\displaystyle \textbf{Question 1: }\text{The students of a class are made to stand}
\displaystyle \text{equally in rows. If 3 students are extra in each row, there would be}
\displaystyle \text{1 row less. If 3 students are less in a row, there would be 2 more rows.}
\displaystyle \text{Find the number of students in the class.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total number of students be }x\text{ and the number of rows be }y.
\displaystyle \text{Number of students in each row}=\frac{x}{y}.
\displaystyle \text{If 3 students are extra in each row, the number of rows becomes }y-1.
\displaystyle \therefore x=\left(\frac{x}{y}+3\right)(y-1)
\displaystyle \Rightarrow x=x-\frac{x}{y}+3y-3
\displaystyle \Rightarrow \frac{x}{y}-3y+3=0\qquad\ldots\text{(i)}
\displaystyle \text{If 3 students are less in each row, the number of rows becomes }y+2.
\displaystyle \therefore x=\left(\frac{x}{y}-3\right)(y+2)
\displaystyle \Rightarrow x=x+\frac{2x}{y}-3y-6
\displaystyle \Rightarrow \frac{2x}{y}-3y-6=0\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{x}{y}=u.
\displaystyle \text{Equations (i) and (ii) become}
\displaystyle u-3y+3=0\qquad\ldots\text{(iii)}
\displaystyle 2u-3y-6=0\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iii) from equation (iv),}
\displaystyle u=9
\displaystyle \text{Substituting }u=9\text{ in equation (iii),}
\displaystyle 9-3y+3=0
\displaystyle \Rightarrow 3y=12
\displaystyle \Rightarrow y=4
\displaystyle \text{Since }\frac{x}{y}=u,\quad\frac{x}{4}=9
\displaystyle \Rightarrow x=36
\displaystyle \therefore \text{The number of students in the class is }36.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The perimeter of a rectangle is }70\text{ cm. The length}
\displaystyle \text{of the rectangle is }5\text{ cm more than twice its breadth. Express the}
\displaystyle \text{given situation as a system of linear equations in two variables and}
\displaystyle \text{hence solve it.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length be }x\text{ cm and the breadth be }y\text{ cm.}
\displaystyle \text{Perimeter of the rectangle}=70\text{ cm.}
\displaystyle \therefore 2(x+y)=70
\displaystyle \Rightarrow x+y=35\qquad\ldots\text{(i)}
\displaystyle \text{The length is }5\text{ cm more than twice the breadth.}
\displaystyle \therefore x=2y+5
\displaystyle \Rightarrow x-2y=5\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, the required system is }x+y=35\text{ and }x-2y=5.
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle 3y=30
\displaystyle \Rightarrow y=10
\displaystyle \text{Substituting }y=10\text{ in equation (i),}
\displaystyle x+10=35
\displaystyle \Rightarrow x=25
\displaystyle \therefore \text{The length is }25\text{ cm and the breadth is }10\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In a rectangle, if the length is increased by}
\displaystyle 3\text{ metres and breadth is decreased by }4\text{ metres, the area of the rectangle}
\displaystyle \text{is reduced by }67\text{ square metres. If length is reduced by }1\text{ metre and}
\displaystyle \text{breadth is increased by }4\text{ metres, the area is increased by }89\text{ sq. metres.}
\displaystyle \text{Find the dimensions of the rectangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the rectangle be }x\text{ m and breadth be }y\text{ m.}
\displaystyle \therefore \text{Area of the rectangle}=xy\text{ sq. m.}
\displaystyle \text{When length is increased by }3\text{ m and breadth is decreased by }4\text{ m,}
\displaystyle (x+3)(y-4)=xy-67
\displaystyle \Rightarrow xy-4x+3y-12=xy-67
\displaystyle \Rightarrow 4x-3y=55\qquad\ldots\text{(i)}
\displaystyle \text{When length is reduced by }1\text{ m and breadth is increased by }4\text{ m,}
\displaystyle (x-1)(y+4)=xy+89
\displaystyle \Rightarrow xy+4x-y-4=xy+89
\displaystyle \Rightarrow 4x-y=93\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 2y=38
\displaystyle \Rightarrow y=19
\displaystyle \text{Substituting }y=19\text{ in equation (ii),}
\displaystyle 4x-19=93
\displaystyle \Rightarrow 4x=112
\displaystyle \Rightarrow x=28
\displaystyle \therefore \text{The length is }28\text{ m and the breadth is }19\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{ABCD is a cyclic quadrilateral such that}
\displaystyle \angle A=(4y+20)^\circ,\ \angle B=(3y-5)^\circ,\ \angle C=(4x)^\circ\text{ and}
