Class 10: Question Bank & Solution (CBSE)

$\displaystyle \text{ARITHMETIC PROGRESSION }$

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

$\displaystyle \text{India is a competitive manufacturing location due to the low cost of manpower and strong}$
$\displaystyle \text{technical and engineering capabilities, which contribute to high-quality production. The}$
$\displaystyle \text{production of TV sets in a factory increases uniformly by a fixed number every year. The factory}$
$\displaystyle \text{produced }16000\text{ sets in the }6^{\text{th}}\text{ year and }22600\text{ sets in the }9^{\text{th}}\text{ year.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Find the production during the first year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the production in the first year}=a\text{ and the common difference}=d.$
$\displaystyle a_n=a+(n-1)d$
$\displaystyle \text{For the }6^{\text{th}}\text{ year,}$
$\displaystyle a+5d=16000\qquad ...(1)$
$\displaystyle \text{For the }9^{\text{th}}\text{ year,}$
$\displaystyle a+8d=22600\qquad ...(2)$
$\displaystyle \text{Subtracting (1) from (2),}$
$\displaystyle 3d=6600$
$\displaystyle d=2200$
$\displaystyle \text{Substituting }d=2200\text{ in (1),}$
$\displaystyle a+5(2200)=16000$
$\displaystyle a=5000$
$\displaystyle \therefore \text{The production during the first year was }5000\text{ sets.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the production during the }8^{\text{th}}\text{ year.}$
$\displaystyle \text{Answer:}$
$\displaystyle a=5000,\qquad d=2200$
$\displaystyle a_8=a+7d$
$\displaystyle =5000+7(2200)$
$\displaystyle =5000+15400$
$\displaystyle =20400$
$\displaystyle \therefore \text{The production during the }8^{\text{th}}\text{ year was }20400\text{ sets.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the total production during the first }3\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle a=5000,\qquad d=2200,\qquad n=3$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle S_3=\frac{3}{2}[2(5000)+(3-1)(2200)]$
$\displaystyle =\frac{3}{2}[10000+4400]$
$\displaystyle =\frac{3}{2}\times14400$
$\displaystyle =21600$
$\displaystyle \therefore \text{The total production during the first }3\text{ years was }21600\text{ sets.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{In which year was the production }29200\text{ sets?}$
$\displaystyle \text{Answer:}$
$\displaystyle a=5000,\qquad d=2200,\qquad a_n=29200$
$\displaystyle a_n=a+(n-1)d$
$\displaystyle 29200=5000+(n-1)(2200)$
$\displaystyle 24200=2200(n-1)$
$\displaystyle n-1=11$
$\displaystyle n=12$
$\displaystyle \therefore \text{The production was }29200\text{ sets in the }12^{\text{th}}\text{ year.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the difference between the production during the }7^{\text{th}}\text{ year and}$
$\displaystyle \text{the }4^{\text{th}}\text{ year.}$
$\displaystyle \text{Answer:}$
$\displaystyle a_7=a+6d$
$\displaystyle a_4=a+3d$
$\displaystyle a_7-a_4=(a+6d)-(a+3d)$
$\displaystyle =3d$
$\displaystyle =3(2200)=6600$
$\displaystyle \therefore \text{The difference in production is }6600\text{ sets.}$
$\displaystyle \\$

