Question 13 Marks
If $p, q$ and $r$ in continued proportion, then prove the following :
$p ^2- q ^2+ r ^2= q ^4\left(\frac{1}{ p ^2}-\frac{1}{ q ^2}-\frac{1}{ r ^2}\right)$
Answer$p: q:: q: r \Rightarrow q^2=p r $
$p^2-q^2+r^2=q^4\left(\frac{1}{p^2}-\frac{1}{q^2}-\frac{1}{r^2}\right) $
$\text { RHS } $
$q^4\left(\frac{1}{p^2}-\frac{1}{q^2}-\frac{1}{r^2}\right) $
$=q^4\left(\frac{q^2 r^2-p^2 r^2+p^2 q^2}{p^2 q^2 r^2}\right) $
$=q^4 q^2\left(\frac{r^2-q^2+p^2}{q^2 q^4}\right) $
$=p^2-q^2+r^2=\text { LHS } $
$\text { LHS }=\text { RHS. Hence, proved. }$
LHS $=$ RHS. Hence, proved.
View full question & answer→Question 23 Marks
If p, q and r in continued proportion, then prove the following:
$(p + q + r )(p - q + r) = p^2 + q^2 + r^2$
Answer$p : q :: q : r \Rightarrow q^2 = pr$
$(p + q + r )(p - q + r) = p^2 + q^2 + r^2$
LHS
$(p + q + r)(p - q + r)$
$= p^2 + pq + pr - pq - q^2 - qr + pr + qr + r^2$
$= p^2 + 2pr - q^2 + r^2$
$= p^2 + 2q^2 - q^2 + r^2$
$= p^2 + q^2 + r^2 = RHS$
LHS = RHS. Hence, proved.
View full question & answer→Question 33 Marks
If $p, q$ and $r$ in continued proportion, then prove the following:
$(p^2 - q^2)(q^2 + r^2) = (q^2 - r^2)(p^2 + q^2)$
Answer$p : q :: q : r => q^2 = pr1$
$LHS (p^2 q^2)( q^2 + r^2)$
$= (p^2 -(pr)^2)(pr)^2 + r^2)$
$= p^3r - p^2r^2 + p^2r^2 - pr^3$
$= pr (p^2 - r^2)$
RHS
$(q^2 - r^2)(p^2 + q^2)$
$= ((pr)^2-r^2 )(p^2+ (pr)^2)$
$= p^3r - p^2r^2 + p^2t^2 - pr^3$
$= pr(p^2 - r^2)$
LHS = RHS. Hence, proved.
View full question & answer→Question 43 Marks
If $(7m +8n)(7p - 8q) = (7m - 8n)(7p + 8q),$ then prove that $m: n = p: q$
Answer$(7 m+8 n)(7 p-8 q)=(7 m-8 n) c$
$\Rightarrow \frac{7 m+8 n}{7 m-8 n}=\frac{7 p+8 q}{7 p-8 q}$
Applying componendo and dividendo,
$\frac{7 m +8 n +7 m -8 n }{7 m +8 m -7 m +8 n }=\frac{7 p +8 q +7 m -8 q }{7 m +8 q -7 m +8 q } $
$\Rightarrow \frac{14 m }{16 n }=\frac{14 p }{16 q }$
Dividing both sides by $\frac{14}{16}$
$\frac{ m }{ n }=\frac{ p }{ q }$
Hence. $m: n=p: q$.
