\displaystyle \textbf{Question 1: }\text{Frame a formula for each of the following statements:}

\displaystyle \text{(i) The area }(A)\text{ of a rectangle is equal to the product of its}
\displaystyle \text{length }(l)\text{ and breadth }(b).
\displaystyle A=l\times b

\displaystyle \text{(ii) The area }(A)\text{ of a triangle is half the product of its base}
\displaystyle (b)\text{ and height }(h).
\displaystyle A=\frac{1}{2}\times b\times h

\displaystyle \text{(iii) The volume }(V)\text{ of a cone is one-third the product of }\pi,
\displaystyle \text{the square of its radius }(r)\text{ and its height }(h).
\displaystyle V=\frac{1}{3}\pi r^2h

\displaystyle \text{(iv) The perimeter }(P)\text{ of a rectangle is twice the sum of its}
\displaystyle \text{length }(l)\text{ and breadth }(b).
\displaystyle P=2(l+b)

\displaystyle \text{(v) The perimeter }(P)\text{ of a square is four times its side }(s).
\displaystyle P=4s

\displaystyle \text{(vi) The distance }(s)\text{ through which a body falls freely from rest}
\displaystyle \text{under gravity is }4.9\text{ times the square of the time }(t).
\displaystyle s=4.9t^2

\displaystyle \text{(vii) The reciprocal of the focal length }(f)\text{ is equal to the sum}
\displaystyle \text{of the reciprocals of the object distance }(u)\text{ and image distance }(v).
\displaystyle \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\displaystyle \text{(viii) Two years ago, a man whose present age is }x\text{ years was three}
\displaystyle \text{times as old as his son, whose present age is }y\text{ years.}
\displaystyle x-2=3(y-2)

\displaystyle \text{(ix) A two-digit number having }x\text{ as the tens digit and }y\text{ as}
\displaystyle \text{the units digit is five times the sum of its digits.}
\displaystyle 10x+y=5(x+y)

\displaystyle \text{(x) The number of diagonals }(d)\text{ that can be drawn from one vertex}
\displaystyle \text{of an }n\text{-sided polygon to all the other vertices is }3\text{ less than }n.
\displaystyle d=n-3
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A workman is paid Rs. }x\text{ for each day he works}
\displaystyle \text{and fined Rs. }y\text{ for each day he is absent. If he works for }N
\displaystyle \text{days in a month of }30\text{ days, find an expression for his total}
\displaystyle \text{earnings }(E)\text{ in rupees.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of days absent}=30-N.
\displaystyle \text{Amount earned}=Nx.
\displaystyle \text{Amount fined}=(30-N)y.
\displaystyle \therefore E=Nx-(30-N)y.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A purse contains }x\text{ notes of Rs. }10\text{ each, } y\text{ notes of}
\displaystyle \text{Rs. }5\text{ each, }z\text{ coins of }50\text{ paise each and }t\text{ coins of } 5\text{ paise each.}
\displaystyle \text{Find the total money }(M)\text{ in rupees.}
\displaystyle \text{Answer:}
\displaystyle 50\text{ paise}=\frac{1}{2}\text{ rupee and } 5\text{ paise}=\frac{1}{20}\text{ rupee}.
\displaystyle \therefore M=10x+5y+\frac{1}{2}z+\frac{1}{20}t.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A shopkeeper buys }m\text{ kg of rice at Rs. }x \text{ per kg and}
\displaystyle \text{another }n\text{ kg of rice at Rs. }y\text{ per kg. He mixes the two}
\displaystyle \text{quantities and sells the mixture at Rs. }z\text{ per kg. Find an expression}
\displaystyle \text{for his (i) total profit and (ii) profit per cent.}
\displaystyle \text{Answer:}
\displaystyle \text{Total cost price}=mx+ny.
\displaystyle \text{Total selling price}=(m+n)z.
\displaystyle \text{(i) Total profit}=(m+n)z-(mx+ny).

