Question 13 Marks
If $(3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y),$ find $x: y.$
Answer$(3x - 4y) : (2x - 3y) = (5x - 6y) : (4x - 5y)$
$\frac{3 x-4 y}{2 x-3 y}=\frac{5 x-6 y}{4 x-5 y}$
Applying componendo and dividendo
$\frac{3 x-4 y+2 x-3 y}{3 x-4 y-2 x+3 y}=\frac{5 x-6 y+4 x-5 y}{5 x-6 y-4 x+5 y} $
$ \frac{5 x-7 y}{x-y 0}=\frac{9 x-11 y}{x-y} $
$ 5 x-7 y=9 x-11 y $
$ 11 y-7 y=9 x-5 x $
$ 4 y=4 x $
$ \frac{x}{y}=\frac{1}{1}$
$x : y = 1 : 1$
View full question & answer→Question 23 Marks
If $\frac{x^2+y^2}{x^2-y^2}=2 \frac{1}{8}$ find $\frac{x^3+y^3}{x^3-y^3}$
Answer$\frac{x^3+y^3}{x^3-y^3}$
$=\frac{\left(\frac{x}{y}\right)^3+1}{\left(\frac{x}{y}\right)^3-1} $
$ =\frac{\left(\frac{5}{3}\right)^2+1}{\left(\frac{5}{3}\right)^3-1}$
$ =\frac{\frac{125}{27}+1}{\frac{125}{27}-1}$
$ =\frac{\frac{125+27}{27}}{\frac{125-27}{27}} $
$ =\frac{125+27}{125-27} $
$=\frac{76}{49}=1 \frac{27}{49}$
View full question & answer→Question 33 Marks
If x, y, z are in continued proportion prove that $\frac{(x+y)^2}{(y+z)^2}=\frac{x}{z}$
Answer∵ x, y, z are in continued proportion
$\therefore \frac{x}{y}=\frac{y}{z}$
$\Rightarrow y^2=z x$ ......(1)
$\Rightarrow \frac{x+y}{y}=\frac{y+z}{z} \quad \ldots($ By componendo $)$
$\Rightarrow \frac{x+y}{y+z}=\frac{y}{z} \quad \ldots$ (By alternendo)
$\Rightarrow \frac{(x+y)^2}{(y+z)^2}=\frac{y^2}{z^2} \quad \ldots$ (squaring both sides)
$\Rightarrow \frac{(x+y)^2}{(y+z)^2}=\frac{z x}{z^2} \quad \ldots[$ from (1)]
$\Rightarrow \frac{(x+y)^2}{(y+z)^2}=\frac{x}{z}$
Hence Proved.
View full question & answer→Question 43 Marks
If $\frac{4 m+3 n}{4 m-3 n}=\frac{7}{4}$ use properties of proportion to find $\frac{2 m^2-11 n^2}{2 m^2+11 n^2}$
Answer$\frac{m}{n}=\frac{11}{4}$
$ \frac{m^2}{n^2}=\frac{121}{16}$
$\frac{2 m^2}{11 n^2}=\frac{2 \times 121}{11 \times 16} \quad$ (Multiplying both sides by $2/11$)
$\frac{2 m^2}{11 n^2}=\frac{11}{8}$
$\frac{2 m^2+11 n^2}{2 m^2-11 n^2}=\frac{11+8}{11-8}$ (Applying componendo and divdendo)
$\frac{2 m^2+11 n^2}{2 m^2-11 n^2}=\frac{19}{3}$
$\frac{2 m^2-11 n^2}{2 m^2+11 n^2}=\frac{3}{19} \quad$ (Applying invertendo)
View full question & answer→Question 53 Marks
If 7x – 15y = 4x + y, find the value of x: y. Hence, use componendo and dividend to find the values of:
$\frac{3 x^2+2 y^2}{3 x^2-2 y^2}$
Answer7x - 15y = 4x + y
7x - 4x = y + 15y
3x = 16y
$\frac{x}{y}=\frac{16}{3}$
$\Rightarrow \frac{x^2}{y^2}=\frac{256}{9}$
$\Rightarrow \frac{3 x^2}{2 y^2}=\frac{768}{18}=\frac{128}{3} \quad$ (Multiplying both sides by $3 / 2$ )
$\Rightarrow \frac{3 x^2+2 y^2}{3 x^2-2 y^2}=\frac{128+3}{128-3} \quad$ (Applying componendo and dividendo)
$\Rightarrow \frac{3 x^2+2 y^2}{3 x^2-2 y^2}=\frac{131}{125}$
View full question & answer→Question 63 Marks
There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3: 1. How any more girls should be added to the council so that the ratio of the number of boys to the number of girls maybe 9: 5?
