POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-CALENDER & CLOCKS

POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-CALENDER & CLOCKS

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.

Calendar :

Some important facts about calendar we must know :

Odd days: Number of days more than the complete number of weeks in a given period is the number of odd days during that period.

Ordinary year has 365 days ;It has 1odd days.

Leap year has 366 days ; It has 2 odd days.

100 year contain 5 odd days.

200 year contain 3 odd days.

300 year contain 1 odd day.

400 year contain 0 odd day.

Sunday -------> '0' odd day.
Monday -------> '1' odd day.

Leap year:

Every Year which is divisible by 4.

Every century year must be divisible by both 4 and 400.

3.One leap year contains '2' odd days.

The years which are mul of '4' are called leap years.

Leap year -------> 366 days (feb --> 29 days).
Ordinary year -------> 365 days.
 MONTHS DAYS

Jan 31
Feb 28 (or) 29
Mar 31
Apr 30
May 31
Jun 30
Jul 31
Aug 31
Sep 30
Nov 30
Dec 31One week = '7' days.

Leap year ------> '52' weeks + '2' odd days.
Ordinary year ------> '52' weeks + '1' odd day.

Steps involved in determining the day for given date.

§  Split the completed years into multiple of 400 and remainder.

§  Split the remainder into multiples of 100 and remainder.

§  Compute the odd day for the remainder years.

§  Compute the odd day for the completed months.

§  Compute the odd day for the date.

§  Add up the number of odd days obtained in (3) to (5).

§  Use the below table to determine the required day.

Odd days		1	2	3	4	5	6
Day	Sun	Mon	Tues	Wed	Thurs	Fri	Sat

                                  Example:

1.What was the day of the week on 16^th November 1989?

1989 = 1600 + 389 [Step 1]

389 = 300+ 89 [Step 2]

89 = 22 leap years + 67 ordinary years [Step 3]

= 22*2 + 67* 1 = 111 days = R{111/7} = 6 odd days.

10 months completed i.e up-to October .

No of days in ten months = 365 – (30+31) = 304 days.

No of odd days in ten months = R{304/7} = 3 odd days. [Step 4]

No of odd days for given date = R{16/7} = 2 odd days. [Step 5]

Sum of odd days from step 3,4,5 = 6 + 3 + 2 = 11 days = 4 odd days.

From the above table,we can find the day, The required day is Thursday.

2.On what dates of November 1989 did Thursday fall ?

These type of problems we need to find the day on 1^st November, 1989.

For years,we have 6 odd days,

For months, we have 3 odd days,

For dates,we have 1 odd day i.e 1^st November.

Sum of odd days = 10 days = 3 odd days.

The day on 1^st November 1989 is Wednesday. So first Thursday on November month is 2 nd November, 9^th ,16^th,23^th and 30^th .

3.The calendar for the year 2012 is same as for the year ?

Count the number of odd days from 2012 onwards to get 0 odd day.

Year	2012	2013	2014	2015	2016
Odd Days	2	1	1	1	2

At the end of 2016, 7 odd days = 0 odd day.

So calendar for the year 2012 is same as that for the year 2017.

4.Today is Saturday,After 65 days , it will be :

R[65/7] = 2 odd days.

This table is for these type of problems alone. If today is Monday we have to take 0 odd day as Monday , 1 odd day as Tuesday and so on . But In this problem,Today is Saturday So we have to starts with Saturday .

Odd days		1	2	3	4	5	6
Day	Sat	Sun	Mon	Tues	Wed	Thur	Fri

So the required day is Monday .

In calendar Problems , Above four models Frequent. Be thorough with the concepts which given. If you have any doubts plz post here.

                                          CLOCKS :

In a clock, the minute hand travels 360° in one hour or 6° in one minute and the hour hand moves 360° in 12 hours or 0.5° in one minute. Since the two hands move in the same direction the relative speed is 5.5°.

