\displaystyle \textbf{Question 1: } \text{The sum }S\text{ of }n\text{ successive odd numbers starting from }3
\displaystyle \text{is given by the relation }S=n(n+2).\text{ Determine }n,\text{ if the sum is }168.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S=n(n+2)
\displaystyle 168=n(n+2)
\displaystyle n^2+2n-168=0
\displaystyle (n-12)(n+14)=0
\displaystyle \Rightarrow n=12\text{ or }n=-14
\displaystyle \text{Since }n\text{ cannot be negative, }n\neq -14.
\displaystyle \therefore n=12
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\displaystyle \textbf{Question 2: } \text{A stone is thrown vertically downwards and the formula}
\displaystyle d=16t^2+4t\text{ gives the distance, }d\text{ meters, that it falls in }t\text{ seconds.}
\displaystyle \text{How long does it take to fall }420\text{ meters?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }d=16t^2+4t
\displaystyle 420=16t^2+4t
\displaystyle 16t^2+4t-420=0
\displaystyle 4t^2+t-105=0
\displaystyle 4t^2+21t-20t-105=0
\displaystyle t(4t+21)-5(4t+21)=0
\displaystyle (t-5)(4t+21)=0
\displaystyle \Rightarrow t=5\text{ or }t=-\frac{21}{4}
\displaystyle \text{Since time cannot be negative, }t\neq -\frac{21}{4}.
\displaystyle \therefore t=5
\displaystyle \therefore \text{The stone takes }5\text{ seconds to fall }420\text{ meters.}
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\displaystyle \textbf{Question 3: } \text{The product of the digits of a two-digit number is }24.
\displaystyle \text{If its unit's digit exceeds twice its ten's digit by }2,\text{ find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the ten's digit be }x\text{ and the unit's digit be }y.
\displaystyle \therefore xy=24
\displaystyle \text{Also, }y=2x+2
\displaystyle \therefore x(2x+2)=24
\displaystyle 2x^2+2x-24=0
\displaystyle x^2+x-12=0
\displaystyle (x+4)(x-3)=0
\displaystyle \Rightarrow x=-4\text{ or }x=3
\displaystyle \text{Since }x\text{ is a digit, }x\neq -4.
\displaystyle \therefore x=3
\displaystyle y=2(3)+2=8
\displaystyle \therefore \text{The required number is }38.
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\displaystyle \textbf{Question 4: } \text{The ages of two sisters are }11\text{ years and }14\text{ years.}
\displaystyle \text{In how many years' time will the product of their ages be }304?
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number of years be }n.
\displaystyle \therefore (11+n)(14+n)=304
\displaystyle 154+25n+n^2=304
\displaystyle n^2+25n-150=0
\displaystyle (n-5)(n+30)=0
\displaystyle \Rightarrow n=5\text{ or }n=-30
\displaystyle \text{Since time cannot be negative, }n\neq -30.
\displaystyle \therefore n=5
\displaystyle \therefore \text{The product of their ages will be }304\text{ in }5\text{ years.}
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\displaystyle \textbf{Question 5: } \text{One year ago, a man was }8\text{ times as old as his son.}
\displaystyle \text{Now his age is equal to the square of his son's age. Find their present ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the present age of the father be }x\text{ years.}
\displaystyle \text{Let the present age of the son be }y\text{ years.}
\displaystyle \text{Given, }x-1=8(y-1)
\displaystyle \text{Also, }x=y^2
\displaystyle \therefore y^2-1=8(y-1)
\displaystyle y^2-1=8y-8
\displaystyle y^2-8y+7=0
\displaystyle (y-7)(y-1)=0
\displaystyle \Rightarrow y=7\text{ or }y=1
\displaystyle \text{If }y=1,\text{ then one year ago son's age was }0,\text{ which is not possible.}
\displaystyle \therefore y=7
\displaystyle x=y^2=7^2=49
\displaystyle \therefore \text{The present ages of the father and son are }49\text{ years and }7\text{ years.}
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\displaystyle \textbf{Question 6: } \text{The age of a father is twice the square of the age of his son.}
\displaystyle \text{Eight years hence, the age of the father will be }4\text{ years more}
\displaystyle \text{than three times the age of the son. Find their present ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the present age of the father be }x\text{ years.}
\displaystyle \text{Let the present age of the son be }y\text{ years.}
\displaystyle \text{Given, }x=2y^2
\displaystyle \text{Also, }x+8=3(y+8)+4
\displaystyle 2y^2+8=3y+24+4
\displaystyle 2y^2+8=3y+28
\displaystyle 2y^2-3y-20=0
\displaystyle 2y^2-8y+5y-20=0
\displaystyle 2y(y-4)+5(y-4)=0
\displaystyle (y-4)(2y+5)=0
\displaystyle \Rightarrow y=4\text{ or }y=-\frac{5}{2}
\displaystyle \text{Since age cannot be negative, }y\neq -\frac{5}{2}.
