\displaystyle \textbf{Question 1: }\text{7 audio cassettes and 3 video cassettes cost}
\displaystyle \text{Rs. }1110\text{, while 5 audio cassettes and 4 video cassettes cost Rs. }1350.
\displaystyle \text{Find the cost of an audio cassette and a video cassette.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost of one audio cassette be Rs. }x\text{ and one video cassette be Rs. }y.
\displaystyle \text{Cost of 7 audio cassettes and 3 video cassettes}= \text{Rs. }1110.
\displaystyle \therefore 7x+3y=1110\qquad\ldots\text{(i)}
\displaystyle \text{Cost of 5 audio cassettes and 4 video cassettes}= \text{Rs. }1350.
\displaystyle \therefore 5x+4y=1350\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }4,
\displaystyle 28x+12y=4440\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,
\displaystyle 15x+12y=4050\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle 13x=390
\displaystyle \Rightarrow x=30
\displaystyle \text{Substituting }x=30\text{ in equation (i),}
\displaystyle 7(30)+3y=1110
\displaystyle \Rightarrow 210+3y=1110
\displaystyle \Rightarrow 3y=900
\displaystyle \Rightarrow y=300
\displaystyle \therefore \text{The cost of one audio cassette is Rs. }30\text{ and one video cassette is Rs. }300.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Reena has pens and pencils which together are }40
\displaystyle \text{in number. If she has 5 more pencils and 5 less pens, the number of pencils}
\displaystyle \text{would become 4 times the number of pens. Find the original number of pens}
\displaystyle \text{and pencils.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original number of pens be }x\text{ and the original number of pencils be }y.
\displaystyle \text{The total number of pens and pencils is }40.
\displaystyle \therefore x+y=40\qquad\ldots\text{(i)}
\displaystyle \text{After the change, the number of pens}=x-5
\displaystyle \text{and the number of pencils}=y+5.
\displaystyle \text{The number of pencils becomes 4 times the number of pens.}
\displaystyle \therefore y+5=4(x-5)
\displaystyle \Rightarrow 4x-y=25\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 5x=65
\displaystyle \Rightarrow x=13
\displaystyle \text{Substituting }x=13\text{ in equation (i),}
\displaystyle 13+y=40
\displaystyle \Rightarrow y=27
\displaystyle \therefore \text{Reena originally had }13\text{ pens and }27\text{ pencils.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The coach of a cricket team buys 7 bats and 6 balls}
\displaystyle \text{for Rs. }3800.\text{ Later, he buys 3 bats and 5 balls for Rs. }1750.\text{ Find the cost}
\displaystyle \text{of each bat and each ball.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost of each bat be Rs. }x\text{ and the cost of each ball be Rs. }y.
\displaystyle \text{Cost of 7 bats and 6 balls}=\text{Rs. }3800.
\displaystyle \therefore 7x+6y=3800\qquad\ldots\text{(i)}
\displaystyle \text{Cost of 3 bats and 5 balls}=\text{Rs. }1750.
\displaystyle \therefore 3x+5y=1750\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 21x+18y=11400\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }7,
\displaystyle 21x+35y=12250\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iii) from equation (iv),}
\displaystyle 17y=850
\displaystyle \Rightarrow y=50
\displaystyle \text{Substituting }y=50\text{ in equation (ii),}
\displaystyle 3x+5(50)=1750
\displaystyle \Rightarrow 3x=1500
\displaystyle \Rightarrow x=500
\displaystyle \therefore \text{The cost of each bat is Rs. }500\text{ and each ball is Rs. }50.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A lending library has a fixed charge for the first}
\displaystyle \text{three days and an additional charge for each day thereafter. Saritha paid}
\displaystyle \text{Rs. }27\text{ for a book kept for seven days, while Susy paid Rs. }21\text{ for the book}
\displaystyle \text{she kept for five days. Find the fixed charge and the charge for each extra day.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the fixed charge for the first three days be Rs. }x\text{ and the charge for each extra day be Rs. }y.