AnswerA ratio of number of boys to the number of girls = 3: 1
Let the number of boys be 3x and the number of girls be x.
3x + x = 36
4x = 36
x = 9
∴ Number of boys = 27
Number of girls = 9
Le n number of girls be added to the council.
From given information, we have:
$\frac{27}{9+n}=\frac{9}{5}$
$135=81+9 n$
9n = 54
n = 6
Thus 6 girl are added to the council
View full question & answer→Question 73 Marks
If $\frac{a}{b}=\frac{c}{d}$ show that $(a+b):(c+d)=\sqrt{a^2+b^2}: \sqrt{c^2+d^2}$
AnswerLet $\frac{a}{b}=\frac{c}{d}=k$ (say)
$`\Rightarrow a = bk , c = dk$
$\text { L.H.S }=\frac{a+b}{c+d}$
$=\frac{b k+b}{d k+d} $
$ =\frac{b(k+1)}{d(k+1)} $
$=\frac{b}{d}$
$\text { R.H.S }=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} $
$ =\frac{\sqrt{(b k)^2+b^2}}{\sqrt{(d k)^2+d^2}} $
$=\frac{\sqrt{b^2\left(k^2+1\right)}}{\sqrt{d^2\left(k^2+1\right)}}$
$ =\frac{\sqrt{b^2}}{\sqrt{d^2}} $
$=\text { b/d } $
$ \therefore \text { L.H.S }=\text { R.H.S }$
View full question & answer→Question 83 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.
Answer$\text { Given } \frac{x}{y}=\frac{(x+z)^2}{(y+z)^2} $
$ x\left(y^2+z^2+2 y z\right)=y\left(x^2+z^2+2 x y\right) $
$y^2+x z^2+2 x y z=x^y+y z^2+2 x y z $
$ x y^2+x z^2=y z^2-x z^2$
$ x y(y-x)=z^2(y-x) $
$x y=z^2$
Hence z is mean proportional between x and y
View full question & answer→Question 93 Marks
Find two numbers such that the mean proportional between them is $14$ and third proportional to them is $112.$
AnswerLet the required number be a and b.
