Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.2 | Set 1
Last Updated : 04 Sep, 2024
Question 1. Consider a tent cylindrical in shape and surmounted by a conical top having a height of 16 m and a radius as common for all the surfaces constituting the whole portion of the tent which is equal to 24 m. The height of the cylindrical portion of the tent is 11 m. Find the area of Canvas required for the tent.
Solution:
According to the question
Diameter of the cylinder = 24 m,
So the radius(r) = 24/2 = 12 m,
Height of the cylindrical part (H1) = 11m
and the total height of the shape = 16 m
So, the height of the cone (h)= 16 - 11 = 5m
Now, first we find the slant height of the cone
So, according to the formula of slant height
l = √r2 + h2
l = √122 + 52 = 13m
Now, we find the curved surface area of the cone
A1 = πrl
= 22/7 × 6 × 13 ......(1)
Now, we find the curved surface area of the cylinder
A2 = 2πrH1
= 2π(12)(11) ......(2)
Now we find the total area of canvas required for tent,
So,
A = A1 + A2
= 22/7 × 12 × 13 + 2 × 22/7 × 12 × 11
= 490 + 829.38 = 1319.8 = 1320 m2
Hence, the total canvas required for tent is 1320 m2
Question 2. Consider a Rocket. Suppose the rocket is in the form of a Circular Cylinder Closed at the lower end with a Cone of the same radius attached to its top. The Cylindrical portion of the rocket has radius say, 2.5 m and the height of that cylindrical portion of the rocket is 21m. The Conical portion of the rocket has a slant height of 8m, then calculate the total surface area of the rocket and also find the volume of the rocket.
Solution:
According to the question
Radius of the cylindrical part of the rocket(r) = 2.5 m,
Height of the cylindrical part of the rocket (h) = 21 m,
Slant Height of the conical surface of the rocket(l) = 8 m,
Now, we find the curved surface area of the cone
A1 = πrl
= π(2.5)(8)
= π x 20 ......(1)
Now, we find the curved surface area of the cone
A2 = 2πrh + πr2
= (2π × 2.5 × 21) + π (2.5)2
= (π × 10.5) + (π × 6.25) ......(2)
Now we find the total curved surface area
A = A1 + A2
= (π x 20) + (π x 10.5) + (π x 6.25)
= 62.83 + 329.86 + 19.63 = 412.3 m2
Hence, the total curved surface area of the conical surface is 412.3 m2
Let us considered H be the height of the conical portion in the rocket,
So, l2 = r2 + H2
H2 = l2 - r2
h = √82 - 2.52 = 23.685 m
Now we find the volume of the conical surface of the rocket
V1 = 1/3πr2H
= 1/3 × 22/7 × (2.5)2 × 23.685 ......(3)
Now we find the volume of the cylindrical part
V2 = πr2h
V2 = 22/7 × 2.52 × 21
Therefore, the total volume of the rocket
V = V1 + V2
V = 461.84 m2
Hence, the total volume of the Rocket is 461.84 m2
Question 3. Take a tent structure in vision being cylindrical in shape with height 77 dm and is being surmounted by a cone at the top having height 44 dm. The diameter of the cylinder is 36 m. Find the curved surface area of the tent.
Solution:
According to the question
Height of the tent = 77 dm,
Height of a surmounted cone = 44 dm,
Diameter of the cylinder = 36 m,
So, the radius of the cylinder(r) = 36/2 = 18 m
Therefore, the height of the cylindrical Portion(h) = 77 - 44 = 33 dm = 3.3 m
Let us considered l be the slant height of the cone,
So,l2 = r2 + h2
l2 = 182 + 3.32
l2 = 324 + 10.89 = 334.89 = 18.3 m
Now we find the curved surface area of the cylinder
A1 = 2πrh
= 2π184.4 m2 ......(1)
Now we find the curved surface area of the cone
A2 = πrh
= π × 18 × 18.3 ......(2)
Hence, the total curved surface of the tent
A = A1 + A2
Put all the values from eq(1) and (2)
A = (2π x 18 × 4.4) + (π x 18 × 18.3) = 1532. 46 m2
Hence, the total Curved Surface Area is 1532.46 m2
Question 4. A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy.
Solution:
According to the question
The height of the cone (h) = 4 cm
Diameter of the cone (d) = 6 cm,
So, radius of the cone (r) = 3
Also radius of the cone = radius of the hemisphere.
