Question 12 Marks
Find the mean proportional to $(x – y)$ and $(x^3 – x^2y).$
AnswerLet the mean proportional to $(x - y)$ and $\left(x^3-x^2 y\right)$ be n
$\Rightarrow(x-y), n,\left(x^3-x^2 y\right)$ are in continued proportional
$\Rightarrow(x-y): n=n:\left(x^3-x^2 y\right) $
$ \Rightarrow n^2=x^2(x-y)(x-y) $
$\Rightarrow n^2=x^2(x-y)^2 $
$ \Rightarrow n=x(x-y)$
View full question & answer→Question 22 Marks
Find the third proportional to $a^2 – b^2$ and $a + b.$
AnswerLet the third proportional to $a^2-b^2$ and $a + b$ be $n$
$\Rightarrow a^2-b^2, a+b$ and $n$ are in continued proportion
$\Rightarrow a^2-b^2: a+b=a_b: n$
$ \Rightarrow n=\frac{(a+b h)^2}{a^2-b^2}=\frac{(a+b)^2}{(a+b)(a-b)}=\frac{a+b}{a-b}$
View full question & answer→Question 32 Marks
Find the fourth proportional to $2xy, x^2$ and $y^2.$
AnswerLet the fourth proportional to $2 x y, x^2$ and $y^2$ be $n$
$\Rightarrow 2 x y: x^2=y^2: n $
$\Rightarrow 2 x y x n=x^2 \times y^2$
$ \Rightarrow n=\frac{x^2 y^2}{2 x y}=\frac{x y}{2}$
View full question & answer→Question 42 Marks
A woman reduces her weight in the ratio $7 : 5.$ What does her weight become if originally it was $84\ kg?$
AnswerLet the reduced weight be x.
Original weight $= 84$ kg
Thus, we have
$84 : x = 7 : 5$
$\Rightarrow \frac{84}{x}=\frac{7}{5}$
$ \Rightarrow 84 \times 5=7 \times x$
$ \Rightarrow x=\frac{84 \times 5}{7} $
$ \Rightarrow x=60$
Thus, her reduced weight is $60$ kg
View full question & answer→Question 52 Marks
What quantity must be added to each term of the ratio x: y so that it may become equal to c: d?
AnswerLet the required quantity which is to be added be p.
Then, we have:
$\frac{x+p}{y+p}=\frac{c}{d}$
$dx+p d=c y+c p$
pd - cp = cy - dx
p(d - c) = cy - dx
$p=\frac{c y-d x}{d-c}$
View full question & answer→Question 62 Marks
Find the value of x, if: (3x – 7): (4x + 3) is the sub-triplicate ratio of 8: 27.
Answer(3x - 7) : (4x + 3) is the sub-triplicate ratio of 8 : 27
Sub-triplicate ratio of 8 : 27 = 2 : 3
$\frac{3 x-7}{4 x+3}=\frac{2}{3}$
9x - 21 = 8x + 6
9x - 8x = 6 + 21
x = 27
View full question & answer→Question 72 Marks
Find the value of x, if: (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25.
Answer(2x + 1): (3x + 13) is the sub-duplicate ration of 9 : 25
Sub-duplicate ratio of 9 : 20 = 3 : 5
$\frac{2 x+1}{3 x+13}=\frac{3}{5}$
10x + 5 = 9x + 39
10x - 9x = 39 - 5
x = 34
View full question & answer→Question 82 Marks
Find the value of x, if: $(2 x+3):(5 x-38)$ is the duplicate ratio of $\sqrt{5}: \sqrt{6}$
Answer$(2 x+3):(5 x-38)$ is the duplicate ratio of $\sqrt{5}: \sqrt{6}$
Duplicate ratio of $\sqrt{5}: \sqrt{6}=5: 6$
$\frac{2 x+3}{5 x-38}=\frac{5}{6} $
$12 x+18=25 x-190$
$ 25 x-12 x=190+18 $
$ 13 x=208 $
$ x=\frac{208}{13}=16$
View full question & answer→Question 92 Marks
Find the ratio compounded of the duplicate ratio of 5: 6, the reciprocal ratio of 25: 42 and the sub-duplicate ratio of 36: 49.