View full question & answer→Question 53 Marks
If $a: b:: c: d$, then prove that
$\frac{4 a+9 b}{4 c+9 d}=\frac{4 a-9 b}{4 c-9 d}$
Answer$\frac{4 a+9 b}{4 c+9 d}=\frac{4 a-9 b}{4 c-9 d} $
$\frac{a}{b}=\frac{c}{d}$
Multiplying both sides by $\frac{4}{9}$
$\Rightarrow \frac{ a }{ b } \times \frac{4}{9}=\frac{ c }{ d } \times \frac{4}{9} $
$\Rightarrow \frac{4 a }{9 b }=\frac{4 c }{9 d }$
Applying componendo and dividendo,
$\frac{4 a+9 b}{4 a-9 b}=\frac{4 c+9 d}{4 c-9 d} $
$\Rightarrow \frac{4 a+9 b}{4 c+9 d}=\frac{4 a-9 b}{4 c-9 d}$
Hence, $4 a+9 b: 4 c+9 d=4 a-9 b: 4 c-9 d$
View full question & answer→Question 63 Marks
If $a : b : : c : d,$ then prove that
$7a+11b : 7a -11b = 7c +11d : 7c - 11d$
Answer$7 a+11 b: 7 a-11 b=7 c+11 d: 7 c-11 d$
$\frac{a}{b}=\frac{c}{d}$
Multiplying both sides by $\frac{7}{11}$
$\Rightarrow \frac{ a }{ b } \times \frac{7}{11}=\frac{ c }{ d } \times \frac{7}{11} $
$\Rightarrow \frac{7 a }{11 b }=\frac{7 c }{11 d }$
Applying componendo and dividendo,
$\frac{7 a +11 b }{7 a -11 b }=\frac{7 c +11 d }{7 c -11 d }$
Hence, $7 a+11 b: 7 a-11 b=7 c+11 d: 7 c-11 d$
View full question & answer→Question 73 Marks
If $a : b : : c : d,$ then prove that
$2a+7b : 2a-7b = 2c+7d : 2c-7d$
Answer$2 a +7 b : 2 a -7 b =2 c +7 d : 2 c -7 d $
$\frac{ a }{ b }=\frac{ c }{ d } $
$\text { Multiplying both sides by } \frac{2}{7} $
$\Rightarrow \frac{ a }{ b } \times \frac{2}{7}=\frac{ c }{ d } \times \frac{2}{7} $
$\Rightarrow \frac{2 a }{7 b }=\frac{2 c }{7 d }$
Applying componendo and dividendo,
$\frac{2 a +7 b }{2 a -7 b }=\frac{2 c +7 d }{2 c -7 d }$
Hence, $2 a+7 b: 2 a-7 b=2 c+7 d: 2 c-7 d$
View full question & answer→Question 83 Marks
If three quantities are in continued proportion, show that the ratio of the first to the third is the duplicate ratio of the first to the second.
AnswerLet x, y and z are the three quantities which are in continued proportion
Then, $x : y :: y : z => y^2 = xz$
Now, we have to prove that
$x : z = x^2 : y^2$
$\Rightarrow xy^2 = x^2z$
$LHS$
$= xy^2 = x(xz) = x^2z = RHS$
$LHS = RHS$
View full question & answer→Question 93 Marks
If x, 12 and 16 are in continued proportion! find x.
AnswerSince x, 12 and 16 are in continued proportion
⇒ x : 12 : : 12 : 16
⇒ 16 x = 12 x 12
⇒ 16 x = 144
⇒ x = 9
x = 9
View full question & answer→Question 103 Marks
Find the mean proportion of the following:
$\frac{28}{3} \text { and } \frac{175}{27}$
AnswerLet $x$ be the mean proportion
$\frac{28}{3}: x :: x : \frac{175}{27} $
$\Rightarrow x \times x =\frac{28}{3} \times \frac{175}{27} $
$\Rightarrow x ^2=\frac{4900}{81} $
$\Rightarrow x =\sqrt{\frac{4900}{81}} $
$\Rightarrow x =\frac{70}{9}$
The mean proportion is $\frac{70}{9}$
View full question & answer→Question 113 Marks
Find the mean proportion of the following :
$ab^3$ and $a^3b$
AnswerLet $x$ be the mean proportion
$a b^3: x:: x: a^3 b $
$\Rightarrow x \times x=a b^3 x a^3 b $
$\Rightarrow x^2=a^4 b^4 $
$\Rightarrow x=\sqrt{a^4 b^4} $
$\Rightarrow x=a^2 b^2$
The mean proportion is $a ^2 b ^2$
View full question & answer→Question 123 Marks
Find the mean proportion of the following :
$0.09$ and $0.25$
AnswerLet $x$ be the mean proportion
$0.09: x:: x: 0.25 $
$\Rightarrow x \times x=0.09 \times 0.25 $
$\Rightarrow x^2=0.0225 $
$\Rightarrow x=\sqrt{0.0225} $
$\Rightarrow x=0.15$
The mean prooortion is $0.15$
View full question & answer→Question 133 Marks
Find the mean proportion of the following :
24 and 6
AnswerLet $x$ be the mean proportion
$24: x:: x: 6$
$
\Rightarrow x \times x=24 \times 6
$
$
\Rightarrow x^2=144
$
$
\Rightarrow x=\sqrt{144}
$
$
\Rightarrow x =12
$
The mean proportion is 12 .