The following results are very useful to answer questions fast.

o The two hands overlap each other every 65 5/11 minutes.

o In a 12-hour gap the two hands would overlap / coincide 11times and hence in a 24-hour gap 22 overlap would take place.

o In a 12-hour gap the two hands would be perpendicular to each other 22 times.

o In a 12-hour gap the two hands would be in a straight line  22 times.

o In a 12-hour gap the two hands would be in a straight line but opposite in direction is 11  times.

o In 60 minutes, the minute hand gains 55 minutes on the hour hand.

Another one shortcut formula for Clocks:

Angle = 30H – (11/2)M

H represents the hour   : M represents the minute

                                           Example :

1.What is the angle between the two hands of a clock at 3:27 p.m.?

As per formula,

H = 3 ; M = 27

Angle = 30*3 – (11/2)27

= 90^o – 148.5^o

= 58.5^o (here no need to represent the negative in that)

2.At what time between 2 a.m. and 3 a.m. is the angle between the two hands of a clock 28°?

The job is to find the time say, t minutes after 2 a.m. when (y – x) = 28° . there are two possibilities – one in which the minute hand is ahead of the hour hand and the other in which the hour hand is ahead of the minute hand.

CASE 1

Now, y = the angular movement of the minute hand in t minutes = 6t°.

Also x = 60 + 0.5t. [since at 2 a.m., 60° is already covered]

So, 6t – 60 – 0.5t = 28 or

5.5t …………………… = 88 or

t ………………………… = 16.

CASE 2

In this case, (x – y) = 28°, y = 6t and x = 60 + 0.5t.

60 + 0.5t – 6t …… = 28 or

5.5t …………………… = 32 or

t ……………………….. = 32/5.5

= 5 9/11.

Thus, there are two points of time at which the angle between the two hands

is 28° and these two points of time are 16 minutes past 2 a.m. and 5 9/11

minutes past 2 a.m.

                                    EXAMPLES

1.Today is Tuesday. After 62 days it will be:

Each day of the week is repeated after 7 days. After 63 days, it will be Tuesday.
After 62 days, it will be Monday.

2.Monday falls on 20th March, 1995. What was the day on 3rd November, 1994?

Counting the number of days after 3rd November, 1994 we have: Nov. Dec. Jan. Feb. March
27+ 31 + 31+ 28 + 20 = 137days =19weeks + 4days . Number of odd days = 4.
The day on 3rd November, 1994 is (7 - 4) days beyond the day on 20th March, 1995.
So, the required day is Thursday.

3.What will be the day of the week on 1st Jan 2001 ?

1600 yeaas contain 0 odd day.
300 years contain 1 odd day.
100 years contain 5 odd days.
2000 years contain (0+ 1 +5) odd days =6 odd days.
1st Jan, 2001 has one odd day.
Total number of odd days upto 1st Jan, 2001 = 7 odd days =Q odd days.
The day will be Sunday.

4.Today is 1st April. The day of the week is Wednesday. This is a leap year. The day of the week on this day after 3 years will be:

This being a leap year, none of the next 3 years is a leap year.
So,the day of the week will be 3 days beyond Wednesday.
The day after 3 years will be Saturday.

5.How many days are there from 2nd January 1995 to 15th March, 1995?

Jan Feb March 30 +28 + 15 = 73 days.

6.January 1, 1995 was a Sunday. What day of the week lies on January 1, 1996?

1995 being on ordinary year, it has 1 odd day. So, the first day of 1996 will be one day beyond Sunday, i.e. it will be Monday

7.On 8th Feb, 1995 it was Wednesday. The day of the week on 8th Feb, 1994 was

1994 being an ordinary year, it has 1 odd day. So, the day on 8th Feb, 1995 is one day beyond the day on 8th Feb, 1994. But, 8th Feb, 1995 was Wednesday. 8th Feb, 1994 was Tuesday.

8.May 6, 1993 was Thursday. What day of the week was on May 6, 1992?

