\displaystyle \textbf{Question 1: }\text{A piece of cloth costs Rs. }35.\text{ If the piece were }4\text{ m longer and each metre}
\displaystyle \text{costs Rs. }1\text{ less, the cost would remain unchanged. How long is the piece?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the piece of cloth be }x\text{ m.}
\displaystyle \therefore \text{Original cost per metre}=\text{Rs. }\frac{35}{x}.
\displaystyle \text{New length of the piece}=(x+4)\text{ m.}
\displaystyle \therefore \text{New cost per metre}=\text{Rs. }\frac{35}{x+4}.
\displaystyle \text{According to the question,}
\displaystyle \frac{35}{x}-\frac{35}{x+4}=1
\displaystyle \Rightarrow \frac{35(x+4)-35x}{x(x+4)}=1
\displaystyle \Rightarrow \frac{140}{x(x+4)}=1
\displaystyle \Rightarrow x(x+4)=140
\displaystyle \Rightarrow x^2+4x-140=0
\displaystyle \Rightarrow (x+14)(x-10)=0
\displaystyle \Rightarrow x=-14\text{ or }x=10
\displaystyle \text{Since length cannot be negative, }x=-14\text{ is rejected.}
\displaystyle \therefore \text{The piece of cloth is }10\text{ m long.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Some students planned a picnic. The budget for food was Rs. }480.\text{ But eight of}
\displaystyle \text{these failed to go and thus the cost of food for each member increased by Rs. }10.\text{ How many}
\displaystyle \text{students attended the picnic?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of students who originally planned to attend be }x.
\displaystyle \therefore \text{Original cost of food per student}=\text{Rs. }\frac{480}{x}.
\displaystyle \text{Number of students who actually attended}=(x-8).
\displaystyle \therefore \text{New cost of food per student}=\text{Rs. }\frac{480}{x-8}.
\displaystyle \text{According to the question,}
\displaystyle \frac{480}{x-8}-\frac{480}{x}=10
\displaystyle \Rightarrow \frac{480x-480(x-8)}{x(x-8)}=10
\displaystyle \Rightarrow \frac{3840}{x(x-8)}=10
\displaystyle \Rightarrow x(x-8)=384
\displaystyle \Rightarrow x^2-8x-384=0
\displaystyle \Rightarrow (x-24)(x+16)=0
\displaystyle \Rightarrow x=24\text{ or }x=-16
\displaystyle \text{Since the number of students cannot be negative, }x=-16\text{ is rejected.}
\displaystyle \therefore \text{Number of students who attended}=24-8=16.
\displaystyle \therefore \text{The number of students who attended the picnic is }16.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the list price of a toy is reduced by Rs. }2,\text{ a person can buy }2\text{ toys more}
\displaystyle \text{for Rs. }360.\text{ Find the original price of the toy.}\hfill\text{[CBSE 2002 C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original price of the toy be Rs. }x.
\displaystyle \therefore \text{Number of toys bought for Rs. }360=\frac{360}{x}.
\displaystyle \text{Reduced price of the toy}=\text{Rs. }(x-2).
\displaystyle \therefore \text{Number of toys bought at reduced price}=\frac{360}{x-2}.
\displaystyle \text{According to the question,}
\displaystyle \frac{360}{x-2}-\frac{360}{x}=2
\displaystyle \Rightarrow \frac{360x-360(x-2)}{x(x-2)}=2
\displaystyle \Rightarrow \frac{720}{x(x-2)}=2
\displaystyle \Rightarrow x(x-2)=360
\displaystyle \Rightarrow x^2-2x-360=0
\displaystyle \Rightarrow (x-20)(x+18)=0
\displaystyle \Rightarrow x=20\text{ or }x=-18
\displaystyle \text{Since price cannot be negative, }x=-18\text{ is rejected.}
\displaystyle \therefore \text{The original price of the toy is Rs. }20.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Rs. }9000\text{ were divided equally among a certain number of persons. Had there been}
\displaystyle 20\text{ more persons, each would have got Rs. }160\text{ less. Find the original number of persons.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original number of persons be }x.
\displaystyle \therefore \text{Original share of each person}=\text{Rs. }\frac{9000}{x}.
\displaystyle \text{If there were }20\text{ more persons, their number would be }x+20.
\displaystyle \therefore \text{New share of each person}=\text{Rs. }\frac{9000}{x+20}.