Answer![]()
Duplicate ratio of 3 : 5 = 5 : 3
Reciprocal ratio of 5 : 6 = 25 : 36
Sub- duplicate ratio of 36 : 49 = 6 : 7
Required compound ratio $=\frac{25 \times 42 \times 6}{36 \times 25 \times 7}=1: 1$ View full question & answer→Question 102 Marks
If $5x + 6y: 8x + 5y = 8: 9,$ find $x: y.$
Answer$(5 x+6 y)(8 x+5 y)=\frac{8}{9}$
$45 x+54 y=64 x=40 y $
$ 64 x-45 x=54 y-40 y$
$ 19 x=14 y $
$ \frac{x}{y}=\frac{14}{19}$
View full question & answer→Question 112 Marks
If $\frac{x^2+y^2}{x^2-y^2}=2 \frac{1}{8}$ find $\frac{x}{y}$
AnswerGiven $(x^2 + y^2)/(x^2 - y^2) =2 \frac{1}{8}$ find $\frac{x}{y}$
$\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$
Applying componendo and dividendo
$\frac{x^2+y^2+x^2-y^2}{x^2+y^2-x^2+y^2}=\frac{17+8}{17-8}$
$\frac{2 x^2}{2 y^3}=\frac{25}{9}$
$\frac{x}{y}=\frac{5}{3}=1 \frac{2}{3}$
View full question & answer→Question 122 Marks
If $a: b = 3: 5,$ find: $(10a + 3b): (5a + 2b)$
AnswerGiven $\frac{a}{b}=\frac{3}{5}$
$\frac{10 a+3 b}{5 a+2 b} $
$=\frac{10\left(\frac{a}{b}\right)+3}{5\left(\frac{a}{b}\right)+2} $
$=\frac{10\left(\frac{3}{5}\right)+3}{5\left(\frac{3}{5}\right)+2} $
$=\frac{6+3}{3+2}$
$=\frac{9}{5}$
View full question & answer→Question 132 Marks
If $\frac{4 m+3 n}{4 m-3 n}=\frac{7}{4}$ use properties of proportion to find $m : n$
AnswerGiven $\frac{4 m+3 n}{4 m-3 n}=\frac{7}{4}$
Applying componendo and dividendo
$\frac{4 m+3 n+4 m-3 n}{4 m+3 n-4 m+3 n}=\frac{7+4}{7-4}$
$\frac{8 m}{6 n}=\frac{11}{3}$
$ \frac{m}{n}=\frac{11}{4}$
View full question & answer→Question 142 Marks
If 7x – 15y = 4x + y, find the value of x: y. Hence, use componendo and dividendo to find the values of: $\frac{9 x+5 y}{9 x-5 y}$
Answer7x – 15y = 4x + y
7x - 4x = y + 15y
3x = 16y
$\frac{x}{y}=\frac{16}{3}$
$\Rightarrow \frac{9 x}{5 y}=\frac{144}{15} \quad$ (Multiplying both sides by $9 / 5$ )
$\Rightarrow \frac{9 x+5 y}{9 x-5 y}=\frac{144+15}{144-15} \quad$ (Applying componendo and dividendo)
$\Rightarrow \frac{9 x+5 y}{9 x-5 y}=\frac{159}{129}=\frac{53}{43}$
View full question & answer→Question 152 Marks
If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that: a: b = c: d.
AnswerGiven $\frac{4 a+9 b}{4 a-9 b}=\frac{4 c+9 d}{4 c-9 d}$
Applying componendo and dividendo
$\frac{4 a+9 b+4 a-9 b}{4 a+9 b-4 a+9 b}=\frac{4 c+9 d+4 c-9 d}{4 c+9 d-4 c+9 d}$
$\frac{8 a}{18 b}=\frac{8 c}{18 d}$
$\frac{a}{b}=\frac{c}{d}$
View full question & answer→Question 162 Marks
If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a: b = c: d.
