161. A car can cover a certain distance in 4	1	hours.
     	2

     If the speed is increased by 5 km/hour, it would take
     

     1	hour less to cover the same distance.
     2

     
     Find the slower speed of the car.

1. 1. 50 km/hour

   
   
   

   2. 40 km/hour

   
   
   

   3. 45 km/hour

   
   
   

   4. 60 km/hour

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the initial speed of the car be x kmph and the distance be y km.
   

   Then, y =	9	x     ...(i)
   	2

   
   and, y = 4 (x + 5)     ...(ii)
   

   ∴	9x	= 4 (x + 5)
   	2

   
   ⇒   9x = 8x + 40
   ⇒   x = 40 kmph

   Correct Option: B

   Let the initial speed of the car be x kmph and the distance be y km.
   

   Then, y =	9	x     ...(i)
   	2

   
   and, y = 4 (x + 5)     ...(ii)
   

   ∴	9x	= 4 (x + 5)
   	2

   
   ⇒   9x = 8x + 40
   ⇒   x = 40 kmph

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162. Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is

1. 1. 20 km

   
   
   

   2. 21 km

   
   
   

   3. 22 km

   
   
   

   4. 24 km

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the distance of office be x km.
   

   ∴	x	−	x	=	11
   	24		30		60

   
   

   ⇒	5x − 4x	=	11
   	120		60

   
   

   ⇒	x	=	11
   	120		60

   
   

   ⇒  x =	11	× 120 = 22 km.
   	60

   
   
   Second Methdod :
   Here, S₁ = 24, t₁ = 5
   S₂ = 30, t₂ = 6
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	24 × 30(5 + 6)
   	30 − 24

   
   

   =	720 × 11	= 22 km
   	6 × 60

   Correct Option: C

   Let the distance of office be x km.
   

   ∴	x	−	x	=	11
   	24		30		60

   
   

   ⇒	5x − 4x	=	11
   	120		60

   
   

   ⇒	x	=	11
   	120		60

   
   

   ⇒  x =	11	× 120 = 22 km.
   	60

   
   
   Second Methdod :
   Here, S₁ = 24, t₁ = 5
   S₂ = 30, t₂ = 6
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	24 × 30(5 + 6)
   	30 − 24

   
   

   =	720 × 11	= 22 km
   	6 × 60

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163. Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?

1. 1. 5 km

   
   
   

   2. 8 km

   
   
   

   3. 3 km

   
   
   

   4. 2 km

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the required distance be x km.
   

   Then,	x	−	x	=	24
   	3		5		60

   
   

   ⇒	5x − 3x	=	2
   	15		5

   
   

   ⇒	2x	= 2
   	3

   
   ⇒  2x = 2 × 3 ⇒ x = 3 km
   
   Second Methdod :
   Here, S₁ = 3, t₁ = 9
   S₂ = 5, t₂ = 15
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	(3 × 5)(9 + 15)
   	5 − 3

   
   

   =	15	×	24
   	2		60

   
   = 3 km

   Correct Option: C

   Let the required distance be x km.
   

   Then,	x	−	x	=	24
   	3		5		60

   
   

   ⇒	5x − 3x	=	2
   	15		5

   
   

   ⇒	2x	= 2
   	3

   
   ⇒  2x = 2 × 3 ⇒ x = 3 km
   
   Second Methdod :
   Here, S₁ = 3, t₁ = 9
   S₂ = 5, t₂ = 15
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	(3 × 5)(9 + 15)
   	5 − 3

   
   

   =	15	×	24
   	2		60

   
   = 3 km

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164. A student goes to school at the rate of

     2	1	km/h and reaches 6 minutes late.
     	2

     If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is

1. 1. 5

   
   
   

   2. 4

   
   
   

   3. 3

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the required distance be x km.
   

   x	−	x	=	16
   5/2		3		60

   
   

   ⇒	2x	−	x	=	4
   	5		3		15

   
   

   ⇒	6x − 5x	=	4
   	15		15

   
   ⇒  x = 4 km.
   
   Second Methdod :
   

   Here, S1 = 2	1	, t1 = 6
   	2

   
   S₂ = 3, t₂ = 10
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	5	× 3(6 + 10)
   	2
   	3 −	5
   		2

   
   

   = 15 ×	16	km = 4 km
   	60

   Correct Option: B

   Let the required distance be x km.
   

   x	−	x	=	16
   5/2		3		60

   
   

   ⇒	2x	−	x	=	4
   	5		3		15

   
   

   ⇒	6x − 5x	=	4
   	15		15

   
   ⇒  x = 4 km.
   
   Second Methdod :
   

   Here, S1 = 2	1	, t1 = 6
   	2

   
   S₂ = 3, t₂ = 10
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	5	× 3(6 + 10)
   	2
   	3 −	5
   		2

   
   

   = 15 ×	16	km = 4 km
   	60

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165. When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 minutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is

1. 1. 6 km

   
   
   

   2. 7 km

   
   
   

   3. 12 km

   
   
   

   4. 16 km

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the distance be x km.
   

   ∴	x	−	x	=	12
   	10		12		60

   
   

   ⇒	6x − 5x	=	1
   	60		5

   
   

   ⇒  x =	1	× 60 = 12 km
   	5

   
   
   Second Methdod :
   Here, S₁ = 10, t₁ = 6
   S₂ = 12, t₂ = 6
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	(10 × 12)(6 + 6)
   	12 − 10

   
   

   =	120 × 12
   	2

   
   

   = 60 ×	12	km = 12 km
   	60

   Correct Option: C

   Let the distance be x km.
   

   ∴	x	−	x	=	12
   	10		12		60

   
   

   ⇒	6x − 5x	=	1
   	60		5

   
   

   ⇒  x =	1	× 60 = 12 km
   	5

   
   
   Second Methdod :
   Here, S₁ = 10, t₁ = 6
   S₂ = 12, t₂ = 6
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	(10 × 12)(6 + 6)
   	12 − 10

   
   

   =	120 × 12
   	2

   
   

   = 60 ×	12	km = 12 km
   	60

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