\displaystyle \text{Saritha kept the book for 7 days, i.e. for }7-3=4\text{ extra days.}
\displaystyle \therefore x+4y=27\qquad\ldots\text{(i)}
\displaystyle \text{Susy kept the book for 5 days, i.e. for }5-3=2\text{ extra days.}
\displaystyle \therefore x+2y=21\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)=21
\displaystyle \Rightarrow x=15
\displaystyle \therefore \text{The fixed charge is Rs. }15\text{ and the charge for each extra day is Rs. }3.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Jamila sold a table and a chair for Rs. }1050\text{, thereby}
\displaystyle \text{making a profit of }10\%\text{ on a table and }25\%\text{ on the chair. If she had}
\displaystyle \text{taken a profit of }25\%\text{ on the table and }10\%\text{ on the chair, she would}
\displaystyle \text{have got Rs. }1065.\text{ Find the cost price of each.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost price of the table be Rs. }x\text{ and that of the chair be Rs. }y.
\displaystyle \text{Selling price of the table at }10\%\text{ profit}=\frac{110x}{100}
\displaystyle \text{Selling price of the chair at }25\%\text{ profit}=\frac{125y}{100}
\displaystyle \therefore \frac{110x}{100}+\frac{125y}{100}=1050
\displaystyle \Rightarrow 110x+125y=105000\qquad\ldots\text{(i)}
\displaystyle \text{Selling price of the table at }25\%\text{ profit}=\frac{125x}{100}
\displaystyle \text{Selling price of the chair at }10\%\text{ profit}=\frac{110y}{100}
\displaystyle \therefore \frac{125x}{100}+\frac{110y}{100}=1065
\displaystyle \Rightarrow 125x+110y=106500\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 15x-15y=1500
\displaystyle \Rightarrow x-y=100\qquad\ldots\text{(iii)}
\displaystyle \Rightarrow x=y+100
\displaystyle \text{Substituting }x=y+100\text{ in equation (i),}
\displaystyle 110(y+100)+125y=105000
\displaystyle \Rightarrow 235y=94000
\displaystyle \Rightarrow y=400
\displaystyle \text{From equation (iii), }x=400+100=500
\displaystyle \therefore \text{The cost price of the table is Rs. }500\text{ and that of the chair is Rs. }400.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Susan invested certain amount of money in two}
\displaystyle \text{schemes }A\text{ and }B\text{, which offer interest at the rate of }8\%\text{ per annum and}
\displaystyle 9\%\text{ per annum, respectively. She received Rs. }1860\text{ as annual interest.}
\displaystyle \text{However, if she had interchanged the amount of investment in the two}
\displaystyle \text{schemes, she would have received Rs. }20\text{ more as annual interest.}
\displaystyle \text{How much money did she invest in each scheme?}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the amount invested in scheme }A\text{ be Rs. }x\text{ and in scheme }B\text{ be Rs. }y.
\displaystyle \text{Annual interest from scheme }A=\frac{8x}{100}
\displaystyle \text{Annual interest from scheme }B=\frac{9y}{100}
\displaystyle \therefore \frac{8x}{100}+\frac{9y}{100}=1860
\displaystyle \Rightarrow 8x+9y=186000\qquad\ldots\text{(i)}
\displaystyle \text{On interchanging the investments, the annual interest would be Rs. }1860+20=\text{Rs. }1880.
\displaystyle \therefore \frac{9x}{100}+\frac{8y}{100}=1880
\displaystyle \Rightarrow 9x+8y=188000\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 17x+17y=374000
\displaystyle \Rightarrow x+y=22000\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle x-y=2000\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=24000
\displaystyle \Rightarrow x=12000
\displaystyle \text{Substituting }x=12000\text{ in equation (iii),}
\displaystyle 12000+y=22000
\displaystyle \Rightarrow y=10000
\displaystyle \therefore \text{Susan invested Rs. }12000\text{ in scheme }A\text{ and Rs. }10000\text{ in scheme }B.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The cost of 4 pens and 4 pencil boxes is Rs. }100.
\displaystyle \text{Three times the cost of a pen is Rs. }15\text{ more than the cost of a}
\displaystyle \text{pencil box. Form the pair of linear equations for the above situation.}
\displaystyle \text{Find the cost of a pen and a pencil box.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost of a pen be Rs. }x\text{ and the cost of a pencil box be Rs. }y.
\displaystyle \text{The cost of 4 pens and 4 pencil boxes is Rs. }100.
