Question 15 Marks
Make $y$ the subject of the formula $\frac{x}{ a }+\frac{y}{ b }=1$. Find $y$, when $a =2, b =8$ and $x =5$.
Answer$\frac{x}{ a }+\frac{y}{ b }=1 $
$\Rightarrow \frac{y}{ b }=1-\frac{x}{ a } $
$\Rightarrow \frac{y}{ b }=\frac{ a -x}{ a } $
$\Rightarrow y = b \left(1-\frac{x}{ a }\right) $
$\Rightarrow y = b -\frac{ b }{ a } x$
Substituting $a=2, b=8$ and $x=5$, we get
$y=8-\frac{8}{2} \times 5$
$=-12$
View full question & answer→Question 25 Marks
Make $x$ the subject of the formula $y =\frac{1-x^2}{1+x^2}$, Find $x$, when $y =\frac{1}{2}$
Answer$y=\frac{1-x^2}{1+x^2} $
$\Rightarrow y\left(1+x^2\right)=1-x^2 $
$\Rightarrow y+y^2=1-x^2 $
$\Rightarrow y^2+x^2=1-y $
$\Rightarrow x^2(1+y)=1-y $
$\Rightarrow x^2=\frac{1-y}{1+y} $
$\Rightarrow x=\sqrt{\frac{1-y}{1+y}}$
Substituting $y =\frac{1}{2}$, we get
$x=\sqrt{\frac{1-\frac{1}{2}}{1+\frac{1}{2}}} $
$=\sqrt{\frac{1}{2}} .$
View full question & answer→Question 35 Marks
Make $h$ the subject of the formula $K=\sqrt{\frac{h g}{d^2}-a^2}$. Find $h$, when $k=-2, a=-3, d=8$ and $g=32$.
Answer$K=\sqrt{\frac{ hg }{ d ^2}- a ^2}$ Squaring both sides, we get
$\Rightarrow K^2=\frac{h g}{d^2}-a^2 $
$\Rightarrow K^2+a^2=\frac{h g}{d^2} $
$\Rightarrow\left(K^2+a^2\right) d^2=h g $
$\Rightarrow h=\frac{\left(K^2+a^2\right) d^2}{g}$
Substituting $k =2, a =-3, d =8$ and $g =32$, we get
$h=\frac{\left((-2)^2+(-3)^2\right)(8)^2}{32} $
$=\frac{(4+9) 64}{32} $
$=26 .$
View full question & answer→Question 45 Marks
Make $x$ the subject of the formula $a=1-\frac{2 b}{c x-b}$. Find $x$, when $a=5, b=12$ and
Answer$a=1-\frac{2 b}{c x-b} $
$\Rightarrow a-1=-\frac{2 b}{c x-b} $
$\Rightarrow(a-1)(c x-b)+2 b=0 $
$\Rightarrow a c x-a b-c x+3 b=0 $
$\Rightarrow x(a c-c)+b(3-a)=0 $
$\Rightarrow x c(a-1)=-b(3-a) $
$\Rightarrow x=\frac{b(a-3)}{c(a-1)}$
Substituting $a=5, b=12$ and $c=2$, we get
$x=\frac{12(5-3)}{2(5-1)} $
$=\frac{12 \times 2}{2 \times 4} $
$=3$
View full question & answer→Question 55 Marks
Make a the subject of the formula $S=\frac{n}{2}\{2 a+(n-1) d\}$. Find $a$ when $S=50, n=10$ and $d=2$.