View full question & answer→Question 143 Marks
Find the third proportion to the following:
$\frac{9}{25} \text { and } \frac{18}{25}$
AnswerLet $x$ be the third proportion
$\frac{9}{25}: \frac{18}{25}:: \frac{18}{25}: x $
$\Rightarrow \frac{9}{25} \times x =\frac{18}{25} \times \frac{18}{25} $
$= x =\frac{18}{25} \times \frac{18}{25} \times \frac{25}{9} $
$= x =\frac{36}{25}$
The third proportion is $\frac{36}{25}$
View full question & answer→Question 153 Marks
Find the third proportion to the following :
$(x - y)$ and $m (x - y)$
AnswerLet z be lhe third proportion
$(x - y) : m (x - y) : : m (x - y) : z$
$\Rightarrow (x -y) x z - m(x - y) x m (x - y)$
$\Rightarrow (x - y) z = m^2 (x - y)^2$
$\Rightarrow z = m^2 (x - y)$
The third proportion is $m^2 (x - y) . $
View full question & answer→Question 163 Marks
Find the third proportion to the following :
$16x^2 $and $24x$
AnswerLet y be !he third proportion
$16x^2 : 24x : : 24 x : y$
$\Rightarrow 16x^2 : 24 x : : 24 x : y$
$\Rightarrow 16x^2 y = 576 x^2$
$\Rightarrow 16 y = 576$
$\Rightarrow y = 36$
The third proporbon is $36.$
View full question & answer→Question 173 Marks
Find the third proportion to the following :
3 and 15
AnswerLet x be the third proportion
3 : 15 :: 15 : x
⇒ 3 × x- 15 x 15
⇒ 3 x = 225
⇒ x = 75
The third proportion is 75.
View full question & answer→Question 183 Marks
Find the fourth proportion to the following :
$(x^2 - y^2),(x^3 + y^3)$ and $(x^3 - xy^2 + x^2y- y^3)$
AnswerLet $z$ be the fourth proportion
$\left(x^2-y^3\right)\left(x^3+y^3\right)::\left(x^3-x y^2+x^2 y-y^3\right): z $
$\Rightarrow\left(x^2-y^2\right) z=\left(x^3+y^3\right)\left(x^3-x y^2+x^2 y-y^3\right) $
$\Rightarrow z=\frac{\left(x^3+y^3\right)\left(x^3-x y^2+x^2 y-y^3\right)}{x^2-y^2} $
$\Rightarrow z=\frac{(x+y)\left(x^2-x y^2+x^2 y-y^3\right)}{(x+y)(x-y)} $
$\Rightarrow z=\left(x^2-x y+y^2\right)(x+y)^2$
Therefore the fourth proportion is $\left(x^2+y^2-x y\right)(x+y)^2$
View full question & answer→Question 193 Marks
Find the fourth proportion to the following:
$0.7, 4.9$ and $1.6$
AnswerLet $x$ be the fourth prcportion .
$0.7: 4.9:: 1.6: x $
$\Rightarrow 0.7 \times X=4.9 \times 1.6 $
$\Rightarrow x=\frac{4.9 \times 1.6}{0.7} $
$\Rightarrow x=11.2$
The fourth proportion is $11.2$
View full question & answer→Question 203 Marks
Find the fourth proportion to the following:
$3,5$ and $15$
AnswerLet $x$ be the fourth proportion to $3,5$ and $15$
$3: 5:: 15: x $
$\Rightarrow 3 \times x=5 \times 15$
$\Rightarrow x=\frac{5 \times 15}{3}$
$\Rightarrow x=25$
The fourth proportion is $25 .$
View full question & answer→Question 213 Marks
Find the value of the unknown in the following proportion :
c
Answer$a : \frac{9}{2}:: \frac{7}{2}: \frac{11}{2}$
$\Rightarrow a \times \frac{11}{2}=\frac{9}{2} \times \frac{7}{2}$
$\Rightarrow a =\frac{9}{2} \times \frac{7}{2} \times \frac{2}{11}$
$\Rightarrow a =\frac{63}{22} $
$a =\frac{63}{22}$
View full question & answer→Question 223 Marks
Find the value of the unknown in the following proportion:
$\frac{1}{2}: m :: \frac{14}{9}: \frac{4}{3}$
Answer$\frac{1}{2}: m :: \frac{14}{9}: \frac{4}{3} $
$ \frac{1}{2}: m :: \frac{14}{9}: \frac{4}{3} $
$\Rightarrow m \times \frac{14}{9}=\frac{1}{2} \times \frac{4}{3} $
$ \Rightarrow m =\frac{1}{2} \times \frac{4}{3} \times \frac{9}{14}=\frac{3}{7}$
$ m =\frac{3}{7}$
View full question & answer→Question 233 Marks
If $a : b = c : d;$ then show that $(ax+ by) : b = (cx+ dy) : d$
Answer$\frac{ a }{ b }=\frac{ c }{ d } \Rightarrow \frac{ ax }{ b }=\frac{ cx }{ d }$
Adding $y$ to both sides
$\Rightarrow \frac{a x}{b}+y=\frac{c x}{d}+y $
$\Rightarrow \frac{a x+b y}{b}=\frac{c x+d y}{d} $
$\Rightarrow(a x+b y): b=(e x+d y): d$
Proved.
