, prove that $\frac{5 a+11 b}{5 c+11 d}=\frac{5 a-11 b}{5 c-11 d}$

Answer

$\begin{aligned}
& \because a : b :: c : d \\
& \therefore \frac{a}{b}=\frac{c}{d} \\
& \Rightarrow \frac{5 a}{11 b}=\frac{5 c}{11 d} \ldots\left(\text { Multiplying by } \frac{5}{11}\right)
\end{aligned}$
Applying componendo and dividendo,
$
\frac{5 a+11 b}{5 a-11 d}=\frac{5 c+11 d}{5 c-11 d}
$
$\Rightarrow \frac{5 a+11 b}{5 c+11 b}=\frac{5 a-11 b}{5 c-11 d} \ldots$ (Applying alternendo).

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Question 113 Marks

If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.

Answer

∵ x + 5 is the mean proportion between x + 2 and x + 9, then
(x + 5)² = (x + 2) (x + 9)
⇒ x² + 10x + 25 = x² + 11x + 18
⇒ x² + 10x – x² – 11x = 18 – 25
⇒ – x = – 7
∵ x = 7.

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Question 123 Marks

Find the mean proportion of $: (a – b)$ and $(a³ – a²b), a> b$

Answer

Let $x$ be the mean proportion to
$(a – b)$ and $(a³ – a²b), a> b$
then $(a - b) : x : : x: (a^3 – a^2b)$
$x^2 = (a – b) (a^3 – a^2b)$
$= (a – b) a^2 (a – b) = a^2 (a – b)^2$
$\therefore x = a (a – b)$
Hence the mean proportion $= a (a – b).$

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Question 133 Marks

Find the third proportional to $5 \frac{1}{4}$ and 7 .

Answer

Let $x$ be the third proportional to $5 \frac{1}{4}$ and 7
then $5 \frac{1}{4}: 7:: 7: x$
$
\begin{aligned}
& \Rightarrow \frac{21}{7}: 7:: 7: x \\
& \therefore \frac{21}{4} \times x=7 \times 7 \\
& x=\frac{7 \times 7 \times 4}{21} \\
& =\frac{28}{3} \\
& =9 \frac{1}{3}
\end{aligned}
$
$\therefore$ Third proportional $=9 \frac{1}{3}$.

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Question 143 Marks

If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.

Answer

$\because b$ is the mean proportional between $a$ and $c _r$ then,
$
b^2=a \times c \Rightarrow b^2=a c \text {... }$
(i) Now $a, c, a^2+b^2$ and $b^2+c^2$ are in proportion
$
\text { if } \frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}
$
if $a\left(b^2+c^2\right)=\left(a^2+b^2\right)$
if $a\left(a c+c^2\right)=c\left(a^2+a c\right)$...[from (i)]
if $a c(a+c)=a^2 c+a c^2$
if $a c(a+c)=a c(a+c)$
which is true.
Hence proved.

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Question 153 Marks

If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D

Answer

Let $A: B=2: 3, B: C=4: 5$ and $C: D=6: 7$
$
\frac{ A }{ B }=\frac{2}{3}, \frac{ B }{ C }=\frac{4}{5}, \frac{ C }{ D }=\frac{6}{7}
$
Multiplying $\frac{ A }{ B } \times \frac{ B }{ C } \times \frac{ C }{ D }=\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7}$
$
\begin{aligned}
& \therefore \frac{A}{D}=\frac{16}{35} \\
& \Rightarrow A: D=16: 35 .
\end{aligned}
$

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Question 163 Marks

If the angles of a triangle are in the ratio 2 : 3 : 4, find the angles.

Answer

Sum of angles of a triangle $=180^{\circ}$
Ratio among angles $=2: 3: 4$
Sum of ratios $=2+3+4=9$
First angle $=180^{\circ} \times \frac{2}{9}=40^{\circ}$
Second angle $=180^{\circ} \times \frac{3}{9}=60^{\circ}$
Third angle $=180^{\circ} \times \frac{4}{9}=80^{\circ}$
$\therefore$ Angles are $40^{\circ}, 60^{\circ}$ and $80^{\circ}$.

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Question 173 Marks

The sides of a triangle are in the ratio 7 : 5 : 3 and its perimeter is 30 cm. Find the lengths of sides.

Answer

Perimeter of a triangle $=30 cm$.
Ratio among sides $=7: 5: 3$
Sum of ratios $7+5+3=15$
Length of the first side $=30 \times \frac{7}{15}=14 cm$
Length of the second side $=30 \times \frac{5}{15}=10 cm$
Length of the third side $=30 \times \frac{3}{15}=6 cm$
$\therefore$ Sides are $14 cm , 10 cm , 6 cm$.

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Question 183 Marks

A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3 : 4. How much money did each receive?

