Class 10: Question Bank & Solution (CBSE)

$\displaystyle \text{LINEAR EQUATIONS INTWO VARIABLES}$

$\displaystyle \textbf{Case Study - 1}$

$\displaystyle \text{A test consists of `True' or `False' questions. One mark is awarded for every correct answer,}$
$\displaystyle \text{while }\frac14\text{ mark is deducted for every wrong answer. A student knew the answers to some}$
$\displaystyle \text{of the questions. The rest of the questions he attempted by guessing. He answered }120$
$\displaystyle \text{questions and scored }90\text{ marks.}$
$\displaystyle \text{For a correct answer, marks awarded}=1.$
$\displaystyle \text{For a wrong answer, marks deducted}=\frac14=0.25.$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{If the answers to all the questions he attempted by guessing were wrong, then}$
$\displaystyle \text{how many questions did he answer correctly?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of questions answered correctly}=x.$
$\displaystyle \therefore \text{Number of questions answered wrongly}=120-x.$
$\displaystyle \text{According to the given marking scheme,}$
$\displaystyle x-\frac14(120-x)=90$
$\displaystyle 4x-120+x=360$
$\displaystyle 5x=480$
$\displaystyle x=96$
$\displaystyle \therefore \text{He answered }96\text{ questions correctly.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{How many questions did he guess?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total number of questions answered}=120.$
$\displaystyle \text{Number of questions answered correctly}=96.$
$\displaystyle \text{Since all the guessed answers were wrong,}$
$\displaystyle \text{Number of questions guessed}=120-96=24.$
$\displaystyle \therefore \text{He guessed }24\text{ questions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If the answers to all the questions he attempted by guessing were wrong and}$
$\displaystyle \text{he answered }80\text{ correctly, then how many marks did he get?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of questions answered correctly}=80.$
$\displaystyle \text{Number of questions answered wrongly}=120-80=40.$
$\displaystyle \text{Marks obtained}=80(1)-40\left(\frac14\right)$
$\displaystyle =80-10=70$
$\displaystyle \therefore \text{He obtained }70\text{ marks.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If the answers to all the questions he attempted by guessing were wrong, then}$
$\displaystyle \text{how many questions should he answer correctly to score }95\text{ marks?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of questions answered correctly}=x.$
$\displaystyle \therefore \text{Number of questions answered wrongly}=120-x.$
$\displaystyle \text{To score }95\text{ marks,}$
$\displaystyle x-\frac14(120-x)=95$
$\displaystyle 4x-120+x=380$
$\displaystyle 5x=500$
$\displaystyle x=100$
$\displaystyle \therefore \text{He should answer }100\text{ questions correctly to score }95\text{ marks.}$
$\displaystyle \\$

$\displaystyle \textbf{Case Study - 2}$

$\displaystyle \text{Amit is planning to buy a house, and its layout is given below. The design and measurements}$
$\displaystyle \text{have been made such that the combined area of the two bedrooms and the kitchen is }95\text{ sq. m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Form the pair of linear equations in two variables from this situation.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{From the horizontal dimensions of the layout,}$
$\displaystyle x+2+y=15$
$\displaystyle \therefore x+y=13\qquad ...(1)$
$\displaystyle \text{Area of Bedroom 1}=5x$
$\displaystyle \text{Area of Bedroom 2}=5x$
$\displaystyle \text{Area of Kitchen}=5y$
$\displaystyle \text{The combined area of the two bedrooms and the kitchen is }95\text{ sq. m.}$
$\displaystyle 5x+5x+5y=95$
$\displaystyle 10x+5y=95$
$\displaystyle \therefore 2x+y=19\qquad ...(2)$
$\displaystyle \therefore \text{The required pair of linear equations is }x+y=13\text{ and }2x+y=19.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the length of the outer boundary of the layout.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of the layout}=15\text{ m}$
$\displaystyle \text{Breadth of the layout}=5+2+5=12\text{ m}$
$\displaystyle \text{Length of the outer boundary}=2(15+12)$
$\displaystyle =2\times27=54\text{ m}$
$\displaystyle \therefore \text{The length of the outer boundary is }54\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the area of each bedroom and the kitchen in the layout.}$
$\displaystyle \text{Answer:}$
$\displaystyle x+y=13\qquad ...(1)$
$\displaystyle 2x+y=19\qquad ...(2)$
$\displaystyle \text{Subtracting (1) from (2),}$
$\displaystyle x=6$
$\displaystyle \text{Substituting }x=6\text{ in }x+y=13,$
$\displaystyle 6+y=13$
$\displaystyle y=7$
$\displaystyle \text{Area of Bedroom 1}=5x=5\times6=30\text{ sq. m.}$
$\displaystyle \text{Area of Bedroom 2}=5x=5\times6=30\text{ sq. m.}$
$\displaystyle \text{Area of Kitchen}=5y=5\times7=35\text{ sq. m.}$
$\displaystyle \therefore \text{The areas of Bedroom 1, Bedroom 2 and the kitchen are }30,\ 30\text{ and }35\text{ sq. m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the area of the living room in the layout.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the horizontal part of the living room}=15\times2=30\text{ sq. m.}$
$\displaystyle \text{Width of the lower right part}=15-x=15-6=9\text{ m}$
$\displaystyle \text{Area of the lower right part}=9\times5=45\text{ sq. m.}$
$\displaystyle \therefore \text{Area of the living room}=30+45=75\text{ sq. m.}$
$\displaystyle \therefore \text{The area of the living room is }75\text{ sq. m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the cost of laying tiles in the kitchen at the rate of Rs. }50\text{ per sq. m.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the kitchen}=35\text{ sq. m.}$
$\displaystyle \text{Rate of laying tiles}=\text{Rs. }50\text{ per sq. m.}$
$\displaystyle \text{Cost of laying tiles}=35\times50$
$\displaystyle =\text{Rs. }1750$
$\displaystyle \therefore \text{The cost of laying tiles in the kitchen is Rs. }1750.$
$\displaystyle \\$