\displaystyle \therefore y=4
\displaystyle x=2y^2=2(4)^2=32
\displaystyle \therefore \text{The present ages of the father and son are }32\text{ years and }4\text{ years.}
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\displaystyle \textbf{Question 7: } \text{The speed of a boat in still water is }15\text{ km/hr.}
\displaystyle \text{It can go }30\text{ km upstream and return downstream to the original point}
\displaystyle \text{in }4\text{ hours }30\text{ minutes. Find the speed of the stream.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the stream be }x\text{ km/hr.}
\displaystyle \text{Speed upstream}=(15-x)\text{ km/hr}
\displaystyle \text{Speed downstream}=(15+x)\text{ km/hr}
\displaystyle 4\text{ hours }30\text{ minutes}=\frac{9}{2}\text{ hours}
\displaystyle \therefore \frac{30}{15-x}+\frac{30}{15+x}=\frac{9}{2}
\displaystyle \frac{30(15+x)+30(15-x)}{(15-x)(15+x)}=\frac{9}{2}
\displaystyle \frac{900}{225-x^2}=\frac{9}{2}
\displaystyle 1800=9(225-x^2)
\displaystyle 200=225-x^2
\displaystyle x^2=25
\displaystyle x=\pm5
\displaystyle \text{Since speed cannot be negative, }x\neq -5.
\displaystyle \therefore x=5
\displaystyle \therefore \text{The speed of the stream is }5\text{ km/hr}.
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\displaystyle \textbf{Question 8: } \text{A person sends his servant to the market to buy oranges}
\displaystyle \text{worth Rs. }15.\text{ The servant having eaten three oranges on the way,}
\displaystyle \text{the person pays }25\text{ paisa per orange more than the market price.}
\displaystyle \text{Find the number of oranges received back by the person.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of oranges bought by the servant be }x.
\displaystyle \therefore \text{Market price of each orange}=\text{Rs. }\frac{15}{x}
\displaystyle \text{Number of oranges received back by the person}=(x-3)
\displaystyle \therefore \text{Actual price of each orange}=\text{Rs. }\frac{15}{x-3}
\displaystyle 25\text{ paisa}=\text{Rs. }\frac{1}{4}
\displaystyle \therefore \frac{15}{x-3}-\frac{15}{x}=\frac{1}{4}
\displaystyle \frac{15x-15(x-3)}{x(x-3)}=\frac{1}{4}
\displaystyle \frac{45}{x(x-3)}=\frac{1}{4}
\displaystyle x(x-3)=180
\displaystyle x^2-3x-180=0
\displaystyle (x-15)(x+12)=0
\displaystyle \Rightarrow x=15\text{ or }x=-12
\displaystyle \text{Since the number of oranges cannot be negative, }x\neq -12.
\displaystyle \therefore x=15
\displaystyle x-3=12
\displaystyle \therefore \text{The person receives }12\text{ oranges back.}
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\displaystyle \textbf{Question 9: } \text{Rs. }250\text{ is divided equally among a certain number}
\displaystyle \text{of children. If there were }25\text{ children more, each would have received}
\displaystyle 50\text{ paisa less. Find the number of children.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of children be }x.
\displaystyle \therefore \text{Amount received by each child}=\text{Rs. }\frac{250}{x}
\displaystyle \text{If there were }25\text{ children more, amount received by each child}=\text{Rs. }\frac{250}{x+25}
\displaystyle 50\text{ paisa}=\text{Rs. }\frac{1}{2}
\displaystyle \therefore \frac{250}{x}-\frac{250}{x+25}=\frac{1}{2}
\displaystyle \frac{250(x+25)-250x}{x(x+25)}=\frac{1}{2}
\displaystyle \frac{6250}{x(x+25)}=\frac{1}{2}
\displaystyle x(x+25)=12500
\displaystyle x^2+25x-12500=0
\displaystyle (x-100)(x+125)=0
\displaystyle \Rightarrow x=100\text{ or }x=-125
\displaystyle \text{Since the number of children cannot be negative, }x\neq -125.