Answer$S=\frac{n}{2}\{2 a+(n-1) d\} $
$\Rightarrow 2 S=n\{2 a+(n-1) d\} $
$\Rightarrow \frac{2 s}{n}=2 a+(n-1) d $
$\Rightarrow \frac{2 S}{n}-(n-1) d=2 a $
$\Rightarrow\left\{\frac{S}{n}-\frac{(n-1) d}{2}\right\}=a$
Substituting $S=50, n=10$ and $d=2$, we get
$a=\left\{\frac{S}{n}-\frac{(n-1) d}{2}\right\} $
$=\left\{\frac{50}{10}-\frac{9 \times 2}{2}\right\} $
$=5-9 $
$=-4 .$
View full question & answer→Question 65 Marks
Make $y$ the subject of the formula $x=\frac{1-y^2}{1+y^2}$. Find $y$ if $x=\frac{3}{5}$
Answer$x=\frac{1-y^2}{1+y^2} $
$\Rightarrow x\left(1+y^2\right)=1-y^2 $
$\Rightarrow x+y^2=1-y^2 $
$\Rightarrow x y^2+y^2=1-x $
$\Rightarrow y^2(x+1)=1-x $
$\Rightarrow y^2=\frac{1-x}{1+x} $
$\Rightarrow y=\sqrt{\frac{1-x}{1+x}}$
Substituting $x =\frac{3}{5}$, we get
$y=\sqrt{\frac{1-\frac{3}{5}}{1+\frac{3}{5}}} $
$=\sqrt{\frac{2}{8}} $
$=\sqrt{\frac{1}{4}} $
$=\frac{1}{2} .$
View full question & answer→Question 75 Marks
If $s=\frac{n}{2}[2 a+(n-1) d]$, the $n$ express $d$ in terms of $s, a$ and $n$, find $d$ if $n=3, a=n+1$ and $s=18$.
Answer$s=\frac{n}{2}[2 a+(n-1) d] $
$\Rightarrow s=a n+\frac{n(n-1) d}{2} $
$\Rightarrow s-a n=\frac{n(n-1) d}{2} $
$\Rightarrow\left(\frac{2 s-2 a n}{n(n-1)}\right)=d $
$\Rightarrow d=\frac{2}{n(n-1)}(s-a n)$
Given that $n=3, a=n+1$ and $s=18$
Since
$a=n+1 $
$\Rightarrow a=3+1 $
$=4$
Substituting we get
$\Rightarrow d=\frac{2}{3(3-1)}(18-(4)(3)) $
$\Rightarrow d=\frac{2}{3(2)}(18-12) $
$\Rightarrow d=\frac{1}{3}(6) $
$\Rightarrow d=2 .$
View full question & answer→Question 85 Marks
The total energy $E$ possess by a body of Mass $' m\ '$, moving with a velocity $' v\ '$ at a height $' h\ '$ is given by: $E=\frac{1}{2} m u^2+m g h$. Find $m$, if $v=2, g=10, h=5$ and $E=104$.
Answer$E=\frac{1}{2} m u^2+m g h $
$\Rightarrow E=m\left(\frac{1}{2} u^2+g h\right) $
$\Rightarrow m=\frac{E}{\frac{1}{2} u^2+g h} $
$\Rightarrow m=\frac{E}{\frac{u^2+2 g h}{2}} $
$\Rightarrow m=\frac{2 E}{u^2+2 g h}$
Given that $v=2, g=10, h=5$ and $E=104$
$\Rightarrow m=\frac{2(104)}{(2)^2+2(10)(5)} $
$\Rightarrow m=\frac{208}{4+100} $
$\Rightarrow m=2$
View full question & answer→Question 95 Marks
$"\ $The volume of a cone $V$ is equal to the product of one third of $\pi$ and square of radius $r$ of the base and the height $h^{\prime \prime}$. Express this statement as a formula. Make $r$ the subject formula. Find$r$, when $V=1232 \ cm ^3, \pi=\frac{22}{7}, h =24 \ cm$.