Given, $14$ is the mean proportional between a and b
$\Rightarrow a: 14=14: b$
$\Rightarrow ab =196 $
$ \Rightarrow a =\frac{196}{ b }......(1)$
Also given third proportional to a and b is $112$
$\Rightarrow a: b=b: 112 $
$ \Rightarrow b^2=112 a \ldots .(2)$
Using $(1)$ we have
$b^2=112 \times \frac{196}{b} $
$ b^3=(14)^3(2)^3$
$ b=28$
From $(1)$
$a =\frac{196}{28}=7$
Thus the two numbers are $7$ and $28.$
View full question & answer→Question 103 Marks
if $\left(a^2+b^2\right)\left(x^2+y^2\right)= (ax + by)^2 ;$ prove that $\frac{a}{x}=\frac{b}{y}$
AnswerGiven $\left(a^2+b^2\right)\left(x^2+y^2\right)=(a x+b y)^2$
$a^2 x^2+a^2 y^2+b^2 x^2+b^2 y^2+b^2 y^2+2 a b x y $
$a^2 y^2+b^2 x^2-2 a b x y=0 \\ (a y-b x)^2=0 $
$ \text { ay - bx }=0 $
$ \text { ay }=\text { bx } $
$ \frac{a}{x}=\frac{b}{y}$
View full question & answer→Question 113 Marks
If $(a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d),$ prove that $a: b = c: d.$
Show, that $a, b, c, d$ are in proportion if:
$(a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d)$
AnswerGiven $\frac{a+b+c+d}{a-b-c+d}=\frac{a+b-c-d}{a-b+c-d}$
Applying componendo and dividendo
$\frac{(a+b+c+d)+(a+b-c-d)}{(a+b+c+d)-(a+b-c-d)}=\frac{(a-b+c-d)+(a-b-c+d)}{(a-b+c-d)-(a-b-c+d)}$
$\frac{2(a+b)}{2(c+d)}=\frac{2(a-b)}{2(c-d)}$
$\frac{a+b}{c+d}=\frac{a-b}{c-d}$
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$
Applying componendo and dividendo
$\frac{a+b+a-b}{a+b-a-b}=\frac{c+d+c-d}{c+d-c+d}$
$\frac{2 a}{2 b}=\frac{2 c}{2 d}$
$\frac{a}{b}=\frac{c}{d}$
View full question & answer→Question 123 Marks
If a : b = c : d, prove that: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).
AnswerGiven $\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{6 a}{7 b}=\frac{6 c}{7 d} \quad$ (Multiplying each side by $6 / 7$ )
$\Rightarrow \frac{6 a+7 b}{7 b}=\frac{6 c+7 d}{7 d}$ (By componendo)
$\Rightarrow \frac{6 a+7 b}{6 c+7 d}=\frac{7 b}{7 d}=\frac{b}{d} \ldots . .(1)$
Also $\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{3 a}{4 b}=\frac{3 c}{4 d} \quad$ (Mutipling each side by $3 / 4$ )
$\Rightarrow \frac{3 a-4 b}{4 b}=\frac{3 c-4 d}{4 d} \quad$ (By dividendo)
$\Rightarrow \frac{3 a-4 b}{3 c-4 d}=\frac{4 b}{4 d}=\frac{b}{d}$ ...... (2)
From (1) and (2)
$\frac{6 a+7 b}{6 c+7 d}=\frac{3 a-4 b}{3 c-4 c}$
(6a+ 7b)(3c - 4d) = (6c + 7d)(3a - 4b)
View full question & answer→Question 133 Marks
If $\frac{x^2+3 x y^2}{3 x^2 y+y^3}=\frac{m^2+3 m n^2}{3 m^2 n+n^3}$ show that nx = my
Answer$\frac{x^2+3 x y^2}{3 x^2 y+y^3}=\frac{m^2+3 m n^2}{3 m^2 n+n^3}$
Applying componendo and dividendo
$\frac{x^3+3 x y^2+3 x^2 y+y^3}{x^3+3 x y^2-3 x^2 y-y^3}=\frac{m^3+3 m n^2+3 m^2 n+n^3}{m^3+3 m n^2-3 m^2 n-n^3} $
$ \frac{(x+y)^3}{(x-y)^3}=\frac{(m+n)^3}{(m-n)^3}$
$ \frac{x+y}{x-y}=\frac{m+n}{m-n}$
Applying componendo and dividendo
$\frac{x+y+x-y}{x+y-x+y}=\frac{m+n+m-n}{m+n-m+n} $
$ \frac{2 x}{2 y}=\frac{2 m}{2 n} $
$\frac{x}{y}=\frac{m}{n} \\ nx = my$
View full question & answer→Question 143 Marks
If x$=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}$ prove that $3 b x^2-2 a x+3 b=0$
AnswerSince $\frac{x}{1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}$
Applying componendo and dividendo we get
$\frac{x+1}{x-1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}+\sqrt{a+3 b}-\sqrt{a}-3 b}{\sqrt{a+3 b}+\sqrt{a-3 b}-\sqrt{a+3 b}+\sqrt{a-3 b}}$
$\frac{x+1}{x-1}=\frac{2 \sqrt{a+3 b}}{2 \sqrt{a-3 b}}$
Squaring both sides.