View full question & answer→Question 243 Marks
Find the duplicate ratio of the following :
$(a+b): (a^2- b^2)$
Answer$(a+b):\left(a^2-b^2\right) $
$=(a+b)^2:\left(a^2-b^2\right)^2$
$=\frac{(a+b)^2}{(a+b)^2(a-b)^2}$
$=\frac{1}{(a-b)^2}$
Duplicate ratio $=1:( a - b )^2$
View full question & answer→Question 253 Marks
Find the compounded ratio of the following:
$\sqrt{8}: 4,3: \sqrt{5}$ and $\sqrt{20}: \sqrt{27}$
AnswerCompounded ratio of $\sqrt{8}: 4,3: \sqrt{5}$ and $\sqrt{20}: \sqrt{27}$
$=\frac{\sqrt{8}}{4} \times \frac{3}{\sqrt{5}} \times \frac{\sqrt{20}}{\sqrt{27}}$
$=\frac{2 \sqrt{2}}{4} \times \frac{3}{\sqrt{5}} \times \frac{2 \sqrt{5}}{3 \sqrt{3}} $
$=\frac{\sqrt{2}}{\sqrt{3}}$
Compounded raoo $=\sqrt{2}: \sqrt{3}$
View full question & answer→Question 263 Marks
If $a : e = 5 : 12, e : i = 8 : 3$ and $i : u = 9 : 16,$ then find $a : u$
Answer$a : e =5: 12 \Rightarrow \frac{ a }{ e }=\frac{5}{12} $
$ e : i =8: 3 \Rightarrow \frac{ e }{ i }=\frac{8}{3} $
$ i : u =9: 16 \Rightarrow \frac{ i }{ u }=\frac{9}{16}$
$\Rightarrow \frac{ a }{ e } \times \frac{ e }{ i } \times \frac{ i }{ u }=\frac{5}{12} \times \frac{8}{3} \times \frac{9}{16} $
$ \Rightarrow \frac{ a }{ u }=\frac{10}{6}=\frac{5}{8} $
$ a : u =5: 8$
View full question & answer→Question 273 Marks
A man's monthly inoome is Rs 5,000. He saves every month a minimum of R~ 800. Find the ratio of his:
monthly savings to monthly expenses.
Answermonthly savings to monthly expenses.
Monthly expenses $=$ Rs 5,000 $-800=$ Rs 4,200
$
\frac{\text { Monthly Savings }}{\text { Monthly expenses }}=\frac{800}{4200}=\frac{8}{42}=\frac{4}{21}
$
Therefore, monthly savings : monthly expenses $=4: 21$
View full question & answer→Question 283 Marks
A man's monthly inoome is Rs 5,000. He saves every month a minimum of R~ 800. Find the ratio of his:
Annual expenses to annual Income.
AnswerAnnual income=monthly income x $12=$ Rs 5,000 x $12=$ Rs 60,000
Monthly expenses $=$ Rs 5,000 $-800=$ Rs 4,200
Annual expenses $=$ monthly expenses $\times 12=$ Rs 4,200 $12=\operatorname{Rs} 50,400$
$\frac{\text { Annual expenses }}{\text { Annual income }}=\frac{50400}{60000}=\frac{504}{600}=\frac{21}{25}$
Therefore, Annual expenses : annual income $=21: 25$
View full question & answer→Question 293 Marks
The ratio of the pocket money saved by a boy and his sister is $5:4.$ If the brother saves $Rs.100$ more, how much more should the sister save in order to keep the ratio of their savings unchanged ?
AnswerLet money saved by the boy and her sister be $S x$ and $4 x$ and let the sister need to save Rs $y$ more
$\frac{5 x+100}{4 x+y}=\frac{5}{4}$
$\Rightarrow 20 x+400=20 x+5 y $
$\Rightarrow 5 y=400 $
$\Rightarrow y=80$
The sister needs to save $Rs.80$ more.
