Rs Aggarwal 2018 for Class 10 Math Chapter 1 - Real Numbers

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Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 10 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 874:

Question 1:

Two cubes each of volume 27 cm³ are joined end to end to form a solid. Find the surface area of the resulting cuboid. [CBSE 2011, 14]

Answer:

$As, volume of a cube = 27 {cm}^{3} \Rightarrow {(edge)}^{3} = 27 \Rightarrow edge = \sqrt[3]{27} \Rightarrow edge = 3 cm The length of the resulting cuboid, l = 3 + 3 = 6 cm, its breadth, b = 3 cm and its height, h = 3 cm Now, the surface area of the resulting cuboid = 2 (l b + b h + h l) = 2 (6 \times 3 + 3 \times 3 + 3 \times 6) = 2 (18 + 9 + 18) = 2 \times 45 = 90 {cm}^{2}$

Match the following columns:

Column I	Column II
(a) A solid metallic sphere of radius 8 cm is melted and the material is used to make solid right cones with height 4 cm and base radius of 8 cm. How many cones are formed?	(p) 18
(b) A 20-m-deep well with diameter 14 m is dug up and the earth from digging is evenly spread out to form a platform 44 m by 14 m. The height of the platform is ...........m.	(q) 8
(c) A sphere of radius 6 cm is melted and recast in the shape of a cylinder of radius 4 cm. Then, the height of the cylinder is ......... cm.	(r) 16 : 9
(d) The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is ....... .	(s) 5

Answer:

(a)
Volume of the sphere $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3} π \times {(8)}^{3}) {cm}^{3}$

Volume of each cone $= \frac{1}{3} π r^{2} h$
$= \frac{1}{3} π \times {(8)}^{2} \times 4 {cm}^{3}$

Number of cones formed $= \frac{Volume of the sphere}{Volume of each cone}$

$= \frac{4 π \times 8 \times 8 \times 8 \times 3}{3 \times π \times 8 \times 8 \times 4} = 8$
Hence, $(a) \Rightarrow (q)$

(b)
Volume of the earth dug out = Volume of the cylinder
$= π r^{2} h = \frac{22}{7} \times 7 \times 7 \times 20$
Let the height of the platform be h.
Then, volume of the platform = volume of the cuboid
$= (44 \times 14 \times h) m^{3}$
Therefore,

$\frac{22}{7} \times 7 \times 7 \times 20 = 44 \times 14 \times h \Rightarrow 3080 = 616 \times h \Rightarrow h = \frac{3080}{616} \Rightarrow h = 5 m$

Hence, $(b) \Rightarrow (s)$

(c)
Volume of the sphere
$= \frac{4}{3} π r^{3}$

$= \frac{4}{3} π \times 6 \times 6 \times 6$

Let h be the height of the cylinder.
Then, volume of the cylinder $= {πr}^{2} h$
$= π \times 4 \times 4 \times h$
Therefore,

$\frac{4}{3} π \times 6 \times 6 \times 6 = π \times 4 \times 4 \times h \Rightarrow \frac{4}{3} \times 6 \times 6 \times 6 = 4 \times 4 \times h \Rightarrow 228 = 16 \times h \Rightarrow h = \frac{228}{16} \Rightarrow h = 18 cm$

Hence, $(c) \Rightarrow (p)$

(d)
Let the radii of the spheres be R and r respectively.
Then, ratio of their volumes $= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}}$
Therefore,
$\frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}} = \frac{64}{27} \Rightarrow \frac{R^{3}}{r^{3}} = \frac{64}{27} \Rightarrow {(\frac{R}{r})}^{3} = {(\frac{4}{3})}^{3} \Rightarrow \frac{R}{r} = \frac{4}{3}$
Hence, the ratio of their surface areas $= \frac{4 π R^{2}}{4 π r^{2}}$
$= \frac{R^{2}}{r^{2}} = {(\frac{R}{r})}^{2} = {(\frac{4}{3})}^{2} = \frac{16}{9} = 16 : 9$
Hence, $(d) \Rightarrow (r)$