Let us considered l be the slant height of the cone,
l = √r2 + h2
l = √32 + 42 = 5 cm
Now we find the curved surface area of the cone
A1 = πrl
= π(3)(5) = 47.1 cm2
Now we find the curved surface area of the hemisphere
A1 = 2πr2
= 2π(3)2 = 56.23 cm2
Hence, the total surface area of toy
A = A1 + A2
= 47.1 + 56.23 = 103.62 cm2
Hence, the curved surface area of the toy is 103.62 cm2
Question 5. A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm, respectively. Find the total surface area of the solid. (Use π = 22/7).
Solution:
According to the question
The radius of the common base (r) = 3.5 cm,
Height of the cylindrical part (h) = 10 cm,
Height of the conical part (h1) = 6 cm
Let us considered l be the slant height of the cone,
So,
l = √r2 + h2 = √3.52 + 62 = 48.25 cm
Now we find the curved surface area of the cone
A1 = πrl
= π(3.5)(48.25) = 76.408 cm2
Now we find the curved surface area of the hemisphere
A2 = 2πrh
= 2π(3.5) (10) = 220 cm2
Therefore, the total surface area of the solid
A = A1 + A2
= 76.408 + 220 = 373.408 cm2
Hence, total surface area of the solid is 373.408 cm2.
Question 6. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical parts are 5cm and 13 cm, respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm.
Solution:
According to the question
The height of the cylindrical part(h1) = 13 cm,
Radius of the cylindrical part(r) = 5 cm,
Height of the whole solid(H) = 30 cm,
The height of the conical part,
h2 = 30 - 13 - 5 = 12 cm
Now we find the slant height of the cone
l = √r2 + h22 = √52 + 122 = 13 cm
Now we find the curved surface area of the cylinder
A1 = 2πrh1
= 2π(5)(13) = 408.2 cm2
Now we find the curved surface area of the cone
A2 = πrl
= π(5)(13) cm2 = 204.1 cm2
Now we find the curved surface area of the hemisphere
A3 = 2πr2
= 2π(5)2 = 157 cm2
Hence, the total curved surface area of the toy
A = A1 + A2 + A3
= (408.2 + 204.1 + 157) = 769.3 cm2
Hence, the surface area of the toy is 769.3 cm2
Question 7. Consider a cylindrical tub having radius as 5 cm and its length 9.8 cm. It is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in tub. If the radius of the hemisphere is 3.5 cm and the height of the cone outside the hemisphere is 5 cm. Find the volume of water left in the tub.
Solution:
According to the question
Radius of the Cylindrical tub(r) = 5 cm,
Height of the Cylindrical tub(h1) = 9.8 cm,
Height of the cone outside the hemisphere (h2) = 5 cm,
Radius of the hemisphere = 5 cm.
Now we find the volume of the cylindrical tub
V1 = πr2h1
= π(5)2 9.8 = 770 cm3
Now we find the volume of the Hemisphere
V2 = 2/3 × π × r3
= 2/3 × 22/7 × 3.53 = 89.79 cm3
Now we find the volume of the cone
V3= 1/3 × π × r2 × h2
= 1/3 × 22/7 × 3.52 × 5 = 64.14 cm3
Hence, The total volume
V = V2 + V3
V = 89.79 + 64.14 = 154 cm3
Hence, the total volume of the solid = 154 cm3
Hence, the volume of water left in the tube V = V1 - V2
V = 770 - 154 = 616 cm3
Hence, the volume of water left in the tube is 616 cm3.
Question 8. A circus tent has a cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20 cm. The height of the cylindrical and conical portions is 4.2 cm and 2.1 cm. Find the volume of that circus tent.
Solution:
According to the question
Radius of the cylindrical part(R) = 20 m,
Height of the cylindrical part(h1) = 4.2 m,
Height of the conical part(h2) = 2.1 m,
Now we find the volume of the cylindrical part
V1 = πr2 h1
= π(20)2 4.2 = 5280 m3
Now we find the volume of the conical part
V2 = 1/3 × π × r2 × h2
= 1/3 × 22/7 × r2 × h2
= 13 × 22/7 × 202 × 2.1 = 880 m3
Hence, the total volume of the circus tent
V = V1 + V2 = 6160 m3
Hence, the volume of the circus tent is 6160 m3
Question 9. A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with the conical ends, each of axis 9 cm. Determine the capacity of the tank.
Solution:
According to the question
Base diameter of the cylinder = 21 cm,
Radius (r) = diameter/2 = 25/2 = 11.5 cm,
Height of the cylindrical part of the tank (h1) = 18 cm,
Height of the conical part of the tank (h2) = 9 cm.
