321. A man starts from a place P and reaches the place Q in 7 hours.

     He travels	1	th of the distance at 10 km/hour
     	4

     and the remaining distance at 12 km/hour. The distance between P and Q is

1. 1. 72 km

   
   
   

   2. 90 km

   
   
   

   3. 80 km

   
   
   

   4. 70 km

   
   
   

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1. View Hint View Answer Discuss in Forum

   Time =	Distance
   	Speed

   
   According to the question,
   

   x	+	3x	= 7
   10		12

   
   

   ⇒	x	+	x	= 7
   	10		4

   
   

   ⇒	2x + 5x	= 7
   	20

   
   ⇒  7x = 7 × 20
   

   ∴  x =	7 × 20	= 20 km.
   	7

   
   ∴  PQ = 4x = 4 × 20 = 80 km.

   Correct Option: C

   Time =	Distance
   	Speed

   
   According to the question,
   

   x	+	3x	= 7
   10		12

   
   

   ⇒	x	+	x	= 7
   	10		4

   
   

   ⇒	2x + 5x	= 7
   	20

   
   ⇒  7x = 7 × 20
   

   ∴  x =	7 × 20	= 20 km.
   	7

   
   ∴  PQ = 4x = 4 × 20 = 80 km.

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322. A student goes to school at the

     rate of 2	1	km/hr and reaches 6 minutes late.
     	2

     If he travels at the speed of 3 km/hr. he is 10 minutes early. What is the distance to the school?

1. 1. 4 km

   
   
   

   2. 3	1	km
      	2

   
   
   

   3. 1 km

   
   
   

   4. 3	1	km
      	4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the distance of school be x km.
   Difference of time = 6 + 10
   

   = 16 minutes =	16	hour
   	60

   
   

   =	4	hour
   	15

   
   

   Time =	Distance
   	Speed

   
   

   ∴	x	−	x	=	4
   	5/2		3		15

   
   

   ⇒	2x	−	x	=	4
   	5		3		15

   
   

   ⇒	6x − 5x	−	4
   	15		15

   
   ⇒  x = 4 km.
   

   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
   	60

   
   

   =	16	= 4 km
   	4

   Correct Option: A

   Let the distance of school be x km.
   Difference of time = 6 + 10
   

   = 16 minutes =	16	hour
   	60

   
   

   =	4	hour
   	15

   
   

   Time =	Distance
   	Speed

   
   

   ∴	x	−	x	=	4
   	5/2		3		15

   
   

   ⇒	2x	−	x	=	4
   	5		3		15

   
   

   ⇒	6x − 5x	−	4
   	15		15

   
   ⇒  x = 4 km.
   

   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
   	60

   
   

   =	16	= 4 km
   	4

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323. A student goes to school at the

     rate of	5	km/hr and reaches 6 minutes late.
     	2

     If he travels at the speed of 3 km/hr, he reaches 10 minutes earlier. The distance of the school is

1. 1. 45 km

   
   
   

   2. 20 km

   
   
   

   3. 10 km

   
   
   

   4. 4 km

   
   
   

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1. View Hint View Answer Discuss in Forum

   Distance of school = x km
   Difference of time
   

   = 16 minutes =	16	hour
   	60

   
   

   ∴	x	−	x	=	16
   	5/2		3		60

   
   

   ⇒	2x	−	x	=	4
   	5		3		15

   
   

   ⇒	6x − 5x	=	4
   	15		15

   
   

   ⇒	x	=	4
   	15		15

   
   

   ⇒ x =	4	× 15 = 4km
   	15

   
   
   Second Method :
   

   Here, S1 =	5	,t1 = 6
   	2

   
   S₂ = 3, t₂ = 10
   

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

   
   

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

   
   

   = 15 ×	16	km = 4 km.
   	60

   Correct Option: D

   Distance of school = x km
   Difference of time
   

   = 16 minutes =	16	hour
   	60

   
   

   ∴	x	−	x	=	16
   	5/2		3		60

   
   

   ⇒	2x	−	x	=	4
   	5		3		15

   
   

   ⇒	6x − 5x	=	4
   	15		15

   
   

   ⇒	x	=	4
   	15		15

   
   

   ⇒ x =	4	× 15 = 4km
   	15

   
   
   Second Method :
   

   Here, S1 =	5	,t1 = 6
   	2

   
   S₂ = 3, t₂ = 10
   

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

   
   

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

   
   

   = 15 ×	16	km = 4 km.
   	60

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324. A car can finish a certain journey in 10 hours at the speed of 42 kmph. In order to cover the same distance in 7 hours, the speed of the car (km/h) must be increased by :

1. 1. 12

   
   
   

   2. 15

   
   
   

   3. 18

   
   
   

   4. 24

   
   
   

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1. View Hint View Answer Discuss in Forum

   Distance covered by car = 42 × 10 = 420 km.
   New time = 7 hours
   

   ∴  Required speed =	420	= 60 kmph.
   	7

   
   ∴  Required increase
   = (60 – 42) kmph
   = 18 kmph

   Correct Option: C

   Distance covered by car = 42 × 10 = 420 km.
   New time = 7 hours
   

   ∴  Required speed =	420	= 60 kmph.
   	7

   
   ∴  Required increase
   = (60 – 42) kmph
   = 18 kmph

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325. A man cycles at the speed of 8 km/hr and reaches office at 11 am and when he cycles at the speed of 12 km/hr he reaches office at 9 am. At what speed should he cycle so that he reaches his office at 10 am?

1. 1. 9.6 kmph.

   
   
   

   2. 10 kmph.

   
   
   

   3. 11.2 kmph.

   
   
   

   4. Cannot be determined

   
   
   

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1. View Hint View Answer Discuss in Forum

   Distance of the office = x km.
   Difference of time = 2 hours
   

   ∴	x	−	x	= 2
   	8		12

   
   

   ⇒	3x − 2x	= 2
   	24

   
   

   ⇒	x	= 2 ⇒ x = 48 km.
   	24

   
   ∴  Time taken at the speed of 8 kmph
   

   =	48	= 6 hours
   	8

   
   ∴  Required time to reach the
   office at 10 a.m. i.e., in 5 hours
   

   =		48		kmph
   		5

   
   = 9.6 kmph

   Correct Option: A

   Distance of the office = x km.
   Difference of time = 2 hours
   

   ∴	x	−	x	= 2
   	8		12

   
   

   ⇒	3x − 2x	= 2
   	24

   
   

   ⇒	x	= 2 ⇒ x = 48 km.
   	24

   
   ∴  Time taken at the speed of 8 kmph
   

   =	48	= 6 hours
   	8

   
   ∴  Required time to reach the
   office at 10 a.m. i.e., in 5 hours
   

   =		48		kmph
   		5

   
   = 9.6 kmph

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