\displaystyle \text{According to the question,}
\displaystyle \frac{9000}{x}-\frac{9000}{x+20}=160
\displaystyle \Rightarrow \frac{9000(x+20)-9000x}{x(x+20)}=160
\displaystyle \Rightarrow \frac{180000}{x(x+20)}=160
\displaystyle \Rightarrow x(x+20)=1125
\displaystyle \Rightarrow x^2+20x-1125=0
\displaystyle \Rightarrow (x-25)(x+45)=0
\displaystyle \Rightarrow x=25\text{ or }x=-45
\displaystyle \text{Since the number of persons cannot be negative, }x=-45\text{ is rejected.}
\displaystyle \therefore \text{The original number of persons is }25.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In a class test, the sum of Shefali's marks in Mathematics and English is }
\displaystyle 30.\text{ Had she got} \ 2\text{ marks more in Mathematics and }3\text{ marks less in English, the}
\displaystyle \text{product of her marks would have been }210.\text{ Find her marks in two subjects.}\hfill\text{[CBSE 2008, 14, 19]}
\displaystyle \text{Answer:}
\displaystyle \text{Let Shefali's marks in Mathematics be }x.
\displaystyle \therefore \text{Her marks in English}=30-x.
\displaystyle \text{According to the question,}
\displaystyle (x+2)(30-x-3)=210
\displaystyle \Rightarrow (x+2)(27-x)=210
\displaystyle \Rightarrow 27x-x^2+54-2x=210
\displaystyle \Rightarrow x^2-25x+156=0
\displaystyle \Rightarrow x^2-12x-13x+156=0
\displaystyle \Rightarrow x(x-12)-13(x-12)=0
\displaystyle \Rightarrow (x-12)(x-13)=0
\displaystyle \Rightarrow x=12\text{ or }x=13
\displaystyle \text{If }x=12,\text{ then English marks}=30-12=18.
\displaystyle \text{If }x=13,\text{ then English marks}=30-13=17.
\displaystyle \therefore \text{Her marks are }12\text{ in Mathematics and }18\text{ in English, or }13\text{ in Mathematics and }17\text{ in English.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A student scored a total of }32\text{ marks in class tests in Mathematics }
\displaystyle \text{and Science. Had he scored }2\text{ marks less in Science and }4\text{ marks more in Mathematics, the}
\displaystyle \text{product of his marks would have been }253.\text{ Find his marks in the two subjects.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the student's marks in Mathematics be }x.
\displaystyle \therefore \text{His marks in Science}=32-x.
\displaystyle \text{According to the question,}
\displaystyle (x+4)(32-x-2)=253
\displaystyle \Rightarrow (x+4)(30-x)=253
\displaystyle \Rightarrow 30x-x^2+120-4x=253
\displaystyle \Rightarrow x^2-26x+133=0
\displaystyle \Rightarrow x^2-7x-19x+133=0
\displaystyle \Rightarrow x(x-7)-19(x-7)=0
\displaystyle \Rightarrow (x-7)(x-19)=0
\displaystyle \Rightarrow x=7\text{ or }x=19
\displaystyle \text{If }x=7,\text{ then Science marks}=32-7=25.
\displaystyle \text{If }x=19,\text{ then Science marks}=32-19=13.
\displaystyle \therefore \text{His marks are }7\text{ in Mathematics and }25\text{ in Science, or }19\text{ in Mathematics and }13\text{ in Science.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A dealer sells an article for Rs. }24\text{ and gains as much percent as the cost price}
\displaystyle \text{of the article. Find the cost price of the article.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost price of the article be Rs. }x.
\displaystyle \therefore \text{Gain percent}=x\%.
\displaystyle \text{Gain}=\text{Rs. }\frac{x}{100}\times x=\text{Rs. }\frac{x^2}{100}.