$\displaystyle \textbf{Case Study - 3}$

$\displaystyle \text{It is common for governments to revise travel fares from time to time based on various factors}$
$\displaystyle \text{such as inflation, which is a general increase in prices and a fall in the purchasing value of}$
$\displaystyle \text{money. Such revisions apply to different types of vehicles such as auto-rickshaws, taxis and}$
$\displaystyle \text{radio cabs. The auto-rickshaw fare in a city consists of a fixed charge together with a charge}$
$\displaystyle \text{for the distance travelled. Study the following situations.}$
$\displaystyle \text{Situation 1: In City A, for a journey of }10\text{ km, the charge paid is Rs. }75,\text{ and for a}$
$\displaystyle \text{journey of }15\text{ km, the charge paid is Rs. }110.$
$\displaystyle \text{Situation 2: In City B, for a journey of }8\text{ km, the charge paid is Rs. }91,\text{ and for a}$
$\displaystyle \text{journey of }14\text{ km, the charge paid is Rs. }145.$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{If the fixed charge of an auto-rickshaw is Rs. }x\text{ and the running charge is}$
$\displaystyle \text{Rs. }y\text{ per km, then the pair of linear equations representing Situation 1 is:}$
$\displaystyle \text{(a) }x+10y=110,\quad x+15y=75$
$\displaystyle \text{(b) }x+10y=75,\quad x+15y=110$
$\displaystyle \text{(c) }10x+y=110,\quad15x+y=75$
$\displaystyle \text{(d) }10x+y=75,\quad15x+y=110$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Fixed charge}=\text{Rs. }x,\qquad\text{Running charge}=\text{Rs. }y\text{ per km.}$
$\displaystyle \text{For a journey of }10\text{ km,}$
$\displaystyle x+10y=75$
$\displaystyle \text{For a journey of }15\text{ km,}$
$\displaystyle x+15y=110$
$\displaystyle \therefore \text{The required equations are }x+10y=75\text{ and }x+15y=110.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A person travels a distance of }50\text{ km. The amount he has to pay is:}$
$\displaystyle \text{(a) Rs. }155\qquad\text{(b) Rs. }255\qquad\text{(c) Rs. }355\qquad\text{(d) Rs. }455$
$\displaystyle \text{Answer:}$
$\displaystyle x+10y=75\qquad ...(1)$
$\displaystyle x+15y=110\qquad ...(2)$
$\displaystyle \text{Subtracting (1) from (2),}$
$\displaystyle 5y=35$
$\displaystyle y=7$
$\displaystyle \text{Substituting }y=7\text{ in (1),}$
$\displaystyle x+10(7)=75$
$\displaystyle x=5$
$\displaystyle \text{Therefore, the fare for travelling }50\text{ km}=x+50y$
$\displaystyle =5+50(7)=355$
$\displaystyle \therefore \text{The amount to be paid is Rs. }355.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{What will a person have to pay for travelling a distance of }30\text{ km in City B?}$
$\displaystyle \text{(a) Rs. }185\qquad\text{(b) Rs. }289\qquad\text{(c) Rs. }275\qquad\text{(d) Rs. }305$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For City B, let the fixed charge be Rs. }x\text{ and running charge be Rs. }y\text{ per km.}$
$\displaystyle x+8y=91\qquad ...(1)$
$\displaystyle x+14y=145\qquad ...(2)$
$\displaystyle \text{Subtracting (1) from (2),}$
$\displaystyle 6y=54$
$\displaystyle y=9$
$\displaystyle \text{Substituting }y=9\text{ in (1),}$
$\displaystyle x+8(9)=91$
$\displaystyle x=19$
$\displaystyle \text{Therefore, the fare for travelling }30\text{ km}=x+30y$
$\displaystyle =19+30(9)=289$
$\displaystyle \therefore \text{The amount to be paid is Rs. }289.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{The graph of the lines representing the conditions in Situation 2 is:}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{For Situation 2, the equations are}$
$\displaystyle x+8y=91\qquad\text{and}\qquad x+14y=145.$
$\displaystyle \text{Solving the equations, we get }x=19\text{ and }y=9.$
$\displaystyle \therefore \text{The two lines intersect at the point }(19,9).$
$\displaystyle \text{Among the given graphs, graph (iii) represents the two lines intersecting at }(19,9).$
$\displaystyle \therefore \text{Graph (iii) is correct.}$
$\displaystyle \\$

