4 Mark Question

Question 14 Marks

If $\frac{2 x-3 y}{3 z+y}=\frac{z-y}{z-x}=\frac{x+3 z}{2 y-3 x}$, then prove that every ratio $=\frac{x}{y}$.

Answer

Let $\frac{2 x-3 y}{3 z+y}=\frac{ z -y}{ z -x}=\frac{x+3 z }{2 y-3 x}= k$
$\therefore \quad k =\frac{2 x-3 y}{3 z +y}=\frac{-3( z -y)}{-3( z -x)}=\frac{x+3 z }{2 y-3 x}$
...[Multiplying numerator and denominator of second ratio by -3 ]
$\therefore \quad k =\frac{(2 x-3 y)-3( z -y)+(x+3 z)}{(3 z +y)-3( z -x)+(2 y-3 x)}$
... [Theorem on equal ratios]
$ =\frac{2 x-3 y-3 z+3 y+x+3 z}{3 z+y-3 z+3 x+2 y-3 x}=\frac{3 x}{3 y}$
$\therefore \quad k =\frac{x}{y}$
$\therefore \quad \text { Each ratio }=\frac{ x }{ y }$

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Question 24 Marks

If $a, b, c, d$ are in proportion, then prove that : $\frac{ a ^2+ ab + b ^2}{ a ^2- ab + b ^2}=\frac{ c ^2+ cd + d ^2}{c^2- cd + d ^2}$

Answer

$\frac{a^2+a b+b^2}{a^2-a b+b^2}=\frac{c^2+c d+d^2}{c^2-c d+d^2}$
$\text { L.H.S. }=\frac{a^2+a b+b^2}{a^2-a b+b^2}$
$=\frac{(b k)^2+(b k) b+b^2}{(b k)^2-(b k) b+b^2}$
$=\frac{b^2 k^2+b^2 k+b^2}{b^2 k^2-b^2 k+b^2}$
$=\frac{b^2\left(k^2+k+1\right)}{b^2\left(k^2-k+1\right)}$
$=\frac{ k ^2+ k +1}{ k ^2- k +1}$
$\text { R.H.S. }=\frac{c^2+c d+d^2}{c^2-c d+d^2}$
$=\frac{(d k)^2+(d k) d+d^2}{(d k)^2-(d k) d+d^2}$
$=\frac{d^2 k^2+d^2 k+d^2}{d^2 k^2-d^2 k+d^2}$
$=\frac{d^2\left(k^2+k+1\right)}{d^2\left(k^2-k+1\right)}$
$\text { 3, }=\frac{k^2+k+1}{k^2-k+1}$
$\therefore \quad \text { L.H.S. }=\text { R.H.S. }$
$\therefore \quad \frac{a^2+a b+b^2}{a^2-a b+b^2}=\frac{c^2+c d+d^2}{c^2-c d+d^2}$

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Question 34 Marks

If $a, b, c, d$ are in proportion, then prove that : $\quad \sqrt{\frac{a^2+5 c^2}{b^2+5 d^2}}=\frac{a}{b}$

Answer

$ \sqrt{\frac{a^2+5 c^2}{b^2+5 d^2}}=\frac{a}{b}$
$\text { L.H.S. }=\sqrt{\frac{a^2+5 c ^2}{b^2+5 d^2}}$
$=\sqrt{\frac{(b k)^2+5(d k)^2}{b^2+5 d^2}}$
$=\sqrt{\frac{b^2 k^2+5 d^2 k^2}{b^2+5 d^2}}$
$=\sqrt{\frac{k^2\left(b^2+5 d^2\right)}{b^2+5 d^2}}$
$=\sqrt{k^2}$
$=k$
$\text { R.H.S. }=\frac{a}{b}=\frac{b k}{b}$
$\text { L.H.S. }=\text { R.H.S. }$
$\therefore \quad \sqrt{\frac{a^2+5 c^2}{b^2+5 d^2}}=\frac{a}{b}$
...[From (i)]
...[From (i)]

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Question 44 Marks

If $a, b, c, d$ are in proportion, then prove that : $\quad \frac{11 a^2+9 a c}{11 b^2+9 b d}=\frac{a^2+3 a c}{b^2+3 b d}$

Answer

$a, b, c$ are in continued proportion. …[Given]
$\therefore \quad \frac{a}{b}=\frac{ c }{ d }$
Let $\frac{ a }{ b }=\frac{ c }{ d }= k$
$\therefore \quad a = bk$ and $c = dk$
i. $ \frac{11 a^2+9 a c}{11 b^2+9 b d}=\frac{a^2+3 a c}{b^2+3 b d}$
$\text { L.H.S. }=\frac{11 a^2+9 a c}{11 b^2+9 b d}$
$=\frac{11(b k)^2+9(b k)(d k)}{11 b^2+9 b d} \quad \ldots[\text { From (i)] }$
$=\frac{11 b^2 k^2+9 b d k^2}{11 b^2+9 b d}$
$=\frac{k^2\left(11 b^2+9 b d\right)}{11 b^2+9 b d}$
$=k^2$
$=\frac{(b k)^2+3(b k)(d k)}{b^2+3 b d}$
$\text { R.H.S. }=\frac{a^2+3 a c}{b^2+3 b d}$
$=\frac{b^2 k^2+3 b d k^2}{b^2+3 b d}$
$=\frac{k^2\left(b^2+3 b d\right)}{b^2+3 b d}$
$\text { L.H.S. }=\text { R.H.S. }$
$11 a^2+9 a c$
$\text { [From (i)] }$
$\text { R }$

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Question 54 Marks

Which number should be subtracted from $7, 12$ and $18$ such that the resultant numbers are in continued proportion?