Answer

Total amount to be disturbed $=₹ 2100$
Ratio between orphanage and a blind school $=3: 4$
Sum of ratios $=3+4=7$
$\therefore$ Orphaange school's share $=₹ 2100 \times \frac{3}{7}=₹ 900$
Blind School's share $=₹ 2100 \times \frac{4}{7}=₹ 1200$.

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Question 193 Marks

A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 91 kg.

Answer

Ratio between the original weight and reduced weight $=$ $7: 5$
Let original weight $=7 x$
then reduced weight $=5 x$
If original weight $=91 kg$
then reduced weight
$=\frac{91 \times 5 x}{7 x}$
$=65 kg$.

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Question 203 Marks

The income of a man is increased in the ratio of 10 : 11. If the increase in his income is Rs 600 per month, find his new income.

Answer

Let the present income = 10x
then increased income = 11x
∴ Increase per month
= 11x - 10x
= x
∴ x = ₹600
Now his new income
= 11x
= 11 x 600
= ₹6600.

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Question 213 Marks

If (x + 2y) : (2x – y) is equal to the duplicate ratio of 3 : 2, find x : y.

Answer

If $(x+2 y):(2 x-y)$ is equal to the duplicate ratio of $3: 2$
$
\begin{aligned}
& \text { then } \frac{x+2 y}{2 x-y}-\frac{(3)^2}{(2)^2}=\frac{9}{4} \\
& \Rightarrow 9(2 x-y)=4(x+2 y) \\
& \Rightarrow 18 x-9 x=4 x+8 y \\
& \Rightarrow 18 x-4 x=8 y+9 y \\
& \Rightarrow 14 x-17 y \\
& \Rightarrow \frac{x}{y}=\frac{17}{14} \\
& \therefore x: y=17: 14 .
\end{aligned}
$

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Question 223 Marks

If (3x + 1) : (5x + 3) is the triplicate ratio of 3 : 4, find the value of x.

Answer

If $(3 x+1):(5 x+3)$ is the triplicate ratio of $3: 4$
$
\begin{aligned}
& \text { then } \frac{3 x+1}{5 x+3}=\frac{(3)^2}{(4)^3}=\frac{27}{64} \\
& \Rightarrow 64(3 x+1)=27(5 x+3) \\
& \Rightarrow 192 x+64=135 x+81 \\
& \Rightarrow 192 x-135 x=81-64 \\
& \Rightarrow 57 x=17 \\
& \Rightarrow x=\frac{17}{57}
\end{aligned}$
Hence $x=\frac{17}{57}$.

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Question 233 Marks

If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x.

Answer

$\begin{aligned} & \frac{x-9}{3 x+6}=\left(\frac{4}{9}\right)^2 \\ & \Rightarrow \frac{x-6}{3 x+6}=\frac{16}{81} \\ & \Rightarrow 81 x-729=48 x+96 \\ & \Rightarrow 81 x-48 x=96+729 \\ & \Rightarrow 33 x=825 \\ & \Rightarrow x=\frac{825}{33} \\ & =25 .\end{aligned}$

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Question 243 Marks

If $\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}$, Find $x : y$

Answer

$
\begin{aligned}
& \frac{3 x+5 y}{3 x-5 y}=\frac{7}{3} \\
& \Rightarrow 9 x +15 y =21 x -35 y \ldots[\text { By cross multiplication] } \\
& \Rightarrow 21 x -9 x =15 y +35 y \\
& \Rightarrow 12 x =50 y \\
& \Rightarrow \frac{x}{y}=\frac{50}{12}=\frac{25}{6}
\end{aligned}$
Hence, $x: y=25: 6$.

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Question 253 Marks

If 3A = 4B = 6C, find A : B : C

Answer

$
\begin{aligned}
& 3 A =4 B \\
& \Rightarrow \frac{ A }{ B }=\frac{4}{3}
\end{aligned}
$
or
$
\begin{aligned}
& A: B=4: 3 \\
& \text { and } 4 B=6 C \\
& \Rightarrow \frac{B}{C}=\frac{6}{4}=\frac{3}{2}
\end{aligned}
$
or
$
\begin{aligned}
& C=3: 2 \\
& \therefore A: B: C=4: 3: 2 .
\end{aligned}
$

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Question 263 Marks

If $A: B=\frac{1}{4}: \frac{1}{5}$ and $B: C=\frac{1}{7}: \frac{1}{6}$, find $A: B: C$.

Answer

$A : B =\frac{1}{4}: \frac{1}{5}$
$
\text { B : C }=\frac{1}{7}: \frac{1}{6}$
LCM of B's terms 4 and $6=12$
Making terms of B's; as 12
$
\begin{aligned}
& \frac{A}{B}=\frac{5 \times 3}{4 \times 3}=\frac{15}{12}=15: 12 \\
& \frac{B}{C}=\frac{6 \times 2}{7 \times 2}=\frac{12}{14}=12: 14 \\
& \therefore A: B: C=15: 12: 14 .
\end{aligned}
$

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Question 273 Marks

The ratio of the pocket money saved by Lokesh and his sister is 5 : 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged?