\displaystyle \text{(ii) Profit per cent} =\frac{\text{Total profit}}{\text{Total cost price}}\times100
\displaystyle =\frac{(m+n)z-(mx+ny)}{mx+ny}\times100.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A man cycles for }p\text{ hours at }x\text{ km per hour} \text{ and for}
\displaystyle \text{another }q\text{ hours at }y\text{ km per hour. Find his average speed }(A)
\displaystyle \text{for the whole journey.}
\displaystyle \text{Answer:}
\displaystyle \text{Total distance}=px+qy.
\displaystyle \text{Total time}=p+q.
\displaystyle \therefore A=\frac{px+qy}{p+q}\text{ km per hour}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The average age of }x\text{ boys in a class is }y \text{ years. A new}
\displaystyle \text{boy of age }z\text{ years joins the class. Find the new average age }(A).
\displaystyle \text{Answer:}
\displaystyle \text{Total age of }x\text{ boys}=xy\text{ years}.
\displaystyle \text{New total age}=xy+z\text{ years}.
\displaystyle \text{New number of boys}=x+1.
\displaystyle \therefore A=\frac{xy+z}{x+1}\text{ years}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A cricketer has an average score of }85 \text{ runs per innings in}
\displaystyle x\text{ innings and an average of }63\text{ runs per innings in }y \text{ innings. Find the}
\displaystyle \text{average score }(A)\text{ per innings.}
\displaystyle \text{Answer:}
\displaystyle \text{Total runs scored in }x\text{ innings}=85x.
\displaystyle \text{Total runs scored in }y\text{ innings}=63y.
\displaystyle \text{Total runs scored}=85x+63y.
\displaystyle \text{Total number of innings}=x+y.
\displaystyle \therefore A=\frac{85x+63y}{x+y}\text{ runs per innings}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In a class of }x\text{ children, each of }y \text{ children pays Rs. }10
\displaystyle \text{and each of the remaining children pays Rs. }6\text{ for a charity show.}
\displaystyle \text{Find the total amount }(C)\text{ collected in rupees.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of remaining children}=x-y.
\displaystyle C=10y+6(x-y)
\displaystyle =10y+6x-6y=6x+4y.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A shopkeeper marks each article at Rs. }m \text{ and gives a }20\%
\displaystyle \text{discount on the marked price. If the cost price of an article is Rs. }C,
\displaystyle \text{find the formula for profit }(P).\text{ Find }P\text{ when }m=300 \text{ and }C=200.
\displaystyle \text{Answer:}
\displaystyle \text{Selling price}=m-\frac{20}{100}m=0.8m.
\displaystyle \therefore P=0.8m-C.
\displaystyle \text{When }m=300\text{ and }C=200,
\displaystyle P=0.8(300)-200=240-200=40.
\displaystyle \therefore \text{profit}=\text{Rs. }40.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The hiring charges }(h)\text{ of a taxi are Rs. }100 \text{ plus Rs. }8
\displaystyle \text{per km for the distance travelled beyond }25\text{ km. If the distance}
\displaystyle \text{travelled is }x\text{ km, write a formula for the hiring charges.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance travelled beyond }25\text{ km}=x-25.
\displaystyle \therefore h=100+8(x-25),\qquad x>25.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Make }b\text{ the subject of the formula } P=2(l+b).\text{ Find }b
\displaystyle \text{when }P=66\text{ and }l=18.
\displaystyle \text{Answer:}
\displaystyle P=2(l+b)
\displaystyle \frac{P}{2}=l+b
\displaystyle \therefore b=\frac{P}{2}-l.
\displaystyle \text{When }P=66\text{ and }l=18,
\displaystyle b=\frac{66}{2}-18=33-18=15.
\displaystyle \therefore b=15.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Given }s=ut-\frac{1}{2}gt^2.
\displaystyle \text{(i) Make }g\text{ the subject of the formula.}
\displaystyle \text{(ii) Find }g\text{ when }t=20,\ s=10\text{ and }u=50.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }s=ut-\frac{1}{2}gt^2
\displaystyle \frac{1}{2}gt^2=ut-s
\displaystyle \therefore g=\frac{2(ut-s)}{t^2}.

\displaystyle \text{(ii) When }t=20,\ s=10\text{ and }u=50,
\displaystyle g=\frac{2\{50(20)-10\}}{20^2}
\displaystyle =\frac{2(1000-10)}{400}
\displaystyle =\frac{1980}{400}=4.95.
\displaystyle \therefore g=4.95.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Given }T=2\pi\sqrt{\frac{l}{g}}.
\displaystyle \text{(i) Make }l\text{ the subject of the formula.}
\displaystyle \text{(ii) Find }l\text{ when }T=2,\ g=9.8\text{ and }\pi=\sqrt{10}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }T=2\pi\sqrt{\frac{l}{g}}
\displaystyle T^2=4\pi^2\frac{l}{g}
\displaystyle \therefore l=\frac{gT^2}{4\pi^2}.