Again applying componendo and dividendo
$\frac{x^2+2 x+1+x^2-2 x+1}{x^2+2 x+1-x^2+2 x-1}=\frac{a+3 b+a-3 b}{a+3 b-a+3 b}$
$\frac{2\left(x^2+1\right)}{2(2 x)}=\frac{2(a)}{2(3 b)}$
$3 b\left(x^2+1\right)=2 a x$
$3 b x^2-2 a x+3 b=0$
View full question & answer→Question 153 Marks
If a, b, c are in continued proportion, prove that
$\frac{a^2+b^2+c^2}{(a+b+c)^2}=\frac{a-b+c}{a+b+c}$
AnswerGiven, a, b and c are in continued proportion.
Therefore,
$\frac{a}{b}$=$\frac{b}{c}$
$a c=b^2$
LHS $=\frac{a^2+b^2+c^2}{(a+b+c)^2}$
$=\frac{a^2+b^2+c^2+2 a b+2 b c+2 a c-(2 a b+2 b c+2 a c)}{(a+b+c)^2} $
$=\frac{(a+b+c)^2-2\left(a b+b c+b^2\right)}{(a+b+c)^2}$
$ =\frac{(a+b+c)^2-2 b(a+c+b)}{(a+b+c)^2} $
$=\frac{(a+b+c)(a+b+c-2 b)}{(a+b+c)^2} $
$ =\frac{a-b+c}{a+b+c}$
Hence proved.
View full question & answer→Question 163 Marks
What least number be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?
AnswerLet the number subtracted be x
∴ (7 - x) : (17 - x) :: (17 - x)(47 - x)
$\frac{7-x}{47-x}=(17-x)^2$
$329-47 x-7 x+x^2=289-34 x=x^2$
$329-289=-34 x+54 x$
20x = 40
x = 2
Thus the required number which should be subtracted is 2
View full question & answer→Question 173 Marks
If a,b,c are in continued proportion, show that: $\frac{a^2+b^2}{b(a+c)}=\frac{b(a+c)}{b^2+c^2}$
AnswerSine a, b, c are in continued proportion,
$\frac{ a }{ b }=\frac{ b }{ c }$
$\Rightarrow b^2=a c$
Now, $\left(a^2+b^2\right)\left(b^2+c^2\right)=\left(a^2+a c\right)\left(a c+c^2\right)$
$=a(a+c) c(a+c)$
$=a c(a+c)^2 $
$ =b^2(a+c)^2$
$ \Rightarrow\left(a^2+b^2\right)\left(b^2+c^2\right)=[b(a+c)][b(a+c)]$
$\Rightarrow \frac{ a ^2+ b ^2}{ b ( a + c )}=\frac{ b ( a + c )}{ b ^2+ c ^2}$
View full question & answer→Question 183 Marks
What least number must be added to each of the numbers $6, 15, 20$ and $43$ to make them proportional?
AnswerLet the number added be x.
$\therefore (6 + x) : (15 + x) :: (20 + x) (43 + x)$
$\Rightarrow \frac{6+x}{15+x}=\frac{20+x}{43+x}$
$\Rightarrow (6 + x)(43 + x) (20 + x)(15 + x)$
$\Rightarrow 258+6 x+43 x+x^2=300+20 x+15 x+x^2 $
$ \Rightarrow 49 x-35 x=300-258$
$ \Rightarrow 14 x=42 $
$ \Rightarrow x=3$
Thus, the required number which should be added is $3.$
View full question & answer→Question 193 Marks
If $x+5$ is the mean proportional between $x+2$ and $x+9$; find the value of $x$.