1992 being a leap year, it has 2 odd days.
So, the day on May, 1993 is 2 days beyond the day on May 6, 1992.
But, on May 6, 1993 it was Thursday.
So, on May 6, 1992 it was Tuesday.

9.The year next to 1996 having the same calender as that of 1996 is:

Starting with 1996, we go on counting the number of odd days till sum is divisible by 7.
Year 1996 1997 1998 1999 2000
Odd days 2 1 1 1 2 =7 odd days
0 odd day.
Calendar for 2001 will be the same as that of 1995.

10.January 1, 1992 was Wednesday. What day of the week was on January 1,1993?

1992 being a leap year, it has 2 odd days. So, the first day of the year
1993 must be two days beyond Wednesday.
So, it was Friday.

11.On what dates of April, 1994 did Sunday fall?

Find the day on 1st April, 1994. 1600 years contain 0 odd day 300 years contain 1 odd day 93 years = (23 leap years + 70 ordinary years) = (46 + 70) odd days =4 odd days. Number of days upto 1St April, 1994 Jan Feb March April 31 + 28 ÷ 31 + 1 =9ldays=Ooddday. Total number of odd days = (0÷ 1 +4+0) =5 Odd day.  Day on 1st April 1994 is ‘Friday’. Sunday was on 3rd April, 1994. Thus, Sunday fell on 3rd, 10th, 17th & 24th

12.The day on 5th March of a year is the same day on what date of thesame year?

Since any date in March is the same day of the week as the corresponding date in November of that year, so the same day falls on 5th November.

13.The calendar for 1990 is the same as for

count the number of days for 1990 onwards to get 0 odd day.
Year 1990 1991 1992 1993 1994 1995
Odd days 1 1 2 1 1 1 = 7 or 0 odd day.
Calendar for 1990 is the same as for the year 1996

14.If the first day of the year 1991 was Tuesday, what day of the week must have been on 1st January, 1998?

Total number of odd days from 1st Jan, 1991 to 1st Jan, 1998
Year 1991 1992 1993 1994 1995 1996 1997
Odd days I + 2 + 1 + 1 + 1 + 2 + 1 = 9 odd days
2 odd days
The day is 2 days beyond the day on 1st Jan, 1991.
i.e. The required day must be Thursday.

Courtesy : http://akulapraveen.blogspot.in

POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-BOATS & STREAMS

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.

BOATS AND STREAMS

1.When a boat is moving in the same direction as the stream or water current, the boat is said to be moving with the stream or moving downstream.

2.Instead of boats in water, it could be a swimmer or a cyclist cycling against or along the wind.

3. When a boat is moving in a direction opposite to that of the stream or water current, the boat is said to be moving against the stream or water current or moving downstream.

4. When the speed of the boat is given, it is the speed of the boat in still water.

5. Speed of the boat against stream or while moving upstream = Speed of the boat in still water - Speed of the stream.

6. Speed of the boat with stream or while moving downstream= Speed of the boat in still water + Speed of the Stream.

7. If 'p' is the speed of the boat down the stream and 'q' is the speed of the boat up the stream, then,

Speed of the boat in still water = (p+q) / 2.

Speed of the boat of the water stream = (p-q) / 2.

8.These problems are governed by the following results:

Downstream (along the current) speed (D) = Boat speed (B) + current (stream) speed (C).         D=B+C

Upstream (against the current) speed (U) = Boat speed – current (stream) speed. U=B–C

Speed of the boat = average of downstream and upstream speeds B = (D + U)/2

Speed of the current = half the difference of downstream and upstream speeds    C = (D – U)/2

                                             Example:

1.A boat takes 5 hours to go from A to B and 8 hours to return to A. If AB distance is 40 km, find the speed of (a) the boat and (b) the current.

Sol: Since B to A takes more time, it is upstream and hence AB is downstream. Downstream speed = 40/5 = 8 kmph.

Upstream speed= 40/8 = 5 kmph.