\displaystyle \text{(ii) When }T=2,\ g=9.8\text{ and }\pi=\sqrt{10},
\displaystyle l=\frac{9.8\times2^2}{4(\sqrt{10})^2}
\displaystyle =\frac{39.2}{40}=0.98.
\displaystyle \therefore l=0.98.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Let }S= \frac{n}{2}\{2a+(n-1)d\}.
\displaystyle \text{(i) Find }a\text{ when }S=185,\ n=10\text{ and }d=3.
\displaystyle \text{(ii) Find }d\text{ when }a=11,\ n=10\text{ and }S=380.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }S=\frac{n}{2}\{2a+(n-1)d\}
\displaystyle \frac{2S}{n}=2a+(n-1)d
\displaystyle 2a=\frac{2S}{n}-(n-1)d
\displaystyle \therefore a=\frac{1}{2} \left\{\frac{2S}{n}-(n-1)d\right\}.
\displaystyle a=\frac{1}{2} \left\{\frac{2(185)}{10}-(10-1)(3)\right\}
\displaystyle =\frac{1}{2}(37-27)=5.
\displaystyle \therefore a=5.

\displaystyle \text{(ii) }S=\frac{n}{2}\{2a+(n-1)d\}
\displaystyle \frac{2S}{n}=2a+(n-1)d
\displaystyle (n-1)d=\frac{2S}{n}-2a
\displaystyle \therefore d=\frac{1}{n-1} \left\{\frac{2S}{n}-2a\right\}.
\displaystyle d=\frac{1}{10-1} \left\{\frac{2(380)}{10}-2(11)\right\}
\displaystyle =\frac{1}{9}(76-22)=6.
\displaystyle \therefore d=6.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Let }v=\sqrt{u^2+2ax}.
\displaystyle \text{(i) Make }x\text{ the subject of the formula.}
\displaystyle \text{(ii) Find }x\text{ when }v=35,\ u=25\text{ and }a=50.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }v=\sqrt{u^2+2ax}
\displaystyle v^2=u^2+2ax
\displaystyle 2ax=v^2-u^2
\displaystyle \therefore x=\frac{v^2-u^2}{2a}.

\displaystyle \text{(ii) When }v=35,\ u=25\text{ and }a=50,
\displaystyle x=\frac{35^2-25^2}{2(50)}
\displaystyle =\frac{1225-625}{100}=\frac{600}{100}=6.
\displaystyle \therefore x=6.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The volume }(V)\text{ of a hollow cylindrical pipe} \text{ with outer radius }(R),
\displaystyle \text{inner radius }(r)\text{ and length }(h)\text{ is given by the formula } V=\pi(R^2-r^2)h.
\displaystyle \text{(i) Make }h\text{ the subject of the formula.}
\displaystyle \text{(ii) Find }h\text{ when }R=2.6,\ r=2.3,\ \pi=\frac{22}{7} \text{ and }V=115.5.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }V=\pi(R^2-r^2)h
\displaystyle \therefore h=\frac{V}{\pi(R^2-r^2)}.

\displaystyle \text{(ii) When }R=2.6,\ r=2.3,\ \pi=\frac{22}{7} \text{ and }V=115.5,
\displaystyle h=\frac{115.5}{\frac{22}{7}(2.6^2-2.3^2)}
\displaystyle =\frac{115.5\times7}{22(6.76-5.29)}
\displaystyle =\frac{808.5}{22(1.47)}=\frac{808.5}{32.34}=25.
\displaystyle \therefore h=25.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Let }x=\frac{m+n}{m-n}.
\displaystyle \text{(i) Make }n\text{ the subject of the formula.}
\displaystyle \text{(ii) Find }n\text{ when }m=36\text{ and }x=2.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }x=\frac{m+n}{m-n}
\displaystyle x(m-n)=m+n
\displaystyle mx-nx=m+n
\displaystyle mx-m=nx+n
\displaystyle m(x-1)=n(x+1)
\displaystyle \therefore n=\frac{m(x-1)}{x+1}.