$\displaystyle \text{QUADRATIC EQUATIONS }$

$\displaystyle \textbf{Case Study - 1}$

$\displaystyle \text{Raj and Ajay are very close friends. Both families decide to travel to Ranikhet in their own cars.}$
$\displaystyle \text{Raj's car travels at a speed of }x\text{ km/h, while Ajay's car travels }5\text{ km/h faster than Raj's car.}$
$\displaystyle \text{Raj takes }4\text{ hours longer than Ajay to complete the journey of }400\text{ km.}$ $\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{What will be the distance covered by Ajay's car in two hours?}$
$\displaystyle \text{(a) }2(x+5)\text{ km}\qquad\text{(b) }(x-5)\text{ km}$
$\displaystyle \text{(c) }2(x+10)\text{ km}\qquad\text{(d) }(2x+5)\text{ km}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Speed of Raj's car}=x\text{ km/h}$
$\displaystyle \therefore \text{Speed of Ajay's car}=(x+5)\text{ km/h}$
$\displaystyle \text{Distance covered}=\text{Speed}\times\text{Time}$
$\displaystyle \therefore \text{Distance covered by Ajay's car in }2\text{ hours}=2(x+5)\text{ km}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Which of the following quadratic equations describes the speed of Raj's car?}$
$\displaystyle \text{(a) }x^2-5x-500=0\qquad\text{(b) }x^2+4x-400=0$
$\displaystyle \text{(c) }x^2+5x-500=0\qquad\text{(d) }x^2-4x+400=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Time taken by Raj}=\frac{400}{x}\text{ hours}$
$\displaystyle \text{Time taken by Ajay}=\frac{400}{x+5}\text{ hours}$
$\displaystyle \text{Since Raj takes }4\text{ hours longer than Ajay,}$
$\displaystyle \frac{400}{x}-\frac{400}{x+5}=4$
$\displaystyle \frac{400(x+5)-400x}{x(x+5)}=4$
$\displaystyle \frac{2000}{x(x+5)}=4$
$\displaystyle x(x+5)=500$
$\displaystyle x^2+5x-500=0$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{What is the speed of Raj's car?}$
$\displaystyle \text{(a) }20\text{ km/h}\qquad\text{(b) }15\text{ km/h}\qquad\text{(c) }25\text{ km/h}\qquad\text{(d) }10\text{ km/h}$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+5x-500=0$
$\displaystyle x^2+25x-20x-500=0$
$\displaystyle x(x+25)-20(x+25)=0$
$\displaystyle (x+25)(x-20)=0$
$\displaystyle x=-25\quad\text{or}\quad x=20$
$\displaystyle \text{Since speed cannot be negative, }x=20\text{ km/h.}$
$\displaystyle \therefore \text{The speed of Raj's car is }20\text{ km/h.}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{How much time did Ajay take to travel }400\text{ km?}$
$\displaystyle \text{(a) }20\text{ hours}\qquad\text{(b) }40\text{ hours}\qquad\text{(c) }25\text{ hours}\qquad\text{(d) }16\text{ hours}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Speed of Raj's car}=20\text{ km/h}$
$\displaystyle \therefore \text{Speed of Ajay's car}=20+5=25\text{ km/h}$
$\displaystyle \text{Time taken by Ajay}=\frac{\text{Distance}}{\text{Speed}}$
$\displaystyle =\frac{400}{25}=16\text{ hours}$
$\displaystyle \therefore \text{Ajay took }16\text{ hours to travel }400\text{ km.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Case Study - 2}$

