Chapter 1 Objective Type Question Answer Q1. Choose the correct option 1. The HCF of `96` and `404` will be a) `120` b) `4` c) `10` d) `3` 2. The HCF of `12` and `15` will be a) `3` b) `4` c) `10` d) `5` 3. Product of  numbers `= 32` and their LCM `= 8`, then their HCF Will be a) `4` b) `8` c) `32` d) `256` 4. The HCF of `4` and `7` will be  a) `1` b) `2` c) `3` d) `4` 5. For some integer m, every sum integer is of the form a) `2m+3` b) `2m + 1` c) `2m` d) `2m+ 5` 6. The rato of LCM and HCF for the numbers `5, 15, 20` Will be a) `9: 1` b) `4:3` c) `11:1` d) `12:1` Answer  1(b), 2.(a), 3.(a), 4.(a), 5.(c), 6.(d) Q2. Fill in the blanks 1) Dividend = divisor×qutient+_____. 2) The value of HCF of the numbers `8,9` and `25` will be______ . Answer 1. Remainder, 2. 1 Q3. Write True/False 1) Two positive integers a and b, there exist whole numbers q and r satisfying `a = bq+r, 0le r ltb.` 2) A prime factorisation of a natural number is unique apart from the order of its factors Q...

NCERT Class 10th solution of Exercise 13.1 NCERT Class 10th solution of Exercise 13.2 NCERT Class 10th Maths Projects Exercise 13.3 Take `pi = 22/7` unless stated otherwise. Q1. A metallic sphere of radius `text{4.2 cm}` is melted and recast into the shape of a cylinder of radius `text{6 cm}`. Find the height of the cylinder. `text{Sol. :}` `text{Given :}` `text{Radius of Sphere }R = 4.2 text{ cm}` `text{Radius of Cylinder }r = 6 text{ cm}` `text{To find :}` `text{Height of the Cylinder}` `text{Solve :}` `text{According to question}` `text{Volume of the Cylinder = Volume of the Sphere}` `pir^2h = 4/3piR^3` `h = (4piR^3)/(3pir^2)` `h = (4R^3)/(3r^2)` `h = (4times(4.2)^3)/(3(6)^2)` `h = (4times4.2times4.2times4.2)/(3times6times6)` `h = (4times4.2times0.7times0.7)/3` `h = (8.232)/3` `h = 2.744text{ cm}` `text{Answer :}` `text{The height of the cylinder 2.744 cm.` ☝  Like,  Share,  and  Subscribe.  Q2. Metallic spheres of radii `text{6 cm, 8 cm}` and `text{10 cm,}` ...

NCERT Class 10th solution of Exercise 13.1 NCERT Class 10th solution of Exercise 13.3 NCERT Class 10th Maths Projects Chapter 13 Surface Areas And Volume Exercise 13.2 Unless stated otherwise, take `pi = 22/7`. Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to `text{1 cm}` and the height of the cone is equal to its radius. Find the volume of the solid in terms of `pi`. `text{Sol. :}` `text{Given :}` `text{Radius of Cone & hemisphere } r = 1text{ cm}` `text{Height of Cone & hemisphere } h = r =1text{ cm}` `text{To find :}` `text{Volume of the solid}` `text{Solve :}` `text{Volume of Solid = Volume of hemisphere}` `text{                           +  Volume of Cone}` `text{V} = 2/3pir^3 + 1/3pir^2h` `text{    }= pir^2(2/3r + 1/3h)` `text{    }= pi(1)^2(2/3times1 + 1/3 times 1)` `text{    }= pi(2/3 + 1/3)` `text{    }= pi...