Now we find the volume of the cylindrical portion
V1 = πr2 h1
= π(11.5)2 18 = 7474.77 cm3
Now we find the volume of the conical portion
V2 = 1/3 × 22/7 × r2 × h2
= 1/3 × 22/7 × 11.52 × 9 = 1245.795 cm3
Hence, the total the capacity of the tank
V = V1 + V2
= 7474.77 + 1245.795
= 8316 cm3
Hence, the capacity of the tank is 8316 cm3
Question 10. A conical hole is drilled in a circular cylinder of height 12 cm and base radius 5 cm. The height and base radius of the cone are also the same. Find the whole surface and volume of the remaining Cylinder.
Solution:
According to the question
Height of the cone = Height of the cylinder = h = 12 cm,
Radius of the cone = Radius of the cylinder = r = 5 cm.
Let us considered l be the slant height of the cone
So,
l = √r2 + h2 = √52 + 122 = 13 cm
Now we find the total surface area of the remaining part in the circular cylinder
A= πr2 + 2πrh + πrl
= π(5)2 + 2π(5)(12) + π(5)(13) = 210 π cm2
Now we find the volume of the remaining part of the circular cylinder
V = πr2h - 1/3 πr2h
= πr2h - 1/3 × 22/7 × r2 × h
= π(5)2(12) - 1/3 × 22/7 × 52 × 12 = 200 π cm2
Hence, the area of the remaining part is 210 π cm2 and the volume is 200 π cm2
Question 11. A tent is in the form of a cylinder of diameter 20 m and height 2.5 m surmounted by a cone of equal base and height 7.5 m. Find the capacity of tent and the cost of canvas as well at a price of Rs.100 per square meter.
Solution:
According to the question
Diameter of the cylinder = 20 m,
So, the radius of the cylinder = 10 m,
Height of the cylinder (h1) = 2.5 m,
Radius of the cone = Radius of the cylinder = r = 15 m,
Height of the Cone (h2) = 7.5 m.
Let us considered l be the slant height of the cone
So,
l = √r2 + h22 = √152 + 7.52 = 12.5 m
Now we find the volume of the cylinder
V1 = πr2h1
= π(10)2 2.5 = 250π m3
Now we find the volume of the Cone
V2 = 1/3πr2h2
= 1/3 × 22/7 × 102 × 7.5 = 250π m3
Hence, the total capacity of the tent
V = V1 + V2
= 250 π + 250 π = 500π m3
Hence, the total capacity of the tent = V = 4478.5714 m3
Now we find the total area of the canvas required for the tent is
S = 2πrh1 + πrl
= 2(π)(10)(2.5) + π(10)(12.5) = 550 m2
Hence, the total cost of the canvas is (100 x 550) = Rs. 55000
Question 12. Consider a boiler which is in the form of a cylinder having length 2 m and there’s a hemispherical ends each of having a diameter of 2 m. Find the volume of the boiler.
Solution:
According to the question
Diameter of the hemisphere = 2 m,
So, the radius of the hemisphere (r) = 1 m,
Height of the cylinder (h) = 2 m,
Now we find the volume of the cylinder
V1 = πr2h
= π(1)2 x 2 = 22/7 × 2 = 44/7m3
Since, at each of the ends of the cylinder, two hemispheres are attached.
so, the volume of two hemispheres
V2 = 2 × 2/3πr3
= 2 × 2/3 × 22/7 × 13 = 22/7 × 4/3 = 88/21 m3
Hence, the volume of the boiler is
V = V1 + V2
V = 44/7 + 88/21 = 220/21 m3
Hence, the volume of the boiler Is 21.3
Summary
Exercise 16.2 Set 1 focuses on problems related to cylinders. It covers calculations involving the curved surface area, total surface area, and volume of cylinders. The questions range from direct application of formulas to more complex problem-solving scenarios. Students are required to work with various dimensions of cylinders, including radius and height, and understand the relationships between these measurements and the cylinder's properties.