AnswerGiven $\frac{7 a+8 b}{7 a-8 b}=\frac{7 c+8 d}{7 c-8 d}$ ...(Invertendo)
Applying componendo and dividendo
$\frac{7 a+8 b+7 a-8 b}{7 a+8 b-7 a+8 b}=\frac{7 c+8 d+7 c-8 d}{7 c+8 d-7 c+8 d}$
$\Rightarrow \frac{14 a}{16 b}=\frac{14 c}{16 d}$
$\Rightarrow \frac{a}{b}=\frac{c}{d}$
Hence a : b = c : d
View full question & answer→Question 172 Marks
If $\frac{5 x+6 y}{5 u+6 v}=\frac{5 x-6 y}{5 u-6 v}$ then prove that x : y = u : v
Answer$\frac{5 x+6 y}{5 u+6 v}=\frac{5 x-6 y}{5 u-6 v}$ (By aletrnendo)
$\frac{5 x+6 y}{5 x-6 y}=\frac{5 u+6 v}{5 u-6 v}$
$\frac{5 x+6 y+5 x-6 y}{5 x+6 y-5 x+6 y}=\frac{5 u+6 v+5 u-6 v}{5 u+6 v-5 u=6 v}$ (By componendo and dividendo)
$\frac{10 x}{12 y}=\frac{10 u}{12 v}$
$\frac{x}{y}=\frac{u}{v}$
View full question & answer→Question 182 Marks
Given $\frac{a}{b}=\frac{c}{d}$ prove that $\frac{3 a-5 b}{3 a+5 b}=\frac{3 c-5 d}{3 c+5 d}$
Answer$\frac{a}{b}=\frac{c}{d}$
$\frac{3 a}{5 b}=\frac{3 c}{5 d}$ ....(Multiplying each side by $\frac{3}{5}$ )
$\frac{3 a+5 b}{3 a-5 b}=\frac{3 c+5 d}{3 c-5 d}$ ...(By componendo and divdendo)
$\frac{3 a-5 b}{3 a+5 b}=\frac{3 c-5 d}{3 c+5 d}$ ...(by Invertendo)
View full question & answer→Question 192 Marks
If $a : b = c : d,$ prove that $: xa + yb : xc + yd = b : d.$
AnswerGiven $\frac{ a }{ b }=\frac{ c }{ d }$
$\Rightarrow \frac{x a }{ yb }=\frac{x c }{ yd }\left(\right.$ Multiplying each side by $\left.\frac{x}{ y }\right.)$
$\Rightarrow \frac{x a + yb }{ yb }=\frac{x c + yd }{ yd }$ (By componendo)
$\Rightarrow \frac{x a + yb }{x c + yd }=\frac{ yb }{ yd } $
$$\Rightarrow \frac{x a + yb }{x c + yd }=\frac{ b }{ d }$
View full question & answer→Question 202 Marks
If a : b = c : d, prove that: (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b).
AnswerGiven $\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{9 a}{13 b}=\frac{9 c}{13 d}$ (Multiplying each side by $9 / 13$ )
$\Rightarrow \frac{9 a+13 b}{13 a-13 b}=\frac{9 c+13 d}{9 c-13 d}$ (By componendo and divdendo)
$\Rightarrow(9 a+13 b)(9 c-13 d)=(9 c+13 d)(9 a-13 b)$
View full question & answer→Question 212 Marks
If a, b, and c are in continued proportion, prove that $\frac{a^2+a b+b^2}{b^2+b c+c^2}=\frac{a}{c}$
AnswerGiven, a, b and c are in continued proportion.
Therefore,
$\frac{a}{b}=\frac{b}{c}$
$ac = b ^2$
Here,
$\frac{a^2+a b+b^2}{b^2+b c+c^2}$
$=\frac{a^2+a b+a c}{a c+b c+c^2}$
$ =\frac{a(a+b+c)}{c(a+b+c)} $
$ =\frac{a}{c}$
Hence proved.
View full question & answer→Question 222 Marks
If $x^2, 4$ and $9$ are in continued proportion, find $x.$
AnswerGiven, $x^2, 4$ and $9$ are in continued proportion.
$\frac{x^2}{4}=\frac{4}{9} $
$ \Rightarrow 9 x^2=16 $
$ \Rightarrow x^2=\frac{16}{9} $
$ \Rightarrow x=\frac{4}{3}$
View full question & answer→Question 232 Marks
Find the mean proportional between $a-b$ and $a^3-a^2 b$
AnswerLet the mean proportional between $a – b$ and $a^3 – a^2b be x.$
$\Rightarrow a – b, x, a^3 – a^2b.$ are in continued proportion.