AnswerGiven, $x+5$ is the mean proportional between $x+2$ and $x+9$.
$\Rightarrow(x+2),(x+5)$ and $(x+9)$ are in continued proportion.
$\Rightarrow(x+2):(x+5)=(x+5):(x+9)$
$\Rightarrow(x+5)^2=(x+2)(x+9)$
$\Rightarrow x^2+25+10 x=x^2+2 x+9 x+18$
$\Rightarrow 25-18=11 x-10 x$
$\Rightarrow x=7$
View full question & answer→Question 203 Marks
Find the third proportional to $a - b$ and $a^2-b^2$
AnswerLet the third proportional to $a - b$ and $a^2-b^2$ be $x$
$\Rightarrow a - b , a ^2- b ^2, x$ are in continued proportion.
$\Rightarrow a-b: a^2-b^2=a^2-b^2: x$
$\Rightarrow \frac{a-b}{a^2-b^2}=\frac{a^2-b^2}{x} $
$ \Rightarrow x=\frac{\left(a^2-b^2\right)^2}{a-b} $
$ \Rightarrow x=\frac{(a+b)(a-b)\left(a^2-b^2\right)}{a-b} $
$ \Rightarrow x=(a+b)\left(a^2-b^2\right)$
View full question & answer→Question 213 Marks
Find two numbers such that the mean proportional between them is $12$ and the third proportional to them is $96.$
AnswerLet a and b be the two numbers whose mean proportional is $12$
$\therefore a b=12^2 \Rightarrow a b=144 \Rightarrow \frac{144}{a} ...... (1)$
Now third proportional is $96$
$\therefore a : b : : b : 96$
$\Rightarrow b^2=96 a $
$ \Rightarrow\left(\frac{144}{a}\right)^2=96 a$
$\Rightarrow a^3=\frac{144 \times 144}{96}$
$ \Rightarrow a^3=216 $
$ \Rightarrow a=6$
$b=\frac{144}{6}=24$
Therefore the number are $6$ and $24.$
View full question & answer→Question 223 Marks
Given four quantities a, b, c and d are in proportion. Show that
$(a-c) b^2:(b-d) c d=\left(a^2-b^2-a b\right):\left(c^2-d^2-c d\right)$
AnswerLet $\frac{a}{b}=\frac{c}{d}=k$
$\Rightarrow a=b k$ and $c=d k$
$L . H . S=\frac{(a-c) b^2}{(b-d) c d}$
$=\frac{(b k-d k) b^2}{(b-d) d^2 k} $
$ =\frac{b^2}{d^2}$
R.H.S $=\frac{a^2-b^2-a b}{c^2-d^2-c d}$
$=\frac{b^2 k^2-b^2-b k b}{d^2 k^2-d^2-d k d}$
$=\frac{b^2\left(k^2-1-k\right)}{d^2\left(k^2-1-k\right)}$
$=\frac{b^2}{d^2}$
$\Rightarrow L.H.S = R.H.S$
Hence proved
View full question & answer→Question 233 Marks
If y is the mean proportional between x and z. prove that
$\frac{x^2-y^2+z^2}{x^{-2}-y^{-2}+z^{-2}}=y^4$
AnswerGiven, y is the mean proportional between x and z.
$\Rightarrow y^2=x z$
$\text { LHS }=\frac{x^2-y^2+z^2}{x^{-2}-y^{-2}+z^{-2}} $
$=\frac{x^2-y^2+z^2}{\frac{1}{x^2}-\frac{1}{y^2}+\frac{1}{z^2}}$
$=\frac{x^2-x z+z^2}{\frac{1}{x^2}-\frac{1}{x z}+\frac{1}{z^2}} \quad\left(\because y^2=x y\right) $
$ =\frac{x^2-x z=z^2}{\frac{z^2-x z+x^2}{x^2 z^2}}$
$=x^2 z^2 $
$ =(x z)^2$
$=\left(y^2\right)^2 \quad\left(\therefore y^2=x z\right)$
$=y^4 $
$ =\text { R.H.S }$
View full question & answer→Question 243 Marks
Divide Rs 1,290 into A,B and C such that A is $\frac{2}{5}$ of $B$ and $B: C=4: 3$.
