Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.1 | Set 2
Last Updated : 30 Aug, 2024
Question 19. How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm x 10 cm x 7 cm?
Solution:
Diameter of the coin = 1.75 cm
Radius = 1.75/2 = 0.875 cm
Thickness or height = 2 mm = 0.2 cm
Volume of the cylinder = πr2h
= π 0.8752 × 0.2
Volume of the cuboid = 11 × 10 × 7 cm3
Let the number of coins needed to be melted be n.
Volume of cuboid = n × Volume of cylinder
11 × 10 × 7 = π 0.8752 × 0.2 x n
11 × 10 × 7 = 22/7 x 0.8752 × 0.2 x n
n = 1600
Therefore, the number of coins required are 1600
Question 20. The surface area of a solid metallic sphere is 616 cm2. It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed.
Solution:
Radius of sphere = r cm
Surface area of the solid metallic sphere = 616 cm3
Surface area of the sphere = 4πr2
4πr2 = 616
r2 = 49
r = 7
Radius of the solid metallic sphere = 7 cm
Let radius of cone be x cm
Volume of the cone = 1/3 πx2h
= 1/3 πx2 (28) ….. (i)
Volume of the sphere = 4/3 πr3
= 4/3 π73 ………. (ii)
(i) and (ii) are equal
1/3 πx2 (28) = 4/3 π73
x2 (28) = 4 x 73
x2 = 49
r =7
Diameter of the cone = 7 x 2 = 14 cm
Therefore, the diameter of the base of the cone is 14 cm
Question 21. A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Height of the cylindrical bucket = 32 cm
Radius of the cylindrical bucket = 18 cm
Volume of cylinder = π × r2 × h
Volume of cone = 1/3 π × r2 × h
Volume of the conical heap = Volume of the cylindrical bucket
1/3 π × r2 × 24 = π × 182 × 32
r2 = 182 x 4
r = 18 x 2 = 36 cm
Let slant height of conical heap be l cm
l = √(h2 + r2)
l = √(242 + 362) = √1872
l = 43.26 cm
Therefore, the radius = 36cm slant height of the conical heap = 43.26 cm
Question 22. A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed.
Solution:
Let the number of cones be n
Radius of metallic sphere = 5.6 cm
Radius of the cone = 2.8 cm
Height of the cone = 3.2 cm
Volume of a sphere = 4/3 π × r3
= 4/3 π × 5.63
Volume of cone = 1/3 π × r2 × h
= 1/3 π × 2.82 × 3.2
Volume of sphere = n * Volume of each cone
Number of cones (n) = Volume of the sphere/ Volume of the cone
n = 4/3 π × 5.63/(1/3 π × 2.82 × 3.2)
n = (4 x 5.63)/(2.82 × 3.2)
n = 28
Therefore, 28 such cones can be formed.
Question 23. A solid cuboid of iron with dimensions 53 cm x 40 cm x 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe.
Solution:
Volume of cuboid = (53 x 40 x 15) cm3
Internal radius of the pipe = 7/2 cm = r
External radius of the pipe = 8/2 = 4 cm = R
Let h be length of pipe
Volume of iron in the pipe = (External Volume) – (Internal Volume)
= πR2h – πr2h
= πh(R2– r2)
= πh(R – r) (R + r)
= π(4 – 7/2) (4 + 7/2) x h
= π(1/2) (15/2) x h
Volume of iron in the pipe = volume of iron in cuboid
π(1/2) (15/2) x h = 53 x 40 x 15
h = (53 x 40 x 15 x 7/22 x 2/15 x 2) cm
h = 2698.18 cm
Therefore, the length of the pipe is 2698.2 cm.
Question 24. The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder.
Solution:
Internal diameter of hollow spherical shell = 6 cm
Internal radius of hollow spherical shell = 6/2 = 3 cm = r
External diameter of hollow spherical shell = 10 cm
External diameter of hollow spherical shell = 10/2 = 5 cm = R
Diameter of the cylinder = 14 cm
Radius of cylinder = 14/2 = 7 cm
Let the height of cylinder be taken as h cm
Volume of cylinder = Volume of spherical shell
π × r2 × h = 4/3 π × (R3 – r3)
π × 72 × h = 4/3 π × (53 – 33)
h = 4/3 x 2
h = 8/3 cm
Therefore, the height of the cylinder = 8/3 cm
Question 25. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone?