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

$\displaystyle \text{Your friend Veer wants to participate in a }200\text{ m race. He can currently complete the race in}$
$\displaystyle 51\text{ seconds, and with each day of practice, his time decreases by }2\text{ seconds. He wants to}$
$\displaystyle \text{complete the race in }31\text{ seconds.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Which of the following terms are in AP for the given situation?}$
$\displaystyle \text{(a) }51,53,55,\ldots\qquad\text{(b) }51,49,47,\ldots$
$\displaystyle \text{(c) }-51,-53,-55,\ldots\qquad\text{(d) }51,55,59,\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Veer's current time}=51\text{ seconds.}$
$\displaystyle \text{His time decreases by }2\text{ seconds with each day of practice.}$
$\displaystyle \therefore \text{The AP is }51,49,47,45,\ldots$
$\displaystyle \text{Here, }a=51\text{ and }d=49-51=-2.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{What is the minimum number of days he needs to practice until his goal is achieved?}$
$\displaystyle \text{(a) }10\qquad\text{(b) }12\qquad\text{(c) }11\qquad\text{(d) }9$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Current time}=51\text{ seconds}$
$\displaystyle \text{Target time}=31\text{ seconds}$
$\displaystyle \text{Reduction in time required}=51-31=20\text{ seconds}$
$\displaystyle \text{Reduction in time per day}=2\text{ seconds}$
$\displaystyle \therefore \text{Number of days of practice}=\frac{20}{2}=10$
$\displaystyle \therefore \text{Veer needs to practice for }10\text{ days to achieve his goal.}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Which of the following terms is not in the AP of the above given situation?}$
$\displaystyle \text{(a) }41\qquad\text{(b) }30\qquad\text{(c) }37\qquad\text{(d) }39$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The AP is }51,49,47,45,43,41,39,37,35,33,31,\ldots$
$\displaystyle \text{All the terms of this AP are odd numbers.}$
$\displaystyle \text{Since }30\text{ is an even number, it cannot be a term of the given AP.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If the }n^{\text{th}}\text{ term of an AP is given by }a_n=2n+3,\text{ then the common}$
$\displaystyle \text{difference of the AP is:}$
$\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }5\qquad\text{(d) }1$
$\displaystyle \text{Answer:}$
$\displaystyle a_n=2n+3$
$\displaystyle a_{n+1}=2(n+1)+3=2n+5$
$\displaystyle d=a_{n+1}-a_n$
$\displaystyle =(2n+5)-(2n+3)=2$
$\displaystyle \therefore \text{The common difference is }2.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the value of }x\text{ for which }2x,\ x+10,\ 3x+2\text{ are three consecutive}$
$\displaystyle \text{terms of an AP.}$
$\displaystyle \text{(a) }6\qquad\text{(b) }-6\qquad\text{(c) }18\qquad\text{(d) }-18$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For three consecutive terms of an AP, twice the middle term equals the sum of}$
$\displaystyle \text{the first and third terms.}$
$\displaystyle 2(x+10)=2x+(3x+2)$
$\displaystyle 2x+20=5x+2$
$\displaystyle 18=3x$
$\displaystyle x=6$
$\displaystyle \therefore \text{The value of }x\text{ is }6.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

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

$\displaystyle \text{Your elder brother wants to buy a car and plans to take a loan of Rs. }118000\text{ from a bank.}$
$\displaystyle \text{He repays the loan through monthly instalments, starting with the first instalment of Rs. }1000.$
$\displaystyle \text{If the amount of each instalment increases by Rs. }100\text{ every month, answer the questions}$
$\displaystyle \text{that follow.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{The amount paid by him in the }30^{\text{th}}\text{ instalment is:}$
$\displaystyle \text{(a) Rs. }3900\qquad\text{(b) Rs. }3500\qquad\text{(c) Rs. }3700\qquad\text{(d) Rs. }3600$
$\displaystyle \text{Answer:}$
$\displaystyle a=1000,\qquad d=100,\qquad n=30$
$\displaystyle a_n=a+(n-1)d$
$\displaystyle a_{30}=1000+(30-1)(100)$
$\displaystyle =1000+2900=3900$
$\displaystyle \therefore \text{The amount paid in the }30^{\text{th}}\text{ instalment is Rs. }3900.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The amount paid by him in the first }30\text{ instalments is:}$
$\displaystyle \text{(a) Rs. }37000\qquad\text{(b) Rs. }73500\qquad\text{(c) Rs. }75300\qquad\text{(d) Rs. }75000$
$\displaystyle \text{Answer:}$
$\displaystyle a=1000,\qquad d=100,\qquad n=30$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle S_{30}=\frac{30}{2}[2(1000)+(30-1)(100)]$
$\displaystyle =15[2000+2900]$
$\displaystyle =15\times4900=73500$
$\displaystyle \therefore \text{The amount paid in the first }30\text{ instalments is Rs. }73500.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{What amount does he still have to pay after the }30^{\text{th}}\text{ instalment?}$
$\displaystyle \text{(a) Rs. }45500\qquad\text{(b) Rs. }49000\qquad\text{(c) Rs. }44500\qquad\text{(d) Rs. }54000$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total loan amount}=\text{Rs. }118000$
$\displaystyle \text{Amount paid in the first }30\text{ instalments}=\text{Rs. }73500$
$\displaystyle \text{Amount still to be paid}=118000-73500$
$\displaystyle =44500$
$\displaystyle \therefore \text{The amount still to be paid is Rs. }44500.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If the total number of instalments is }40,\text{ then the amount paid in the last}$
$\displaystyle \text{instalment is:}$
$\displaystyle \text{(a) Rs. }4900\qquad\text{(b) Rs. }3900\qquad\text{(c) Rs. }5900\qquad\text{(d) Rs. }9400$
$\displaystyle \text{Answer:}$
$\displaystyle a=1000,\qquad d=100,\qquad n=40$
$\displaystyle a_{40}=a+(40-1)d$
$\displaystyle =1000+39(100)$
$\displaystyle =1000+3900=4900$
$\displaystyle \therefore \text{The amount paid in the last instalment is Rs. }4900.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The ratio of the first instalment to the last instalment is:}$
$\displaystyle \text{(a) }1:49\qquad\text{(b) }10:49\qquad\text{(c) }10:39\qquad\text{(d) }39:10$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First instalment}=\text{Rs. }1000$
$\displaystyle \text{Last instalment}=\text{Rs. }4900$
$\displaystyle \text{Required ratio}=1000:4900$
$\displaystyle =10:49$
$\displaystyle \therefore \text{The ratio of the first instalment to the last instalment is }10:49.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \text{SIMILAR TRIANGLES }$