\displaystyle \angle D=(7x+5)^\circ.\text{ Find the four angles.}
\displaystyle \text{Answer:}
\displaystyle \text{In a cyclic quadrilateral, opposite angles are supplementary.}
\displaystyle \therefore \angle A+\angle C=180^\circ
\displaystyle (4y+20)+4x=180
\displaystyle \Rightarrow 4x+4y=160
\displaystyle \Rightarrow x+y=40\qquad\ldots\text{(i)}
\displaystyle \text{Also, }\angle B+\angle D=180^\circ
\displaystyle (3y-5)+(7x+5)=180
\displaystyle \Rightarrow 7x+3y=180\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 3x+3y=120\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (ii),}
\displaystyle 4x=60
\displaystyle \Rightarrow x=15
\displaystyle \text{Substituting }x=15\text{ in equation (i),}
\displaystyle 15+y=40
\displaystyle \Rightarrow y=25
\displaystyle \angle A=4(25)+20=120^\circ
\displaystyle \angle B=3(25)-5=70^\circ
\displaystyle \angle C=4(15)=60^\circ
\displaystyle \angle D=7(15)+5=110^\circ
\displaystyle \therefore \angle A=120^\circ,\ \angle B=70^\circ,\ \angle C=60^\circ,\ \angle D=110^\circ.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Yash scored 40 marks in a test, getting 3 marks}
\displaystyle \text{for each right answer and losing 1 mark for each wrong answer. Had 4 marks}
\displaystyle \text{been awarded for each correct answer and 2 marks been deducted for each}
\displaystyle \text{incorrect answer, then Yash would have scored 50 marks. How many questions}
\displaystyle \text{were there in the test?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of correct answers be }x\text{ and wrong answers be }y.
\displaystyle \text{Yash gets 3 marks for each correct answer and loses 1 mark for each wrong answer.}
\displaystyle \therefore 3x-y=40\qquad\ldots\text{(i)}
\displaystyle \text{If 4 marks are awarded for each correct answer and 2 marks are deducted}
\displaystyle \text{for each wrong answer, his score would be 50 marks.}
\displaystyle \therefore 4x-2y=50
\displaystyle \Rightarrow 2x-y=25\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle x=15
\displaystyle \text{Substituting }x=15\text{ in equation (i),}
\displaystyle 3(15)-y=40
\displaystyle \Rightarrow 45-y=40
\displaystyle \Rightarrow y=5
\displaystyle \therefore \text{Total number of questions}=x+y=15+5=20.
\displaystyle \therefore \text{There were }20\text{ questions in the test.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The car hire charges in a city comprise of fixed charges}
\displaystyle \text{together with the charge for the distance covered. For a journey of 12 km,}
\displaystyle \text{the charge paid is Rs. }89\text{ and for a journey of 20 km, the charge paid}
\displaystyle \text{is Rs. }145.\text{ What will a person have to pay for travelling a distance}
\displaystyle \text{of 30 km?}\hfill\text{[CBSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the fixed charge be Rs. }x\text{ and the charge per km be Rs. }y.
\displaystyle \text{For a journey of }12\text{ km, the total charge is Rs. }89.
\displaystyle \therefore x+12y=89\qquad\ldots\text{(i)}
\displaystyle \text{For a journey of }20\text{ km, the total charge is Rs. }145.
\displaystyle \therefore x+20y=145\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 8y=56
\displaystyle \Rightarrow y=7
\displaystyle \text{Substituting }y=7\text{ in equation (i),}
\displaystyle x+12(7)=89
\displaystyle \Rightarrow x=5
\displaystyle \therefore \text{Fixed charge}=\text{Rs. }5\text{ and charge per km}=\text{Rs. }7.
\displaystyle \text{Charge for travelling }30\text{ km}=x+30y
\displaystyle =5+30(7)=5+210=215
\displaystyle \therefore \text{The person will have to pay Rs. }215.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A part of monthly hostel charges in a college are}
\displaystyle \text{fixed and the remaining depend on the number of days one has taken food}
\displaystyle \text{in the mess. When a student }A\text{ takes food for 20 days, he has to pay}
\displaystyle \text{Rs. }1000\text{ as hostel charges whereas a student }B\text{, who takes food for}
\displaystyle 26\text{ days, pays Rs. }1180\text{ as hostel charges. Find the fixed charge and}
\displaystyle \text{the cost of food per day.}\hfill\text{[CBSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the fixed monthly charge be Rs. }x\text{ and cost of food per day be Rs. }y.