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 1Chapter 7 of RD Sharma’s Class 10 textbook focuses on Statistics a fundamental area in mathematics that deals with data collection, analysis, interpretation, and presentation. Exercise 7.3 | Set 1 of this chapter covers the various problems designed to enhance students' understanding and application
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 2Question 14. Find the mean of the following frequency distribution:Class interval:25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59Frequency:1422166534 Solution: Let’s consider the assumed mean (A) = 42 Class intervalMid – value xidi = xi – 42ui = (xi – 42)/5fifiui25 – 2927-15-314-4230 – 3432-10-222
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Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 1In this article, we will explore the solutions to Exercise 7.4 from Chapter 7 of RD Sharma's Class 10 Mathematics textbook which focuses on Statistics. This chapter is crucial as it lays the foundation for understanding how data is collected, organized, analyzed, and interpreted in the various field
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Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 2Question 11. An incomplete distribution is given below :Variable10-2020-3030-4040-5050-6060-7070-80Frequency1230-65-2518You are given that the median value is 46 and the total number of items is 230. (i) Using the median formula fill up missing frequencies. (ii) Calculate the AM of the completed dis
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 1Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It provides methods for summarizing and drawing conclusions from the data which is essential in various fields such as economics, sociology, and science. Chapter 7 of RD Sharma's Class 10
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 2Question 11. Find the mean, median, and mode of the following data:Classes0-5050-100100-150150-200200-250250-300300-350Frequency2356531 Solution: Let mean (A) = 175 Classes Class Marks (x) Frequency (f) c.f. di = x - A A = 175 fi * di0-502522-150-30050-1007535-100-300100-150125510-50-250150-200175-A
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.6Chapter 7 of RD Sharma's Class 10 textbook focuses on Statistics a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of data. This chapter is crucial for understanding how to summarize and make sense of numerical data in various contexts. The aim is to
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Chapter 8: Quadratic Equations
Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.1Question 1. Which of the following are quadratic equations?(i) x2 + 6x - 4 = 0 Solution: As we know that, quadratic equation is in form of: ax2 + bx + c = 0 Here, given equation is x2 + 6x - 4 =0 in quadratic equation. So, It is a quadratic equations. (ii) √3x2 − 2x + ½= 0 Solution: As we know that,
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.2Quadratic equations are a fundamental concept in algebra that involves a polynomial of degree two. They play a crucial role in various fields of mathematics, physics, engineering, and many other disciplines. Understanding quadratic equations is essential for solving problems related to motion, area,
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1Solve the following quadratic equations by factorizationQuestion 1. (x - 4) (x + 2) = 0 Solution: We have equation, (x - 4) (x + 2) = 0 Implies that either x - 4 = 0 or x + 2 = 0 Therefore, roots of the equation are 4 or -2. Question 2. (2x + 3) (3x - 7) = 0 Solution: We have equation, (2x + 3) (3x
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 2The Quadratic equations are fundamental to algebra and appear frequently in various mathematical contexts. They play a crucial role in understanding polynomial functions and solving real-world problems. In this article, we will delve into Exercise 8.3 | Set 2 from RD Sharma’s Class 10 Mathematics bo
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.4Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: x^2-4\sqrt{2x}+6=0 . Solution: Given: x^2-4\sqrt{2x}+6=0 We have to make the equation a perfect square. => x^2-4\sqrt{2x} = -6 => x^2 - 2\times2\sqrt{2x} + (2\sqrt{2})^2 = -6 + (2\sqrt
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.5Question 1. Find the discriminant of the following quadratic equations:(i) 2x2 - 5x + 3 = 0 Solution: Given quadratic equation: 2x2 - 5x + 3 = 0 ....(1) As we know that the general form of quadratic equation is ax2 + bx + c = 0 ....(2) On comparing eq(1) and (2), we get Here, a = 2, b = -5 and c = 3
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 1Question 1. Determine the nature of the roots of following quadratic equations : (i) 2x² – 3x + 5 = 0 (ii) 2x² – 6x + 3 = 0 (iii) 3/5 x² – 2/3 x + 1 = 0 (iv) 3x² – 4√3 x + 4 = 0 (v) 3x² – 2√6 x + 2 = 0 Solution: (i) 2x² – 3x + 5 = 0 Here a=2, b=-3, c=5 D=b2-4ac=(-3)2-4*2*5 =-9-40=-31 D<0 Roots ar
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 2Question 11. If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k. Solution: 2x² + px – 15 = 0 -5 is its one root It will satisfy it 2(-5)2+p(-5)-15=0 2*25-5p-15=0 50-15-5p=0 -5p+35=0 -5p=-35 p=-35/-5=7 Now in
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Class 10 RD Sharma Solutions- Chapter 8 Quadratic Equations - Exercise 8.7 | Set 1Question 1: Find two consecutive numbers whose squares have the sum 85. Solution: Let first number is = x ⇒ Second number = (x+1) Now according to given condition— ⇒ Sum of squares of the numbers = 85 ⇒ x2 + (x+1)2 = 85 ⇒ x2 + x2 + 2x + 1 = 85 [ because (a+b)2 = a2 + 2ab + b2] ⇒ 2x2 + 2x + 1 - 85 =