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

$\displaystyle \text{Vijay is trying to find the height of a tower near his house. He uses the properties of similar}$
$\displaystyle \text{triangles. Vijay's house is }20\text{ m high and casts a shadow }10\text{ m long on the ground. At the}$
$\displaystyle \text{same time, the tower casts a shadow }50\text{ m long, while Ajay's house casts a shadow }20\text{ m}$
$\displaystyle \text{long on the ground.}$ $\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{What is the height of the tower?}$
$\displaystyle \text{(a) }20\text{ m}\qquad\text{(b) }50\text{ m}\qquad\text{(c) }100\text{ m}\qquad\text{(d) }200\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower}=h\text{ m.}$
$\displaystyle \text{Since the sun's rays make the same angle, the corresponding triangles are similar.}$
$\displaystyle \therefore \frac{\text{Height of Vijay's house}}{\text{Shadow of Vijay's house}}=\frac{\text{Height of tower}}{\text{Shadow of tower}}$
$\displaystyle \frac{20}{10}=\frac{h}{50}$
$\displaystyle 2=\frac{h}{50}$
$\displaystyle h=100\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }100\text{ m.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{What will be the length of the shadow of the tower when Vijay's house casts a}$
$\displaystyle \text{shadow of }12\text{ m?}$
$\displaystyle \text{(a) }75\text{ m}\qquad\text{(b) }50\text{ m}\qquad\text{(c) }45\text{ m}\qquad\text{(d) }60\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the tower's shadow}=x\text{ m.}$
$\displaystyle \text{Height of Vijay's house}=20\text{ m},\qquad\text{Height of tower}=100\text{ m}$
$\displaystyle \text{By similarity of triangles,}$
$\displaystyle \frac{20}{12}=\frac{100}{x}$
$\displaystyle 20x=1200$
$\displaystyle x=60\text{ m}$
$\displaystyle \therefore \text{The length of the tower's shadow is }60\text{ m.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{What is the height of Ajay's house?}$
$\displaystyle \text{(a) }30\text{ m}\qquad\text{(b) }40\text{ m}\qquad\text{(c) }50\text{ m}\qquad\text{(d) }20\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of Ajay's house}=h\text{ m.}$
$\displaystyle \text{By similarity of triangles,}$
$\displaystyle \frac{20}{10}=\frac{h}{20}$
$\displaystyle 2=\frac{h}{20}$
$\displaystyle h=40\text{ m}$
$\displaystyle \therefore \text{The height of Ajay's house is }40\text{ m.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{When the tower casts a shadow of }40\text{ m, what will be the length of the shadow}$
$\displaystyle \text{of Ajay's house at the same time?}$
$\displaystyle \text{(a) }16\text{ m}\qquad\text{(b) }32\text{ m}\qquad\text{(c) }20\text{ m}\qquad\text{(d) }8\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the shadow of Ajay's house}=x\text{ m.}$
$\displaystyle \text{Height of tower}=100\text{ m},\qquad\text{Height of Ajay's house}=40\text{ m}$
$\displaystyle \text{By similarity of triangles,}$
$\displaystyle \frac{100}{40}=\frac{40}{x}$
$\displaystyle 100x=1600$
$\displaystyle x=16\text{ m}$
$\displaystyle \therefore \text{The length of the shadow of Ajay's house is }16\text{ m.}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{When the tower casts a shadow of }40\text{ m, what will be the length of the shadow}$
$\displaystyle \text{of Vijay's house at the same time?}$
$\displaystyle \text{(a) }15\text{ m}\qquad\text{(b) }32\text{ m}\qquad\text{(c) }16\text{ m}\qquad\text{(d) }8\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the shadow of Vijay's house}=x\text{ m.}$
$\displaystyle \text{Height of tower}=100\text{ m},\qquad\text{Height of Vijay's house}=20\text{ m}$
$\displaystyle \text{By similarity of triangles,}$
$\displaystyle \frac{100}{40}=\frac{20}{x}$
$\displaystyle 100x=800$
$\displaystyle x=8\text{ m}$
$\displaystyle \therefore \text{The length of the shadow of Vijay's house is }8\text{ m.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