\displaystyle \text{For student }A\text{, who takes food for }20\text{ days,}
\displaystyle x+20y=1000\qquad\ldots\text{(i)}
\displaystyle \text{For student }B\text{, who takes food for }26\text{ days,}
\displaystyle x+26y=1180\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 6y=180
\displaystyle \Rightarrow y=30
\displaystyle \text{Substituting }y=30\text{ in equation (i),}
\displaystyle x+20(30)=1000
\displaystyle \Rightarrow x+600=1000
\displaystyle \Rightarrow x=400
\displaystyle \therefore \text{The fixed charge is Rs. }400\text{ and cost of food is Rs. }30\text{ per day.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The larger of two supplementary angles exceeds the}
\displaystyle \text{smaller by }18\text{ degrees. Find the two angles.}\hfill\text{[CBSE 2019, 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the larger angle be }x^\circ\text{ and the smaller angle be }y^\circ.
\displaystyle \text{Since the two angles are supplementary,}
\displaystyle x+y=180\qquad\ldots\text{(i)}
\displaystyle \text{The larger angle exceeds the smaller angle by }18^\circ.
\displaystyle \therefore x-y=18\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2x=198
\displaystyle \Rightarrow x=99
\displaystyle \text{Substituting }x=99\text{ in equation (i),}
\displaystyle 99+y=180
\displaystyle \Rightarrow y=81
\displaystyle \therefore \text{The two angles are }99^\circ\text{ and }81^\circ.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Meena went to a bank to withdraw Rs. }2000.
\displaystyle \text{She asked the cashier to give her Rs. }50\text{ and Rs. }100\text{ notes only. Meena}
\displaystyle \text{got 25 notes in all. Find how many notes of Rs. }50\text{ and Rs. }100\text{ she}
\displaystyle \text{received.}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of Rs. }50\text{ notes be }x\text{ and Rs. }100\text{ notes be }y.
\displaystyle \text{The total number of notes is }25.
\displaystyle \therefore x+y=25\qquad\ldots\text{(i)}
\displaystyle \text{The total amount withdrawn is Rs. }2000.
\displaystyle \therefore 50x+100y=2000
\displaystyle \Rightarrow x+2y=40\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle y=15
\displaystyle \text{Substituting }y=15\text{ in equation (i),}
\displaystyle x+15=25
\displaystyle \Rightarrow x=10
\displaystyle \therefore \text{Meena received }10\text{ notes of Rs. }50\text{ and }15\text{ notes of Rs. }100.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A shopkeeper gives books on rent for reading. She}
\displaystyle \text{takes a fixed charge for the first two days, and an additional charge for}
\displaystyle \text{each day thereafter. Latika paid Rs. }22\text{ for a book kept for 6 days,}
\displaystyle \text{while Anand paid Rs. }16\text{ for the book kept for four days. Find the}
\displaystyle \text{fixed charges and charge for each extra day.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the fixed charge for the first two days be Rs. }x.
\displaystyle \text{Let the additional charge for each extra day be Rs. }y.
\displaystyle \text{Latika keeps the book for }6\text{ days, i.e. for }4\text{ extra days.}
\displaystyle \therefore x+4y=22\qquad\ldots\text{(i)}
\displaystyle \text{Anand keeps the book for }4\text{ days, i.e. for }2\text{ extra days.}
\displaystyle \therefore x+2y=16\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle 2y=6
\displaystyle \Rightarrow y=3
\displaystyle \text{Substituting }y=3\text{ in equation (ii),}
\displaystyle x+2(3)=16
\displaystyle \Rightarrow x=10
\displaystyle \therefore \text{The fixed charge is Rs. }10\text{ and the charge for each extra day}
\displaystyle \text{is Rs. }3.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In a pair of supplementary angles, the greater}
\displaystyle \text{angle exceeds the smaller by }50^\circ.\text{ Express the given situation as a}
\displaystyle \text{system of linear equations in two variables and hence obtain the measure}
\displaystyle \text{of each angle.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the greater angle be }x^\circ\text{ and the smaller angle be }y^\circ.
\displaystyle \text{Since the two angles are supplementary,}
\displaystyle x+y=180\qquad\ldots\text{(i)}
\displaystyle \text{The greater angle exceeds the smaller angle by }50^\circ.