Solution:
Internal diameter of hollow sphere = 4 cm
Internal radius of hollow sphere = 2 cm
External diameter of hollow sphere = 8 cm
External radius of hollow sphere = 4 cm
Volume of the hollow sphere = 4/3 π × (43 – 23) … (i)
Diameter of the cone = 8 cm
Radius of the cone = 4 cm
Let the height of the cone be x cm
Volume of the cone =1/3 π × 42 × (x) ….. (ii)
(i) and (ii) are equal
4/3 π × (43 – 23) = 1/3 π × 42 × h
4 x (64 – 8) = 16 x h
h = 14
Therefore, the height of the cone = 14 cm
Question 26. A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Solution:
The internal radius of hollow sphere = 2 cm
The external radius of hollow sphere = 4 cm
Volume of the hollow sphere = 4/3 π × (43 – 23) … (i)
The base radius of the cone = 4 cm
Let the height of the cone be x cm
Volume of the cone = 1/3 π × 42 × h ….. (ii)
4/3 π × (43 – 23) = 1/3 π × 42 × h
4 x (64 – 8) = 16 x h
h = 14
Let Slant height of the cone be l
l = √(h2 + r2)
l = √(142 + 42) = √212
l = 14.56 cm
Therefore, the height = 14 cm and slant height = 14.56 cm
Question 27. A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.
Solution:
Radius of the spherical ball = 3 cm
Volume of the sphere = 4/3 πr3
Volume = 4/3 π33
Volume of first ball = 4/3 π 1.53
Volume of second ball = 4/3 π23
Let the radius of the third ball = r cm
Volume of third ball = 4/3 πr3
Volume of the spherical ball is equal to sum of volumes of three balls
4/3 π33 = 4/3 π 1.53+ 4/3 π23 + 4/3 πr3
(3)3 = (2)3 + (1.5)3 + r3
r3 = 33– 23– 1.53 cm3
r3 = 15.6 cm3
r = (15.625)1/3 cm
r = 2.5 cm
Diameter = 2 x radius = 2 x 2.5 cm
= 5.0 cm.
Therefore, the diameter of the third ball = 5 cm
Question 28. A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic meters of gravel are required to grave the path to a depth of 20 cm?
Solution:
Diameter of the circular pond = 40 m
Radius of the pond = 40/2 = 20 m = r
Thickness (width of the path) = 2 m
Height = 20 cm = 0.2 m
Thickness (t) = R – r
2 = R – 20
R = 22 m
Volume of the hollow cylinder = π (R2– r2) × h
= π (222– 202) × 0.2
= 52.8 m3
Therefore, the volume of the hollow cylinder = 52.8 m3
Question 29. A 16 m deep well with diameter 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7m. Find the height of the platform?
Solution:
Radius(r) of the cylinder = 3.5/2 m = 1.75 m
Depth of the well or height of the cylinder (h) = 16 m
Volume of the cylinder= πr2h
= π × 1.752 × 16
The length of the platform (l) = 27.5 m
Breadth of the platform (b) =7 m
Let the height of the platform be x m
Volume of the cuboid = l×b×h
= 27.5×7×(x)
Volume of cylinder = Volume of cuboid
π × 1.75 × 1.75 × 16 = 27.5 × 7 × x
x = 0.8 m = 80 cm
Therefore, the height of the platform = 80 cm.
Question 30. A well of diameter 2 m is dug 14 m deep. The earth taken out of it is evenly spread all around it to form an embankment of height 40 cm. Find the width of the embankment?
Solution:
Radius of the circular cylinder (r) = 2/2 m = 1 m
Height of the well (h) = 14 m
Volume of the solid circular cylinder = π r2h
= π × 12× 14 …. (i)
The height of the embankment = 40 cm = 0.4 m
Let the width of the embankment be (x) m.