Answer

Let $x$ be subtracted from $7, 12$ and $18$ such that resultant numbers are in continued proportion.
$ (7-x),(12-x),(18-x) \text { are in continued proportion. }$
$\therefore (12-x)^2=(7-x)(18-x)$
$\therefore 144-24 x+x^2=126-25 x+x^2$
$\therefore-24 x+25 x=126-144$
$\therefore \quad x=-18$
$\quad\quad\text{tally}$
$(7-x)=7-(-18)=25$
$(12-x)=12-(-18)=30$
$(18-x)=18-(-18)=36$
$30^2=900 \text { and } 25 \times 36=900$
$25,30,36$ are in continued proportion

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Question 64 Marks

Solve. $\frac{\sqrt{x+7}+\sqrt{x-2}}{\sqrt{x+7}-\sqrt{x-2}}=\frac{5}{1}$

Answer

$\frac{(\sqrt{x+7}+\sqrt{x-2})+(\sqrt{x+7}-\sqrt{x-2})}{(\sqrt{x+7}+\sqrt{x-2})-(\sqrt{x+7}-\sqrt{x-2})}=\frac{5+1}{5-1} \quad \ldots$ (using componendo-dividendo)
$
\begin{array}{}
\therefore \quad \frac{2 \sqrt{x+7}}{2 \sqrt{x-2}}=\frac{6}{4} \\
\therefore \frac{\sqrt{x+7}}{\sqrt{x-2}}=\frac{3}{2} \\
\therefore \quad \frac{x+7}{x-2}=\frac{9}{4} \\
\therefore \quad 4 x+28=9 x-18 \\
\therefore 28+18=9 x-4 x \\
\therefore \quad 46=5 x \\
\therefore \quad \frac{46}{5}=x \\
\end{array}
$
...(squaring both sides of the equation)
$\therefore \quad x=\frac{46}{5}$ is the solution of the given equation

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Question 74 Marks

Solve the equation. $\frac{3 x^2+5 x+7}{10 x+14}=\frac{3 x^2+4 x+3}{8 x+6}$

Answer

$\quad \frac{3 x^2+5 x+7}{10 x+14}=\frac{3 x^2+4 x+3}{8 x+6}$
$\frac{\left(6 x^2+10 x+14\right)}{10 x+14}=\frac{\left(6 x^2+8 x+6\right)}{8 x+6} \quad \ldots .($ multiplying both sides by $2)$
$ \frac{\left(6 x^2+10 x+14\right)-(10 x+14)}{10 x+14}=\frac{\left(6 x^2+8 x+6\right)-(8 x+6)}{8 x+6} \quad \ldots \text { (using dividendo) }$
$\therefore \quad \frac{6 x^2}{10 x+14}=\frac{6 x^2}{8 x+6}$
This equation is true for $x=0 $
$ \therefore x=0$ is a solution of the given equation.
If $x \neq 0$ then $x^2 \neq 0, $
$\therefore$ dividing by $6 x^2, \quad \frac{1}{10 x+14}=\frac{1}{8 x+6}$
$ \therefore 8 x+6=10 x+14$
$\therefore 6-14=10 x-8 x$
$\therefore \quad-8=2 x$
$\therefore \quad x=-4$
$\therefore x=-4$ or $x=0$ are the solutions of the given equation.

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Question 84 Marks

If $a$ and $b$ are integers and $a<b, b \neq \pm 1$ then compare $\frac{a-1}{b-1}, \frac{a+1}{b+1}$.

Answer

$a<b \quad \therefore a-1<b-1$
Now consider the subtraction $\frac{a-1}{b-1}-\frac{a+1}{b+1}$
$
\begin{aligned}
\frac{a-1}{b-1}-\frac{a+1}{b+1} & =\frac{(a-1)(b+1)-(a+1)(b-1)}{(b-1)(b+1)} \\
& =\frac{(a b-b+a-1)-(a b+b-a-1)}{b^2-1} \\
& =\frac{a b-b+a-1-a b-b+a+1}{b^2-1} \\
& =\frac{2 a-2 b}{b^2-1} \\
& =\frac{2(a-b)}{b^2-1} \ldots \ldots \ldots . .(1)
\end{aligned}
$
Now $a<b \quad \therefore a-b<0$
also $b^2-1>0$ because $b \neq \pm 1$
$\frac{2(a-b)}{b^2-1}<0$.........(2)
$\frac{a-1}{b-1}-\frac{a+1}{b+1}<0$ ...... from (1) \& (2)
$
\frac{a-1}{b-1}<\frac{a+1}{b+1}
$

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Question 94 Marks

The ratio of two numbers is 5 : 7. If 40 is added in each number, then the ratio becomes
25 : 31, Find the numbers.