\displaystyle \therefore x-y=50\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, the required system is }x+y=180\text{ and }x-y=50.
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2x=230
\displaystyle \Rightarrow x=115
\displaystyle \text{Substituting }x=115\text{ in equation (i),}
\displaystyle 115+y=180
\displaystyle \Rightarrow y=65
\displaystyle \therefore \text{The two angles are }115^\circ\text{ and }65^\circ.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A bag contains some red and blue balls. Ten percent}
\displaystyle \text{of the red balls, when added to twenty percent of the blue balls, give a}
\displaystyle \text{total of }24.\text{ If three times the number of red balls exceeds the number}
\displaystyle \text{of blue balls by }20,\text{ find the number of red and blue balls.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of red balls be }x\text{ and the number of blue balls be }y.
\displaystyle \text{Ten percent of red balls plus twenty percent of blue balls is }24.
\displaystyle \therefore \frac{10x}{100}+\frac{20y}{100}=24
\displaystyle \Rightarrow x+2y=240\qquad\ldots\text{(i)}
\displaystyle \text{Three times the number of red balls exceeds the blue balls by }20.
\displaystyle \therefore 3x-y=20\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (ii) by }2,
\displaystyle 6x-2y=40\qquad\ldots\text{(iii)}
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle 7x=280
\displaystyle \Rightarrow x=40
\displaystyle \text{Substituting }x=40\text{ in equation (ii),}
\displaystyle 3(40)-y=20
\displaystyle \Rightarrow 120-y=20
\displaystyle \Rightarrow y=100
\displaystyle \therefore \text{There are }40\text{ red balls and }100\text{ blue balls.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{2 women and 5 men can together finish a piece}
\displaystyle \text{of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days.}
\displaystyle \text{Find the time taken by 1 woman alone to finish the embroidery, and that}
\displaystyle \text{taken by 1 man alone.}
\displaystyle \text{Answer:}
\displaystyle \text{Let one woman's one-day work be }x\text{ and one man's one-day work be }y.
\displaystyle \text{2 women and 5 men finish the work in 4 days.}
\displaystyle \therefore 2x+5y=\frac{1}{4}\qquad\ldots\text{(i)}
\displaystyle \text{3 women and 6 men finish the work in 3 days.}
\displaystyle \therefore 3x+6y=\frac{1}{3}\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }12,
\displaystyle 24x+60y=3\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }12,
\displaystyle 36x+72y=4\qquad\ldots\text{(iv)}
\displaystyle \text{Multiplying equation (iii) by }3,
\displaystyle 72x+180y=9\qquad\ldots\text{(v)}
\displaystyle \text{Multiplying equation (iv) by }2,
\displaystyle 72x+144y=8\qquad\ldots\text{(vi)}
\displaystyle \text{Subtracting equation (vi) from equation (v),}
\displaystyle 36y=1
\displaystyle \Rightarrow y=\frac{1}{36}
\displaystyle \text{Substituting }y=\frac{1}{36}\text{ in equation (ii),}
\displaystyle 3x+6\left(\frac{1}{36}\right)=\frac{1}{3}
\displaystyle \Rightarrow 3x+\frac{1}{6}=\frac{1}{3}
\displaystyle \Rightarrow 3x=\frac{1}{6}
\displaystyle \Rightarrow x=\frac{1}{18}
\displaystyle \therefore \text{Time taken by 1 woman alone}=\frac{1}{x}=18\text{ days.}
\displaystyle \text{Time taken by 1 man alone}=\frac{1}{y}=36\text{ days.}
\displaystyle \therefore \text{One woman takes }18\text{ days and one man takes }36\text{ days.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{There are two examination rooms }A\text{ and }B.\text{ If 10}
\displaystyle \text{candidates are sent from }A\text{ to }B,\text{ the number of students in each room}
\displaystyle \text{is same. If 20 candidates are sent from }B\text{ to }A,\text{ the number of students}
\displaystyle \text{in }A\text{ is double the number of students in }B.\text{ Find the number of}
\displaystyle \text{students in each room.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of students in rooms }A\text{ and }B\text{ be }x\text{ and }y.