\displaystyle \therefore 4x+4y=100
\displaystyle \Rightarrow x+y=25\qquad\ldots\text{(i)}
\displaystyle \text{Three times the cost of a pen is Rs. }15\text{ more than the cost of a pencil box.}
\displaystyle \therefore 3x=y+15
\displaystyle \Rightarrow 3x-y=15\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, the required pair of linear equations is }x+y=25\text{ and }3x-y=15.
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 4x=40
\displaystyle \Rightarrow x=10
\displaystyle \text{Substituting }x=10\text{ in equation (i),}
\displaystyle 10+y=25
\displaystyle \Rightarrow y=15
\displaystyle \therefore \text{The cost of a pen is Rs. }10\text{ and the cost of a pencil box is Rs. }15.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Vijay had some bananas, and he divided them into}
\displaystyle \text{two lots }A\text{ and }B.\text{ He sold first lot at the rate of Rs. }2\text{ for 3 bananas}
\displaystyle \text{and the second lot at the rate of Rs. }1\text{ per banana and got a total}
\displaystyle \text{of Rs. }400.\text{ If he had sold the first lot at the rate of Rs. }1\text{ per banana}
\displaystyle \text{and the second lot at the rate of Rs. }4\text{ per five bananas, his total}
\displaystyle \text{collection would have been Rs. }460.\text{ Find the total number of bananas}
\displaystyle \text{he had.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of bananas in lots }A\text{ and }B\text{ be }x\text{ and }y\text{ respectively.}
\displaystyle \text{Collection from lot }A\text{ at Rs. }2\text{ for 3 bananas}=\frac{2x}{3}.
\displaystyle \text{Collection from lot }B\text{ at Rs. }1\text{ per banana}=y.
\displaystyle \therefore \frac{2x}{3}+y=400
\displaystyle \Rightarrow 2x+3y=1200\qquad\ldots\text{(i)}
\displaystyle \text{At the changed rates, collection from lot }A=x
\displaystyle \text{and collection from lot }B=\frac{4y}{5}.
\displaystyle \therefore x+\frac{4y}{5}=460
\displaystyle \Rightarrow 5x+4y=2300\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }5,
\displaystyle 10x+15y=6000\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }2,
\displaystyle 10x+8y=4600\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle 7y=1400
\displaystyle \Rightarrow y=200
\displaystyle \text{Substituting }y=200\text{ in equation (i),}
\displaystyle 2x+3(200)=1200
\displaystyle \Rightarrow 2x=600
\displaystyle \Rightarrow x=300
\displaystyle \therefore \text{Total number of bananas}=x+y=300+200=500.
\displaystyle \therefore \text{Vijay had }500\text{ bananas.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{(a) The cost of 2 kg apples and 1 kg of grapes}
\displaystyle \text{on a day was found to be Rs. }320.\text{ The cost of 4 kg apples and 2 kg}
\displaystyle \text{grapes was found to be Rs. }600.\text{ If cost of 1 kg of apples and 1 kg}
\displaystyle \text{of grapes is Rs. }x\text{ and Rs. }y\text{ respectively, represent the given}
\displaystyle \text{situation algebraically as a system of equations and check whether the}
\displaystyle \text{system so obtained is consistent or not.}\hfill\text{[CBSE 2025]}
\displaystyle \text{OR}
\displaystyle \text{(b) Solve for }x\text{ and }y:
\displaystyle \sqrt{2}x+\sqrt{3}y=5,\qquad \sqrt{3}x+\sqrt{8}y=-\sqrt{6}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let the cost of 1 kg of apples be Rs. }x\text{ and that of 1 kg of grapes be Rs. }y.
\displaystyle \text{Cost of 2 kg apples and 1 kg grapes is Rs. }320.
\displaystyle \therefore 2x+y=320
\displaystyle \Rightarrow 2x+y-320=0\qquad\ldots\text{(i)}
\displaystyle \text{Cost of 4 kg apples and 2 kg grapes is Rs. }600.
\displaystyle \therefore 4x+2y=600
\displaystyle \Rightarrow 4x+2y-600=0\qquad\ldots\text{(ii)}
\displaystyle \text{Here, }a_1=2,\ b_1=1,\ c_1=-320,\quad a_2=4,\ b_2=2,\ c_2=-600.
