Ch 14 EOC Class 9 Maths Ganita Manjari | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 14 End of Chapter (EOC) Exercises. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this comprehensive review walkthrough covers advanced guesstimate problems, spherical and cylindrical volume relations, scaling principles, optimization challenges, and practical real-world geometric puzzles entirely clear.
🔑 Key Concepts & Advanced Principles:
- Scaling & Dimensional Changes: Analyzing how volumes and surface areas scale when linear dimensions are modified.
- Composite Solids & Displacements: Calculating fluid level changes using Archimedes' principle and marble drop volumes.
NCERT Class 9 Maths Ganita Manjari Chapter 14 End of Chapter Solutions
Question 1
Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions \(10\text{ cm} \times 15\text{ cm} \times 20\text{ cm}\).
Solution:
1. Volume of the cuboidal container = \(10 \times 15 \times 20 = 3000\text{ cm}^3\).
2. Assuming an average ice cream scoop is roughly spherical with a radius of approximately \(2\text{ cm}\), its volume is \(V_{\text{scoop}} = \frac{4}{3}\pi (2)^3 = \frac{32\pi}{3} \approx 33.5\text{ cm}^3\).
3. Accounting for packing efficiency and air gaps (packing factor \(\approx 0.64\)), usable ice cream volume \(= 3000 \times 0.64 = 1920\text{ cm}^3\).
4. Estimated number of scoops = \(\frac{1920}{33.5} \approx \mathbf{55 \text{ to } 60 \text{ scoops}}\).
Question 2
A cube of integer side length \(a\) is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length \(a + 1\).
Solution:
1. Volume of the initial cube of side \(a\) = \(a^3\).
2. Volume of the larger cube of side \(a + 1\) = \((a + 1)^3\).
3. Number of unit cubes to be added = \((a + 1)^3 - a^3 = a^3 + 3a^2 + 3a + 1 - a^3 = \mathbf{3a^2 + 3a + 1}\).
Question 3
Solve the following:
(i) Could a person drink enough water in a lifetime to fill an entire room the size of your classroom?
(ii) Estimate the number of bricks used to build the walls of your classroom.
Solution:
* (i) Classroom water volume: A typical classroom volume is around \(150\text{ m}^3 = 150,000\text{ litres}\). Since a human drinks about \(2\text{ litres}\) per day over 80 years (\(\approx 30,000\text{ days}\)), total lifetime water intake is roughly \(60,000\text{ litres}\). Thus, \(\mathbf{\text{no}}\), a single person's lifetime drinking water is less than half a classroom's volume.
* (ii) Brick estimation: For a classroom with wall dimensions totaling \(40\text{ m}\) perimeter, \(3\text{ m}\) height, and \(0.2\text{ m}\) thickness (\(\text{wall volume} \approx 24\text{ m}^3\)), using standard brick size (\(20\text{ cm} \times 10\text{ cm} \times 10\text{ cm} = 0.002\text{ m}^3\)), the estimated number of bricks is \(\frac{24}{0.002} \approx \mathbf{12,000 \text{ bricks}}\).
Question 4
You are given two options: (A) one chocolate cube of side \(50\text{ mm}\), (B) hundred chocolate cubes of side \(10\text{ mm}\). Which option gives you more chocolate?
Solution:
1. Volume of Option A (one cube of side \(50\text{ mm} = 5\text{ cm}\)):
\(V_A = 5^3 = 125\text{ cm}^3\).
2. Volume of Option B (hundred cubes of side \(10\text{ mm} = 1\text{ cm}\)):
\(V_B = 100 \times (1^3) = 100 \times 1 = 100\text{ cm}^3\).
3. Comparing volumes, \(125\text{ cm}^3 > 100\text{ cm}^3\), so \(\mathbf{\text{Option A}}\) gives you more chocolate.
Question 5
If all the people in the world are crowded together in one location, how much area would it cover?
Solution:
1. World population \(\approx 8\text{ billion}\) people.
2. Assuming standing space per person is approximately \(0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2\).
3. Total area = \(8,000,000,000 \times 0.25 = 2,000,000,000\text{ m}^2 = 2000\text{ km}^2\), which easily fits inside a small state or a region like Luxemburg/Tokyo.
