Question 514 Marks
If $a + c = mb$ and $\frac{1}{b}+\frac{1}{d}=\frac{m}{c}$, prove that $a , b , c$ and $d$ are in proportion.
Answer
$
\begin{aligned}
& a+c=m b \text { and } \frac{1}{b}+\frac{1}{d}=\frac{m}{c} \\
& a+c=m b \\
& \frac{a}{c}+\frac{c}{d}=m \quad \text { (Dividing by b)...(i) }
\end{aligned}
$
and $\frac{1}{b}+\frac{1}{d}=\frac{m}{c} \frac{c}{b}+\frac{c}{d}=m \quad$ (Multiplying by $C$ ) From (i) and (ii)
$
\begin{aligned}
& \frac{a}{b}+\frac{c}{b} \\
& =\frac{c}{b}+\frac{c}{d} \\
& \Rightarrow \frac{a}{b}=\frac{c}{d}
\end{aligned}$
Hence, a, b, c and d are proportional.
View full question & answer→Question 524 Marks
If $y$ is mean proportional between $x$ and $z,$ prove that $xyz (x + y + z)^3 = (xy + yz + zx)^3.$
Answer$\because y$ is the mean proportional between $x$ and $z,$ then
$y^2 = xz …(i)$
$ \text{L.H.S.} = xyz (x + y + z)^3$
$= xzy (x + y+ z)^3$
$= y^2 y(x + y+ z)^3 ...[from (i)]$
$= [y (x + y + z)]^3$
$= [xy + y^2 + yz]^{3 } ...[from (i)]$
$= (xy + xz + yz)^3$
$= (xy + yz + zx)^3$
$= \text{R.H.S.}$
$\because \text{L.H.S = R.H.S.}$ hence proved.
View full question & answer→Question 534 Marks
If $b$ is the mean proportional between $a$ and $c,$ prove that $(ab + bc)$ is the mean proportional between $(a^2 + b^2)$ and $(b^2 + c^2).$
Answer$b$ is the mean proportional between $a$ and $c$ then
$b^2 = ac …(i)$
Now if $(ab + bc)$ is the mean proportional
between $(a^2 + b^2)$ and $(b^2 + c^2),$ then
$(ab + bc)^2 = (a^2 + b^2) (b^2 + c^2)$
Now $\text{L.H.S.} = (ab + bc)^2 = a^2b + b^2c^2 + 2ab^2c$
$= a^2(ac) + ac(c)^2+ 2a ac.c ...[$from $(i)]$
$= ac(a^2 + c^2 + 2ac) = ac(a + c)^2$
$\text{R.H.S.} = (a^2 + b^2)(b^2 + c^2)$
$= (a^2 + ac)(ac + c^2) ...[$from $(i)]$
$= a(a + c) \times (a + c) = ac(a + c)2$
$\therefore \text{L.H.S.} = \text{R.H.S.}$
View full question & answer→Question 544 Marks
Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
AnswerLet the two numbers are $a$ and $b$.
$\because 28$ is the mean proportional
$
\because a: 28:: 28: b
$
$
\begin{aligned}
& \therefore ab =(28)^2=784 \\
& \Rightarrow a =\frac{784}{b} ...(i)
\end{aligned}
$
$\because 224$ is the third proportional
$
\begin{aligned}
& \therefore a : b :: b : 224 \\
& \Rightarrow b ^2=224 a ...(ii)
\end{aligned}$
Substituting the value of a in (ii)
$
\begin{aligned}
& b ^2=24 \times \frac{784}{b} \\
& \Rightarrow b ^3=224 \times 784 \\
& \Rightarrow b ^2=175616=(56) 3 \\
& \therefore b =56
\end{aligned}$
Now substituting the value of $b$ in (i)
$
a =\frac{784}{56}=14$
Hence numbers are 14,56 .
View full question & answer→Question 554 Marks
What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?
AnswerLet $x$ be added to each number then
$
16+x, 26+x \text { and } 40+x
$
are in continued proportion.
$
\Rightarrow \frac{16+x}{26+x}=\frac{26+x}{40+x}
$Cross Multiplying
$
\begin{aligned}
& (16+x)(40+x)=(26+x)(26+x) \\
& \Rightarrow 640+16 x+40 x+x^2=676+26 x+26 x+x^2 \\
& \Rightarrow 640+56 x+x^2=676+52 x+x^2 \\
& \Rightarrow 56 x+x^2-52 x-x^2=676-640 \\
& \Rightarrow 4 x=36 \\
& \Rightarrow x=\frac{36}{4}=9
\end{aligned}
$
$\therefore 9$ is to be added.
View full question & answer→Question 564 Marks
Arrange the following ratios in ascending order of magnitude: 2 : 3, 17 : 21, 11 : 14 and 5 : 7
AnswerWriting the given ratios in fraction
$
\frac{2}{3}, \frac{17}{21}, \frac{11}{14}, \frac{5}{7}$
LCM of $3,21,14,7=42$
Converting the given ratio as equivalent
$
\begin{aligned}
& \frac{2}{3}=\frac{2 \times 14}{3 \times 14}=\frac{28}{42} ; \frac{17}{21}=\frac{17 \times 2}{21 \times 2}=\frac{34}{42} \\
& \frac{11}{14}=\frac{11 \times 3}{14 \times 3}=\frac{33}{42} ; \frac{5}{7}=\frac{5 \times 6}{7 \times 6}=\frac{30}{42}
\end{aligned}$
From above, writing in asecnding order,
$
\begin{aligned}
& \frac{28}{42}, \frac{30}{42}, \frac{33}{42}, \frac{34}{42} \\
& \text { or } \\
& \frac{2}{3}, \frac{5}{7}, \frac{11}{14}, \frac{17}{21}
\end{aligned}
$
or
$2: 3 ; 5: 7 ; 11: 14$ and $17: 21$.
