311. If the radius of the base, and the height of a right circular cone are increased by 20%, what is the approximate percentage increase in volume?
     

1. 1. 60
      

   
   
   

   2. 68.8
      

   
   
   

   3. 72.8
      

   
   
   

   4. 75

   
   
   

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1. View Hint View Answer Discuss in Forum

   Volume of cone =	1	πr²h
   	3

   
   

   Single equivalent increase in radius =		20 + 20 +	20 × 20		% = 44%
   			100

   
   Single equivalent percentage effect for 44% and 20%
   

   =		44 + 20 +	44 × 20		% = (64 + 8.8)%
   			100

   
   = 72.8%
   = Increase in volume

   Correct Option: C

   Volume of cone =	1	πr²h
   	3

   
   

   Single equivalent increase in radius =		20 + 20 +	20 × 20		% = 44%
   			100

   
   Single equivalent percentage effect for 44% and 20%
   

   =		44 + 20 +	44 × 20		% = (64 + 8.8)%
   			100

   
   = 72.8%
   = Increase in volume

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312. Which of the the following statements is not correct?
     

1. 1. For a given radius and height, a right circular cone has the lesser volume among a right circular cone and a right circular cylinder.

   
   
   

   2. If side of a cube is increased by 10%, the volume will increase by 33.1%.

   
   
   

   3. If the radius of a sphere is increased by 20%, the surface area will increase by 40%.

   
   
   

   4. Cutting a sphere into 2 parts does not change the total volume.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Percentage increase in surface area of sphere =		20 + 20 +	20 × 20		% = 44%
   			100

   Correct Option: C

   Percentage increase in surface area of sphere =		20 + 20 +	20 × 20		% = 44%
   			100

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313. There is a 4% increase in volume when a liquid freezes to its solid state. The percentage decrease when solid melts to liquid again,is

1. 1. 3	3	%
      	13

   
   
   

   2. 4%

   
   
   

   3. 4	1	%
      	13

   
   
   

   4. 3	11	%
      	13

   
   
   

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1. View Hint View Answer Discuss in Forum

   Required percentage decrease =	4	× 100
   	(100 + 4)

   
   

   =	400	=	50	= 3	11	%
   	104		13		13

   Correct Option: D

   Required percentage decrease =	4	× 100
   	(100 + 4)

   
   

   =	400	=	50	= 3	11	%
   	104		13		13

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314. A circle is inscribed in a square. An equilateral triangle of side 4√3 cm is inscribed in that circle. The length of the diagonal of the square (in centimetres) is
     

1. 1. 4√2
      

   
   
   

   2. 8
      

   
   
   

   3. 8√2
      

   
   
   

   4. 16

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Area of equilateral triangle ABC
   

   =	√3	× (4√3)²	48√3	= 12√3 cm²
   	4		4

   
   Again, AD is the height and O is the centre of the circle
   ∴ Area of ∆ ABC
   

   =	1	× BC × AD
   	2

   
   

   ⇒ 12√3 =	1	× 4√3 × AD
   	2

   
   

   ⇒ AD =	12√3	= 6
   	2√3

   
   

   ∴ OD =	1	AD = 2 cm
   	3

   
   ∴ OB = √BD² + OD²
   = (2√3)² + 2²
   = √16 = 4 cm.
   ∴ Side of square = 2 × OB = 2 × 4 = 8 cm.
   ∴ Diagonal of square = √2 × Side = 8√2 cm

   Correct Option: C

   
   Area of equilateral triangle ABC
   

   =	√3	× (4√3)²	48√3	= 12√3 cm²
   	4		4

   
   Again, AD is the height and O is the centre of the circle
   ∴ Area of ∆ ABC
   

   =	1	× BC × AD
   	2

   
   

   ⇒ 12√3 =	1	× 4√3 × AD
   	2

   
   

   ⇒ AD =	12√3	= 6
   	2√3

   
   

   ∴ OD =	1	AD = 2 cm
   	3

   
   ∴ OB = √BD² + OD²
   = (2√3)² + 2²
   = √16 = 4 cm.
   ∴ Side of square = 2 × OB = 2 × 4 = 8 cm.
   ∴ Diagonal of square = √2 × Side = 8√2 cm

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315. The radius and the height of a cone are each increased by 20%. Then the volume of the cone increases by
     

1. 1. 20%
      

   
   
   

   2. 20.5%
      

   
   
   

   3. 62%
      

   
   
   

   4. 72.8%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Volume of cone =	1	πr²h
   	3

   
   

   ∴ Resultant increase in radius² =		20 + 20 +	20 × 20		% = 44%
   			100

   
   

   Resultant increase in r² and h =		44 + 20 +	44 × 20		% = (64 + 8.8)%
   			100

   
   = 72.8%

   Correct Option: D

   Volume of cone =	1	πr²h
   	3

   
   

   ∴ Resultant increase in radius² =		20 + 20 +	20 × 20		% = 44%
   			100

   
   

   Resultant increase in r² and h =		44 + 20 +	44 × 20		% = (64 + 8.8)%
   			100

   
   = 72.8%

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