Question 6
A school provides milk to its students in cylindrical glasses, each with a diameter of \(7\text{ cm}\). If each glass is filled with milk to a height of \(12\text{ cm}\), how many litres of milk are needed to serve 1600 students? Use \(1000\text{ cm}^3 = 1\text{ litre}\).
Solution:
1. Radius of glass \(r = \frac{7}{2} = 3.5\text{ cm}\), height \(h = 12\text{ cm}\).
2. Volume of milk in one glass = \(\pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 12 = \frac{22}{7} \times 12.25 \times 12 = 462\text{ cm}^3\).
3. Total volume for 1600 students = \(1600 \times 462 = 739,200\text{ cm}^3\).
4. Converting to litres (\(1000\text{ cm}^3 = 1\text{ litre}\)): \(\frac{739,200}{1000} = \mathbf{739.2\text{ litres}}\).
Question 7
The surface area of a sphere of radius \(5\text{ cm}\) is five times the area of the curved surface of a cone of radius \(4\text{ cm}\). Find the height and the volume of the cone.
Solution:
1. Surface area of sphere (\(r = 5\text{ cm}\)) = \(4\pi (5)^2 = 100\pi\text{ cm}^2\).
2. Given \(100\pi = 5 \times (\text{CSA of cone})\) \(\implies \text{CSA of cone} = 20\pi\text{ cm}^2\).
3. \(\text{CSA of cone} = \pi r l \implies \pi (4) l = 20\pi \implies l = 5\text{ cm}\).
4. Height of cone \(h = \sqrt{l^2 - r^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = \mathbf{3\text{ cm}}\).
5. Volume of cone = \(\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (4)^2 (3) = \mathbf{16\pi\text{ cm}^3} \approx 50.24\text{ cm}^3\).
Question 8
Take Earth to be a perfect sphere with radius \(6370\text{ km}\). Take Jupiter to be a perfect sphere with radius \(69,900\text{ km}\). Take the Sun to be a perfect sphere with radius \(695,700\text{ km}\). Compute approximately:
(i) the ratio of the volume of Jupiter to the volume of the Earth;
(ii) the ratio of the volume of the Sun to the volume of the Earth.
Solution:
* (i) Ratio of Jupiter to Earth volume:
\(\left(\frac{69,900}{6370}\right)^3 \approx (10.97)^3 \approx \mathbf{1320 : 1}\).
* (ii) Ratio of Sun to Earth volume:
\(\left(\frac{695,700}{6370}\right)^3 \approx (109.2)^3 \approx \mathbf{1,300,000 : 1}\).
Question 9
Show that the volume of a sphere is equal to \(\frac{2}{3}\) of the volume of the smallest cylinder which encloses it.
Solution:
1. For a sphere of radius \(r\), the smallest enclosing cylinder has radius \(r\) and height \(h = 2r\).
2. Volume of sphere = \(\frac{4}{3}\pi r^3\).
3. Volume of cylinder = \(\pi r^2 h = \pi r^2 (2r) = 2\pi r^3\).
4. Ratio = \(\frac{\frac{4}{3}\pi r^3}{2\pi r^3} = \frac{4}{6} = \mathbf{\frac{2}{3}}\). Hence proved.
Question 10
Suppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, \(1\text{ metre}\) longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere. How high above the ground is the second string? Repeat for: (i) Moon, (ii) Jupiter, (iii) a volleyball. Are you surprised by the three answers? Why or why not?
Solution:
1. Let original radius be \(R\). Original circumference \(C = 2\pi R\).
2. New circumference \(C' = 2\pi R + 1 = 2\pi R'\).
3. \(2\pi (R' - R) = 1 \implies R' - R = \frac{1}{2\pi} \approx \frac{1}{2 \times 3.1416} \approx \mathbf{0.159\text{ m}}\) (or \(15.9\text{ cm}\)).
4. Surprising conclusion: The radius increase depends solely on the added length (\(1\text{ m}\)), not on the initial size of the sphere! Thus, the height above ground is \(\approx 15.9\text{ cm}\) for the Earth, Moon, Jupiter, and volleyball alike.
Question 11
We have a cylinder with a base radius of \(r\text{ cm}\) and height \(h\text{ cm}\). A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.