Answer

Let the first number be $5 x$ and the second number be $7 x$.
From the given condition, $\frac{5 x+40}{7 x+40}=\frac{25}{31}$
$
\begin{aligned}
31(5 x+40) & =25(7 x+40) \\
155 x+1240 & =175 x+1000 \\
1240-1000 & =175 x-155 x \\
240 & =20 x \\
x & =12
\end{aligned}
$
$\therefore$ first number $=5 \times 12=60$ and second number $=7 \times 12=84$
$\, therefore, $ given numbers are 60 and 84

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Question 104 Marks

$5m – n = 3m + 4n,$ then find the values of the following expressions.
i. $\quad \frac{m^2+n^2}{m^2-n^2}$
ii. $\frac{3 m+4 n}{3 m-4 n}$

Answer

$5m – n = 3m + 4n …$ [Given]
$\therefore 5m – 3m = 4n + n$
$\therefore 2m = 5n$
$\text { i. } \quad \frac{ m ^2}{ n ^2}=\frac{25}{4} \quad \ldots \text { [Squaring both sides] }$
$\therefore \quad \frac{m^2+n^2}{ m ^2- n ^2}=\frac{25+4}{25-4}$
$\therefore \quad \frac{m^2+n^2}{m^2-n^2}=\frac{29}{21}$
$\therefore \quad \frac{m^2+n^2}{m^2-n^2}=29: 21$
$\text { ii. } \quad \frac{m}{n}=\frac{5}{2}$
$\therefore \quad \frac{m}{n} \times \frac{3}{4}=\frac{5}{2} \times \frac{3}{4}$
$\quad \ldots\left[\text { Multiplying both sides by } \frac{3}{4}\right]$
$\therefore \quad \frac{3 m}{4 n}=\frac{15}{8}$
$\therefore \quad \frac{3 m+4 n}{3 m-4 n}=\frac{15+8}{15-8}$
$\therefore \quad \frac{3 m+4 n}{3 m-4 n}=\frac{23}{7}$
$\therefore \quad \frac{3 m+4 n}{3 m-4 n}=23: 7$

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Question 114 Marks

Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?

Answer

Present age of Vatsala = 14 years
Present age of Sara = 10 years
After x years,
Vatsala’s age = (14 + x) years
Sara’s age = (10 + x) years
According to the given condition,
After x years the ratio of their ages will become 5 : 4
$\therefore \frac{14+x}{10+x}=\frac{5}{4}$
∴ 4(14 + x) = 5(10 + x)
∴ 56 + 4x = 50 + 5x
∴ 56 – 50 = 5x – 4x
∴ 6 = x
∴ x = 6
∴ After 6 years, the ratio of their ages will become 5 : 4.

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Question 124 Marks

The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha’s birth her mothers age was 27 years. Find the present ages of Abha and her mother.

Answer

The ratio of ages of Abha and her mother is 2 : 5.
Let the common multiple be x.
∴ Present age of Abha = 2x years and
Present age of Abha’s mother = 5x years
According to the given condition, the age of Abha’s mother at the time of Abha’s birth = 27 years
∴ 5x – 2x = 27
∴ 3x = 27
∴ x = 9
∴ Present age of Abha = 2x = 2 x 9 = 18 years
∴ Present age of Abha’s mother = 5x =5 x 9 = 45 years
The present ages of Abha and her mother are 18 years and 45 years respectively.

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Question 134 Marks

From the following pairs of numbers, find the reduced form of ratio of first number to second number.
i. $72,60$
ii. $38,57$
iii. $52,78$

Answer

$ \text { i. } 72,60$
$\text { Ratio }=\frac{72}{60}=\frac{12 \times 6}{12 \times 5}=\frac{6}{5}=6: 5 $
ii. $38,57$
$\text { Ratio }=\frac{38}{57}=\frac{19 \times 2}{19 \times 3}=\frac{2}{3}=2: 3$
iii. $52,78$
$\text { Ratio }=\frac{52}{78}=\frac{26 \times 2}{26 \times 3}=\frac{2}{3}=2: 3$

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Question 144 Marks

The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana’s present age ?

Answer

The ratio of present ages of Rehana and her mother is 2 : 7
Let the common multiple be x.
∴ Present age of Rehana = 2x years and Present age of Rehana’s mother = 7x years
After 2 years,
Rehana’s age = (2x + 2) years
Age of Rehana’s mother = (7x + 2) years
According to the given condition,
After 2 years, the ratio of their ages will be 1 : 3
$\therefore \frac{2 x+2}{7 x+2}=\frac{1}{3}$
∴ 3(2x + 2) = 1(7x + 2)
∴ 6x + 6 = 7x + 2
∴ 6 – 2 = 7x – 6x
∴ 4 = x
∴ x = 4
∴ Rehana’s present age = 2x = 2 x 4 = 8 years
∴ Rehana’s present age is 8 years.

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