366. If average of two numbers x and	1	(where x ≠ 0) is A, what will be the average of x3	1	?
     	x		x3

     
     

1. 1. 4A³ - 2A
      

   
   
   

   2. 4A³ - 3A
      

   
   
   

   3. 4A³ - 4A
      

   
   
   

   4. 4A³ - A

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to the question ,
   

   A =	1 +	1
   		x
   	2

   
   

   ⇒ x +	1	= 2A
   	x

   
   On cubing both sides,
   

   ⇒		x +	1		3	= (2A)3 = 8A3
   			x

   
   

   ⇒ x3 +	1	+ 3		x +	1		= 8A3
   	x3				x

   
   

   ⇒ x3 +	1	+ 3 × 2A = 8A3
   	x3

   
   

   ⇒ x3 +	1	= 8A3 - 4A
   	x3

   
   

   ∴ Required average =	x3 +	1
   		x3
   	2

   
   

   Required average =	8A3 - 4A	= 4A3 - 2A
   	2

   
   

   Correct Option: B

   According to the question ,
   

   A =	1 +	1
   		x
   	2

   
   

   ⇒ x +	1	= 2A
   	x

   
   On cubing both sides,
   

   ⇒		x +	1		3	= (2A)3 = 8A3
   			x

   
   

   ⇒ x3 +	1	+ 3		x +	1		= 8A3
   	x3				x

   
   

   ⇒ x3 +	1	+ 3 × 2A = 8A3
   	x3

   
   

   ⇒ x3 +	1	= 8A3 - 4A
   	x3

   
   

   ∴ Required average =	x3 +	1
   		x3
   	2

   
   

   Required average =	8A3 - 4A	= 4A3 - 2A
   	2

   
   

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367. If p + m = 6 and p³ + m³ = 72, then the value of pm is
     

1. 1. 6
      

   
   
   

   2. 12
      

   
   
   

   3. 9
      

   
   
   

   4. 8

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 8,
   p + m = 6
   On cubing both sides,
   (p + m)³ = p³ + m³ + 3pm(p + m)
   ⇒ (6)³ = 72 + 3pm × 6
   ⇒ 216 – 72 = 18pm
   ⇒ 18pm = 144
   ⇒ pm = 144 ÷ 18 = 8

   Correct Option: D

   Using Rule 8,
   p + m = 6
   On cubing both sides,
   (p + m)³ = p³ + m³ + 3pm(p + m)
   ⇒ (6)³ = 72 + 3pm × 6
   ⇒ 216 – 72 = 18pm
   ⇒ 18pm = 144
   ⇒ pm = 144 ÷ 18 = 8

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368. If x = 5, y = 6 and z = –11, then the value of x³ + y³ + z³ is
     

1. 1. –890
      

   
   
   

   2. –970
      

   
   
   

   3. –870
      

   
   
   

   4. –990

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 21,
   x + y + z = 5 + 6 – 11 = 0
   x³ + y³ + z³ = = 3xyz
   = 3 × 5 × 6 × (–11) = –990

   Correct Option: D

   Using Rule 21,
   x + y + z = 5 + 6 – 11 = 0
   x³ + y³ + z³ = = 3xyz
   = 3 × 5 × 6 × (–11) = –990

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369. If un =	1	-	1	, then the value of u1 + u2 + u3 + u4 + u5 is
     	n		n + 1

     
     

1. 1. 	1
      	2

      
      

   
   
   

   2. 	1
      	3

      
      

   
   
   

   3. 	2
      	5

      
      

   
   
   

   4. 	5
      	6

   
   
   

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1. View Hint View Answer Discuss in Forum

   un =	1	-	1
   	n		n + 1

   
   

   ∴ u1 =	1	-	1
   	1		1 + 1

   
   

   ⇒ u1 = 1 -	1	;
   	2

   
   

   Similarlly ,u2 =	1	-	1	;
   	2		3

   
   

   u3 =	1	-	1	;
   	3		4

   
   

   u4 =	1	-	1	;
   	4		5

   
   

   u5 =	1	-	1
   	5		6

   
   

   ∴ u1 + u2 + u3 + u4 + u5 = 1 -	1	+	1	-	1	+	1	-	1	+	1	-	1	+	1	-	1
   	2		2		3		3		4		4		5		5		6

   
   

   u1 + u2 + u3 + u4 + u5 = 1 -	1	=	6 - 1
   	6		6

   
   

   Hence , u1 + u2 + u3 + u4 + u5 =	5
   	6

   
   

   Correct Option: D

   un =	1	-	1
   	n		n + 1

   
   

   ∴ u1 =	1	-	1
   	1		1 + 1

   
   

   ⇒ u1 = 1 -	1	;
   	2

   
   

   Similarlly ,u2 =	1	-	1	;
   	2		3

   
   

   u3 =	1	-	1	;
   	3		4

   
   

   u4 =	1	-	1	;
   	4		5

   
   

   u5 =	1	-	1
   	5		6

   
   

   ∴ u1 + u2 + u3 + u4 + u5 = 1 -	1	+	1	-	1	+	1	-	1	+	1	-	1	+	1	-	1
   	2		2		3		3		4		4		5		5		6

   
   

   u1 + u2 + u3 + u4 + u5 = 1 -	1	=	6 - 1
   	6		6

   
   

   Hence , u1 + u2 + u3 + u4 + u5 =	5
   	6

   
   

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370. If m + n = – 2, then the value of m³ + n³ - 6mn is
     

1. 1. 8
      

   
   
   

   2. 4
      

   
   
   

   3. –8
      

   
   
   

   4. –4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 8,
   m + n = –2
   On cubing both sides,
   (m + n)³ = (-2)³ = -8
   ⇒ m³ + n³ + 3mn(m + n) = -8
   ⇒ m³ + n³ - 6mn = -8

   Correct Option: C

   Using Rule 8,
   m + n = –2
   On cubing both sides,
   (m + n)³ = (-2)³ = -8
   ⇒ m³ + n³ + 3mn(m + n) = -8
   ⇒ m³ + n³ - 6mn = -8

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