AnswerVolume of cone $= V$
Product of one third of $\pi$ and square of radius $r$ of the base and the height $h=\frac{1}{3} \pi r^2 h$
So, $V=\frac{1}{3} \pi r^2 h$
$\Rightarrow \frac{3 V }{\pi h }= r ^2 $
$\Rightarrow r =\sqrt{\frac{3 V }{\pi h }}$
Substituting $V =1232 \ cm ^3, \pi=\frac{22}{7}, h =24 \ cm$
$\Rightarrow r =\sqrt{\frac{3 \times 1232}{\frac{22}{7} \times 24}} $
$=\sqrt{49}$
$=7\ cm .$
View full question & answer→Question 105 Marks
$"$The volume of a cylinder $V$ is equal to the product of $\pi$ and square of radius $r$ and the height $h "$. Express this statement as a formula. Make $r$ the subject formula. Find $r _{ x }$, when $V =44 \ cm ^3, \pi=\frac{22}{7}, h =14 \ cm$.
AnswerVolume of cylinder $= V$
Product of $\pi$ and Square of radius $r$ and the height $h=\pi r^2 h$
i.e. $V=\pi r^2 h$
$V=\pi r^2 h$
$\Rightarrow \frac{V}{h \pi}=r^2$
$\Rightarrow r =\sqrt{\frac{ V }{\pi h }}$
When $V=44 \ cm ^3, \pi=\frac{22}{7}, h=14 \ cm$
$\Rightarrow r =\sqrt{\frac{44}{\frac{22}{7} \times 14}} $
$=\sqrt{1}$
View full question & answer→Question 115 Marks
Make $z$ the subject of the formula $y =\frac{2 z+1}{2 z-1}$. If $x =\frac{y+1}{y-1}$, express $z$ in terms of $x _{ r }$ and find its value when $x=34$.
Answer$y=\frac{2 z+1}{2 z-1} $
$\Rightarrow(2 z-1) y=2 z+1 $
$\Rightarrow 2 z y-y=2 z+1 $
$\Rightarrow 2 z y-2 z=1+y $
$\Rightarrow z(2 y-1)=1+y $
$\Rightarrow z=\frac{1+y}{2 y-1} $
$\Rightarrow x=\frac{y+1}{y-1} $
$\Rightarrow x=\frac{\left(\frac{2 z+1}{2 z-1}\right)+1}{\left(\frac{2 z+1}{2 z-1}\right)-1} $
$=\frac{2 z+1+2 z-1}{2 z+1-2 z+1} $
$=\frac{4 z}{2} $
$=2 z $
$\Rightarrow z=\frac{x}{2}$
Substituting $x=34$, we get
$z=\frac{34}{2} $
$=17 .$
View full question & answer→Question 125 Marks
Make $f$ the subject of the formula $D=\sqrt{\left(\frac{f+p}{f-p}\right)}$. Find $f_r$ when $D=13$ and $P=21$.
Answer$=\sqrt{\left(\frac{f+p}{f-p}\right)}$
squaring both sides, we get
$\Rightarrow D^2=\left(\frac{f+p}{f-p}\right) $
$\Rightarrow D^2(f-p)=(f+p) $
$\Rightarrow D^2 f-D^2 p=f+p $
$\Rightarrow D^2 f-f=p+D^2 p $
$\Rightarrow f\left(D^2-1\right)=p\left(D^2+1\right) $
$\Rightarrow f=\frac{p\left(D^2+1\right)}{\left(D^2-1\right)}$
Substituting the values of $D=13$ and $p=21$
$=\frac{21\left(13^{\ } 2+1\right)}{\left(13^2-1\right)} $
$=\frac{21 \times 170}{168}$
$=21.25$
View full question & answer→Question 135 Marks
Make $I$ the subject of the following $M=\frac{L}{F}\left(\frac{1}{2} N-C\right) \times I$. Find $I$, If $M=44, L=20, F=15, N=50$ and $C=13$.