$\Rightarrow a – b : x = x : a^3 – a^2b$
$\Rightarrow x \times x = (a – b) (a^3 – a^2b)$
$\Rightarrow x^2 = (a – b) a^2(a – b) = [a(a – b)]^2$
$\Rightarrow x = a(a – b)$
View full question & answer→Question 242 Marks
Find the mean proportional between $6+3 \sqrt{3}$ and $8-4 \sqrt{3}$
AnswerLet the mean proportional between $6 + 3\sqrt3$ and $8 – 4\sqrt3 be x.\Rightarrow 6 + 3\sqrt3, x$ and $8 – 4\sqrt3$ are in continued proportion.
$\Rightarrow 6 + 3\sqrt3 : x = x : 8 – 4\sqrt3$
$\Rightarrow x \times x = (6 + 3\sqrt3) (8 – 4\sqrt3)$
$\Rightarrow x^2= 48 + 24\sqrt3- 24\sqrt3 – 36$
$\Rightarrow x^2= 12$
$\Rightarrow x = 2\sqrt3$
View full question & answer→Question 252 Marks
Find the third proportional to $2 \frac{2}{3}$ and 4
AnswerLet the third proportional to $2 \frac{2}{3}$ and 4 be $x$
$=>2 \frac{2}{3}, 4, x$ are in continued proportion.
$\Rightarrow 2 \frac{2}{3}: 4=4: x$
$\Rightarrow \frac{\frac{8}{3}}{4}=\frac{4}{x}$
$\Rightarrow x=16 x \frac{3}{8}=6$
View full question & answer→Question 262 Marks
If $p + r = mq$ and $\frac{1}{q}+\frac{1}{s}=\frac{m}{r}$ then prove that $p : q = r : s$
Answer$\frac{1}{q}+\frac{1}{s}=\frac{m}{r}$
$\frac{s+q}{q s}=\frac{m}{r}$
$\frac{s+q}{s}=\frac{m q}{r}$
$\frac{s+q}{s}=\frac{p+r}{r} \quad(\because p + r = mq )$
$1+\frac{q}{s}=\frac{p}{r}+1$
$\frac{q}{s}=\frac{p}{r}$
$\frac{p}{q}=\frac{r}{s}$
Hence proved
View full question & answer→Question 272 Marks
If p: q = r: s; then show that: mp + nq : q = mr + ns : s.
Answer$\frac{p}{q}=\frac{r}{s}$
$ \begin{aligned} & \Rightarrow \frac{m p}{q}=\frac{m r}{s} \\ & \Rightarrow \frac{m p}{q}+n=\frac{m r}{s}+n \\ & \Rightarrow \frac{m p+n q}{q}=\frac{m r+n s}{s} \end{aligned} $
Hence, $m p+n q: q=m r+n s: s$
View full question & answer→Question 282 Marks
Find the third proportional to $\frac{x}{y}+\frac{y}{x}$ and $\sqrt{x^2+y^2}$
AnswerLet the required third proportional be p
$\Rightarrow \frac{x}{y}+\frac{y}{x}, \sqrt{x^2+y^2}, p$ are in contined proportion
$\Rightarrow \frac{x}{y}+\frac{y}{x}: \sqrt{x^2+y^2}=\sqrt{x^2+y^2}: p $
$ \Rightarrow p\left(\frac{x}{y}+\frac{y}{x}\right)=\left(\sqrt{x^2+y^2}\right)^2$
$ \Rightarrow p\left(\frac{x^2+y^2}{x y}\right) 0=x^2+y^2 $
$ \Rightarrow p=x y$
View full question & answer→Question 292 Marks
Find the fourth proportional to $3a, 6a^2$ and $2ab^2$
AnswerLet the fourth proportional to $3a, 6a^2$ and $2ab^2$ be $x.$
$\Rightarrow 3a : 6a^2 = 2ab^2 : x$
$\Rightarrow 3a \times x = 2ab^2 6a^2$
$\Rightarrow 3a \times x = 12a^3b^2$
$\Rightarrow x = 4a^2b^2$
View full question & answer→Question 302 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 $z$ be the three quantities which are in continued proportion.