Answer

Let the savings of Lokesh and his sister are $5 x$ and $6 x$. and the Lokesh should save Rs y more Now, according to the problem,
$
\begin{aligned}
& \Rightarrow \frac{5 x+y}{6 x+30}=\frac{5}{6} \\
& \Rightarrow 30 x+6 y=30 x+150 \\
& \Rightarrow 30 x+6 y-30 x=150 \\
& \Rightarrow 6 y=150 \\
& \therefore y=\frac{150}{6}=25
\end{aligned}$
Hence, Lokesh should save ₹25 more.

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Question 283 Marks

If a : b = 3 : 5, find (3a + 5b): (7a – 2b).

Answer

$
\begin{aligned}
& a: b=3: 5 \\
& \Rightarrow \frac{a}{b}=\frac{3}{5} \\
& \Rightarrow 3 a+5 n: 7 a-2 b
\end{aligned}$
Dividing each term by $b$
$
\begin{aligned}
& 3 \frac{a}{b}+5: 3 \frac{a}{b}-2 \\
& \Rightarrow 3 \times \frac{3}{5}+5: 7 \times \frac{3}{5}-2 \\
& \Rightarrow\left(\frac{9}{5}+5\right):\left(\frac{21}{5}-2\right) \\
& \Rightarrow \frac{9+25}{5}: \frac{21-10}{5} \\
& \Rightarrow \frac{34}{5}: \frac{11}{5} \\
& =34: 11 .
\end{aligned}
$

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Question 293 Marks

If (7 p + 3 q) : (3 p – 2 q) = 43 : 2 find p : q

Answer

$\begin{aligned} & (7 p+3 q):(3 p-2 q)=43: 2 \\ & \Rightarrow \frac{7 p+3 q}{3 p-2 q}=\frac{43}{2} \\ & \Rightarrow 129 p-86 q=4 p+6 q \\ & \Rightarrow 129 p-14 p=6 q+86 q \\ & \Rightarrow 115 p=92 q \\ & \Rightarrow \frac{p}{q}=\frac{92}{115}=\frac{4}{5} \\ & \therefore p: q=4: 5 .\end{aligned}$

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Question 303 Marks

If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}$ prove that each ratio's equal to : $\frac{x+y+z}{a+b+c}$

Answer

$
\begin{aligned}
& \frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}= k \text { (say) } \\
& x = k ( b + c - a ), \\
& y = k ( c + a - b ), \\
& z = k ( a + b - c ) \\
& \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 .
\end{aligned}$
Hence proved.

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Question 313 Marks

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$, prove that each ratio is $\left[\frac{2 a^3+5 c^3+7 e^3}{2 b^3+5 d^3+7 f^3}\right]^{\frac{1}{3}}$

Answer

$
\begin{aligned}
& \frac{a}{b}=\frac{c}{d}=\frac{e}{f}= k \text { (say) } \\
& \therefore a = k , c = dk _{ l } e = fk \\
& {\left[\frac{2 a^3+5 c^3+7 e^3}{2 b^3+5 d^3+7 f^3}\right]^{\frac{1}{3}}}
\end{aligned}
$
$
\begin{aligned}
& =\left[\frac{2 b^3 k^3+5 d^3 k^3+7 f^3 k^3}{2 b^3+5 d^3+7 f^3}\right]^{\frac{1}{3}} \\
& =k\left[\frac{2 b^3+5 d^3+7 f^3}{2 b^3+5 d^3+7 f^3}\right]^{\frac{1}{3}}= k
\end{aligned}$
Hence proved.

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Question 323 Marks

If $\frac{a}{b}=\frac{c}{d}=\frac{c}{f}$, prove that each ratio is $\sqrt{\frac{3 a^2-5 c^2+7 e^2}{3 b^2-5 d^2+7 f^2}}$

Answer

$
\begin{aligned}
& \frac{a}{b}=\frac{c}{d}=\frac{c}{f}= k \text { (say) } \\
& \therefore a = k , c = dk _{ l } e = fk \\
& \sqrt{\frac{3 a^2-5 c^2+7 e^2}{3 b^2-5 d^2+7 f^2}}
\end{aligned}
$
$
=\sqrt{\frac{3 b^2 k^2-5 d^2 k^2+7 f^2 k^2}{3 b^2-5 d^2-7 f^2}}
$
$
= k \sqrt{\frac{3 b^2-5 d^2+7 f^2}{3 b^2-5 d^2+7 f^2}}= k$
Hence proved.

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[3 marks sum] - Mathematics STD 10 Questions - Vidyadip