View full question & answer→Question 574 Marks
In an examination, the ratio of passes to failures was 4 : 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination
Answer
$
\begin{aligned}
& \text { Let the number of passes }=4 x \\
& \text { and number of failures }=x \\
& \text { The total number of students appeared }=4 x+x=5 x \\
& \text { In the second case, the number of students appeared }=5 x \\
& -30 \\
& \text { and number of passes }=4 x-20 \\
& \therefore \text { No. of failure } \\
& =(5 x-30)-(4 x-20) \\
& =5 x-30-4 x+20 \\
& =x-10
\end{aligned}
$According to the condition
$
\begin{aligned}
& \frac{4 x-20}{x-10}=\frac{5}{1} \\
& \Rightarrow 5 x-50=4 x-20 \\
& \Rightarrow 5 x-4 x=-20+50 \\
& \Rightarrow x=30 \\
& \therefore \text { Number of students appeared }=5 x \\
& =5 \times 30 \\
& =150 .
\end{aligned}
$
View full question & answer→Question 584 Marks
In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1, How many students were there in the class?
AnswerLet the number of students in the class $=x$
Ratio of boys and girls $=4: 3$
$\therefore$ No. of boys $=\frac{4 x}{7}$ and no. of girls $=\frac{3 x}{7}$
According to the problem,
$
\begin{aligned}
& \left(\frac{4 x}{7}+20\right):\left(\frac{3 x}{7}-12\right)=2: 1 \\
& \frac{4 x+140}{7}: \frac{3 x-84}{7}: 2: 1 \\
& \Rightarrow \frac{4 x+140}{7} \times \frac{7}{3 x-84}=\frac{2}{1} \\
& \Rightarrow \frac{4 x+140}{3 x-84}=\frac{2}{1} \\
& \Rightarrow 6 x-168=4 x+140 \\
& \Rightarrow 6 x-4 x=140+148 \\
& \Rightarrow 2 x=308 \\
& \Rightarrow x=\frac{308}{2}=154
\end{aligned}$
Hence, number of students $=154$.
View full question & answer→Question 594 Marks
The monthly pocket money of Ravi and Sanjeev are in the ratio 5:7. Their expenditures are in the ratio 3:5. If each saves Rs. 80 every month, find their monthly pocket money.
AnswerLet the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.
Let their expenditures be 3y and 5y respectively.
Ravi’s savings = 5x – 3y
Sanjeev’s savings = 7x – 5y
By the given information,
5x – 3y = 80 … (1)
7x – 5y = 80 … (2)
From (1) and (2), we have
5x – 3y = 7x – 5y
⇒ x = y
From equation (1),
5x – 3x = 80
⇒ 2x = 80
⇒ x = 40
Hence,
Monthly pocket money of Ravi = 5 × 40 = Rs. 200
Monthly pocket money of Sanjeev = 7 × 40 = Rs. 280
View full question & answer→Question 604 Marks
The ratio of the number of boys to the number of girls in a school of 560 pupils is 5 : 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3 : 2.
AnswerRatio between boys and girls $=5: 3$
No. of pupils $=560$
Sum of ratios $=5+3=8$
$\therefore$ No of boys $=\frac{5}{8} \times 560=350$
and no. of girl $=\frac{3}{8} \times 560=350$
No. of new boys admitted $=10$
$\therefore$ Total number of boys $=350+10=360$
Let the no. of girl admitted $= x$
$\therefore$ Total number of girls $=210+x$
Now according to the condition,
$
\begin{aligned}
& 360: 210+x=3: 2 \\
& \Rightarrow \frac{360}{210+x}=\frac{3}{2} \\
& \Rightarrow 630+3 x=720 \\
& \Rightarrow 3 x=720-630=90 \\
& \therefore x=\frac{90}{3}=30
\end{aligned}
$
$\therefore$ No of girls to be admitted $=30$.
View full question & answer→Question 614 Marks
In a mixture of 45 litres, the ratio of milk to water is 13 : 2. How much water must be added to this mixture to make the ratio of milk to water as 3 : 1 ?
AnswerMixture of milk and water $=45$ litres
Ratio of milk and water $=13: 2$
Sum of ratio $=13+2=15$
$\therefore$ Quantity of milk $=\frac{45 \times 13}{15}=39$ litres
and quantity of water $=45 \times \frac{2}{15}=6$ litres
Let $x$ litre of water be added, then water $=(6+x)$ litres
Now nrew ratio $=3: 1$
$
\begin{aligned}
& \therefore 39:(6+ x )=3: 1 \\
& \frac{39}{6+x}-\frac{3}{1} \\
& \Rightarrow 39=18+3 x \\
& \Rightarrow 3 x =39-18=21
\end{aligned}
$
$\therefore x=\frac{21}{3}=7$ litres
$\therefore 7$ litres of water is to be added.
View full question & answer→Question 624 Marks
In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C, Rs 80000 for 5 months. If they together earn Rs 18800 find the share of each.
AnswerA's 6 months investment $=\text ₹ 50000$
$\therefore$ A's 1 month investment
$
\begin{aligned}
& =\text ₹ 50000 \times 6 \\
& =\text ₹ 300000
\end{aligned}$
B's 4 months investment $=\text ₹ 60000$
$\therefore$ B's 1 month investment
$=\text ₹ 60000 \times 4$
= ₹ 240000
C's 5 months investment $=\text ₹ 80000$
$\therefore$ C's 1 month investment
$=\text ₹ 80000 \times 5$
$=\text ₹ 400000$
$\therefore$ Ratio between their investments
$
=300000: 240000: 400000
$
$
=30: 24: 40$
Sum of ratios $=30+24+40=94$
Total earnings $=\text ₹ 18800$
$\therefore$ A's share $=\frac{30}{94} \times 18800=\text ₹ 6000$
B's share $=\frac{24}{94} \times 18800=\text ₹ 4800$
C's share $=\frac{40}{94} \times 18800=\text ₹ 8000$.
View full question & answer→Question 634 Marks
A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got Rs 500 more than C, find the total sum divided.
