Question 13 Marks
If a, b, c are in continued proportion, prove that $a : c = (a^2 + b^2) : (b^2 + c^2).$
Answer$a , b$ and $c$ are the continued proportion
$a: b=b: c$
$\Rightarrow \frac{a}{b}=\frac{b}{c} $
$\Rightarrow b ^2= ac$
Now $\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}$
$=a\left(b^2+c^2\right)=c\left(a^2+b^2\right)$
$\text{L.H.S.}$
$\Rightarrow a\left(b^2+c^2\right) $
$\Rightarrow a\left(a c+c^2\right) $
$\Rightarrow a c(a+c)$
$\text { R.H.S. } $
$\Rightarrow c \left( a ^2+ b ^2\right) $
$\Rightarrow c \left( a ^2+ ac \right) $
$\Rightarrow ac ( a + c ) $
$\text { L.H.S. = R.H.S. }$
Hence proved.
View full question & answer→Question 23 Marks
If $\frac{b y+c z}{b^2+c^2}=\frac{c z+a x}{c^2+a^2}=\frac{a x+b y}{a^2+b^2}$ then show that each ratio is equal to $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
AnswerEach f the given ratio
$=\frac{(b y+c z)}{\left(b^2+c^2\right)}+\frac{(c z+a x)}{\left(c^2+a^2\right)}+\frac{(a x+b y)}{\left(a^2+b^2\right)}$
$=\frac{a x+b y+c z}{a^2+b^2+c^2}$
Now $\frac{b y+c z}{b^2+c^2}=\frac{a x+b y+c z}{a^2+b^2+c^2}$
$\Rightarrow \frac{a^2+b^2+c z}{b^2+c^2}=\frac{a x+b y+c z}{b y+c z}$
$\Rightarrow \frac{a^2}{b^2+c^2}=\frac{z x}{b y+c z} \quad \ldots$ (App. dividendo)
$\Rightarrow \frac{b^2+c^2}{a^2}=\frac{b y+c^2}{a x} \quad \ldots$ (App. invertends)
$\Rightarrow \frac{a^2+b^2+c^2}{a^2}=\frac{a x+b y+c z}{a x} \quad \ldots \text { (by componendo) } $
$\Rightarrow \frac{x}{a}=\frac{a x+b y+c z}{a^2+b^2+c^2} \quad \ldots \text { (similarly) } $
$\therefore \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{a x+b y+c z}{a^2+b^2+c^2}$
Hence proved.
View full question & answer→Question 33 Marks
If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that 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 43 Marks
If $\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d}$, prove that $\frac{a}{b}=\frac{c}{d}$.
Answer$\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d}$
Applying alternendo
$\frac{8 a-5 b}{8 a+5 d}=\frac{8 c+5 d}{8 c+5 d}$
Applying componendo and Dividendo
$\frac{8 a-5 b+8 a+5 d}{8 a-5 b-8 a-5 d}=\frac{8 c-5 d+8 c+5 d}{8 c-5 d-8 c-5 d} $
$\frac{16 a}{-\frac{a}{a}-10 b}=\frac{16 c}{-10 d} $
$\frac{b}{b}=\frac{c}{d}$
Hence Proved.
View full question & answer→Question 53 Marks
If $a=\frac{b+c}{2}, c=\frac{a+b}{2}$ and b is mean proportional between a and c, prove that $\frac{1}{a}+\frac{1}{c}=\frac{1}{b}$.
Answerb is the mean proportional of a and c
$ b ^2= ac$
$ b ^2=\left(\frac{b+c}{2}\right) \cdot\left(\frac{a+b}{2}\right) $
$ \Rightarrow 4 b^2=a b+a c+b^2+b c \quad \ldots\left[\because b^2=a c\right]$
$ \Rightarrow 4 b^2=a b+2 b^2+b c \\ 2 b^2-a b+b c,$
[Dividing both sides by abc]
$\Rightarrow \frac{2 b^2}{a b c}=\frac{a b}{a b c}+\frac{b c}{a b c} $
$\Rightarrow \frac{2}{b}=\frac{1}{c}+\frac{1}{a}, \quad \ldots\left[\because b^2= ac \right]$
Hence proved.
View full question & answer→Question 63 Marks
If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$, prove that $
\frac{x^3}{a^2}+\frac{y^2}{b^2}+\frac{z^3}{c^2}=\frac{(x+y+z)^3}{(a+b++c)^2}
$
AnswerLet $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k \quad$...[By k method $]$
$x=a k, y=b k$ and $z=c k$.
L.H.S.