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

$\displaystyle \text{Rohan wants to measure the distance across a pond during a visit to his native place. He marks}$
$\displaystyle \text{points }A\text{ and }B\text{ on opposite edges of the pond, as shown in the figure. To determine the}$
$\displaystyle \text{distance between the points, he marks another point }C\text{ such that }BC=12\text{ m. He then}$
$\displaystyle \text{marks point }D\text{ such that }CD=40\text{ m and }AD=30\text{ m, with }\angle ADC=90^\circ.$ $\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Which property of geometry will be used to find the distance }AC\text{?}$
$\displaystyle \text{(a) Similarity of triangles}\qquad\text{(b) Thales Theorem}$
$\displaystyle \text{(c) Pythagoras Theorem}\qquad\text{(d) Area of similar triangles}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ADC,\ \angle ADC=90^\circ.$
$\displaystyle \text{Since }AD\text{ and }DC\text{ are known, }AC\text{ can be found using the Pythagoras Theorem.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{What is the distance }AC\text{?}$
$\displaystyle \text{(a) }50\text{ m}\qquad\text{(b) }12\text{ m}\qquad\text{(c) }100\text{ m}\qquad\text{(d) }70\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In right-angled }\triangle ADC,$
$\displaystyle AC^2=AD^2+DC^2$
$\displaystyle =30^2+40^2$
$\displaystyle =900+1600=2500$
$\displaystyle AC=\sqrt{2500}=50\text{ m}$
$\displaystyle \therefore \text{The distance }AC\text{ is }50\text{ m.}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Which of the following does not form a Pythagorean triplet?}$
$\displaystyle \text{(a) }(7,24,25)\qquad\text{(b) }(15,8,17)$
$\displaystyle \text{(c) }(5,12,13)\qquad\text{(d) }(21,20,28)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a Pythagorean triplet, the square of the largest number must equal the sum of}$
$\displaystyle \text{the squares of the other two numbers.}$
$\displaystyle 21^2+20^2=441+400=841$
$\displaystyle 28^2=784$
$\displaystyle \text{Since }841\ne784,\ (21,20,28)\text{ is not a Pythagorean triplet.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the length }AB\text{.}$
$\displaystyle \text{(a) }12\text{ m}\qquad\text{(b) }38\text{ m}\qquad\text{(c) }50\text{ m}\qquad\text{(d) }100\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle AC=50\text{ m},\qquad BC=12\text{ m}$
$\displaystyle \text{Since }A,B,C\text{ are collinear and }B\text{ lies between }A\text{ and }C,$
$\displaystyle AC=AB+BC$
$\displaystyle 50=AB+12$
$\displaystyle AB=38\text{ m}$
$\displaystyle \therefore \text{The length }AB\text{ is }38\text{ m.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the total length of the rope used.}$
$\displaystyle \text{(a) }120\text{ m}\qquad\text{(b) }70\text{ m}\qquad\text{(c) }82\text{ m}\qquad\text{(d) }22\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of rope used}=BC+CD+DA$
$\displaystyle =12+40+30$
$\displaystyle =82\text{ m}$
$\displaystyle \therefore \text{The total length of the rope used is }82\text{ m.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$