View full question & answer→Question 303 Marks
If $r^2 = pq ,$ show that $p : q$ is the duplicate ratio of $(p+r) : ( q+r)$
Answer$\frac{p}{q}=\frac{(p+q)^2}{(q+r)^2} $
$\Rightarrow \frac{p}{q}=\frac{p^2+r^2+2 p r}{q^2+r^2+2 q r} $
$\Rightarrow \frac{p}{q}=\frac{p^2+p q+2 p r}{q^2+p q+2 q r} $
$\Rightarrow p q^2+p^2 q+2 p q r-p^2 q+p q^2+2 p q r$
Since LHS $=$ RHS, therefore $p: q 1 s$ the duplicate ratio of $(p+r):(q+r)$.
View full question & answer→Question 313 Marks
If $(5x+3) : (3x+ 1)$ is the triplicate ratio of $4 : 3 ,$ find $x.$
Answer$\frac{5 x+3}{3 x+1}=\frac{(4)^3}{(3)^3} $
$ \Rightarrow \frac{5 x+3}{3 x+1}=\frac{64}{27}$
$ \Rightarrow 135 x+81=192 x+64 $
$\Rightarrow 57 x=17$
$ \Rightarrow x=\frac{17}{57}$
$ x=\frac{17}{57}$
View full question & answer→Question 323 Marks
If $(3x-4) : (2x+5)$ is the duplicate ratio of $3 : 4 ,$ find $x.$
Answer$\frac{3 x-4}{2 x+5}=\frac{(3)^2}{(4)^2} $
$ \Rightarrow \frac{3 x-4}{2 x+5}=(3)=\frac{9}{16} $
$\Rightarrow 48 x-64-18 x+45 $
$\Rightarrow 30 x-109 $
$ \Rightarrow x=\frac{109}{30} $
$ x=\frac{109}{30}$
View full question & answer→Question 333 Marks
What quantity must be added to each term of the ratio $19 : 51$ to make it equal to $3 : 7 ?$
AnswerLet $x$ be added from each term such that
$\frac{19+x}{51+x}=\frac{3}{7} $
$\Rightarrow 133+7 X=153+3 X $
$\Rightarrow 4 x=20 $
$\Rightarrow x=5$
$5$ should be added to each term.
View full question & answer→Question 343 Marks
What quantity must be subtracted from each term of the ratio $39 : 89$ to make it equal to $2 : 5 ?$
AnswerLet $x$ be subtracted from each term such that
$\frac{39-x}{89-x}=\frac{2}{5} $
$\Rightarrow 195-5 x=178-2 x $
$\Rightarrow 3 x=17 $
$\Rightarrow x=\frac{17}{3}$
$\frac{17}{3}$ should be subtracted from each term.
View full question & answer→Question 353 Marks
Which of the following ratio is greater ?
$5 : 8$ and $7 : 10$
Answer$5: 8 \text { and } 7: 10 $
$\frac{5}{8}: \frac{7}{10}$
$ \Rightarrow 5 \times 10<8 \times 7$
$ \Rightarrow 50<56 $
$ \Rightarrow 5: 8<7: 10$
$ 7: 10 \text { is greater. }$
View full question & answer→Question 363 Marks
which of the following ratio is greater?
$2 : 3$ and $13 : 19$
Answer$2: 3$ and $13: 19$
$\frac{2}{3}: \frac{13}{19} $
$\Rightarrow 2 \times 19<3 \times 13 $
$\Rightarrow 38<39 $
$\Rightarrow 2: 3<13: 19$
$13: 19$ is greater.
View full question & answer→Question 373 Marks
Which of the Following Ratio is Greater ?
$3 : 5$ and $2 : 11$
Answer$3: 5$ and $2: 11$
$\frac{3}{5}: \frac{2}{11} $
$\Rightarrow 3 \times 11>2 \times 5 $
$\Rightarrow 33>10$
$\Rightarrow 3: 5>2: 11$
$3: 5$ is greater.