Then, $x : y :: y : z 3$
$\Rightarrow y^2 = xz ….(1)$
Now, we have to prove that
$x : z = x^2: y^2$
That is we need to prove that
$xy^2= x^2z$
$LHS = xy^2 = x(xz) = x^2z = RHS$ [Using (1)]
Hence, proved.
View full question & answer→Question 312 Marks
Find the fourth proportional to 1.5, 4.5 and 3.5
AnswerLet the fourth proportional to 1.5, 4.5 and 3.5 be x.
⇒ 1.5 : 4.5 = 3.5 : x
⇒ 1.5 × x = 3.5 x 4.5
⇒ x = 10.5
View full question & answer→Question 322 Marks
If q is the mean proportional between p and r, show that: pqr $(p + q + r)^3 = (pq + qr + rp)^3$
AnswerGiven, q is the mean proportional between p and r.
$\Rightarrow q^2 = pr$
$L . HS =p q r(p+q+r)^3$
$=q q^2(p+q+r)^3 \quad\left[\because q^2=p r\right] $
$ =[q(p+q+r)]^3$
$ =\left(p q+q^2+q r\right)^3$
$=(p q+p r+q r)^3 \quad\left[\because q^2=p r\right]$
$ =\text { R.H.S }$
View full question & answer→Question 332 Marks
If the ratio between $8$ and $11$ is the same as the ratio of $2x - y$ to $x + 2y,$ find the value of $\frac{7 x}{9 y}$
Answer$\frac{2 x-y}{x+2 y}=\frac{8}{11} $
$ 22 x-11 y=8 x+16 y$
$ 14 x=27 y $
$ \frac{x}{y}=\frac{27}{14}$
Given, $\frac{7 x }{9 y }=\frac{7 \times 27}{9 \times 14}=\frac{3}{2}$
View full question & answer→Question 342 Marks
If $\frac{m+n}{m+3 n}=\frac{2}{3}$ find $\frac{2 n^2}{3 m^2+m n}$
Answer$\frac{m+n}{m+3 n}=\frac{2}{3}$
$\Rightarrow 3m + 3n = 2m + 6n$
$\Rightarrow m = 3n$
$\Rightarrow \frac{m}{n}=\frac{3}{1}$
$\frac{2 n^2}{3 m^2+m n}=\frac{2}{3\left(\frac{m}{n}\right)^2+\left(\frac{m}{n}\right)}\left(\right.$ Dividing each term by $\left.n^2\right)$
$=\frac{2}{3\left(\frac{3}{1}\right)^2+\left(\frac{3}{1}\right)}$
$=\frac{2}{27+3}=\frac{1}{15}$
View full question & answer→Question 352 Marks
Find the number which bears the same ratio to $\frac{7}{33}$ that $\frac{8}{21}$ does to $\frac{4}{9}$.
AnswerLet the required number be $\frac{x}{y}$
Now, ratio of $\frac{8}{21}$ to $\frac{4}{9}=\frac{\frac{8}{21}}{\frac{4}{9}}=\frac{8}{21} \times \frac{9}{4}=\frac{6}{7}$
thus we have
$\frac{\frac{x}{y}}{\frac{7}{33}}=\frac{6}{7} $
$ \Rightarrow \frac{x}{y}=\frac{\frac{6}{7}}{\frac{7}{33}}$
$ \Rightarrow \frac{x}{y}=\frac{6}{7} \times \frac{7}{33}$
$ \Rightarrow \frac{x}{y}=\frac{2}{11}$
Hence the required number is $\frac{2}{11}$
View full question & answer→Question 362 Marks
If $a: b=3: 8$. Find the value of $\frac{4 a+3 b}{6 a-b}$
Answer$a : b = 3 : 8$
$\Rightarrow \frac{a}{b}=\frac{3}{8}$
$\frac{4 a+3 b}{6 a-b}=\frac{4\left(\frac{a}{b}\right)+3}{6\left(\frac{a}{b}\right)-1}$ (Dividing each term by b)
$=\frac{4\left(\frac{3}{8}\right)+3}{6\left(\frac{3}{8}\right)-1} $
$ =\frac{\frac{3}{2}+3}{\frac{9}{4}-1} $
$ =\frac{\frac{9}{2}}{\frac{5}{4}} $
$ =\frac{18}{5}$
View full question & answer→Question 372 Marks
If $x: y = 4: 7,$ find the value of $(3x + 2y): (5x + y).$
Answer$x : y = 4 : 7$
$\Rightarrow \frac{x}{y}=\frac{4}{7}$
$\frac{3 x+2 y}{5 x+y}=\frac{3\left(\frac{x}{y}\right)+2}{5\left(\frac{x}{y}\right)+1} \quad$ (Dividing each term by y )
$=\frac{3\left(\frac{4}{7}\right)+2}{5\left(\frac{4}{7}\right)+1} $
$ =\frac{\frac{12}{7}+2}{\frac{20}{7}+1} $
$ =\frac{12+14}{20+7} $
$=\frac{26}{27}$
View full question & answer→Question 382 Marks
Find the ratio compounded of the reciprocal ratio of 15: 28, the sub-duplicate ratio of 36: 49 and the triplicate ratio of 5: 4.