$=\frac{a^3 k^3}{a^2}+\frac{b^3 k^3}{b^2}+\frac{c^3 k^3}{c^2} $
$\Rightarrow k^3[a+b+c]$
R.H.S.
$=\frac{[a k+b k+c k]^3}{[a+b+c]^2} $
$\Rightarrow \frac{k^3[a+b+c]^3}{[a+b+c]^2} $
$=k^3(a+b+c) $
$\text { L.H.S. }=\text { R.H.S. }$
Hence proved.
View full question & answer→Question 73 Marks
If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$, show that $\frac{x^3}{a^3}-\frac{y^3}{b^3}=\frac{z^3}{c^3}=\frac{x y z}{z b c}$.
AnswerLet $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$
$x=a k, y=b k, z=c k$
L.H.S.
$=\frac{x^3}{a^3}-\frac{y^3}{b^3}+\frac{z^3}{c^3}$
$\Rightarrow \frac{a^3 k^3}{a^3}-\frac{b^3 k^3}{b^3}-\frac{c^3 k^3}{c^3}$
$\Rightarrow k ^3- k ^3+ k ^3= k ^3$
$\text { R.H.S. } $
$=\frac{x y z}{a b c} $
$\Rightarrow \frac{a k \cdot b k \cdot c k}{a b c} $
$\Rightarrow \frac{k^3 a b c}{a b c} $
$=\text { k }^3 $
$\text { L.H.S. }=\text { R.H.S. }$
Hence proved.
View full question & answer→Question 83 Marks
If $\frac{a}{b}=\frac{c}{d}=\frac{r}{f}$, prove that $\left(\frac{a^2 b^2+c^2 d^2+e^2 f^2}{a b^3+c d^3+e f^3}\right)^{\frac{3}{2}}=\sqrt{\frac{a c e}{b d f}}$
Answer$\text { Let } \frac{a}{b}=\frac{c}{d}=\frac{r}{f}=k $
$\therefore a = bk , c = dk \text {, e }= fk $
$\text { L.H.S. } $
$=\left(\frac{a^2 b^2+c^2 d^2+e^2 f^2}{a b^3+c d^3+e f^3}\right)^{\frac{3}{2}} $
$=\left(\frac{b^2 k^2 \cdot b^2+d^2 k^2 \cdot d^2+f^2 k^2 \cdot f^2}{b k \cdot b^3+d k \cdot d^3+f k \cdot f^3}\right)^{\frac{3}{2}} $
$=\left[\frac{k^2\left(b^4+d^4+f^4\right)}{k\left(b^4+d^4+f^4\right)}\right]^{\frac{3}{2}} $
$=k^{\frac{3}{2}} $
$\text { R.H.S. }=\sqrt{\frac{a c e}{b d f}}=\sqrt{\frac{b k \cdot d k \cdot f k}{b d f}}=k^{\frac{3}{2}} $
$\text { L.H.S. }=\text { R.H.S. }$
Hence proved.
View full question & answer→Question 93 Marks
If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$, prove that$(a b+c d+e f)^2=\left(a^2+c^2+e^2\right)\left(b^2+d^2+f^2\right)$.
AnswerLet $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$ then
$a = bk , c = dk$ and $e = fk$
L.H.S.
$=(a b+c d+e f)^2 $
$=(b k \cdot b+d k \cdot d+f k \cdot f)^2 $
$=k^2\left(b^2+d^2+f^2\right)^2$
R.H.S.
$=\left(a^2+c^2+e^2\right)\left(b^2+d^2+f^2\right) $
$=\left(b^2 k^2+d^2 k^2+f^2 k^2\right)\left(b^2+d^2+f^2\right) $
$=k^2\left(b^2+d^2+f^2\right)\left(b^2+d^2+f^2\right) $
$=k^2\left(b^2+d^2+f^2\right)^2$
L.H.S. = R.H.S.
Hence proved.
View full question & answer→Question 103 Marks
If $a : b = c : d$, show that $(a - c) b^2 : (b - d) cd = (a^2 - b^2 - ab) : (c^2 - d^2 - cd)$.
AnswerLet $\frac{a}{b}=\frac{c}{d}=k$
$\Rightarrow a = bk$ and $c = dk$
L.H.S.
$=\frac{(a-c) b^2}{(b-d) c d}=\frac{(b k-d k) b^2}{(b-d) d k \cdot d}$
$=\frac{b^2 k(b-d)}{d 2 k(b-d)}=\frac{b^2}{d^2}$
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 \cdot b}{d^2 k^2-d^2-d k \cdot d} $
$=\frac{b^2\left(k^2-k-1\right)}{d^2\left(k^2-k-1\right)} $
$=\frac{b^2}{d^2}$
L.H.S. = R.H.S.
