Q 1-Two men starting from the same place walk at the rate of 5 kmph and 5.5 kmph respectively . What time will they take to be 8.5 km apart , if they walk in the same direction ?
A - 4 hrs 15 min
B - 8 hrs 30 min
C - 16 hrs
D - 17 hrs
Answer - D Explanation "To be 0.5 km apart , hey take 1 hour To be 8.5 km apart , they take (1/0.5×8.5) hrs= 17 hrs. "
Q 2-A is twice as fast as B and B is thrice as fast as C is. The journey covered by C in 54 minutes will be covered by B in :
A - 18 min
B - 27 min
C - 38 min
D - 9 min
Answer - A Explanation "Let C's speed = x km/hr, Then , B's speed = 3x km/hr and A's speed = 6x km/hr ∴ Ratio of speed of A, B, C = 6x:3x:x = 6 : 3 : 1 Ratio of time taken = 1/6:1/3:1 = 1 : 3 : 6 If C takes 6 min., then B takes 2 min. If C takes 54 min, then B takes (2/6×54) min = 18 min. "
Q 3-It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car . It takes 20 minutes more ,if 200 km is done by train and the rest by car . The ratio of the speed of the train to that of the car is :
A - 2 : 3
B - 3 : 2
C - 3 : 4
D - 4 : 3
Answer - C Explanation "Let the speed of the train be x km/hr and hat of the car be y km/hr. Then , 120/x+480/y=8 or 1/ x+4/y=1/15 ................................................(I) And 200/x+400/y=25/3 or 1/x+2/y=1/24 .............................................(II) Solving (I) and (II) , we get x=60andy=80 . ∴ Ratio of speed = 60 : 80 = 3 : 4. "
Q 4-If a person walk at 14 km/hr instead of 10 km/hr , he would have walked 20 km more . The actual distance travelled by him is
A - 50 km
B - 56 km
C - 70 km
D - 80 km
Answer - A Explanation "Let the actual distance travelled be x km. then , x/10=x+20/14 ⇔ 14x= 10x+200 ⇔4x=200 ⇔ x= 50 km. "
Q 5-Two men start together to walk to a certain destination , one a 3 kmph and another a 3.75 kmph, he latter arrives half an hour before The former. he distance is :
A - 6 km
B - 7.5 km
C - 8 km
D - 9.5 km
Answer - B Explanation "Let the distance be x km. Then , x/3−x/3.75=1/2 ⇔ 2.5x−2= 3.75 ⇔x=3.75/0.50=15/2 = 7.5 km "
Q 6-With a uniform speed a car covers the distance in 8 hours. Had the speed been increased by 4 km/hr, the same distance could have been covered in 71/2 hours. What is the distance covered ?
A - 420 km
B - 480 km
C - 640 km
D - cannot be determined
Answer - B Explanation "Let the distance be x km. Then , x/71/2−x/8=4 ⇔ 2x/15−x/8=4 ⇔ x= 480 km. "
Q 7-Three persons are walking from a place A to another place B. Their speeds are in the ratio of 4 : 3 : 5 . The time ratio to reach B by these persons will be :
A - 4 : 3 : 5
B - 5 : 3 : 4
C - 15 : 9 20
D - 15 : 20 : 12
Answer - D Explanation "Ratio of speeds = 4 : 3 : 5 ∴ Ratio of times taken = 1/4:1/3:1/5 = 15 : 20 : 12 "
Q 8-In covering a distance of 30 km. Abhay takes 2 hours more than Sameer . If Abhay doubles his speed ,then he would take 1 hour less than Saneer , Abhay's speed is :
A - 5 kmph
B - 6 kmph
C - 6.25 kmph
D - 7.5 kmph
Answer - A Explanation "Let Abhay's speed be x km/hr. Then, 30/x−30/2x=3 ⇔ 6x= 30 ⇔ x= 5 km/hr "
Q 9-In covering a certain distance ,the speeds of A and B are in the ratio of 3 : 4 . A takes 30 minutes more than B to reach the destination . The time taken by A to reach the destination is :
A - 1 hour
B - 11 /2hours
C - 2 hours
D - 2 1/2hours.
Answer - C Explanation "Ratio of speeds = 3 : 4 Ratio of times taken = 4 : 3 Suppose A takes 4xx hrs and B takes 3x hrs to reach the destination. Then , 4x−3x=30/60=1/2orx= ½ ∴ ime taken by A = 4x hrs.= (4×1/2) hrs = 2 hrs. "
Q 10-Excluding stoppages , the speed of a bus is 54 kmph and including stoppages , it is 45 mph. For how many minutes does the bus stop per hour ?
A - 9
B - 10
C - 12
D - 20
Answer - B Explanation "Due to stoppages , it covers 9 km less. Time taken to cover 9 km = (9/54×60) min = 10 min. "
Q 11-A train can travel 50% faster than a car , Both start from point A at the same time and reach point B 75 kms away from A at the same time . On the way however , the train lost about 12.5 minutes while stopping at the stations . The speed of the car is :
A - 100 kmph
B - 110 kmph.
C - 120 kmph
D - 130 kmph
Answer - C Explanation "Let speed of the car be x kmph. Then , speed of the train = 150/100x=(33/2x) kmph ∴ 75/x−75/3/2x= 125/10×60 ⇔ 75/x −50/x= 5/24 ⇔x=(25×24/5) = 120 kmph. "
Q 12-A car travels from P to Q at a constant speed . If its speed were increased by 10 km/hr it would have taken one hour lesser o cover the distance . It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities. ?
A - 420 km
B - 540 km
C - 600 km
D - 650 km.
Answer - A Explanation "Let distance = x km and usual rate = y kmph. Then x/y−x/y+10=1or y(y+10)= 10x .........................................(I) And , x/y−x/y+20=7/4 ory(y+20)= 80x/7 ..............................(II) On dividing (I) by (II) , we get y = 60. Substituting y = 60 in (I) , we get x = 420 km. "
Q 13-A man covered a certain distance at some speed . Had he moved 3 kmph faster , he would have taken 40 minutes less . If he had moved 2 kmph slower , he would have taken 40 minutes more . The distance (in km) is
A - 35
B - 36 2/3
C - 37 1/2
D - 40
Answer - D Explanation "Let distance = x km and usual rate = y kmph. x/y−x/y+3=40/60 or 2y(y+3)=9x .............................................(I) and ,x/y−2−x/y=40/60 or y/(y−2)= 3x .........................................(II) On dividing (I) by (II), we get x = 40 km. "
Q 14-If a train runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph , it is late by 5 minutes only . The correct time for the rain to complete its journey is :
A - 13 min.
B - 15 min.
C - 19 min.
D - 21 min.
Answer - C Explanation "Let the correct time to complete the journey be x min. Distance covered in (x+11) min. at 40 kmph = Distance covered in (x+5) min. at 50 kmph. ∴ (x+11)/60×40=(x+5)/60×50 ⇔ x = 19 min. "
Q 15-Robert is travelling on his cycle and has calculated to reach point A at 2 p.m. . if he ravels at 10 kmph ; he will reach here at 12 noon if he travels at 15 mph at what speed must he travel to reach A at 1 p.m.
A - 8 kmph
B - 11 kmph
C - 12 kmph
D - 14 kmph
Answer - C Explanation "Let the distance travelled be x km Then , x/10−x/15=2 ⇔ 3x−2x=60 ⇔ x=60 Time taken to travel 60 km at 10 km/hr = (60/10)hrs=6hrs So, Rebert started 6 hours before 2 P.M. i.e. at 8 A.M. ∴ Required speed = (60/5) kmph = 12 kmph. "