AnswerReciprocal ratio of 15 : 28 = 28 : 15
Sub-duplicate ratio of 36 : 49 = $\sqrt{36}: \sqrt{49}=6: 7$
Triplicate ratio of $5: 4=5^3: 4^3=125: 64$
Required compound ratio
$=\frac{28 \times 6 \times 125}{15 \times 7 \times 64}=\frac{25}{8}=25: 8$
View full question & answer→Question 392 Marks
If $(x + 3) : (4x + 1)$ is the duplicate ratio of $3 : 5,$ find the value of $x.$
AnswerIf $(x + 3) : (4x + 1)$ is the duplicate ratio of $3 : 5$
Find the value of $x.$
We have,
$\frac{x+3}{4 x+1}=\frac{3^2}{5^2}$
$\Rightarrow \frac{x+3}{4 x+1}=\frac{9}{25} $
$\Rightarrow 25 x+75=36 x+9 $
$ \Rightarrow 11 x=66$
$ \Rightarrow x=6$
View full question & answer→Question 402 Marks
Find the compound ratio of $\sqrt{2}: 1,3: \sqrt{5}$ and $\sqrt{20}: 9$
AnswerRequired compound ratio = $\sqrt{2} \times 3 \times \sqrt{20}: 1 \times \sqrt{5} \times 9$
$=\frac{\sqrt{2} \times 3 \times \sqrt{20}}{1 \times \sqrt{5} \times 9} $
$ =\frac{\sqrt{2} \times \sqrt{4}}{3}$
$=\frac{2 \sqrt{2}}{3}=2 \sqrt{2}: 3$
View full question & answer→Question 412 Marks
if $a : b =5: 3$ find $\frac{5 a-3 b}{5 a+3 b}$
Answer$a : b = 5 : 3$
$\Rightarrow \frac{a}{b}=\frac{5}{3}$
$\frac{5 a-3 b}{5 a+3 b}=\frac{5\left(\frac{a}{b}\right)-3}{5\left(\frac{a}{b}\right)+3} ($ Dividing each term by $3)$
$=\frac{5\left(\frac{5}{3}\right)-3}{5\left(\frac{5}{3}\right)+\{3)}$
$=\frac{\frac{25}{3}-3}{\frac{25}{3}+3}$
$=\frac{25-9}{25+9}$
$=\frac{16}{34}=\frac{8}{17}$
View full question & answer→Question 422 Marks
If 2a = 3b and 4b = 5c, find: a : c.
AnswerWe have,
$2 a=3 b \Rightarrow \frac{a}{b}=\frac{3}{2}$
And $4 b=5 c \Rightarrow \frac{b}{c}=\frac{5}{4}$
Now, $\frac{ a }{ b }=\frac{3}{2}=\frac{3 \times 5}{2 \times 5}=\frac{15}{10}$ and $\frac{ b }{ c }=\frac{5}{4}=\frac{5 \times 2}{4 \times 2}=\frac{10}{8}$
⇒ a : b : c = 15 : 10 : 8
⇒ a : c = 15 : 8
View full question & answer→Question 432 Marks
If 3A = 4B = 6C; find A: B: C.