\displaystyle \text{Selling price}=\text{Cost price}+\text{Gain}
\displaystyle \therefore x+\frac{x^2}{100}=24
\displaystyle \Rightarrow 100x+x^2=2400
\displaystyle \Rightarrow x^2+100x-2400=0
\displaystyle \Rightarrow x^2+120x-20x-2400=0
\displaystyle \Rightarrow x(x+120)-20(x+120)=0
\displaystyle \Rightarrow (x+120)(x-20)=0
\displaystyle \Rightarrow x=-120\text{ or }x=20
\displaystyle \text{Since cost price cannot be negative, }x=-120\text{ is rejected.}
\displaystyle \therefore \text{The cost price of the article is Rs. }20.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A pole has to be erected at a point on the boundary of a circular}
\displaystyle \text{park of diameter 13 metres in such a way that the difference of its distances from two}
\displaystyle \text{diametrically opposite fixed gates }A\text{ and }B\text{ on the boundary is }7\text{ metres. Is it possible to}
\displaystyle \text{do so? If yes, at what distances from the two gates should the pole be erected?}\hfill\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the pole be erected at point }P\text{ on the boundary of the circular park.}
\displaystyle \text{Since }AB\text{ is a diameter, }\angle APB=90^\circ.
\displaystyle \text{Let the shorter distance from the pole to a gate be }x\text{ m.}
\displaystyle \therefore \text{The longer distance}=(x+7)\text{ m.}
\displaystyle \text{Also, }AB=13\text{ m.}
\displaystyle \text{By Pythagoras theorem,}
\displaystyle x^2+(x+7)^2=13^2
\displaystyle \Rightarrow x^2+x^2+14x+49=169
\displaystyle \Rightarrow 2x^2+14x-120=0
\displaystyle \Rightarrow x^2+7x-60=0
\displaystyle \Rightarrow (x+12)(x-5)=0
\displaystyle \Rightarrow x=-12\text{ or }x=5
\displaystyle \text{Since distance cannot be negative, }x=-12\text{ is rejected.}
\displaystyle \therefore x=5,\quad x+7=12.
\displaystyle \therefore \text{Yes, it is possible to erect the pole.}
\displaystyle \therefore \text{It should be }5\text{ m from one gate and }12\text{ m from the other gate.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A cottage industry produces a certain number of pottery articles in a day. It was}
\displaystyle \text{observed on a particular day that the cost of production of each article (in rupees) was }3\text{ more}
\displaystyle \text{than twice the number of articles produced on that day. If the total cost of production on}
\displaystyle \text{that day was Rs. }90,\text{ find} \ \text{the number of articles produced and the cost of each article.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of pottery articles produced on that day be }x.
\displaystyle \therefore \text{Cost of production of each article}=\text{Rs. }(2x+3).
\displaystyle \text{Total cost of production}=\text{Number of articles}\times\text{Cost of each article}
\displaystyle \therefore x(2x+3)=90
\displaystyle \Rightarrow 2x^2+3x-90=0
\displaystyle \Rightarrow 2x^2+15x-12x-90=0
\displaystyle \Rightarrow x(2x+15)-6(2x+15)=0
\displaystyle \Rightarrow (2x+15)(x-6)=0
\displaystyle \Rightarrow x=-\frac{15}{2}\text{ or }x=6
\displaystyle \text{Since the number of articles cannot be negative, }x=-\frac{15}{2}\text{ is rejected.}
\displaystyle \therefore \text{Number of articles produced}=6.
\displaystyle \therefore \text{Cost of production of each article}=2(6)+3=\text{Rs. }15.
\displaystyle \therefore \text{The number of articles produced is }6\text{ and the cost of each article is Rs. }15.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{At }t\text{ minutes past }2\text{ p.m., the time needed by the minute hand}
\displaystyle \text{of a clock to show }3\text{ p.m.} \ \text{was found to be }3\text{ minutes less than }\frac{t^2}{4}\text{ minutes. Find }t.
\displaystyle \text{Answer:}
\displaystyle \text{At }t\text{ minutes past }2\text{ p.m., the time remaining to reach }3\text{ p.m.}=(60-t)\text{ minutes.}
\displaystyle \text{According to the question,}
\displaystyle 60-t=\frac{t^2}{4}-3
\displaystyle \Rightarrow 240-4t=t^2-12
\displaystyle \Rightarrow t^2+4t-252=0
\displaystyle \Rightarrow t^2+18t-14t-252=0
\displaystyle \Rightarrow t(t+18)-14(t+18)=0
\displaystyle \Rightarrow (t+18)(t-14)=0
\displaystyle \Rightarrow t=-18\text{ or }t=14
\displaystyle \text{Since }t\text{ represents the number of minutes past }2\text{ p.m., }t=-18\text{ is rejected.}
\displaystyle \therefore t=14\text{ minutes.}
\displaystyle \\

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