Boat speed = (8 + 5)/2 = 6.5 kmph.

Current speed = (8 – 5)/2 = 1.5 kmph.

2.A man cn row a boat at 20 kmph in still water.If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream ?

Sol: Speed of downstream = boat speed + stream speed = 20 + 6 = 26 kmph

Time required to cover 60 km downstream = d/s = 60/26 = (30/13) hours.

3.The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 42 kmph, find the speed of the stream ?

Sol: The time taken to row his boat upstream is twice the time taken by him to row the same distance downstream. Therefore, the ratio of the time taken is (2:1). So, the ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3:1 .Thus, Speed of the stream = (42)/3 = 14 kmph.

4. A boat travels 36 km upstream in 9 hours and 42 km downstream in 7 hours. Find the speed of the boat in still water and the speed of the water current ?

Sol:  Upstream speed of the boat = 36/9 = 4 kmph

Downstream speed of the boat = 42/ 7 = 6kmph.

Speed of the boat in still water = (6+4) / 2.

= 5 kmph

Speed of the water current = (6-4) /2

= 1 kmph

5.A man can row at 10 kmph in still water. If it takes a total of 5 hours for him to go to a place 24 km away and return, then find the speed of the water current ?

Sol:  Let the speed of the water current be y kmph.

Upstream speed = (10- y) kmph

Downstream speed = (10+y) kmph

Total time = (24/ 10+y) + ( 24/10-y) = 5

Hence, 480/ (100-y2 ) = 5

480= 500-5y2, 5y2= 20

y2= 4, y = 2 kmph.

6.A man can row x kmph in still waters. If in a stream which is flowing at y kmph, it takes him z hrs to row from A to B and back (to a place and back), then

Sol: The distance between A and B = z ( x2 - y2) / 2x.

7. A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and back. How far is the place?

Sol:  Required distance = 1 x ( 62 - ( 1.2)2) kmph

= (36 - 1.44) / 12

= 2.88 km.

In the above case, If distance between A and B, time taken by the boat to go upstream and back again to the starting point, speed of the stream are given; then the speed of the boat in still waters can be obtained using the above given formula.

8. A man rows a certain distance downstream in x hours and returns the same distance in y hrs. If the stream flows at the rate of z kmph then,

Sol: The speed of the man in still water = z(x+y) / ( y-x) kmph.

9.Ramesh can row a certain distance downstream in 6 hours and return the same distance in 9 hours. If the stream flows at the rate of 3 kmph. Find the speed of Ramesh in still water?

Sol:  Ramesh's speed in still water = 3 (9+6) / (9-6)

= 15 kmph.

10.A man rows a certain distance downstream in x hours and returns the same distance in y hours. If the speed of the man in still water z kmph, then

Sol: Speed of the stream = z (y-x) / (x+y) kmph.

11. Abhinay can row a certain distance downstream in x hours and returns the same distance in y hours. If the speed of Abhinay in still water is 12 kmph. Find the speed of the stream?

Sol: Speed of the stream = 12 ( 9-6) / (9+6)

= 2.4 kmph.

12. If a man can swim downstream at 6 kmph and upstream at 2 kmph, his speed in still water is 

Speed in still water = (1/2) * (6 + 2) km/hr = 4 km/hr

13.If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 3 kmph, then the speed of the boat in still water is

Sol: Rate upstream = (7/42)*60 kmh = 10 kmph. Speed of stream = 3 kmph.
Let speed in sttil water is x km/hr
Then, speed upstream = (x —3) km/hr.
x-3 = 10 or x = 13 kmph.