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Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.7 | Set 2Question 11: The sum of a number and its square is 63/4 , find the numbers. Solution: Let the number is = x so it's square is = x2 Now according to condition- ⇒(number)+(number)2=63/4 ⇒ x+x2 = 63/4 ⇒ x2+x-63/4=0 Multiplying by 4- ⇒ 4x2+4x-63=0 ⇒ 4x2 +(18-14)x - 63 =0 [because 63*4=252 so 18*14=252
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.8Chapter 8 of RD Sharma's Class 10 textbook focuses on the Quadratic Equations an essential topic in mathematics that lays the foundation for understanding more advanced algebraic concepts. In this chapter, students will learn how to solve quadratic equations using various methods such as factorizati
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.9The Quadratic equations are a fundamental topic in algebra often introduced at the Class 10 level. They are equations of the form ax2 +bx+c=0 where a, b, and c are constants and x is the variable. Understanding quadratic equations is crucial as they form the basis for many mathematical concepts and
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.10Question 1. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides. Solution: Let the length of other two sides of the triangle be x and y. Therefore, according to the question, x - y = 5 or, x =
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.11Question 1: The perimeter of the rectangular field is 82 m and its area is 400 m2. Find the breadth of the rectangle? Solution: Given: Perimeter = 82 m and its area = 400 m2 Let the breadth of the rectangle be 'b' m. As we know, Perimeter of a rectangle = 2×(length + breadth) 82 = 2×(length + b) 41
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.12Question 1: A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work. Solution: Let us assume that B takes 'x' days to complete a piece of work. So, B’s 1 day work = 1/x Now, A takes
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.13Question 1. A piece of cloth costs ₹ 35. If the piece were 4 m longer and each meter costs ₹ 1 less, the cost would remain unchanged. How long is the piece? Solution: Let us considered the length of piece of cloth = x m Given: The total cost = ₹ 35 So, the cost of 1 m cloth = ₹ 35/x According to the
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Chapter 9: Arithmetic Progressions
Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progression Exercise 9.1Problem 1: Write the first terms of each of the following sequences whose nth term are:(i) an = 3n + 2 Solution: Given: an = 3n + 2 By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence a1 = (3 × 1) + 2 = 3 + 2 = 5 a2 = (3 × 2) + 2 = 6 + 2 = 8 a3 = (3 × 3) + 2 = 9 + 2 = 11 a4 = (3 × 4)
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.2Problem 1: Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference. Solution: Given: an = 5n – 7 Now putting n = 1, 2, 3, 4,5 we get, a1 = 5.1 – 7 = 5 – 7 = -2 a2 = 5.2 – 7 = 10 – 7 = 3 a3 = 5.3 – 7 = 15 – 7 = 8 a4 = 5.4 – 7 = 20 – 7 = 13 We can see that, a2 – a1 = 3 – (
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.3Problem 1: For the following arithmetic progressions write the first term a and the common difference d:(i) -5, -1, 3, 7, ………… Solution: Given sequence is -5, -1, 3, 7, ………… ∴ First term is -5 And, common difference = a2 - a1 = -1 - (-5) = 4 ∴ Common difference is 4 (ii) 1/5, 3/5, 5/5, 7/5, …… Solut
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 1In arithmetic progressions (AP), the terms follow a consistent pattern where each term is derived from the previous one by adding a fixed value, known as the common difference. To analyze various aspects of APs, such as finding specific terms or determining the number of terms, certain formulas and
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 2Question 14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference. Solution: Let’s assume that the first term and the common difference of the A.P to be a and d respectively. Given that, 4th term of an A.P.
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Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions - Exercise 9.5Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P. Solution: Since (8x + 4), (6x – 2) and (2x + 7) are in A.P. Now we know that condition for three number being in A.P.- ⇒2*(Middle Term)=(First Term)+(Last Term) ⇒2(6x-2)=(8x+4)+(2x+7) ⇒12x-4=10x+11 ⇒(12x-10x)=11+4
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 1Class 10 RD Sharma Solutions are comprehensive guides to the exercises and problems presented in the RD Sharma Mathematics textbook, which is a popular reference for students studying mathematics at the 10th class level. These solutions provide step-by-step explanations and answers to the questions
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 2Question 25. In an A.P. the first term is 22, nth term is –11 and the sum of first n term is 66. Find n and the d, the common difference. Solution: Given A.P. has first term(a) = 22, nth term(an) = –11 and sum(Sn) = 123. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 =>
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 3Question 49. Find the sum of first n odd natural numbers. Solution: First odd natural numbers are 1, 3, 5, 7, . . .2n – 1. First term(a) = 1, common difference(d) = 3 – 1 = 2 and nth term(an) = 2n – 1. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 So, = n[1 + 2n – 1] / 2 =
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