AnswerRatio between A, B and C = 7 : 5 : 4
Let A’s share = 7x
B’s share = 5x
and C’s share = 4x
Total sum
= 7x + 5x + 4x
= 16x
Now according to the condition,
5x – 4x = 500
⇒ x = 500
∴ Total sum = 16x
= 16 x 500
= ₹8000.
View full question & answer→Question 644 Marks
Three numbers are in the ratio $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ If the sum of their squares is 244 , find the numbers.
AnswerThe ratio of three numbers $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$
$
\begin{aligned}
& =\frac{16: 4: 3}{12} \\
& =6: 4: 3
\end{aligned}$
Let first number $6 x$, second $4 x$ and third $3 x$
$\therefore$ According to the condition
$
\begin{aligned}
& (6 x)^2+(4 x)^2+(3 x)^2=244 \\
& \Rightarrow 36 x^2+16 x^2+9 x^2=244 \\
& \Rightarrow 61 x 2=244 \\
& \Rightarrow x 2=\frac{244}{61} \\
& =4=(2)^2 \\
& \therefore x=2
\end{aligned}
$
$
\begin{aligned}
& \therefore \text { first number }=6 x=6 \times 2=12 \\
& \text { second number }=4 x=4 \times 2=8 \\
& \text { and third number }=3 x=3 \times 2=6 .
\end{aligned}
$
View full question & answer→Question 654 Marks
Find two numbers in the ratio of 8 : 7 such that when each is decreased by , they are in the ratio 11 : 9.
AnswerThe ratio is $8: 7$
Let the numbers be $8 x$ and $7 x$, According to condition,
$
\begin{aligned}
& \frac{8 x-\frac{25}{2}}{7 x-\frac{25}{2}}=\frac{11}{9} \\
& \Rightarrow \frac{16 x-\frac{25}{2}}{14 x-\frac{25}{2}}=\frac{11}{9} \\
& \Rightarrow \frac{(16 x-25) \times 2}{2(14 x-25)}=\frac{11}{9} \\
& \Rightarrow 154 x-275=144 x-225 \\
& \Rightarrow 154 x-144 x=275-225 \\
& \Rightarrow 10 x=50 \\
& \therefore x=\frac{50}{10}=5 \\
\end{aligned}
$
$\therefore$ Numbers are $8 x=8 \times 5=40$ and $7 x=7 \times 5=35$.
View full question & answer→Question 664 Marks
If $y (3x – y) : x (4x + y) = 5 : 12.$ Find $(x^2 + y^2) : (x + y)^2.$
AnswerIf $y(3 x-y): x(4 x+y)=5: 12$
Find $\left(x^2+y^2\right):(x+y)^2$
$ \frac{3 x y-y^2}{4 x^2+x y}=\frac{5}{12}$
$\Rightarrow 36 x y-12 y^2=20 x^2+5 x y$
$\Rightarrow 20 x^2+5 x y-36 x y+12 y^2=0$
$\Rightarrow 20 x^2-31 x y+12 y^2=0$
$\Rightarrow 20 \frac{x^2}{y^2}-31 x \frac{y}{y^2}+\frac{12 y^2}{y^2}=0 \ldots\left(\text { Dividing by } y ^2\right)$
$\Rightarrow 20\left(\frac{x^2}{y^2}\right)-31\left(\frac{x y}{y^2}\right)+12=0$
$\Rightarrow 20\left(\frac{x}{y}\right)^2-15\left(\frac{x}{y}\right)-16\left(\frac{x}{y}\right)+12=0$
$\Rightarrow 5\left(\frac{x}{y}\right)\left[4\left(\frac{x}{y}\right)-3\right]-4\left[4\left(\frac{x}{y}\right)-3\right]=0$
$\Rightarrow\left[4\left(\frac{x}{y}\right)-3\right]\left[5\left(\frac{x}{y}\right)-4\right]=0$
Either $\left[4\left(\frac{x}{y}\right)-3\right]=0$,
then $4\left(\frac{x}{y}\right)=3$
$ \Rightarrow \frac{x}{y}=\frac{3}{4}$
$\text { or }\left[5\left(\frac{x}{y}\right)-4\right]=0 $
then $5\left(\frac{x}{y}\right)=4$
$ \Rightarrow \frac{x}{y}=\frac{4}{5} $
$(a)$ When $\frac{x}{y}=\frac{3}{4}$
then $\left(x^2: y^2\right):(x+y)^2$
$ =\frac{x^2+y^2}{(x+y)^2}$
$ =\frac{\frac{x^2}{y^2}+\frac{y^2}{y^2}}{\frac{y^2}{\frac{1}{y^2}(x+y)^2}} ($Dividing by $y ^2 ) $
$=\frac{\frac{x^2}{y^2}+1}{\left(\frac{x}{y}+1\right)}$
$=\frac{\left(\frac{3}{4}\right)^2}{\left(\frac{3}{4}+1\right)^2}$
$=\frac{\frac{9}{16}+1}{\left(\frac{7}{4}\right)^2}$
$=\frac{\frac{25}{16}}{\frac{49}{16}}$
$=\frac{25}{16} \times \frac{16}{49}$
$=\frac{25}{49}$
$\because\left(x^2+y^2\right):(x+y)^2=25: 49$
$(b)$ When $\frac{x}{y}=\frac{4}{5}$, then
$ \frac{x^2 y^2}{(x+y)^2}$
$=\frac{\frac{x^2}{y^2}+1}{\left(\frac{x}{y}+1\right)^2}$
$=\frac{\left(\frac{x}{y}\right)^2+1}{\left(\frac{x}{y}+1\right)^2}$
$=\frac{\left(\frac{4}{5}\right)^2+1}{\left(\frac{4}{5}+1\right)^2}$
$=\frac{\frac{16}{25}+1}{\left(\frac{9}{5}\right)^2}$
$=\frac{\frac{41}{25}}{\frac{81}{25}}$
$=\frac{41}{25} \times \frac{25}{81}$
$=\frac{41}{81} $
$\because\left(x^2+y^2\right):(x+y)^2=41: 81$.
View full question & answer→Question 674 Marks
If (4x² + xy) : (3xy – y²) = 12 : 5, find (x + 2y) : (2x + y).