Page No 925:

Question 75:

Match the following columns:

Column I	Column II
(a) The radii of the circular ends of a bucket, in the form of the frustum of a cone of height 30 cm, are 20 cm and 10 cm respectively. The capacity of the bucket is ........cm³.	(p) 2418π
(b) The radii of the circular ends of a conical bucket of height 15 cm are 20 and 12 cm respectively. The slant height of the bucket is ........ cm.	(q) 22000
(c) The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is .........cm².	(r) 12
(d) Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is ........ cm.	(s) 17

Answer:

$We have, Radius of the upper end of the frustum, R = \frac{45}{2} cm, Radius of the lower end of the frustum = Radius of the cylinder = r = \frac{25}{2} cm, Height of the cylinder, h = 6 cm and Total height of the bucket = 40 cm And, the height of the frustum, H = 40 - 6 = 34 cm Also, the slant height of frustum, l = \sqrt{{(R - r)}^{2} + H^{2}} = \sqrt{{(\frac{45}{2} - \frac{25}{2})}^{2} + 34^{2}} = \sqrt{10^{2} + 34^{2}} = \sqrt{100 + 1156} = \sqrt{1256} \approx 35.44 cm Now, The area of the metallic sheet used to make the bucket = CSA of the frustum + CSA of the cylinder + Area of the base of the cylinder = π (R + r) l + 2 πrh + {πr}^{2} = \frac{22}{7} \times (\frac{45}{2} + \frac{25}{2}) \times 35.44 + 2 \times \frac{22}{7} \times \frac{25}{2} \times 6 + \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2} = \frac{22}{7} \times 35 \times 35.44 + \frac{22}{7} \times 150 + \frac{22}{7} \times \frac{625}{4} = \frac{22}{7} \times (1240.4 + 150 + 156.25) = \frac{22}{7} \times 1546.65 = 4860.9 {cm}^{2} Also, The volume of the water that the bucket can hold = Volume of the frustum = \frac{1}{3} π H (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 34 \times [{(\frac{45}{2})}^{2} + {(\frac{25}{2})}^{2} + (\frac{45}{2}) (\frac{25}{2})] = \frac{1}{3} \times \frac{22}{7} \times 34 \times [{(\frac{45}{2})}^{2} + {(\frac{25}{2})}^{2} + (\frac{45}{2}) (\frac{25}{2})] = \frac{748}{21} \times (\frac{2025}{4} + \frac{625}{4} + \frac{1125}{4}) = \frac{748}{21} \times \frac{3775}{4} = 33615.48 {cm}^{3} = 33.61548 L (As, 1000 {cm}^{3} = 1 L) \approx 33.61 L$

Page No 938:

Question 20:

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, then in how much time will the tank be filled completely?

Answer:

$We have, Internal radius of the pipe, r = \frac{20}{2} = 10 cm = 0.1 m, Radius of the cylindrical tank, R = \frac{10}{2} = 5 m and Height of the cylindrical tank, H = 2 m Also, the speed of the water flow in the pipe, h = 4 km / hr = \frac{4 \times 1000 m}{1 hr} = 4000 m / hr Now, The volume of the water flowing out of the pipe in a hour = π r^{2} h = \frac{22}{7} \times 0.1 \times 0.1 \times 4000 = \frac{880}{7} m^{3} And, The volume of the cylindrical tank = π R^{2} H = \frac{22}{7} \times 5 \times 5 \times 2 = \frac{1100}{7} m^{3} So, The time taken to fill the tank = \frac{Volume of the cylindrical tank}{Volume of water flowing out of the pipe in a hour} = \frac{(\frac{1100}{7})}{(\frac{880}{7})} = \frac{1100}{880} = \frac{5}{4} hr = 1 \frac{1}{4} hr = 1 hr and \frac{1}{4} \times 60 \min = 1 hr 15 \min$

So, the tank will be completely filled in 1 hour 15 minutes.

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