\displaystyle \therefore x=100
\displaystyle \therefore \text{The number of children is }100.
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\displaystyle \textbf{Question 10: } \text{An employer finds that if he increases the weekly wages}
\displaystyle \text{of each worker by Rs. }5\text{ and employs five workers less, he increases}
\displaystyle \text{his weekly wage bill from Rs. }3150\text{ to Rs. }3250.\text{ Find the weekly wages}
\displaystyle \text{of the workers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of workers be }x.
\displaystyle \text{Let the weekly wage of each worker be Rs. }y.
\displaystyle xy=3150\qquad \ldots \ldots \text{(i)}
\displaystyle (x-5)(y+5)=3250
\displaystyle xy+5x-5y-25=3250
\displaystyle 3150+5x-5y-25=3250
\displaystyle 5x-5y=125
\displaystyle x-y=25
\displaystyle x=y+25\qquad \ldots \ldots  \text{(ii)}
\displaystyle \text{Substituting (ii) in (i),}
\displaystyle y(y+25)=3150
\displaystyle y^2+25y-3150=0
\displaystyle (y-45)(y+70)=0
\displaystyle \Rightarrow y=45\text{ or }y=-70
\displaystyle \text{Since wage cannot be negative, }y\neq -70.
\displaystyle \therefore y=45
\displaystyle \therefore \text{The weekly wage of each worker is Rs. }45.
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\displaystyle \textbf{Question 11: } \text{A trader bought a number of articles for Rs. }1200.
\displaystyle \text{Ten were damaged and he sold each of the remaining articles at Rs. }2
\displaystyle \text{more than what he paid for it, thus getting a profit of Rs. }60
\displaystyle \text{on the whole transaction. Find the number of articles bought.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of articles bought be }x.
\displaystyle \text{Cost price of one article}=\text{Rs. }\frac{1200}{x}
\displaystyle \text{Number of articles sold}=(x-10)
\displaystyle \text{Selling price of one article}=\text{Rs. }\left(\frac{1200}{x}+2\right)
\displaystyle \text{Total selling price}=\text{Rs. }(x-10)\left(\frac{1200}{x}+2\right)
\displaystyle \text{Since profit is Rs. }60,\text{ total selling price}=\text{Rs. }1260
\displaystyle (x-10)\left(\frac{1200}{x}+2\right)=1260
\displaystyle (x-10)(1200+2x)=1260x
\displaystyle 1200x-12000+2x^2-20x=1260x
\displaystyle 2x^2-80x-12000=0
\displaystyle x^2-40x-6000=0
\displaystyle (x-100)(x+60)=0
\displaystyle \Rightarrow x=100\text{ or }x=-60
\displaystyle \text{Since the number of articles cannot be negative, }x\neq -60.
\displaystyle \therefore x=100
\displaystyle \therefore \text{The number of articles bought is }100.
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\displaystyle \textbf{Question 12: } \text{The total cost price of a certain number of identical}
\displaystyle \text{articles is Rs. }4800.\text{ By selling the articles at Rs. }100\text{ each,}
\displaystyle \text{a profit equal to the cost price of }15\text{ articles is made. Find the}
\displaystyle \text{number of articles bought.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of articles bought be }x.
\displaystyle \text{Cost price of one article}=\text{Rs. }\frac{4800}{x}
\displaystyle \text{Profit}=\text{Cost price of }15\text{ articles}=\text{Rs. }\frac{4800}{x}\times15
\displaystyle \text{Total selling price}=100x
\displaystyle \therefore 100x-4800=\frac{4800}{x}\times15
\displaystyle 100x^2-4800x=72000
\displaystyle 100x^2-4800x-72000=0
\displaystyle x^2-48x-720=0
\displaystyle (x-60)(x+12)=0
\displaystyle \Rightarrow x=60\text{ or }x=-12
\displaystyle \text{Since the number of articles cannot be negative, }x\neq -12.
\displaystyle \therefore x=60
\displaystyle \therefore \text{The number of articles bought is }60.
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