$\displaystyle \text{The speed of a motorboat in still water is }20\text{ km/h. For a distance of }15\text{ km, the}$
$\displaystyle \text{boat takes }1\text{ hour longer to travel upstream than downstream.}$ $\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Let the speed of the stream be }x\text{ km/h. Then the speed of the motorboat}$
$\displaystyle \text{upstream will be:}$
$\displaystyle \text{(a) }20\text{ km/h}\qquad\text{(b) }(20+x)\text{ km/h}$
$\displaystyle \text{(c) }(20-x)\text{ km/h}\qquad\text{(d) }2\text{ km/h}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Speed of the motorboat in still water}=20\text{ km/h}$
$\displaystyle \text{Speed of the stream}=x\text{ km/h}$
$\displaystyle \text{Upstream speed}=\text{Speed in still water}-\text{Speed of stream}$
$\displaystyle =20-x\text{ km/h}$
$\displaystyle \therefore \text{The upstream speed is }(20-x)\text{ km/h.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{What is the relation between speed, distance and time?}$
$\displaystyle \text{(a) Speed}=\frac{\text{Distance}}{\text{Time}}\qquad\text{(b) Distance}=\frac{\text{Speed}}{\text{Time}}$
$\displaystyle \text{(c) Time}=\text{Speed}\times\text{Distance}\qquad\text{(d) Speed}=\text{Distance}\times\text{Time}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The relation between speed, distance and time is}$
$\displaystyle \text{Speed}=\frac{\text{Distance}}{\text{Time}}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Which is the correct quadratic equation for the speed of the current?}$
$\displaystyle \text{(a) }x^2+30x-200=0\qquad\text{(b) }x^2+20x-400=0$
$\displaystyle \text{(c) }x^2+30x-400=0\qquad\text{(d) }x^2+20x-400=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Upstream speed}=(20-x)\text{ km/h}$
$\displaystyle \text{Downstream speed}=(20+x)\text{ km/h}$
$\displaystyle \text{Time taken upstream}=\frac{15}{20-x}\text{ hours}$
$\displaystyle \text{Time taken downstream}=\frac{15}{20+x}\text{ hours}$
$\displaystyle \text{Since the upstream journey takes }1\text{ hour longer,}$
$\displaystyle \frac{15}{20-x}-\frac{15}{20+x}=1$
$\displaystyle \frac{15[(20+x)-(20-x)]}{(20-x)(20+x)}=1$
$\displaystyle \frac{30x}{400-x^2}=1$
$\displaystyle 30x=400-x^2$
$\displaystyle x^2+30x-400=0$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{What is the speed of the current?}$
$\displaystyle \text{(a) }20\text{ km/h}\qquad\text{(b) }10\text{ km/h}\qquad\text{(c) }15\text{ km/h}\qquad\text{(d) }25\text{ km/h}$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+30x-400=0$
$\displaystyle x^2+40x-10x-400=0$
$\displaystyle x(x+40)-10(x+40)=0$
$\displaystyle (x+40)(x-10)=0$
$\displaystyle x=-40\quad\text{or}\quad x=10$
$\displaystyle \text{Since speed cannot be negative, }x=10\text{ km/h.}$
$\displaystyle \therefore \text{The speed of the current is }10\text{ km/h.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{How much time did the boat take to travel downstream?}$
$\displaystyle \text{(a) }90\text{ minutes}\qquad\text{(b) }15\text{ minutes}\qquad\text{(c) }30\text{ minutes}\qquad\text{(d) }45\text{ minutes}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Speed of the current}=10\text{ km/h}$
$\displaystyle \therefore \text{Downstream speed}=20+10=30\text{ km/h}$
$\displaystyle \text{Time taken downstream}=\frac{\text{Distance}}{\text{Speed}}$
$\displaystyle =\frac{15}{30}=\frac12\text{ hour}$
$\displaystyle =30\text{ minutes}$
$\displaystyle \therefore \text{The boat took }30\text{ minutes to travel downstream.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$