Class 6 chapter 10 math solution discuss about the Story of three dimensional objects. Firstly, we see here introduction of Class 6 chapter 10 math solution. Sir Charles Wheatstone was an English scientist and inventor of many scientific advances of the Victorian era, Which was stereoscopic technology and the technique of displaying three-dimensional objects .
He invented and see the technique in 1838. In the SIR CHARLES WHEATSTONE research, he discuss about the phenomenon of two different images being projected onto the retina of the eye while viewing a three-dimensional object. Leonardo da Vinci was an Italian painter, sculptor, draftsman, architect, and engineer. whose skill and intelligence probably exceeded of any other man. “The Last Supper” and “Mona Lisa” are among his most popular and influential paintings. Da Vinci was able to create three-dimensional models of several inventions, including the flying machine, a convex lens grinding machine, and the hydraulic machine. His model later led to many incredible discoveries.
We are use in Daily life many things of two and three dimensional objects. Like as fhe Chair, table, football, box, book, and others. This article is for class 6 students and teacher. We are solve it by the English version system. Many students of class six searching of English version math solution but they can not found it. So, if you are a english version students, this article helpful for you. Here, you continue the reading and get your chapter 10 solution.
Class 6 chapter 10 Number 1 Math Solution
1) Cut out some paper according to the measurement shown in the picture and then fold and 6, y-10 attach it with scotch tape to form a rectangular solid Volume of the boo object. What will be the volume of the rectangular solid object?
Solution :
Cutting out some paper according to the measurement shown in the picture and then folding. Attaching it with scotch tape and will be a rectangular solid object.
Here, length 7 cm, breadth-2 cm and height-5cm.
Now, the volume of the rectangular solid =(length x breadth x height) cubic unit
= (7x2x5) cubic cm.
= 70 cubic cm. (ans)
Class 6 chapter 10 Number 2 Math Solution
2) The following diagram is an open rectangular box. The measurements are given in centimeter units.
a) Find values of a, x, y
(b) Find the volume of the box.
Solution :
a) The given diagram is an open rectangular box. If we fold and attach them, it will look like
From the diagram of the open rectangular box. So, we see that length, x=10 cm, breadth, y=6 cm and height a=4 cm
So, a = 4, x = 6, y = 10 (Ans.)
b) From ‘a’, we get, a = 4, x = 6, y = 10
Volume of the box
= (length x breadth x height) Cubic Cm
So, = (10x6x4)Cubic Cm
= 240 Cubic Cm (Ans.
Read more
CHAPTER 1 : The Story of Numbers
CHAPTER 2 : The Story of Two Dimensional Objects
3 Chapter : Information Investigation and Analysis
CHAPTER 4 : Trees of Prime Factors
CHAPTER 5 : Measurement of Length
The CHAPTER 6 : The World of Integers
CHAPTER 7 : The Game of Fractions
CHAPTER 8: World of Unknown expressions
The CHAPTER 9: Linear Equation
CHAPTER 10 : Story of Three Dimensional Objects
CHAPTER 11 : Unitary Method, Percentages and Ratio
The Chapter 12 : Set the Formula, get the Formula
3 number math solution chapter 10
3) How many small cubic-size pieces will be needed to make each of the shapes shown in the picture ?
Solution :
The figure shown in the picture is a Rubik’s cube it needs 3 small cubes along the direction of its length, breadth and height.
Number of small cubes requires to form the Rubik’s cube = (3x3x3)-27 (Ans.)
Chapter 10 number 4 math
4) Find out how many Mathematics Books from class 6 will be needed to fill up one shelf of the Bookshelf of the library of your school.
Solution :
For the Math book of class six,
The length is 10, the breadth is 6 and height is 3.
For the Bookshelf of the library, the length is 50
Breadth 15 and the height is 12
The volume of the Math book is
= (10 x 6 × 3)
= 180
The volume of the shelf
= Length × breadth × height
= 50 × 15 × 12
Answer, = 9000
The number of books required to fill up one shelf of the bookshelf of our school library
= Volume of the shelf/ Volume of the Math book
= 9000 /180
Answer = 50
Therefore, the required number of Math books is 50.
Chapter 10 number 5 math solution
5) A truck has a space of 12 feet x 6 feet x Observe what’s written on the packet and according feet, to fill up with cartons to carry. If the size of each carton is 2 feet x 2 feet x 1 foot, how many cartons may be possible for the truck to carry?
Solution:
Here, the length, breadth and height of the space of the truck are 12 ft. 6 ft. and 8 ft.
The volume is =
= (12 x 6 x 8) cubic feet
= 576 Cubic ft.
Again, the length, breadth and height of the cartoon are 2
ft., 2 ft. and I ft.
The volume of each carton
= (2 x 2 x 1) cubic feet
= 4 Cubic ft
The number of cartons may be possible for the truck to carry (576 ÷ 4) =144 (Ans.)
6 number math solution class 6
6) A pile of pages was made by putting 200 pieces of paper, like the one in the picture, with one on top of another.
a) What will be the volume of the pile of pages?
b) What is the thickness of one page?
Solution :
a)
Length of the pile of pages = 7 cm breadth = 4 cm and height=3cm
So, Volume of the pile of pages
=(7×4 x 3) Cubic Cm
= 84 Cubic Cm
b)
From the figure, Thickness of 200 pages = 3 cm
So, Thickness of 1 page = 3 ÷ 200 cm
= 0.015 cm ( Ans)
Number 7 math solution
Now answer the following questions:
a) What is the weight of 1 page?
b) What is the weight of the whole packet?
c) By measuring what is the height of the packet, can you find the thickness
of one page?
Solution :
A) Length of 1 page=297mm
= 297 ÷ 1000 m
=0.297 m
and breadth of 1 page 210 mm
= 210 ÷ 1000 m
=. 210 m
Area of 1 page = (0.297 x 210) m²
=0.06237 m²
weight of paper per square meter = 80 gm
So, weight of paper 0.06237 square meter
=(80 x 0.06237) gm
= 4.9896 gm (Ans.)
B) From ‘a’, the weight of 1 sheet is 4.9896 gm
So, Weight of 500 sheets is (4.9896 x 500) gm
= 2494.80 gm
= 2.4948 kg
Weight of the whole packet is 2494.80 gm (Ans.)
C) Yes, I can measure the thickness of a paper by measuring the height of the packet. In this case, let us ignore the thickness of the wrapping paper of the packet.
Let, the height of the packet he 48 mm.
Now, So, the height of 500 sheets is 48 mm
The height of 1 sheet is
= ( 48 ÷ 500) mm
= 0.096 mm ( Ans)
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