Answer
$\begin{aligned} & \left(4 x^2+x y\right):\left(3 x y-y^2\right)=12: 5 \\ & \Rightarrow \frac{4 x^2+x y}{3 x y-y^2}=\frac{12}{5} \\ & \Rightarrow 20 x^2+5 x y=36 x y-12 y^2 \\ & \Rightarrow 20 x^2+5 x y-36 x y+12 y^2=0 \\ & \Rightarrow \frac{20 x^2}{y^2}-\frac{31 x y}{y^2}+\frac{12 y^2}{y^2}=0 \ldots\left(\text { Dividing by } y^2\right) \\ & \Rightarrow 20\left(\frac{x}{y}\right)^2-31\left(\frac{x}{y}\right)+12=0 \\ & \Rightarrow 20\left(\frac{x}{y}\right)^2-15\left(\frac{x}{y}\right)-16\left(\frac{x}{y}\right)+12=0 \\ & \Rightarrow 5\left(\frac{x}{y}\right)\left[4\left(\frac{x}{y}\right)-3\right]-4\left[4\left(\frac{x}{y}\right)-3\right]=0 \\ & \Rightarrow\left[4\left(\frac{x}{y}\right)-3\right]\left[5\left(\frac{x}{y}\right)-4\right]=0\end{aligned}$Either $4\left(\frac{x}{y}\right)-3=0$, then $4\left(\frac{x}{y}\right)=3$
$
\Rightarrow \frac{x}{y}=\frac{3}{4}
$
or $5\left(\frac{x}{y}\right)-4=0$,
then $5\left(\frac{x}{y}\right)=4$
$
\Rightarrow \frac{x}{y}=\frac{4}{5}
$Now $\frac{x+2 y}{2 x+y}=\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1} \ldots($ Dividing by $y )$
(a) When $\frac{x}{y}=\frac{3}{4}$, then
$
\begin{aligned}
& =\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1} \\
& =\frac{\frac{3}{4}+2}{2 \times \frac{3}{4}+1} \\
& =\frac{\frac{11}{4}}{\frac{3}{2}+1} \\
& =\frac{\frac{11}{4}}{\frac{5}{2}} \\
& =\frac{11}{4} \times \frac{2}{5} \\
& =\frac{11}{10} \\
& \therefore(x+2 y):(2 x+y)=11: 10
\end{aligned}
$
(b) When $\frac{x}{y}=\frac{4}{5}$, then
$
\begin{aligned}
& \frac{x+2 y}{2 x+y} \\
& =\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1} \\
& =\frac{\frac{4}{5}+2}{2 x \frac{4}{5}+1} \\
& =\frac{\frac{14}{5}}{\frac{8}{5}+1} \ldots \text { (Dividing by y) } \\
& =\frac{\frac{14}{5}}{\frac{13}{5}} \\
& =\frac{14}{5} \times \frac{5}{13} \\
& =\frac{14}{13}
\end{aligned}
$
...(Dividing by y)
$\begin{aligned} & \text { Hence } \frac{x+2 y}{x 2+y} \\ & =\frac{11}{10} \text { or } \frac{14}{13} \\ & \therefore \frac{x+2 y}{2 x+y} \\ & =11: 10 \text { or } 14: 13 .\end{aligned}$
View full question & answer→Question 684 Marks
If a : b = 3 : 11, find (15a – 3b) : (9a + 5b). a
Answer$
a: b=3: 11 \text { or } \frac{a}{b}=\frac{3}{11}$
Now $\frac{15 a-3 b}{9 a+5 b}$
$=\frac{\frac{15 a}{b}-\frac{3 b}{11}}{\frac{9 a}{b}+\frac{5 b}{b}} \ldots($ Dividing by b)
$
=\frac{\frac{15 a}{b}-3}{9 \frac{a}{b}+5}
$
$=\frac{15 \times \frac{3}{110-3}}{9 \times \frac{3}{11}+5} \ldots\left(\right.$ Substituting the value of $\left.\frac{a}{b}\right)$
$
=\frac{\frac{45}{11}-3}{\frac{27}{11}+5}
$
$
=\frac{\frac{45-33}{11}}{\frac{27+55}{11}}
$
$\begin{aligned} & =\frac{\frac{12}{11}}{\frac{82}{11}} \\ & =\frac{12}{11} \times \frac{11}{82} \\ & =\frac{12}{82} \\ & =\frac{6}{41} \\ & \therefore(15 a-3 b):(9 a+5 b)=6: 41 .\end{aligned}$
View full question & answer→Question 694 Marks
An alloy consists of $27 \frac{1}{2} kg$ of copper and $2 \frac{3}{4} kg$ of tin. Find the ratio by weight of tin to the alloy
Answer
$
\begin{aligned}
& \text { Copper }=27 \frac{1}{2} kg =\frac{55}{4} kg , \\
& \text { Tin }=2 \frac{3}{4} kg =\frac{11}{4} kg
\end{aligned}$
Total alloy
$
\begin{aligned}
& =\frac{55}{2}+\frac{11}{4} \\
& =\frac{110+11}{4} \\
& =\frac{121}{4} kg
\end{aligned}$
Now Ratio between tin and alloy
$
\begin{aligned}
& =\frac{11}{4} kg : \frac{121}{4} kg \\
& =11: 121 \\
& =1: 11 .
\end{aligned}
$
View full question & answer→Question 704 Marks
If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.
Answer$2,6, p, 54$ and $q$ are in continued proportional then
$
\Rightarrow \frac{2}{6}=\frac{6}{p}=\frac{p}{54}=\frac{54}{q}
$
(i) $\because \frac{2}{6}=\frac{6}{p}$ then $2 p =36$
$
\Rightarrow p =18
$
(ii) $\frac{p}{54}=\frac{54}{q}$
$
\begin{aligned}
& \Rightarrow pq =54 \times 54 \\
& \Rightarrow 18 q =54 \times 54 \\
& \Rightarrow q =\frac{54 \times 54}{18} \\
& =162
\end{aligned}$
Hence $p=18, q=162$.