Answer$M=\frac{L}{F}\left(\frac{1}{2} N-C\right) \times I $
$\Rightarrow M-L=\frac{1}{F}\left(\frac{1}{2} N-C\right) \times I $
$\Rightarrow F(M-L)=\left(\frac{1}{2} N-C\right) \times I $
$\Rightarrow F(M-L)=\left(\frac{N-2 C}{2}\right) \times I $
$\Rightarrow \frac{2 F(M-L)}{(N-2 C)}=I$
Substituting the values of $M=44, L=0, F=15, N=50$ and $C=30$, we get
$I=\frac{2 F(M-L)}{(N-2 C)} $
$=\frac{2 \times 15(44-20)}{50-2 \times 13} $
$=\frac{30 \times 24}{24} $
$=30 .$
View full question & answer→Question 145 Marks
Make $a$ the subject of formula $x=\sqrt{\frac{a+b}{a-b}}$
Answer$x=\sqrt{\frac{a+b}{a-b}}$
Squaring both sides
$\Rightarrow x^2=\frac{a+b}{a-b} $
$\Rightarrow x^2(a-b)=a+b $
$\Rightarrow x^2 a-x^2 b=a+b $
$\Rightarrow x^2 a-a=b+x^2 b $
$\Rightarrow a\left(x^2-1\right)=b\left(x^2+1\right) $
$\Rightarrow a=\frac{b\left(x^2+1\right)}{\left(x^2-1\right)} .$
View full question & answer→Question 155 Marks
If $b=\frac{2 a}{a-2}$, and $c=\frac{4 b-3}{3 b+4}$, then express $c$ in terms of $a$.
AnswerGiven $b =\frac{2 a }{ a -2}$, and $c =\frac{4 b -3}{3 b +4}$
Substituting $b =\frac{2 a }{ a -2}$ in $c =\frac{4 b -3}{3 b +4}$
$c=\frac{4\left(\frac{2 a}{a-2}\right)-3}{3\left(\frac{2 a}{a-2}\right)+4} $
$\Rightarrow c=\frac{\frac{8 a}{a-2}-3}{\frac{6 a}{a-2}+4} $
$\Rightarrow c=\frac{\frac{8 a-3(a-2)}{a-2}}{\frac{6 a+4(a-2)}{a-2}} $
$\Rightarrow c=\frac{8 a-3(a-2)}{6 a+4(a-2)} $
$\Rightarrow c=\frac{8 a-3 a+6}{6 a+4 a-8} $
$\Rightarrow c=\frac{5 a+6}{10 a-8} .$
View full question & answer→Question 165 Marks
If $V = pr^2h$ and $S = 2pr^2 + 2$prh, then express $V$ in terms of $S, p$ and $r.$
Answer$V=\pi r^2 h$ and $S=2 \pi r^2+2 \pi r h $
$S=2 \pi r^2+2 \pi r h $
$\Rightarrow 2 r h=S-2 \pi r^2 $
$\Rightarrow h=\frac{S-2 \pi r^2}{2 \pi r}$
Substitute $h$ in $V=\pi r 2 h$
$\Rightarrow V =\pi r ^2\left(\frac{ S -2 \pi r ^2}{2 \pi r }\right) $
$\Rightarrow V = r \left(\frac{ S -2 \pi r ^2}{2}\right) $
$\Rightarrow V =\frac{ Sr }{2}-\pi r ^3 .$
View full question & answer→Question 175 Marks
Apple cost $x$ rupees per dozen and mangoes cost $y$ rupees per score. Write a formula to find the total cost $C$ in rupees of $20$ apples and $30$ mangoes.
AnswerCost of $2$ apples $=x$ rupees $(1$ dozen $=12)$
Cost of $1$ apple $=\frac{x}{12}$ rupees
Cost of $20$ apple $=\frac{20 x }{12}$ rupees
Cost of $20$ mangoes $=y$ rupees $(1$ score $=20$ )
Cost of $1$ mango $=\frac{y}{20}$ rupees
Cost of $30$ mango $=\frac{30 y }{20}$ rupees
Total cost
$ =\frac{20 x}{12}+\frac{30 y}{20}$
$=\frac{20 x}{12}+\frac{3 y}{2}$
$=\frac{20 x+18 y}{12}$
$=\frac{10 x+9 y}{6} $
As per the data: $C=\frac{10 x +9 y }{6}$.
View full question & answer→