Hence proved.
View full question & answer→Question 113 Marks
If $\frac{a}{b}=\frac{c}{d}$ Show that a + b : c + d =$\sqrt{a^2+b^2}: \sqrt{c^2+d^2}$.
Answer$\text { Let } \frac{a}{b}=\frac{c}{d}=k $
$ \Rightarrow a = bk \text { and } 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}$
R.H.S.
$=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}=\frac{\sqrt{b^2 k^2+b^2}}{\sqrt{d^2 k^2+d^2}}$
$ =\frac{b\left(\sqrt{k^2+1}\right)}{d\left(\sqrt{k^2+1}\right)}=\frac{b}{d}$
L.H.S. = R.H.S.
Hence proved.
View full question & answer→Question 123 Marks
If q is the mean proportional between p and r, prove that
$p^2-q^2+r^2=q^4\left[\frac{1}{p^2}-\frac{1}{q^2}+\frac{1}{r^2}\right]$.
AnswerSince, q is the mean proportional of p and r.
Hence, $q_2=$ pr.
$\text { R.H.S. }=q^4\left[\frac{1}{p^2}-\frac{1}{q^2}+\frac{1}{r^2}\right]$
$=q^4\left[\frac{1}{p^2}-\frac{1}{p r}+\frac{1}{r^2}\right] $
$ =q^4\left[\frac{r^2-p r+p^2}{p^2 r^2}\right]$
$ =q^4\left[\frac{p^2-p r+r^2}{(p r)^2}\right]$
$ =q^4\left[\frac{p^2-p r+r^2}{q^4}\right] $
$=p^2-p r+r^2$
$ =p^2-q^2+r^2=\text { L.H.S. }$
Hence proved.
View full question & answer→Question 133 Marks
Given : x = $\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}$
Use componendo and dividendo to prove that $b^2=\frac{2 a^2 x}{x^2+1}$.
Answer$x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}$
By componendo and dividendo,
$\frac{x+1}{x-1}=\frac{2 \sqrt{a^2+b^2}}{2 \sqrt{a^2-b^2}}$
Squaring both sides,
$\frac{x^2+2 x+1}{x^2-2 x+1}=\frac{a^2+b^2}{a^2-b^2}$
By componendo and dividendo.
$\frac{2\left(x^2+1\right)}{4 x}=\frac{2 a^2}{2 b^2}$
$\Rightarrow \frac{x^2+1}{2 x}=\frac{a^2}{b^2}$
$\Rightarrow b ^2=\frac{2 a^2 x}{x^2+1} .$
Hence proved.
View full question & answer→Question 143 Marks
Divide $Rs. 720$ between Sunil, Sbhil and Akhil. So that Sunil gets $4/5$ of Sohil’s and Akhil’s share together and Sohil gets $2/3$ of Akhil’s share.
AnswerLet Akhil's share be ₹ x then
Sohil's share = ₹$\frac{2}{3} x$
and Sunil's share = ₹$\frac{4}{5}\left(\frac{2}{3} x+x\right)$
= ₹$\frac{4}{3} x$.
According to the question
$\frac{4}{3} x+\frac{2}{3} x+x=720$
$ \Rightarrow \frac{4 x+2 x+3 x}{3}=720 $
$ \Rightarrow 9 x=720 \times 3 $
$ \Rightarrow x=\frac{720 \times 3}{9}=240$
$\therefore$ Their respective shares are $₹320, ₹160$ and $₹240.$
View full question & answer→Question 153 Marks
If b is the mean proportional between a and c, prove that $\frac{a^2-b^2+c^2}{a^{-2}-b^{-2}+c^{-2}}$ = $b^4$.
AnswerSince, b is the mean proportional between a and c. So, $b^2$ = ac.
L.H.S. $=\frac{a^2-b^2+c^2}{a^{-2}-b^{-2}+c^{-2}}$
$=\frac{a^2-b^2+c^2}{\frac{1}{a^2}-\frac{1}{b^2}+\frac{1}{c^2}} $
$ =\frac{\left(a^2-b^2+c^2\right)}{\frac{b^2 c^2-a^2 c^2+a^2 b^2}{a^2 b^2 c^2}} $
$ =\frac{a^2 b^2 c^2\left(a^2-b^2+c^2\right)}{b^2 c^2-b^4+a^2 b^2}$
$=\frac{b^4 \times b^2\left(a^2-b^2+c^2\right)}{b^2\left(c^2-b^2+a^2\right)} $
$b^4=\text { R.H.S. }$
Hence proved.