View full question & answer→Question 714 Marks
If (a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion, prove that b is the mean proportional between a and c.
Answer$(a+2 b+c),(a-c)$ and $(a-2 b+c)$ are in continued proportion
$
\begin{aligned}
& \Rightarrow \frac{a+2 b+c}{a-c}=\frac{a-c}{a-2 b+c} \\
& \therefore \frac{a+2 b+c}{a-c}=\frac{a-c}{a-2 b+c}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow(a+2 b+c)(a-2 b+c)=(a-c)^2 \\
& \Rightarrow a^2-2 a b+a c+2 a b-4 b^2+2 b c+a c-2 b c+c^2=a^2- \\
& 2 a c+c^2 \\
& \Rightarrow a^2-2 a b+a c+2 a b-4 b^2+2 b c+a c-2 b c+c^2-a^2+ \\
& 2 a c-c^2=0 \\
& \Rightarrow 4 a c-4 b^2=0 \\
& \Rightarrow a c-b^2=0 \\
& \Rightarrow b^2=a c
\end{aligned}
$Hence $b$ is the mean proportional between $a$ and $c$.
View full question & answer→Question 724 Marks
What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional?
AnswerLet $x$ be added to each number, then numbers will be $15+x_l 17+x_l 34+x_t$ and $38+x$.
Now according to the condition
$
\begin{aligned}
& \frac{15+x}{17+x}=\frac{34+x}{38+x} \\
& \Rightarrow(15- x )(38+ x )=(34+ x )(17+ x ) \\
& \Rightarrow 570+53 x + x ^2=578+51 x + x ^2 \\
& \Rightarrow x ^2+53 x - x ^2-51 x =578-570 \\
& \Rightarrow 2 x =8 \\
& \Rightarrow x =4
\end{aligned}
$
$\therefore 4$ is to be added.
View full question & answer→Question 734 Marks
In an examination, the number of those who passed and the number of those who failed were in the ratio of 3 : 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2 : 1. Find the number of candidates who appeared.
Answer
$\begin{aligned} & \text { Let number of passed }=3 x \\ & \text { and failed }= x \\ & \text { Total candidates appeared }=3 x + x =4 x \text {. } \\ & \text { In second case } \\ & \text { No. of candidates appeared }=4 x +8 \\ & \text { and No. of passed }=3 x -6 \\ & \text { and failed }=4 x +8-3 x +6= x +14 \\ & \text { then ratio will be }=2: 1 \\ & \text { Now according to the condition } \\ & \frac{3 x-6}{x+14}=\frac{2}{1} \\ & \Rightarrow 3 x-6=2 x +28 \\ & \Rightarrow 3 x -2 x =28+6 \\ & \Rightarrow x =34 \\ & \therefore \text { No. of candidates appeared } \\ & =4 x \\ & =4 x 34 \\ & =136 .\end{aligned}$
View full question & answer→Question 744 Marks
The ratio of the shorter sides of a right angled triangle is 5 : 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.
AnswerLet the two shorter sides of a right-angled triangle be $5 x$ and $12 x$.
Third (longest side)
$
\begin{aligned}
& =\sqrt{(5 x)^2+(12 x)^2} \\
& =\sqrt{25 x^2+144 x^2} \\
& =\sqrt{169 x^2} \\
& =13 x
\end{aligned}$
But $5 x +12 x +13 x =360 cm$
$
\Rightarrow 30 x=360
$
$
\Rightarrow x =\frac{360}{30}=12
$
$\therefore$ Length of the longest side $=13 x$
$
\begin{aligned}
& =13 \times 12 cm \\
& =156 cm .
\end{aligned}
$
View full question & answer→Question 754 Marks
If $\frac{b y+c z}{b^2+c^2}=\frac{c z+a x}{c^2+a^2}=\frac{a x+b y}{a_z^2+b^2}$, prove that each of these ratio is equal to $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$
Answer
$
\begin{aligned}
& \frac{b y+c z}{b^2+c^2}=\frac{c z+a x}{c^2+a^2}=\frac{a x+b y}{a^2+b^2} \\
& =\frac{2(a x+b y+c z)}{2\left(a^2+b^2+c^2\right)} \\
& =\frac{a x+b y+c z}{a^2+b^2+c^2} \ldots \text { (Adding) }
\end{aligned}$
Now $\frac{b y+c z}{b^2+c^2}=\frac{a x+b y+c z}{a^2+b^2+c^2}$
$\Rightarrow \frac{b y+c z}{a x+b y+c z}=\frac{b^2+c^2}{a^2+b^2+c^2} \ldots$ (By alternendo)
$
\Rightarrow \frac{b y+c z-a x-b y-c z}{a x+b y+c z}
$
$
=\frac{b^2+c^2-a^2-b^2-c^2}{a^2+b^2+c^2}
$
$
\Rightarrow \frac{-a x}{a x+b y+c z}=\frac{-a}{a^2+b^2+c^2}
$
$
\Rightarrow \frac{x}{a x+b y+c z}=\frac{a}{a^2+b^2+c^2}
$
$
\Rightarrow \frac{x}{a}=\frac{a x+b y+c z}{a^2+b^2+c^2}$
Similarly we can prove that
$
\frac{y}{b}=\frac{ ax + by + cz }{a^2+b^2+c^2}
$and $\frac{z}{c}=\frac{ ax + by + cz }{a^2+b^2+c^2}$...(iii)
from (i), (ii) and (iii)
Hence $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
View full question & answer→Question 764 Marks
If $x =\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}}{\sqrt[3]{a+1}-\sqrt[3]{a-1}}$, prove that : $x^3-3 a x^2+3 x-a=0$
Answer$
x=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}}{\sqrt[3]{a+1}-\sqrt[3]{a-1}}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}+\sqrt[3]{a+1}-\sqrt[3]{a-1}}{\sqrt[3]{a+1}+\sqrt[3]{a-1}-\sqrt[3]{a+1}+\sqrt[3]{a-1}} \\
& \frac{x+1}{x-1}=\frac{2 \sqrt[3]{a+1}}{2 \sqrt[3]{a-1}} \\
& \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}}{\sqrt[3]{a-1}}
\end{aligned}
$Cubing both sides
$
\frac{(x+1)^3}{(x-1)^3}=\frac{a+1}{a-1}$
Again apply onendo and dividendo,
$
\begin{aligned}
& \frac{(x+1)^2+(x-1)^3}{(x+1)^3-(x-1)^3}=\frac{a+1+a-1}{a+1-a+1} \\
& \Rightarrow \frac{2\left(x^3+3 x\right)}{2\left(3 x^2+1\right)}=\frac{2 a}{2} \\
& \Rightarrow \frac{x^3+3 x}{3 x^2+1}=\frac{a}{1} \\
& \Rightarrow x^3+3 x=3 a x^2+a \\
& \Rightarrow x^3-3 a x^2+3 x-a=0
\end{aligned}$
Hence proved.