AnswerGiven, $B: C=4: 3 c \frac{B}{C}=\frac{4}{3}$
And, $A =\frac{2}{5} B \Rightarrow \frac{ A }{ B }=\frac{2}{5}$
Now $\frac{ A }{ B }=\frac{2}{5}=\frac{2 \times 4}{5 \times 4}=\frac{8}{20}$ and $\frac{ B }{ C }=\frac{4 \times 5}{3 \times 5}=\frac{20}{15}$
⇒ A : B : C = 8 : 20 : 15
⇒ A = 8x, B = 20x and C = 15x
8x + 20x + 15x = 1290
⇒ 43x = 1290
⇒ x = 30
A's share = 8x = 8 x 30 = Rs 240
B's share = 20x = 20 x 30 = RS 600
C's share = 15x = 15 x 30 = Rs 450
View full question & answer→Question 253 Marks
Find $x / y$ when $x^2+6 y^2=5 x y$
Answer$x^2+6 y^2=5 x y$
Dividing both sides by $y^2$ we get
$\frac{x^2}{y^2}+\frac{6 y^2}{y^2}=\frac{5 x y}{y^2}$
$\left(\frac{x}{y}\right)^2+6=5\left(\frac{x}{y}\right) $
$ \left(\frac{x}{y}\right)^2-5\left(\frac{x}{y}\right)+6=0$
$\text { Let } \frac{x}{y}=a $
$a^2-5 a+6=0$
$\Rightarrow (a-2)(a-3)=0 $
$ \Rightarrow a=2,3 $
$ \text { hence } \frac{x}{y}=2,3$
View full question & answer→Question 263 Marks
If $(a – b): (a + b) = 1: 11,$ find the ratio $(5a + 4b + 15): (5a – 4b + 3).$
Answer$\frac{a-b}{a+b}=\frac{1}{11}$
$11a - 11b = a = b$
$10a = 12b$
$\frac{a}{b}=\frac{12}{10}=\frac{6}{5}$
So let a = 6k and b = 5k
$\frac{5 a+4 b+15}{5 a-4 b+3}=\frac{5(6 k)+4(5 k)+15}{5(6 k)-4(5 k)+3}$
$=\frac{30 k+20 k+15}{30 k-20 k+3} $
$=\frac{50 k+15}{10 k+3} $
$ =\frac{5(10 k+3)}{10 k+3} $
$=5$
Hence $(5a + 4b + 15):(5a - 4b + 3) = 5 : 1$
View full question & answer→Question 273 Marks
If $(p - x) : (q - x)$ be the duplicate ratio of $p : q$ then show that :
$\frac{1}{p}+\frac{1}{q}=\frac{1}{x}$
AnswerWe have,
$\frac{p-x}{q-x}=\frac{p^2}{q^2}$
$\Rightarrow q^2(p - x) = p^2(q - x)$
$\Rightarrow pq2 - q2x = p2q - p2x$
$\Rightarrow p2x - q2x = p2q - pq2$
$\Rightarrow x(p2 - q2) =pq( p -q)$
$\Rightarrow x(p - q)(p + q) = pq(p - q)$
$\Rightarrow x=\frac{p q}{p+q}$
$\Rightarrow \frac{p q}{p+q}=\frac{1}{x}$
$\Rightarrow \frac{p}{p q}+\frac{q}{p q}=\frac{1}{x}$
$\Rightarrow \frac{1}{q}+\frac{1}{p}=\frac{1}{x}$
$\Rightarrow \frac{1}{p}+\frac{1}{q}=\frac{1}{x}$
View full question & answer→Question 283 Marks
If $r^2_=pq$, show that $p : q$ is the duplicate ratio of $(p + r) : (q + r)$.