Answer3A = 4B = 6C
$3 A=4 B \Rightarrow \frac{A}{B}=\frac{4}{3}$
This means, A:B=4:3
$4 B=6 C \Rightarrow \frac{B}{C}=\frac{6}{4}=\frac{3}{2}$
=>B:C=3:2
The value of B in 4:3 is equal to the value of B in 3:2
Hence A : B : C = 4 : 3 : 2
View full question & answer→Question 442 Marks
If A : B = 2 : 5 and A : C = 3 : 4, find A : B : C
AnswerTo compare 3 ratios, the consquent of first ratio and the antecedent of 2nd ratio must be made equal.
Given that A : B = 2.5 and A:C = 3:4
Interchanging the first ratio we have
B : A = 5:2 and A : C= 3 : 4
L.C.M of 2 and 3 is 6
=> B : A = 5 x 3 : 2 x 3 and A : C = 3 x 2 : 4 x 2
=> B : A = 15 : 6 and A : C = 6 : 8
=> B : A : C = 15 : 6 : 8
=> A : B : C = 6 : 15 : 8
View full question & answer→Question 452 Marks
If A: B $= 3: 4$ and B: C $= 6: 7,$ find A: B: C find:
(i) A: B: C (ii) A: C
Answer(i) A: B: C
$\frac{A}{B}=\frac{3}{4}=\frac{3}{4} \times \frac{3}{3}=\frac{9}{12} $
$ \frac{B}{C}=\frac{6}{7}=\frac{6}{7} \times \frac{2}{2}=\frac{12}{14}$
A : B : C $= 9 : 12 : 14$
(ii) A: C
$\frac{A}{B}=\frac{3}{4} $
$\frac{B}{C}=\frac{6}{7}$
$\therefore \frac{A}{C}=\frac{\frac{A}{B}}{\frac{C}{B}}=\frac{\frac{3}{4}}{\frac{7}{6}}==\frac{9}{14}$
$\therefore A: C=9: 14$
View full question & answer→Question 462 Marks
In a basket, the ratio between the number of oranges and the number of apples is 7: 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes 1: 2. Find the original number of oranges and the original number of apples in the basket.
AnswerLet the original number of oranges and apples be 7x and 13x.
According to the given information,
$\frac{7 x-8}{13 x-11}=\frac{1}{2}$
14x - 16 = 13x - 11
x = 5
Thus the original number of oranges and apples are 7 x 5 = 35 and 13 x 5 = 65 respectively
View full question & answer→Question 472 Marks
By increasing the cost of entry ticket to a fair in the ratio 10: 13, the number of visitors to the fair has decreased in the ratio 6: 5. In what ratio has the total collection increased or decreased?
AnswerLet the cost of the entry ticket initially and at present be 10 x and 13x respectively.
Let the number of visitors initially and at present be 6y and 5y respectively.
Initially, total collection = 10x × 6y = 60 xy
At present, total collection = 13x × 5y = 65 xy
Ratio of total collection = 60 xy: 65 xy = 12: 13
Thus, the total collection has increased in the ratio 12: 13.
View full question & answer→Question 482 Marks
The bus fare between two cities is increased in the ratio 7: 9. Find the increase in the fare, if:
(i) the original fare is Rs 245;
(ii) the increased fare is Rs 207.
AnswerAccording to the given information,
Increased (new) bus fare $=\frac{9}{7}$ original bus fare
(i) We have:
Increased (new) bus fare $=\frac{9}{7} \times \operatorname{Rs} 245=\operatorname{Rs} 315$
Increase in fare = Rs 315 - Rs 245 = Rs 70
(ii) We have:
Rs $207=\frac{9}{7} \times$ original bus fare
Original bus fare $=\operatorname{Rs} 207 \times \frac{7}{9}=$ Rs 161
Increase in fare = Rs 207 - Rs 161 = Rs 46
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What quantity must be subtracted from each term of the ratio 9: 17 to make it equal to 1: 3?
AnswerLet x be subtracted from each term of the ratio 9: 17.
$\frac{9-x}{17-x}=\frac{1}{3}$
27 - 3x = 17 - x
10 = 2x
x = 5
Thus, the required number which should be subtracted is 5.
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