View full question & answer→Question 774 Marks
Find $x$ from the equation $\frac{a+x+\sqrt{a^2 x^2}}{a+x-\sqrt{a^2-x^2}}=\frac{b}{x}$
Answer$
\frac{a+x+\sqrt{a^2 x^2}}{a+x-\sqrt{a^2-x^2}}=\frac{b}{x}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{a+x+\sqrt{a^2-x^2}+a+x-\sqrt{a^2-x^2}}{a+x+\sqrt{a^2-x^2}-a-x+\sqrt{a^2-x^2}}=\frac{b+x}{b-x} \\
& \Rightarrow \frac{2(a+x)}{2 \sqrt{a^2-x^2}}=\frac{b+x}{b-x} \\
& \Rightarrow \frac{a+x}{\sqrt{a^2-x^2}}=\frac{b+x}{b-x}
\end{aligned}
$Squaring both sides,
$
\begin{aligned}
& \frac{(a+x)^2}{a^2-x^2}=\frac{(b+x)^2}{(b-x)^2} \\
& \Rightarrow \frac{(a+x)^2}{(a+x)(a-x)}=\frac{(b+x)^2}{(b-x)^2} \\
& \Rightarrow \frac{a+x}{a-x}=\frac{(b+x)^2}{(b-x)^2}
\end{aligned}$
Again applying componendo and dividendo,
$
\begin{aligned}
& \frac{a+x+a-x}{a+x-a+x} \\
& =\frac{(b+x)^2+(b-x)^2}{(b+x)^2-(b-x)^2}
\end{aligned}
$
$\begin{aligned} & \Rightarrow \frac{2 a}{2 x}=\frac{2\left(b^2+x^2\right)}{4 b x} \\ & \Rightarrow \frac{a}{x}=\frac{b^2+x^2}{2 b x} \\ & 2 a b x=x\left(b^2+x^2\right) \\ & \Rightarrow 2 a b=b^2+x^2 \\ & \Rightarrow x^2=2 a b-b^2 \\ & x=\sqrt{2 a b-b^2} .\end{aligned}$
View full question & answer→Question 784 Marks
If $x =\frac{ pab }{a+b}$, provee that $\frac{x+p a}{x-p a}-\frac{x+p b}{x-p b}=\frac{2\left(a^2-b^2\right)}{a b}$
Answer
$
\begin{aligned}
& x=\frac{p a b}{a+b} \\
& \Rightarrow \frac{x}{p a}+\frac{b}{a+b}
\end{aligned}$
Applying componendo and dividendo
$
\begin{aligned}
& \frac{x+p a}{x-p a} \\
& =\frac{b+a+b}{b-a-b} \\
& =\frac{a+2 b}{-a} \ldots .(i)
\end{aligned}
$Again, $\frac{x}{p b}$
$
=\frac{a}{a+b}$
Applying componendo and dividendo,
$
\frac{x+p b}{x-p b}
$
$\begin{aligned} & =\frac{a+a+b}{a-a-b} \\ & =\frac{2 a+b}{-b} \\ & \text { L.H.S. }=\frac{x+p a}{x-p a}-\frac{x+p b}{x-p b} \\ & =\frac{a+2 b}{-a}-\frac{2 a+b}{-b} \\ & =\frac{a+2 b}{-a}+\frac{2 a+b}{b} \\ & =\frac{a b+2 b^2 2 a^2-a b}{-a b} \\ & =\frac{2 b^2-2 a^2}{-a b} \\ & =\frac{-2 a^2+2 b^2}{-a b} \\ & =\frac{-2\left(a^2-b^2\right)}{-a b} \\ & =\frac{2\left(a^2-b^2\right)}{a b} \\ & =\text { R.H.S. } \\ & \end{aligned}$
View full question & answer→Question 794 Marks
If $x =\frac{2 m a b}{a+b}$, find the value of $\frac{x+m a}{x-m a}+\frac{x+m b}{x-m b}$
Answer
$
\begin{aligned}
& x=\frac{2 m a b}{a+b} \\
& \Rightarrow \frac{x}{m a}+\frac{2 b}{a+b}
\end{aligned}$
Applying componendo and dividendo
$
\begin{aligned}
& \frac{x+m a}{x-m a} \\
& =\frac{2 b+a+b}{2 b-a-b} \\
& =\frac{3 b+a}{b-a}
\end{aligned}
$Again $\frac{x}{m b}$
$
=\frac{2 a}{a+b}$
Applying componendo and dividendo,
$
\frac{x+m b}{x-m b}$
$\begin{aligned}
& =\frac{2 a+a+b}{2 a-a-b} \\
& =\frac{3 a+b}{a-b} \ldots \text { (ii) }
\end{aligned}$
Adding (i) and (ii)
$
\begin{aligned}
& \frac{x+m a}{x-m a}+\frac{x+m b}{x-m b} \\
& =\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b} \\
& =-\frac{3 b+a}{a-b}+\frac{3 a+b}{a-b} \\
& =\frac{-3 b-a+3 a+b}{a-b} \\
& =\frac{2 a-2 b}{a-b} \\
& =\frac{2(a-b)}{a-b} \\
& =2
\end{aligned}
$
View full question & answer→Question 804 Marks
If $\left(3 x^2+2 y^2\right):\left(3 x^2-2 y^2\right)=11: 9$, find the value of $\frac{3 x^4+5 y^4}{3 x^4-5 y^4}$
Answer$
\frac{3 x^4+5 y^4}{3 x^4-5 y^4}=\frac{11}{9}$
Applying componendo and dividendo
$
\begin{aligned}
& \frac{3 x^2+2 y^2+3 x^2-2 y^2}{3 x^2+2 y^2-3 x^2+2 y^2}=\frac{11+9}{11-9} \\