View full question & answer→Question 163 Marks
Given that $\frac{a^3+3 a b^2}{b^2+3 a^2 b}=\frac{63}{62}$.
Using Componendo and Dividendo find $a : b.$
AnswerWe have
$\frac{a^3+3 a b^2}{b^2+3 a^2 b}=\frac{63}{62}$
App. compoenedo and dividendo
$\frac{a^3+3 a b^2+b^3+3 a^2 b}{a^3+3 a b^2-b^3-3 a^2 b}=\frac{63+62}{63-62} $
$ \frac{a^3+3 a b^2+b^3+3 a^2 b}{a^3+3 a b^2-b^3-3 a^2 b}=\frac{125}{1}$
$ \frac{(a+b)^3}{(a-b)^3}=\frac{125}{1} $
$ \frac{a+b}{a-b}=\frac{5}{1}$
Again Applying Componendo & Dividendo
$\frac{a+b+a-b}{a+b-a+b}=\frac{5+1}{5-1} $
$ \frac{2 a}{2 b}=\frac{6}{4}$
$ a : b =3: 2$
View full question & answer→Question 173 Marks
Using the properties of proportion, solve for x, given.
$\frac{x^4+1}{2 x^2}=\frac{17}{8}$
Answer$\frac{x^4+1}{2 x^2}=\frac{17}{8}$
Using Componendo and Dividendo
$\frac{x^2+1+2 x^2}{x^4+1-2 x^2}=\frac{17+8}{17-8} $
$ \Rightarrow \frac{\left(x^2+1\right)^2}{\left(x^2-1\right)^2}=\frac{25}{9}$
$\Rightarrow \frac{x^2+1}{x^2-1}=\frac{5}{3} \ldots$ (taking square root on both the sides)
Again applying Componendo and Dividendo
$\frac{x^2+1+x^2-1}{x^2+1-x^2+1}=\frac{5+3}{5-3}$
$ \frac{2 x^2}{2}=\frac{8}{2} $
$ \Rightarrow x^2-4 \\ \Rightarrow x= \pm 2 .\end{array}$
View full question & answer→Question 183 Marks
Solve for x: $\frac{3 x^2+5 x+18}{5 x^2+6 x+12}=\frac{3 x+5}{5 x-6}$.
AnswerMultiplying the Numerator and Denominator of R.H.S. by -x
$\frac{3 x^2+5 x+18}{5 x^2+6 x+12}=\frac{-3 x^2+5}{-5 x^2-6}$
Since, each ratio $=\frac{\text { Sum of antecedents }}{\text { Sum of consequents }}$
So $\frac{3 x^2+5 x+18-3 x^2-5 x}{5 x^2+6 x+12-5 x^2-6 x}$
$=\frac{-3 x^2-5 x}{-5 x^2-6 x} $
$ \frac{18}{12}=\frac{-3 x^2-5 x}{-5 x^2-6 x}$
$ \Rightarrow \frac{3}{2}=\frac{-3 x^2-5 x}{-5 x^2-6 x} $
$ \Rightarrow \frac{3}{2}=\frac{3 x+5}{5 x+6} $
$\Rightarrow 15 x+18=+x+10$
$ \Rightarrow 9 x=-8$
$ \Rightarrow x=\frac{-8}{9} .$
View full question & answer→Question 193 Marks
Arrange $5: 6, 8: 9, 13: 18$ and $7: 12$ in ascending order of magnitude.