\displaystyle \text{When 10 candidates are sent from }A\text{ to }B,
\displaystyle x-10=y+10
\displaystyle \Rightarrow x-y=20\qquad\ldots\text{(i)}
\displaystyle \text{When 20 candidates are sent from }B\text{ to }A,
\displaystyle x+20=2(y-20)
\displaystyle \Rightarrow x+20=2y-40
\displaystyle \Rightarrow x-2y=-60\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle y=80
\displaystyle \text{Substituting }y=80\text{ in equation (i),}
\displaystyle x-80=20
\displaystyle \Rightarrow x=100
\displaystyle \therefore \text{Room }A\text{ has }100\text{ students and room }B\text{ has }80\text{ students.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A railway half ticket costs half the full fare and}
\displaystyle \text{the reservation charge is the same on half ticket as on full ticket. One}
\displaystyle \text{reserved first class ticket from Mumbai to Ahmedabad costs Rs. }216\text{ and}
\displaystyle \text{one full and one half reserved first class tickets cost Rs. }327.\text{ What is}
\displaystyle \text{the basic first class full fare and what is the reservation charge?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the basic first class full fare be Rs. }x\text{ and reservation charge be Rs. }y.
\displaystyle \text{Cost of one reserved full ticket}=x+y.
\displaystyle \therefore x+y=216\qquad\ldots\text{(i)}
\displaystyle \text{Basic fare of a half ticket}=\frac{x}{2}.
\displaystyle \text{Since the reservation charge is the same,}
\displaystyle \text{cost of one reserved half ticket}=\frac{x}{2}+y.
\displaystyle \text{One full and one half reserved tickets together cost Rs. }327.
\displaystyle \therefore (x+y)+\left(\frac{x}{2}+y\right)=327
\displaystyle \Rightarrow \frac{3x}{2}+2y=327
\displaystyle \Rightarrow 3x+4y=654\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 3x+3y=648\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (ii),}
\displaystyle y=6
\displaystyle \text{Substituting }y=6\text{ in equation (i),}
\displaystyle x+6=216
\displaystyle \Rightarrow x=210
\displaystyle \therefore \text{The basic first class full fare is Rs. }210\text{ and the reservation}
\displaystyle \text{charge is Rs. }6.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A shopkeeper sells a saree at }8\%\text{ profit and a}
\displaystyle \text{sweater at }10\%\text{ discount, thereby getting a sum of Rs. }1008.\text{ If she had}
\displaystyle \text{sold the saree at }10\%\text{ profit and the sweater at }8\%\text{ discount, she}
\displaystyle \text{would have got Rs. }1028.\text{ Find the cost price of the saree and the list}
\displaystyle \text{price of the sweater.} 
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost price of the saree be Rs. }x.
\displaystyle \text{Let the list price of the sweater be Rs. }y.
\displaystyle \text{Selling price of the saree at }8\%\text{ profit}=\frac{108x}{100}.
\displaystyle \text{Selling price of the sweater at }10\%\text{ discount}=\frac{90y}{100}.
\displaystyle \therefore \frac{108x}{100}+\frac{90y}{100}=1008
\displaystyle \Rightarrow 108x+90y=100800\qquad\ldots\text{(i)}
\displaystyle \text{Selling price of the saree at }10\%\text{ profit}=\frac{110x}{100}.
\displaystyle \text{Selling price of the sweater at }8\%\text{ discount}=\frac{92y}{100}.
\displaystyle \therefore \frac{110x}{100}+\frac{92y}{100}=1028
\displaystyle \Rightarrow 110x+92y=102800\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 2x+2y=2000
\displaystyle \Rightarrow x+y=1000\qquad\ldots\text{(iii)}
\displaystyle \Rightarrow y=1000-x
\displaystyle \text{Substituting }y=1000-x\text{ in equation (i),}
\displaystyle 108x+90(1000-x)=100800
\displaystyle \Rightarrow 18x=10800
\displaystyle \Rightarrow x=600
\displaystyle \text{From equation (iii), }y=1000-600=400.
\displaystyle \therefore \text{The cost price of the saree is Rs. }600\text{ and the list price}
\displaystyle \text{of the sweater is Rs. }400.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In a competitive examination, one mark is awarded}
\displaystyle \text{for each correct answer while }\frac{1}{2}\text{ mark is deducted for every wrong answer.}
\displaystyle \text{Jayanti answered }120\text{ questions and got }90\text{ marks. How many questions}
\displaystyle \text{did she answer correctly?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of correct answers be }x\text{ and wrong answers be }y.