& \Rightarrow \frac{6 x^2}{4 y^2}=\frac{20}{2} \\
& \Rightarrow \frac{3 x^2}{2 y^2}= \\
& \Rightarrow \frac{x^2}{y^2}=10 \times \frac{2}{3} \\
& =\frac{20}{3} \\
& \frac{3 x^4+5 y^4}{3 x^4-5 y^4}
\end{aligned}
$
$\begin{aligned} & =\frac{\frac{3 x^4}{y^4}+\frac{25 y^4}{y^4}}{\frac{3 x^4}{y^4}-\frac{25 y^4}{y^4}} \\ & =\frac{3\left(\frac{x^2}{y^2}\right)^2+25}{3\left(\frac{x^2}{y^2}\right)^2-25} \\ & =\frac{3 \times\left(\frac{2}{3}\right)^2+5}{3\left(\frac{20}{3}\right)^2-25} \\ & =\frac{3 \times \frac{400}{9}+25}{3 \times \frac{400}{9}-25} \\ & =\frac{\frac{400}{3}+\frac{25}{1}}{\frac{400}{3}-\frac{25}{1}} \\ & =\frac{\frac{400+75}{3}}{\frac{400-75}{3}} \\ & =\frac{475}{3} \times \frac{3}{325}\end{aligned}$
$=\frac{19}{13}$.
View full question & answer→Question 814 Marks
If $a: b=9: 10$, find the value of $\frac{2 a^2-3 b^2}{2 a^2+3 b^2}$
Answer
$\begin{aligned} & a : b =9: 10 \\ & \Rightarrow \frac{a}{b}=\frac{9}{10} \\ & \frac{2 a^2-3 b^2}{2 a^2+3 b^2} \\ & =\frac{\frac{2 a^2}{b^2}-\frac{3 b^2}{b^2}}{\frac{2 a^2}{b^2}-\frac{3 b^2}{b^2}} \ldots\left(\text { Dividing by } b ^2\right) \\ & =\frac{2\left(\frac{a}{b}\right)^2-3}{2\left(\frac{a}{b}\right)^2+3} \\ & =\frac{2\left(\frac{9}{10}\right)^2-3}{2\left(\frac{9}{10}\right)^2+3} \\ & =\frac{2 \times \frac{81}{100}-3}{2 \times \frac{81}{100}+3} \\ & =\frac{\frac{81}{50}-3}{\frac{81}{50}+3}\end{aligned}$
$\begin{aligned} & =\frac{\frac{81-150}{50}}{\frac{81+150}{50}} \\ & =\frac{-69}{50} \times \frac{50}{231} \\ & =\frac{-69}{231} \\ & =\frac{-23}{77} .\end{aligned}$
View full question & answer→Question 824 Marks
If $a: b=9: 10$, find the value of $\frac{5 a+3 b}{5 a-3 b}$
Answer
$
\begin{aligned}
& a: b=9: 10 \\
& \Rightarrow \frac{a}{b}=\frac{9}{10}
\end{aligned}
$
$
\begin{aligned}
& \frac{5 a+3 b}{5 a-3 b} \\
& =\frac{\frac{5 a}{b}+\frac{3 b}{b}}{\frac{5 a}{b}-\frac{3 b}{b}}
\end{aligned}
$
$=\frac{\frac{5 a}{b}+3}{\frac{5 a}{b}-3} \ldots($ Dividing by b)
$
=\frac{5 \times \frac{9}{10}+3}{5 \times \frac{9}{10}-3}
$
$
=\frac{\frac{9}{2}+3}{\frac{9}{2}-3}
$
$=\frac{\frac{15}{2}}{\frac{3}{2}} \ldots\left(\right.$ Substituting the value of $\left.\frac{a}{b}\right)$
$
\begin{aligned}
& =\frac{15}{2} \times \frac{2}{3} \\
& =5
\end{aligned}
$
View full question & answer→Question 834 Marks
If $x: a=y: b$, prove that
$
\frac{x^4+a^4}{x^3+a^3}+\frac{y^4+b^4}{y^3+b^3}=\frac{(x+y)^4+(a+b)^4}{(x+y)^3+(a+b)^3}
$
Answer
$\begin{aligned} & \frac{x}{a}=\frac{y}{b}= k \text { (say) } \\ & x = ak , y = bk \\ & \text { L.H.S. }=\frac{x^4+a^4}{x^3+a^3}+\frac{y^4+b^4}{y^3+b^3} \\ & =\frac{a^4 k^4+a^4}{a^3 k^3+a^3}+\frac{b^4 k^4+b^4}{b^3 k^3+b^3} \\ & =\frac{a^4\left(k^4+1\right)}{a^3\left(k^3+1\right)}+\frac{b^4\left(k^4+1\right)}{b^3\left(k^3+1\right)} \\ & =\frac{a\left(k^4+1\right)}{k^3+1}+\frac{b\left(k^4+1\right)}{k^3+1} \\ & =\frac{a\left(k^4+1\right)+b\left(k^4+1\right)}{k^3+1} \\ & =\frac{\left(k^4+1\right)(a+b)}{k^3+1} \\ & \text { R.H.S. }=\frac{(x+y)^4+(a+b)^4}{(x+y)^3+(a+b)^3}\end{aligned}$$
\begin{aligned}
& =\frac{(a k+b k)^4+(a+b)^4}{(a k+b k)^3+(a+b)^3} \\
& =\frac{k^4(a+b)^4+(a-b)^4}{k^3(a+b)^3(a+b)^3} \\
& =\frac{(a+b)^4\left(k^4+1\right)}{(a+b)^3\left(k^3+1\right)} \\
& =\frac{(a+b)\left(k^4+1\right)}{k^3+1} \\
& =\frac{\left(k^4+1\right)(a+b)}{k^3+1} \\
& \therefore \text { L.H.S. = R.H.S. }
\end{aligned}$
Hence proved.