AnswerThe given ratios are $5: 6, 8: 9, 13: 18$ and $7: 12.$
Now L.C.M., of $6, 9, 18$ and $12 = 36$
Now $\frac{5}{6}=\frac{5 \times 6}{6 \times 6}=\frac{30}{36}$
$\frac{8}{9}=\frac{8 \times 4}{9 \times 4}=\frac{32}{36}$
$\frac{13}{18}=\frac{13 \times 2}{18 \times 2}=\frac{26}{36}$
Also $\frac{7}{12}=\frac{7 \times 3}{12 \times 3}=\frac{21}{36}$
Here $\frac{21}{36}<\frac{26}{36}<\frac{30}{36}<\frac{32}{36}$
Hence $\frac{21}{36}<\frac{26}{36}<\frac{30}{36}<\frac{32}{36}$
i.e., Ascending order
$7 : 12 < 13 : 18 < 5 : 6 < 8 : 9.$
View full question & answer→Question 203 Marks
Using componendo and idendo, find the value of x
$\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+4}-\sqrt{3 x-5}}=9$
Answer$\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+4}-\sqrt{3 x-5}}=\frac{9}{1}$
Using componendo and dividendo
$\frac{\sqrt{3 x+4}+\sqrt{3 x-5}+\sqrt{3 x+4}-\sqrt{3 x-5}}{\sqrt{3 x+4}+\sqrt{3 x-5}-\sqrt{3 x+4}+\sqrt{3 x-5}}$
$=\frac{9+1}{9-1}=\frac{10}{8}=\frac{5}{4} $
$ \frac{2 \sqrt{3 x+4}}{2 \sqrt{3 x-5}}=\frac{5}{4}$
$\Rightarrow \frac{3 x+4}{3 x-5}=\frac{25}{16} \ldots$ (squaring both sides)
$48 x+64=75 x-125 $
$ \Rightarrow 75 x-48 x=125+64$
$ 27 x=189 $
$\Rightarrow x=\frac{189}{27}$
$ =7.$
View full question & answer→Question 213 Marks
Find the two numbers such that their mean proprtional is $24$ and the third proportinal is $1,536.$
AnswerLet x and y be two numbers
Mean proportional $= 24$
$\Rightarrow \sqrt{x y}=24 $
$ \Rightarrow x y=24 \times 24=576$
$\Rightarrow x=\frac{576}{y} ...(1)$
Also $1536$ is the third proportional then
$x : y = y : 1,536$
$\Rightarrow \frac{x}{y}=\frac{y}{1,536} $
$ \text { From(1), } y ^2=1,536 \times \frac{576}{y} $
$ \Rightarrow y ^3=1,536 \times 576 $
$ \Rightarrow y ^3=24 \times 24 \times 24 \times 24$
$\Rightarrow y=24 \times 24 $
$ \Rightarrow y=96$
Again form $(1),$ we get
$x=\frac{576}{96} $
$=6.$
Hence, the required numbers are $6$ and $$96.$
View full question & answer→Question 223 Marks
Find the fourth proportional to:
$x^3 - y^2, x^4 + x^2y^2+ y^4, x - y$.
AnswerLet A be the fourth proportional then
$ x ^3- y ^2: x ^4+ x ^2 y ^2+ y ^4= x - y : A $
$\Rightarrow \frac{x^3-y^3}{x^4+x^2 y^2+y^4}=\frac{x-y}{ A } $
$ \Rightarrow A \left( x ^3- y ^3\right)=( x - y )\left( x ^4+ x ^2 y ^2+ y ^4\right) $
$\Rightarrow A =\frac{(x-y)\left(x^4+x^2 y^2+y^4\right)}{x^3-y^3} $
$ \Rightarrow A =\frac{(x-y)\left(x^2+y^2+x y\right)\left(x^2+y^2-x y\right)}{(x-y)\left(x^2+x y+y^2\right)} $
$\Rightarrow A = x ^2+ y ^2- xy .$
View full question & answer→Question 233 Marks
The work done by (x – 3) men in (2x + 1) days and the work done by (2x + 1) men in (x + 4) days are in the ratio of 3: 10. Find the value of x.
Answer(x - 3) men do a work in (2x + 1) day
∴ 1 man does it in (2x + 1) (x - 3) days
(2x + 1) men do a work in (x + 4) days
∴ 1 man does it in (x + 4) (2x + 1) days
$\therefore \frac{(2 x+1)(x-3)}{(x+4)(2 x+1)}=\frac{3}{10}$
$\frac{x-3}{x+4}=\frac{3}{10}$
$10 x-30=3 x+12$
$10-3 x=12+30$
$7 x=42$
$x=\frac{42}{7}$
$x=6 .$
View full question & answer→Question 243 Marks
What quantity must be added to each term of the ratio $a + b: a - b$ to make it equal to $(a + b)^2 : (a - b)^2 $?
AnswerLet the quantity to be added be x.
Then
$\frac{(a+b)+x}{(a-b)+x}=\frac{(a+b)^2}{(a-b)^2} $
$ \Rightarrow(a+b)(a-b)^2+(a-b)^2 \cdot x$
$ =(a+b)^2(a-b)+(a+b)^2 \cdot x $
$ \Rightarrow\left[(a+b)^2-(a-b)^2\right] x $
$ =\left(a^2-b^2\right)(a-b)-\left(a^2-b^2\right)(a+b)$
$ \Rightarrow(4 a b x)=\left(a^2-b^2\right)[(a-b)-(a+b)]$
$\Rightarrow x=\frac{-2 b\left(a^2-b^2\right)}{4 a b}=\frac{b^2-a^2}{2 a} .$
View full question & answer→