AnswerGiven, $r^2$= pqDuplicate ratio of $( p + r) : ( q + r) = (p + r)^2: (q + r)^2$
$= (p2 + r2 + 2pr) : (q2 + r2 + 2qr)$
$= (p2 + pq + 2pr) : (q2 + pq + 2qr)$
$= p(p + q + 2r) : q(q + p + 2r)$
$= p : q$
Thus, p : q is the duplicate ratio of $(p + r) : (q + r)$
View full question & answer→Question 293 Marks
If $(3x - 9) : (5x + 4)$ is the triplicate ratio of $3 : 4,$ find the value of x.
AnswerWe have,
$\frac{3 x-9}{5 x+4}=\frac{3^3}{4^3}$
$\Rightarrow \frac{3 x-9}{5 x+4}=\frac{27}{64} $
$ \Rightarrow \frac{3(x-3)}{5 x+4}=\frac{27}{64} $
$ \Rightarrow \frac{x-3}{5 x+4}=\frac{9}{64}$
$ \Rightarrow 64 x-192=45 x+36$
$\Rightarrow 19 x=228$
$ \Rightarrow x=12$
View full question & answer→Question 303 Marks
If $m: n$ is the duplicate ratio of $m + x: n + x$; show that $x^2 = mn$.
Answer$\frac{m}{n}=\frac{(m+x)^2}{(n+x)^2}$
$\frac{m}{n}=\frac{m^2+x^2+2 m x}{n^2+x^2+2 n x}$
$m n^2+m x^2+2 m n x=m^2 n+n x^2+2 m n x $
$ m n^2-m^2 n+m x^2-n x^2=0$
$m n(m-n)-x^2(m-n)=0 $
$x^2(m-n)=m n(m-n) $
$ x^2=m n$
View full question & answer→Question 313 Marks
In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?
AnswerQuantity of milk : Quantity of water = 5 : 2
Quantity of milk $=126 \times \frac{5}{7}=90 kg$
⇒ Quantity of water = 126 - 90 = 36kg
New ratio = 3 : 2
Let the quantity of water to be added be x kg
Then, milk : water $=\frac{90}{36+ x }$
$\frac{90}{36+x}=\frac{3}{2}$
⇒ 180 = 108 + 3x
⇒ 3x = 72
⇒ x = 24
Thus, quantity of water to be added is 24 kg.
View full question & answer→Question 323 Marks
The work done by (x – 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x – 3) days are in the ratio 3: 8. Find the value of x.
AnswerAssuming that all the men do the same amount of work in one day and one day work of each man = 1 units, we have,
Amount of work done by (x – 2) men in (4x + 1) days
= Amount of work done by (x – 2)(4x + 1) men in one day
= (x – 2)(4x + 1) units of work
Similarly,
Amount of work done by (4x + 1) men in (2x – 3) days
= (4x + 1)(2x – 3) units of work
According to the given information,
$\frac{(x-2)(4 x+1)}{(4 x+1)(2 x-3)}=\frac{3}{8}$
$\frac{x-2}{2 x-3}=\frac{3}{8}$
8x - 16 = 6x - 9
2x = 7
$x=\frac{7}{2}=3 . .5$
View full question & answer→Question 333 Marks
A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.
AnswerLet the number of boys be 3x.
Then, number of girls = 2x
3x + 2x = 630
⇒ 5x = 630
⇒ x = 126
⇒ Number of boys = 3x = 3 x 126 = 378
And, Number of girls = 2x = 2 x 126 = 252
After admission of 90 new students, we have
Total number of students = 630 + 90 = 720
Now, let the number of boys be 7x.
then, number of girls = 5x
7x + 5x = 720
⇒ 12x = 720
⇒ x = 60
⇒ Number of boys after new admission = 7x = 7 x 60 = 420
And Number of girls after new admission = 5x = 5 x 60 = 300
Number of newly admitted boys = 420 - 378 = 42
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