\displaystyle \text{(ii) When }m=36\text{ and }x=2,
\displaystyle n=\frac{36(2-1)}{2+1}
\displaystyle =\frac{36}{3}=12.
\displaystyle \therefore n=12.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Let }x=\frac{3-4p}{p+2q}.
\displaystyle \text{(i) Make }p\text{ the subject of the formula.}
\displaystyle \text{(ii) Find }p\text{ when }q=\frac{1}{2}\text{ and }x=\frac{1}{5}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }x=\frac{3-4p}{p+2q}
\displaystyle x(p+2q)=3-4p
\displaystyle px+2qx=3-4p
\displaystyle px+4p=3-2qx
\displaystyle p(x+4)=3-2qx
\displaystyle \therefore p=\frac{3-2qx}{x+4}.

\displaystyle \text{(ii) When }q=\frac{1}{2}\text{ and }x=\frac{1}{5},
\displaystyle p= \frac{3-2\left(\frac{1}{2}\right)\left(\frac{1}{5}\right)} {\frac{1}{5}+4}
\displaystyle =\frac{3-\frac{1}{5}}{\frac{21}{5}} =\frac{\frac{14}{5}}{\frac{21}{5}}
\displaystyle =\frac{14}{21}=\frac{2}{3}.
\displaystyle \therefore p=\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Let }R=\sqrt{\frac{3V}{\pi h}}.
\displaystyle \text{(i) Make }h\text{ the subject of the formula.}
\displaystyle \text{(ii) Find }h\text{ when }V=13.5,\ R=2.5\text{ and }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }R=\sqrt{\frac{3V}{\pi h}}
\displaystyle R^2=\frac{3V}{\pi h}
\displaystyle \pi R^2h=3V
\displaystyle \therefore h=\frac{3V}{\pi R^2}.

\displaystyle \text{(ii) When }V=13.5,\ R=2.5\text{ and }\pi=\frac{22}{7},
\displaystyle h=\frac{3(13.5)}{\frac{22}{7}(2.5)^2}
\displaystyle =\frac{40.5\times7}{22\times6.25}
\displaystyle =\frac{283.5}{137.5}\approx2.06.
\displaystyle \therefore h\approx2.06.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }\frac{1}{f}=\frac{1}{u}+\frac{1}{v}, \text{ find }u\text{ in terms of }v\text{ and }f.
\displaystyle \text{Find }u\text{ when }v=32\text{ and }f=24.
\displaystyle \text{Answer:}
\displaystyle \frac{1}{f}=\frac{1}{u}+\frac{1}{v}
\displaystyle \frac{1}{u}=\frac{1}{f}-\frac{1}{v}
\displaystyle \frac{1}{u}=\frac{v-f}{fv}
\displaystyle \therefore u=\frac{fv}{v-f}.
\displaystyle \text{When }v=32\text{ and }f=24,
\displaystyle u=\frac{24(32)}{32-24}
\displaystyle =\frac{768}{8}=96.
\displaystyle \therefore u=96.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }r=\sqrt{x^2+y^2}, \text{ express }y\text{ in terms of }r\text{ and }x.
\displaystyle \text{Find }y\text{ when }r=17\text{ and }x=8.
\displaystyle \text{Answer:}
\displaystyle r=\sqrt{x^2+y^2}
\displaystyle r^2=x^2+y^2
\displaystyle y^2=r^2-x^2
\displaystyle \therefore y=\pm\sqrt{r^2-x^2}.
\displaystyle \text{Taking the positive value, }y=\sqrt{r^2-x^2}.
\displaystyle \text{When }r=17\text{ and }x=8,
\displaystyle y=\sqrt{17^2-8^2}
\displaystyle =\sqrt{289-64}=\sqrt{225}=15.
\displaystyle \therefore y=15.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Make }b\text{ the subject of } x=\sqrt{\frac{a-b}{a+b}}.
\displaystyle \text{Answer:}
\displaystyle x=\sqrt{\frac{a-b}{a+b}}
\displaystyle x^2=\frac{a-b}{a+b}
\displaystyle x^2(a+b)=a-b
\displaystyle ax^2+bx^2=a-b
\displaystyle bx^2+b=a-ax^2
\displaystyle b(x^2+1)=a(1-x^2)
\displaystyle \therefore b=\frac{a(1-x^2)}{1+x^2}.
\displaystyle \\

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