14. A man rows 13 km upstream in 5 hours and also 28 km downstream in 5 hours. The velpciy of the stream is   

Sol: speed upstream = (13/5) kmph
speed downstream (28/5) kmph
Velocity of stream = (1/2)[(28/5) - (13/5)] = 1.5 kmph

15. A man can row a boat at 10 kmph in still water. If the speed of the stream is 6 kmph, the time taken to row a distance of 80 km down thestream is

Sol; Speed downstream (10+6) km/hr 16 km/hr.
Time taken to cover 80 km downstream = (80/16) hrs = 5 hrs

16.A man can row 9 and 1/3 kmph in still water and finds that it takes him thrice as much time to row up than as to row, down the same distance in the river. The speed of the current is

Sol: Let speed upstream is x kmph. Then, speed downstream = 3x kmph. Speed in still water = (1/2)(3x + x) kmph = 2x kmph. 2x = 28/3 x = 14/3 Speed upstream = 14/3 km/hr, Speed downstream 14 km/hr. speed of the current = (1/2)[14 - (14/3)] = 14/3 = 4 and 2/3 kmph

17.A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water is

Sol: Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec.
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec. = 25/18 m/sec
= (25/18)*(18/5) kmPh = 5 kmph

18.If anshul rows 15 km upstream and 21 km downstream taking 3 hours each time, th’en the speed of the stream is : .

Sol: Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 - 5)kmph = 1 kmph.

19.A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:

Sol: Suppose he move 4 km downstream in x hours. Then,

Speed downstream =		4		km/hr.
		x

Speed upstream =		3		km/hr.
		x

	48	+	48	= 14 or x =	1	.
	(4/x)		(3/x)		2

So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.

Rate of the stream =	1	(8 - 6) km/hr = 1 km/hr.
	2

20. A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is:

Sol: Let man's rate upstream be x kmph.

Then, his rate downstream = 2x kmph.

(Speed in still water) : (Speed of stream) =		2x + x		:		2x - x
		2				2

=	3x	:	x
	2		2

   = 3 : 1.

Courtesy : http://akulapraveen.blogspot.in/

POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-TIME & DISTANCE

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.

TIME AND DISTANCE

1. Speed, Time and Distance:

Speed =		Distance		,	Time =		Distance		,	Distance = (Speed x Time).
		Time					Speed

2. km/hr to m/sec conversion:

x km/hr =		x x	5		m/sec.
			18

3. m/sec to km/hr conversion:

x m/sec =		x x	18		km/hr.
			5

4. If the ratio of the speeds of A and B is a : b, then the ratio of the

the times taken by then to cover the same distance is	1	:	1	or b : a.
	a		b

5. Suppose a man covers a certain distance at x km/hr and an equal distance at ykm/hr. Then,

the average speed during the whole journey is		2xy		km/hr.
		x + y

You have to know only two conversions when needed

1) kilometre/hour into metre/sec use 5/18 from (1000/3600)

2) metre/sec into kilometre/hour use 18/5 from (3600/1000)

Fastest multiplication and division knowledge will help you to solve time and distance easily

 For example;

1.A person crosses a 50 m long street in 5sec. What is his speed in km per hour? Here we came to know he tooks 5 sec to cross 50 metre long So, he goes 10 metre per second, here use our conversion (2)   10*18/5 2*18=36 That person crosses in the speed of 36 km/hr 2.An aeroplane covers a certain distance at a speed of 240 km/hr in 5 hours. To cover the same distance in 1 2/3 hours, it must travel at a speed of? Just listen the question ; It flies 240 kmph and it covers the distance in 5 hours So the total distance covered by aeroplane is (5*240=1200 km) In 300 min(5 hours) it covers 1200 Our question is within 1 2/3hours( 100 min) it need to cover same distance(1200 km)  what is the speed  300 min in the speed 240 km/hr 100 min in the speed? Time and distance is opposite proportion If time increases  to cover same distance speed is decreased If time decreases  to cover same distance speed is increased 300 min                                   240 km/hr 100 min(300/3)                        720 km/hr(240*3) It must travel to cover 1200 km in 1 2/3 hour is 720 km/hr

TIME AND DISTANCE SHORTCUTS

1.If a person travels half a distance in a journey at x km/hr and remaining half at the speed of y km/hr, the average speed of the whole journey is given by:

Example 1

A person travels half of journey at the speed of 30 km/hr and the next half at a speed of 15 km/hr. What is the average speed of the person during the whole journey?