Solution:
1. The cylinder has radius \(r\) and height \(h\). Volume of cylinder = \(\pi r^2 h\).
2. The square base of the pyramid is inscribed in the cylinder's circular base of radius \(r\). The diagonal of the square equals the diameter of the circle (\(2r\)).
3. Side of the square base \(s = \frac{2r}{\sqrt{2}} = r\sqrt{2}\). Area of square base = \(s^2 = (r\sqrt{2})^2 = 2r^2\).
4. Height of the pyramid = \(h\). Volume of pyramid = \(\frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} (2r^2) h = \frac{2}{3}r^2 h\).
5. Ratio of pyramid volume to cylinder volume = \(\frac{\frac{2}{3}r^2 h}{\pi r^2 h} = \mathbf{\frac{2}{3\pi}}\).
Question 12
What is the change in volume when:
(i) the length of a cuboid with dimensions \(l, w, h\) is increased by 1 unit?
(ii) the radius of a cylinder with dimensions \(r, h\) is increased by 1 unit?
(iii) the radius of a sphere is decreased by 1 unit?
Solution:
* (i) Cuboid length increased by 1:
New volume \(= (l + 1)wh = lwh + wh\). Change in volume = \(\mathbf{wh \text{ cubic units}}\) (Option c).
* (ii) Cylinder radius increased by 1:
New volume \(= \pi(r + 1)^2 h = \pi(r^2 + 2r + 1)h = \pi r^2 h + \pi(2r + 1)h\). Change in volume = \(\mathbf{2\pi rh + \pi h \text{ cubic units}}\) (Option e).
* (iii) Sphere radius decreased by 1:
Initial volume \(V = \frac{4}{3}\pi r^3\). New volume \(V' = \frac{4}{3}\pi (r - 1)^3\). Change in volume = \(\frac{4}{3}\pi [r^3 - (r - 1)^3]\).
Question 13
Given a cube with volume \(V\), express its total surface area \(S\) in terms of \(V\).
Solution:
1. Let cube side be \(a\). Volume \(V = a^3 \implies a = V^{1/3}\).
2. Total surface area \(S = 6a^2 = \mathbf{6V^{2/3}}\).
Question 14
Given a cube with total surface area \(S\), express its volume \(V\) in terms of \(S\).
Solution:
1. Total surface area \(S = 6a^2 \implies a^2 = \frac{S}{6} \implies a = \sqrt{\frac{S}{6}}\).
2. Volume \(V = a^3 = \left(\sqrt{\frac{S}{6}}\right)^3 = \mathbf{\left(\frac{S}{6}\right)^{3/2}}\).
Question 15
A cylindrical glass of height \(25\text{ cm}\) and radius \(4\text{ cm}\) has water up to a height of \(16\text{ cm}\). A crow wants to drink water from this glass. The water must be at a height of \(20\text{ cm}\) for the crow to reach it. There are some marbles lying around. How many marbles of radius \(1\text{ cm}\) should the crow drop into the glass to make the water reach the required height?
Solution:
1. Radius of glass \(R = 4\text{ cm}\). Initial water height \(h_1 = 16\text{ cm}\), target height \(h_2 = 20\text{ cm}\).
2. Required rise in water level \(\Delta h = 20 - 16 = 4\text{ cm}\).
3. Volume of water to be displaced (volume of rise) = \(\pi R^2 \Delta h = \pi (4)^2 (4) = 64\pi\text{ cm}^3\).
4. Volume of one spherical marble of radius \(r = 1\text{ cm}\) = \(\frac{4}{3}\pi (1)^3 = \frac{4}{3}\pi\text{ cm}^3\).
5. Number of marbles required = \(\frac{64\pi}{\frac{4}{3}\pi} = 64 \times \frac{3}{4} = \mathbf{48 \text{ marbles}}\).
Question 16
Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.
Solution:
1. The ink reservoir in a ballpoint pen is a thin cylindrical tube of length \(\approx 10\text{ cm}\) and inner radius \(\approx 0.15\text{ cm}\).
2. Volume \(V = \pi r^2 h = \frac{22}{7} \times (0.15)^2 \times 10 \approx \frac{22}{7} \times 0.0225 \times 10 \approx \mathbf{0.7\text{ cm}^3}\) (or \(0.7\text{ ml}\)).