\displaystyle \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\qquad\frac{b_1}{b_2}=\frac{1}{2}
\displaystyle \frac{c_1}{c_2}=\frac{-320}{-600}=\frac{8}{15}
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \text{The lines are parallel and distinct. Hence, the system is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \text{(b) }\sqrt{2}x+\sqrt{3}y=5\qquad\ldots\text{(i)}
\displaystyle \sqrt{3}x+\sqrt{8}y=-\sqrt{6}\qquad\ldots\text{(ii)}
\displaystyle \text{Since }\sqrt{8}=2\sqrt{2},\text{ equation (ii) becomes}
\displaystyle \sqrt{3}x+2\sqrt{2}y=-\sqrt{6}\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (i) by }\sqrt{3},
\displaystyle \sqrt{6}x+3y=5\sqrt{3}\qquad\ldots\text{(iv)}
\displaystyle \text{Multiplying equation (iii) by }\sqrt{2},
\displaystyle \sqrt{6}x+4y=-2\sqrt{3}\qquad\ldots\text{(v)}
\displaystyle \text{Subtracting equation (iv) from equation (v),}
\displaystyle y=-7\sqrt{3}
\displaystyle \text{Substituting }y=-7\sqrt{3}\text{ in equation (i),}
\displaystyle \sqrt{2}x+\sqrt{3}(-7\sqrt{3})=5
\displaystyle \Rightarrow \sqrt{2}x-21=5
\displaystyle \Rightarrow \sqrt{2}x=26
\displaystyle \Rightarrow x=13\sqrt{2}
\displaystyle \therefore x=13\sqrt{2},\qquad y=-7\sqrt{3}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{One says, ``Give me a hundred, friend! I shall}
\displaystyle \text{then become twice as rich as you.'' The other replies, ``If you give me ten,}
\displaystyle \text{I shall be six times as rich as you.'' Tell me what is the amount of their}
\displaystyle \text{respective capital.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the amounts with the first and second persons be Rs. }x\text{ and Rs. }y\text{ respectively.}
\displaystyle \text{If the second person gives Rs. }100\text{ to the first person, their amounts become Rs. }(x+100)\text{ and Rs. }(y-100).
\displaystyle \therefore x+100=2(y-100)
\displaystyle \Rightarrow x-2y=-300\qquad\ldots\text{(i)}
\displaystyle \text{If the first person gives Rs. }10\text{ to the second person, their amounts become Rs. }(x-10)\text{ and Rs. }(y+10).
\displaystyle \therefore y+10=6(x-10)
\displaystyle \Rightarrow 6x-y=70\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (ii) by }2,
\displaystyle 12x-2y=140\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (i) from equation (iii),}
\displaystyle 11x=440
\displaystyle \Rightarrow x=40
\displaystyle \text{Substituting }x=40\text{ in equation (ii),}
\displaystyle 6(40)-y=70
\displaystyle \Rightarrow y=170
\displaystyle \therefore \text{The first person has Rs. }40\text{ and the second person has Rs. }170.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A and }B\text{ each have a certain number of mangoes.}
\displaystyle A\text{ says to }B\text{, ``If you give 30 of your mangoes, I will have twice}
\displaystyle \text{as many as left with you.'' }B\text{ replies, ``If you give me 10, I will}
\displaystyle \text{have thrice as many as left with you.'' How many mangoes does each have?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of mangoes with }A\text{ and }B\text{ be }x\text{ and }y\text{ respectively.}
\displaystyle \text{If }B\text{ gives 30 mangoes to }A,\text{ they are left with }x+30\text{ and }y-30\text{ mangoes respectively.}
\displaystyle \therefore x+30=2(y-30)
\displaystyle \Rightarrow x-2y=-90\qquad\ldots\text{(i)}
\displaystyle \text{If }A\text{ gives 10 mangoes to }B,\text{ they are left with }x-10\text{ and }y+10\text{ mangoes respectively.}
\displaystyle \therefore y+10=3(x-10)
\displaystyle \Rightarrow 3x-y=40\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 3x-6y=-270\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (ii),}
\displaystyle 5y=310
\displaystyle \Rightarrow y=62
\displaystyle \text{Substituting }y=62\text{ in equation (ii),}
\displaystyle 3x-62=40
\displaystyle \Rightarrow 3x=102
\displaystyle \Rightarrow x=34
\displaystyle \therefore A\text{ has }34\text{ mangoes and }B\text{ has }62\text{ mangoes.}
\displaystyle \\

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