\displaystyle \text{Jayanti answered }120\text{ questions.}
\displaystyle \therefore x+y=120\qquad\ldots\text{(i)}
\displaystyle \text{She gets 1 mark for each correct answer and loses }\frac{1}{2}\text{ mark}
\displaystyle \text{for each wrong answer.}
\displaystyle \therefore x-\frac{y}{2}=90
\displaystyle \Rightarrow 2x-y=180\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 3x=300
\displaystyle \Rightarrow x=100
\displaystyle \text{Substituting }x=100\text{ in equation (i),}
\displaystyle 100+y=120
\displaystyle \Rightarrow y=20
\displaystyle \therefore \text{Jayanti answered }100\text{ questions correctly.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Tara scored 40 marks in a test, getting 3 marks}
\displaystyle \text{for each right answer and losing 1 mark for each wrong answer. Had 4 marks}
\displaystyle \text{been awarded for each correct answer and 2 marks been deducted for each}
\displaystyle \text{wrong answer, then Tara would have scored 50 marks. Assuming that Tara}
\displaystyle \text{attempted all questions, find the total number of questions in the test.}
\displaystyle \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of correct answers be }x\text{ and wrong answers be }y.
\displaystyle \text{Tara gets 3 marks for each correct answer and loses 1 mark for each wrong answer.}
\displaystyle \therefore 3x-y=40\qquad\ldots\text{(i)}
\displaystyle \text{If 4 marks are awarded for each correct answer and 2 marks are deducted}
\displaystyle \text{for each wrong answer, her score would be }50\text{ marks.}
\displaystyle \therefore 4x-2y=50
\displaystyle \Rightarrow 2x-y=25\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle x=15
\displaystyle \text{Substituting }x=15\text{ in equation (i),}
\displaystyle 3(15)-y=40
\displaystyle \Rightarrow 45-y=40
\displaystyle \Rightarrow y=5
\displaystyle \therefore \text{Total number of questions}=x+y=15+5=20.
\displaystyle \therefore \text{There were }20\text{ questions in the test.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The two angles of a right angled triangle other}
\displaystyle \text{than }90^\circ\text{ are in the ratio }2:3.\text{ Express the given situation algebraically}
\displaystyle \text{as a system of linear equations in two variables and hence solve it.}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two acute angles be }x^\circ\text{ and }y^\circ.
\displaystyle \text{The sum of the two acute angles of a right angled triangle is }90^\circ.
\displaystyle \therefore x+y=90\qquad\ldots\text{(i)}
\displaystyle \text{The two angles are in the ratio }2:3.
\displaystyle \therefore \frac{x}{y}=\frac{2}{3}
\displaystyle \Rightarrow 3x-2y=0\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, the required system is }x+y=90\text{ and }3x-2y=0.
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 2x+2y=180\qquad\ldots\text{(iii)}
\displaystyle \text{Adding equations (ii) and (iii),}
\displaystyle 5x=180
\displaystyle \Rightarrow x=36
\displaystyle \text{Substituting }x=36\text{ in equation (i),}
\displaystyle 36+y=90
\displaystyle \Rightarrow y=54
\displaystyle \therefore \text{The two acute angles are }36^\circ\text{ and }54^\circ.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The perimeter of an isosceles triangle is }32\text{ cm.}
\displaystyle \text{If each equal side is }\left(\frac{5}{6}\right)^{\text{th}}\text{ of the base, find the area of}
\displaystyle \text{the triangle.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the base be }x\text{ cm and each equal side be }y\text{ cm.}
\displaystyle \text{The perimeter of the triangle is }32\text{ cm.}
\displaystyle \therefore x+2y=32\qquad\ldots\text{(i)}
\displaystyle \text{Each equal side is }\frac{5}{6}\text{ of the base.}
\displaystyle \therefore y=\frac{5}{6}x
\displaystyle \Rightarrow 5x-6y=0\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 3x+6y=96\qquad\ldots\text{(iii)}
\displaystyle \text{Adding equations (ii) and (iii),}
\displaystyle 8x=96
\displaystyle \Rightarrow x=12
\displaystyle \text{Substituting }x=12\text{ in equation (i),}
\displaystyle 12+2y=32
\displaystyle \Rightarrow y=10
\displaystyle \therefore \text{The base is }12\text{ cm and each equal side is }10\text{ cm.}
\displaystyle \text{The altitude bisects the base into two parts of }6\text{ cm each.}
\displaystyle \text{Let the altitude be }h\text{ cm.}
\displaystyle h^2+6^2=10^2
\displaystyle \Rightarrow h^2=100-36=64
\displaystyle \Rightarrow h=8\text{ cm}
\displaystyle \text{Area of the triangle}=\frac{1}{2}\times12\times8=48\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }48\text{ cm}^2.
\displaystyle \\

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