External radius = (1 + x)m
Internal radius = 1 m
Volume of the embankment = π × r2 × h
= π × [(1 + x)2 – (1)2]× 0.4 ….. (ii)
(i) and (ii) are equal
π × 12 × 14 = π × [(1 + x)2 – (1)2] x 0.4
14/0.4 = 1 + x2 + 2x – 1
35 = x2 + 2x
x2 + 2x – 35 = 0
(x + 7) (x – 5) = 0
x = 5 or -7
Therefore, the width of the embankment = 5 m.
Question 31. A well with inner radius 4 m is dug up and 14 m deep. Earth taken out of it has spread evenly all around a width of 3 m it to form an embankment. Find the height of the embankment?
Solution:
Inner radius of the well = 4 m
Depth of the well = 14 m
Volume of the cylinder = π r2h
= π × 42 × 14 …. (i)
Width of the embankment = 3 m
Outer radius of the well = 3 + 4 m = 7 m
Volume of the hollow embankment = π (R2 – r2) × h
= π × (72 – 42) × h …… (ii)
(i) and (ii) are equal
π × 42 × 14 = π × (72 – 42) × h
h = 42 × 14 / (33)
h = 6.78 m
Therefore, the height of the embankment = 6.78 m.
Question 32. A well of diameter 3 m is dug up to 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 3 m
Radius of the well = 3/2 m = 1.5 m
Depth of the well (h) = 14 m
Width of the embankment (thickness) = 4 m
Radius of the outer surface of the embankment = (4 + 1.5) m = 5.5 m
Volume of the embankment = π(R2 – r2) × h
= π(5.52 – 1.52) × h ….. (i)
Volume of earth dug out = π × r2 × h
= π × (3/2)2 × 14 ….. (ii)
(i) and (ii) are equal
π(5.52 – 1.52) × h = π × (3/2)2 × 14
(5.5+1.5)(5.5-1.5) x h = 9 × 14/ 4
h = 9 × 14/ (4 × 28)
h = 9/8 m
= 1.125m
Therefore, the height of the embankment =1.125 m
Question 33. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.
Solution:
The side of the cube = 9 cm
Diameter of the cone = Side of cube
2r = 9
Radius of cone = 9/2 cm = 4.5 cm
Height of cone = side of cube
Height of cone (h) = 9 cm
Volume of the largest cone = 1/3 π × r2 × h
= 1/3 π × 4.52 × 9
= 190.93 cm3
Therefore, the volume of the largest cone to fit in the cube has a volume of 190.93 cm3
Question 34. A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Height of the cylindrical bucket = 32 cm
Radius of the cylindrical bucket = 18 cm
Height of conical heap = 24 cm
Volume of cylinder = π × r2 × h
Volume of cone = 1/3 π × r2 × h
Volume of the conical heap = Volume of the cylindrical bucket
1/3 π × r2 × 24 = π × 182 × 32
r2 = 182 X 4
r = 18 × 2 = 36 cm
Slant height (l) = √(h2 + r2)
l = √(242 + 362) = √576+1296
= √1872
= 43.26 cm
Therefore, the radius =36 cm and slant height = 43.26 cm
Question 35. Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm . What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen?
Solution:
Length of the rectangular surface = 6 m = 600 cm
Breadth of the rectangular surface = 4 m = 400 cm
Height of the rain = 1 cm
Volume of the rectangular surface = length * breadth * height
= 600×400×1 cm3
= 240000 cm3 …………….. (i)
Radius of the cylindrical vessel = 20 cm
Let the height of the cylindrical vessel be h cm
Volume of the cylindrical vessel = π × r2 × h
= π × 202 × h ……….. (ii)
(i) is equal to (ii)
240000 = π × 202 × h
h = 190.9 cm
Therefore, the height of the cylindrical vessel = 191
Summary
Exercise 16.1 Set 2 focuses on problems related to cuboids and cubes. It covers calculations involving surface area (lateral and total) and volume of these three-dimensional shapes. The questions range from straightforward applications of formulas to more complex scenarios involving real-world situations. Students are required to apply their knowledge of geometry, use appropriate formulas, and demonstrate problem-solving skills.
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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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