View full question & answer→Question 844 Marks
If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$, prove that
$
\frac{3 x^3-5 y^3+4 c^3}{3 a^3-5 b^3+4 c^3}=\left(\frac{3 x-5 y+4 c}{3 a-5 b+a c}\right)^3
$
Answer
$\begin{aligned} & \frac{x}{a}=\frac{y}{b}=\frac{z}{c}= k (\text { say) } \\ & x = ak , y = bk , z = ck \\ & \text { L.H.S. }=\frac{3 x^3 5 y^3+4 z^3}{3 a^3 5 b^3+4 c^3} \\ & =\frac{3 a^3 k^3-5 b^3 k^3+4 c^3 k^3}{3 a^3-5 b^3+4 a c^3} \\ & =\frac{k^3\left(3 a^3-5 b^3+4 c^3\right)}{3 a^3-5 b^3+4 c^3} \\ & = k ^3 \\ & \text { R.H.S. }=\left(\frac{3 x-5 y+4 z}{3 a-5 b+4 c}\right)^3 \\ & =\left(\frac{3 a k-5 b k+4 c k}{3 a-5 b+a c}\right)^3 \\ & =\left(\frac{k(3 a-5 b+4 c)}{3 a-5 b+4 c}\right)^3 \\ & =( k )^3 \\ & = k ^3 \\ & \therefore \text { L.H.S. = R.H.S. }\end{aligned}$
View full question & answer→Question 854 Marks
If $q$ is the mean proportional between $p$ and $r$, prove that: $p^3-$ $3 q^2+r^2=q^4\left(\frac{1}{p^2}-\frac{3}{q^2}+\frac{1}{r^2}\right)$
Answer$q$ is mean proportional between $p$ and $r$
$
\begin{aligned}
& q^2=p r \\
& \text { L.H.S. }=p^2-{ }^{3 q}{ }^2+r^2=p^2-3 p r+r^2
\end{aligned}
$
$
\begin{aligned}
& \text { R.H.S. }=q^4\left(\frac{1}{p^2}-\frac{3}{q^2}+\frac{1}{r^2}\right) \\
& =\left(q^2\right)^2\left(\frac{1}{p^2}-\frac{3}{q^2}+\frac{1}{r^2}\right) \\
& =(p r)^2\left(\frac{1}{p^2}-\frac{3}{q^2}+\frac{1}{r^2}\right) \\
& =(p r)^2\left(\frac{1}{p^2}-\frac{3}{ pr }+\frac{1}{r^2}\right) \\
& =p^2 r^2\left(\frac{r^2-3 p r+p^2}{p^2 r^2}\right) \\
& =r^2-3 p r+p^2 \\
& \therefore \text { L.H.S. = R.H.S. }
\end{aligned}$
Hence proved.
View full question & answer→Question 864 Marks
Find two numbers whose mean proportional is 16 and the third proportional is 128.
AnswerLet $x$ and $y$ be two numbers
Their mean proportion $=16$
and third proportion $=128$
$
\begin{aligned}
& \therefore \sqrt{x y}=16 \\
& \Rightarrow xy =256 \\
& \Rightarrow x =\frac{256}{y}
\end{aligned}
$
$
\text { and } \frac{y^2}{x}=128
$
$
\Rightarrow x =\frac{y^2}{128}
$From (i) and (ii)
$
\begin{aligned}
& \frac{256}{y}=\frac{y^2}{128} \\
& \Rightarrow y ^3=256 \times 128 \\
&= 32768 \\
& \Rightarrow y ^3=(32)^3 \\
& \Rightarrow y =32 \\
& \therefore x =\frac{256}{y} \\
& \frac{256}{32} \\
&= 8
\end{aligned}
$
$\therefore$ Numbers are 8,32 .
View full question & answer→Question 874 Marks
If $a, b, c, d, e$ are in continued proportion, prove that: $a : e = a^4 : b^4.$
Answer$a, b, c, d, e$ are in continued proportion
$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{c}= k \text { (say) }$
$ d = ek _{ c } \ c = ek ^2, b = ek ^3 \text { and } a = ek ^4$
Now $ \text{L.H.S.} =\frac{a}{e}$
$=\frac{e k^4}{e}$
$ =k^4$
$ \text{R.H.S.}$ $\frac{a^4}{b^4}$
$=\frac{\left(e k^4\right)^4}{\left(e k^3\right)^4}$
$ =\frac{e^4 k^{16}}{e^4 k^{12}}$
$= k ^{16-12}$
$ = k ^4$
$ \therefore \text { L.H.S. = R.H.S. }$
View full question & answer→Question 884 Marks
Find the compound ratio of:
$(a + b)^2 : (a – b )^2 ,$
$(a^2 – b^2) : (a^2 + b^2),$
$(a^4 – b^4) : (a + b)^4$
Answer$ (a+b)^2:(a-b)^2,$
$ \left(a^2-b^2\right):\left(a 2+b^2\right),$
$ \left(a^4-b^4\right):(a+b)^4$
$ =\frac{(a+b)^2}{(a-b)^2} \times \frac{a^2-b^2}{a^2+b^2} \times \frac{a^4-b^4}{(a+b)^4}$
$ = \frac{(a+b)^2}{(a-b)^2} \times \frac{(a+b)(a-b)}{a^2+b^2} \times \frac{\left(a^2+b^2\right)(a+b)(a-b)}{(a+b)^4}$
$ =\frac{1}{1}$
$ =1: 1 .$
View full question & answer→