Question 17
Solve:
(i) A ball of chapati dough of radius \(6\text{ cm}\) is prepared. Estimate how many chapatis can be made from it?
(ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
Solution:
* (i) Chapati dough estimation:
- Volume of dough ball (\(r = 6\text{ cm}\)) = \(\frac{4}{3}\pi (6)^3 = 288\pi \approx 904.78\text{ cm}^3\).
- A rolled-out chapati is a cylinder of radius \(\approx 7\text{ cm}\) and thickness \(\approx 0.2\text{ cm}\), volume \(\approx \pi (7)^2 (0.2) = 9.8\pi \approx 30.78\text{ cm}^3\).
- Number of chapatis = \(\frac{904.78}{30.78} \approx \mathbf{25 \text{ to } 30 \text{ chapatis}}\).
* (ii) Edible coconut/muskmelon volume: Measured via shell thickness and outer/inner radii using hemispherical shell volume formula \(\frac{2}{3}\pi (R^3 - r^3)\).
Question 18
Looking at the Ganita Manjari, Grade 9, Part 2 textbook, Sheela wonders:
(i) If all the pages of this textbook were laid out side-by-side on the floor would they cover the entire classroom floor?
(ii) What is the maximum number of textbooks that can fit in an empty storeroom of dimensions \(15\text{ ft} \times 20\text{ ft} \times 30\text{ ft}\)?
Solution:
* (i) A typical textbook has about 150 pages, each around \(20\text{ cm} \times 25\text{ cm} = 500\text{ cm}^2 = 0.05\text{ m}^2\). Total page area \(\approx 150 \times 0.05 = 7.5\text{ m}^2\). Since a classroom floor is usually \(30\text{ m}^2\) to \(50\text{ m}^2\), \(\mathbf{\text{no}}\), it would not cover the entire floor.
* (ii) Storeroom capacity: Storeroom volume = \(15 \times 20 \times 30 = 9000\text{ cubic feet}\). A textbook volume \(\approx 0.05\text{ cubic feet}\). Max textbooks \(\approx \frac{9000}{0.05} \approx \mathbf{180,000 \text{ textbooks}}\) (theoretical maximum).
Question 19
If the entire human population decided to climb into one giant cube, how long would the side have to be?
Solution:
1. World population \(\approx 8\text{ billion}\) people. Assuming each person occupies an average volume of \(0.06\text{ m}^3\) (\(\approx 60\text{ litres}\)).
2. Total volume needed = \(8,000,000,000 \times 0.06 = 480,000,000\text{ m}^3\).
3. Side length of cube \(a = \sqrt[3]{480,000,000} \approx \mathbf{783\text{ metres}}\).
Question 20
The Earth's surface has an estimated volume of \(1.38\text{ billion km}^3\) of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering the Earth's surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth's radius is \(\sim 6371\text{ km}\).
(i) Write an expression that gives the thickness of this water layer.
(ii) Simplify the expression in (i) using a calculator.
Solution:
* (i) Expression for thickness (\(t\)):
Volume of water \(V = \frac{4}{3}\pi (R + t)^3 - \frac{4}{3}\pi R^3 \approx 4\pi R^2 t \implies t = \frac{V}{4\pi R^2}\).
* (ii) Calculation:
\(V = 1.38 \times 10^9\text{ km}^3\), \(R = 6371\text{ km}\).
\(t = \frac{1.38 \times 10^9}{4 \times 3.1416 \times (6371)^2} \approx \frac{1.38 \times 10^9}{510,064,472} \approx \mathbf{2.7\text{ km}}\).
Question 21
Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
Solution:
By testing scaling factors on cuboid dimensions \((l, w, h)\), scaling dimensions by a specific ratio (or analyzing reciprocal proportion properties) yields valid solutions such as dimensions scaled inversely.
Question 22
Project: Find the volume of your house making necessary approximations. Present how you solved it.
Solution:
Measured room dimensions (length, width, ceiling height) to compute room volumes and summed them up alongside roof approximations.
Frequently Asked Questions (FAQs)
Q1: How do guesstimate problems help in practical mathematics?
A: They develop order-of-magnitude estimation skills and connect mathematical formulas to real-world physical dimensions.
Q2: How do you calculate liquid displacement when objects are dropped into containers?
A: The volume of the displaced liquid equals the total volume of the submerged objects.
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