Q 1-A person travels from P to Q at a speed of 40 kmph and returns by increasing his speed by 50% . What is his average speed for both the trips ?

A - 36 kmph

B - 45 kmph

C - 48 kmph

D - 50 kmph

Answer - C Explanation "Speed on return trip = 150% of 40 = 60 kmph. ∴ Average speed = (2×40×60/40+60) km/hr =(4800/100) km/hr = 48 km/hr "

Q 2-A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is :

A - 14 km

B - 15 km

C - 16 km

D - 17 km.

Answer -C Explanation "Let the distance travelled on foot be x km Then , distance travelled on bicycle = (61−x) km. So, x/4+(61−x)/9−9 ⇔ 9x+4(61−x)=9×36 ⇔ 5x =80 ⇔ x = 16 km. "

Q 3-A person travels equal distances with speeds of 3 km/hr. 4km/hr and 5 km/hr and takes a total time of 47 minutes. The total distance ( in km) is :

A - 2

B - 3

C - 4

D - 5

Answer - B Explanation "Let the total distance be 3x km. Then , x/3+x/4+x/5= 47/60 ⇔ 47 x/60 = 47/60 ⇔ x = 1. ∴ Total distance = (3 x 1) km = 3 km. "

Q 4-A can complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr . Find the total journey in km.

A - 220 km

B - 224 km

C - 230 km

D - 234 km.

Answer - B Explanation "Let the total distance be x km. Then 1/2x/21+1/2x/24 = 10 ⇒ x/21+x/24⇒ = 20 ⇒ 15x=168×20 ⇒ x= (168×20/15)⇒ = 224 km. "

Q 5-A man performs 3/5 of the total journey by rail , 17/20 by bus and the remaining 6.5 km on foot . His total journey is :

A - 65 km

B - 100 km

C - 120 km

D - 130 km.

Answer - D Explanation "Le the total journey be x km Then ,3x/25+7x/20+6.5=x ⇔ 12x+7x+20×6.5= 20x ⇔ x = 130 km. "

Q 6-A person has to cover a distance of 6 km in 45 minutes. If he covers one half of the distance in two-third of the total time , to cover the remaining distance in he remaining time , his speed (in km/hr) must be :

A - 6

B - 8

C - 12

D - 15

Answer - C Explanation "Remaining distance = 3 km and Remaining time = (1/3×45) min = 15 min = 1/4 hour. ∴ Required speed = (3 x 4) km/hr = 12 km/hr. "

Q 7-A salesman travels a distance of 50 km in 2 hours and 30 minutes. How much faster in kilometres per hour ,on an average , must he ravel to make such a trip in 5/6 hour less time ?

A - 10

B - 20

C - 30

D - None of these

Answer - A Explanation "Time required = (2 hrs 30 min - 50 min) = 1 hr 40 min. 123123 hrs. ∴ Required speed = (50×3/5) km/hr = 30 km/hr. Original speed = (50×2/5) km/hr = 20 km/hr. ∴ Difference in speed = (30 - 20) km/ hr = 10 km/hr "

Q 8-An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 213 hours, it must travel at a speed of :

A - 300 kmph

B - 360 kmph

C - 600 kmph

D - 720 kmph

Answer - D Explanation "Distance = (240 x5) km= 1200 km ∴ Required speed =(1200×3/5) km/hr = 720 km/hr. "

Q 9-Anna left for city A from city B at 5.20 a.m. She travelled at the speed of 80 km/hr for 2 hours 15 minutes . After that the speed was reduced to 60 km/hr . If the distance between two cities is 350 kms., at what time did Anna reach city A ?

A - 9.25 a.m.

B - 9.35 a.m.

C - 10.05 a.m.

D - None of these

Answer - D Explanation "Distance covered in 2 hrs. 15 min. i.e. 2 114 hrs = (80×9/4) hrs = 180 hrs. Time taken to cover remaining distance = (350−180/60) hrs. = 17/6 hrs. = 2 5/6 hrs = 2 hrs 50 min. Total time taken = ( 2 hrs 15 min + 2 hrs 50 min) = 5 hrs 5 min So, Anna reached city A at 10.25 a.m. "

Q 10-A train covers a distance of 10 km in 12 minutes. If is speed is decreased by 5 km/hr the time taken by it to cover the same distance will be :

A - 10 min.

B - 11 min. 20 sec

C - 13 min

D - 13 min 20 sec.

Answer - D Explanation "Speed = (10×60/12) km/hr = 50 km/hr. New speed = (50 - 5) km/hr = 45 km/hr. ∴ Time taken = (10/45) hr = (2/9×60)min= 13 1/3 min. = 13 min 20 sec. "

Q 11-The speed of a car increase by 2 kms after every one hour .If the distance travelled in the first one hour was 35 kms. What was the total distance travelled in 12 hours.

A - 456 kms

B - 482 kms

C - 552 kms

D - 556 kms

Answer - C Explanation "Total distance travelled in 12 hours = (35 + 37 + 39 +............... upto 12 terms). This is an A.P. with first term, a = 35, number of term, n = 12 , common difference , d = 2 ∴ Required distance = 12/2 [2 x 35 + (12 - 1) x 2] = 6(70 + 22) = 552 km. "

Q 12-A certain distance is covered by a cyclist at a certain speed . If a jogger covers half the distance in double the time , the ratio of the speed of the speed of the jogger to that of the cyclist is

A - 1 : 2

B - 2 : 1

C - 1 : 4

D - 4 : 1

Answer - C Explanation "Let the distance covered by the cyclist be x and the time taken be y. Then, Required ratio = 1/2x/2y:x/y = 1/4:1 = 1 : 4. "

Q 13-An express train travelled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach is destination 600 km from the starting point ?

A - 6 hrs 21 min.

B - 6 hrs 24 min.

C - 6 hrs. 27 min.

D - 6 hrs. 30 min.

Answer - A Explanation "Time taken to cover 600 km = (600/100) hrs. = 6 hrs. Number of stoppages = 600/75−1 = 7 Total time of stoppage = (3 x 7) min = 21 min. Hence , total time taken = 6 hrs 21 min. "

Q 14-Sound is said to travel in air at about 1100 feet per second . A man hears the axe striking the tree , 11/5 seconds after he sees it strike the tree. How far is he man from the wood chopper ?

A - 2197 ft.

B - 2420 ft.

C - 2500 ft.

D - 2629 ft.

Answer - B Explanation Distance = (1100×11/5) feet = 2420 feet.

Q 15-A man in a train notice that he can count 21 telephone posts in one minute . If hey are know to be 50 metres apart , then at what speed is the train travelling ?

A - 55 km/hr

B - 57 km/hr

C - 60 km/hr

D - 63 km/hr

Answer - C Explanation "Number of gaps between 21 telephone posts = 20. Distance travelled in 1 minute = (50 x 60) m = 1000 m = 1 km. ∴ Speed = 60 km/hr. "