View full question & answer→Question 383 Marks
Find the reciprocal ratio of e following :
$81pq^2 : 54p^2q$
Answer$81 p q^2: 54 p^2 q$
$=\frac{1}{81 p q^2}: \frac{1}{54 p^2 q}$
$=\frac{1}{81 p q^2} \times \frac{54 p^2 q}{1}$
$=\frac{2 p}{3 q}$
Reciprocal ratio $=2 p: 3 q$
View full question & answer→Question 393 Marks
Find the reciprocal ratio of e following :
$a^3b^2 : a^2b^3$
Answer$a^3 b^2: a^2 b^3 $
$ =\frac{1}{a^3 b^2}: \frac{1}{a^2 b^3}$
$ =\frac{1}{a^3 b^2} \times \frac{a^2 b^3}{1}$
$ =b: a $
$\text { Reciprocal ratio }=b: a$
View full question & answer→Question 403 Marks
Find the reciprocal ratio of e following:
$\frac{17}{45}: \frac{51}{27}$
Answer$\frac{17}{45}: \frac{51}{27} $
$=\frac{45}{17}: \frac{27}{51} $
$=\frac{45}{17} \times \frac{51}{27}=\frac{45}{1} \times \frac{3}{27}=\frac{45}{1} \times \frac{1}{9} $
$=\frac{5}{1}$
Reciprocal ratio $=5: 1$
View full question & answer→Question 413 Marks
Find the sub-triplicate ratio of the following :
$125a^3 : 343b^6$
Answer$125 a^3: 343 b^6$
$=\left(125 a^3\right)^{\frac{1}{3}}:\left(343 b^6\right)^{\frac{1}{3}}$
$=5 a: 7 b^2$
Sub-triplicate ratio $=5 a: 7 b^2$
View full question & answer→Question 423 Marks
Find the sub-triplicate ratio of the following :
$m^3n^6 : m^6n^3$
Answer$m^3 n^6: m^6 n^3 $
$=\left(m^3 n^6\right)^{\frac{1}{3}}:\left(m^6 n^3\right)^{\frac{1}{3}} $
$=m n^2: m^2 n $
$=m: n$
Sub-triplicate ratio $=m: n$
View full question & answer→Question 433 Marks
Find the sub-duplicate ratio of the following:
$\frac{9 a ^2}{5}: \frac{25 a ^2}{3}$
Answer$\frac{9 a ^2}{5}: \frac{25 a ^2}{3}$
$=\sqrt{\frac{9 a ^2}{5}}: \sqrt{\frac{25 a ^2}{3}} $
$=3 a \frac{1}{\sqrt{5}}: 5 a \frac{1}{\sqrt{3}} $
$=3 \sqrt{3}: 5 \sqrt{5}$
Sub-duplicate ratio $=3 \sqrt{3}: 5 \sqrt{5}$
View full question & answer→Question 443 Marks
Find the sub-duplicate ratio of the following
c
Answer$63 m ^2: 28 n ^2 $
$=\sqrt{63 m ^2}: \sqrt{28 n ^2}$
$=3 \sqrt{7} m : 2 \sqrt{7} n$
$=3 m : 2 n$
Sub-duplicate ratio $=3 m: 2 n$
View full question & answer→Question 453 Marks
Find the triplicate ratio of the following :
$\sqrt[3]{a b^2}: \sqrt[3]{a^2 b}$
Answer$\sqrt[3]{a b^2}: \sqrt[3]{a^2 b} $
$=\left(\sqrt[3]{a b^2}\right)^3:\left(\sqrt[3]{a^2 b}\right)^3 $
$=a b^2: a^2 b $
$=b: a$
Triplicate ratio $=b: a$
View full question & answer→Question 463 Marks
Find the triplicate ratio of the following:
$\sqrt{15}: \sqrt{18}$
Answer$\sqrt{15}: \sqrt{18} $
$=(\sqrt{15})^3:(\sqrt{18})^3 $
$=15 \sqrt{15}: 18 \times 3 \sqrt{2} $
$=5 \sqrt{15}: 18 \sqrt{2} $
$\text { Triplicate ratio }=5 \sqrt{15}: 18 \sqrt{2}$
Triplicate ratio $=5 \sqrt{15}: 18 \sqrt{2}$
View full question & answer→Question 473 Marks
Find the triplicate ratio of the following:
$2 \sqrt{5}: 5 \sqrt{2}$
Answer$2 \sqrt{5}: 5 \sqrt{2} $
$=(2 \sqrt{5})^3:(5 \sqrt{2})^3 $
$=\frac{8 \times 5 \sqrt{5}}{125 \times 2 \sqrt{2}} $
$=4 \sqrt{5}: 25 \sqrt{2}$
Triplicate ratio $=4 \sqrt{5}: 25 \sqrt{2}$
View full question & answer→