Sol:

Here; x= 30 and y= 15

Putting the values we get:

Answer= 20 km/hr

Short Cut 2

If a person travels three equal distances in a journey at a speed of x km/hr, y km/hr and z km/hr. The average speed is given by:

Example 2

Ravi starts from Delhi to Agra a distance of about 300 kms. He divided his journey into a distance of thee equal parts in terms of distance and covered them with the speed of 30 km/hr, 60 km/hr and 90 km/hr, calculate his average speed during the journey:

Sol:

Here: x= 30, y= 60, z= 90.

Putting the values we get:

Answer= 49.09 km/hr

Short Cut-3

If a person travels Mth part of a distance at x km/hr, Nth part at y km/hr and the remaining Pth part at z km/hr,

Then his average speed in km/hr is given by:

If instead of proportion, the parts of the distance are given in percentage i.e. M%, N% and P% respectively, then the formula becomes:

Example 3

Hari travels 20% of a distance at an average speed of 20 km/hr, next 30% distance at an average speed of 30 km/hr and the remaining 50% of distance at the average speed of 50 km/hr. Find his average speed during the whole journey.

Sol:

Here M=20, N=30 and P= 50 also x= 20, y= 30 and z= 50

Putting the values we get:

Answer =33.33 km/hr

4If a car does a journey in 'T' hrs, the first half at 'a' km/hr and the second half at 'b' km/hr. The total distance covered by the car:

                     (2 x Time x a x b ) / (a + b).

 A motorcar does a journey in 10 hrs, the first half at 21 kmph and the second half at 24 kmph. Find the distance?Sol: Distance = (2 x 10 x 21 x 24) / (21+24)
                      = 10080 / 45
                      = 224 km.5.If a person goes from 'A' to 'B' at a speed of 'a' kmph and returns at a speed of 'b' kmph and takes 'T' hours in all, then the distance between the A and B:

 Total time taken x (Product of the two Speeds / Addition of the two speeds)

 A boy goes to school at a speed of 3 kmph and returns to the village at a speed of 2 kmph. If he takes 5 hrs in all, what is the distance between the village and the school?Sol: Let the required distance be x km.
Then time taken during the first journey = x/3 hr.
and time taken during the second journey = x/2 hr.
x/3 + x/2 = 5 => (2x + 3x) / 6 = 5
=> 5x = 30.
=> x = 6
required distance = 6 km.

                                            EXAMPLES

1.An athlete runs 200 meters race in 24 seconds. His speed is:

Sol:

Speed =

200 24

m/sec =

25 3

m/sec =

25 3

x

18 5

km/hr = 30 km/hr.

2.Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?

Sol:

In the same time, they cover 110 km and 90 km respectively.

Ratio of their speeds = 110 : 90 = 11 : 9

3.How long will a boy take to run round a square field of side 35 meters, If he runs at the rate of 9 km/hr?

Sol:

Speed = 9 km/hr = 9 x 5 18  m/sec = 5 2  m/sec.

Distance = (35 x 4) m = 140 m.

Time taken = 140 x 2 5  sec = 56 sec.

4.The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet

Sol:

Suppose they meet x hrs after 8 a.m. Then,

(Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330

60x + 75(x – 1) = 330   x  =  3

So, they meet at (8 + 3), i.e. 11 a.m.

5.A person crosses a 600 m long street in 5 minutes, What is his speed in km per hour

Sol:

Speed = 600 5 x 60  m/sec = 2 m/sec = 2 x 18 5  km/hr = 7.2 km/hr.

6.Two men start together to walk to a certain destination, one at 3 kmph and together at 3.75 kmph. The latter arrives half an hour before the former. The distance is:

Sol: Let the distance be x km. Then,

x 3 - x 3.75 = 1 2 2.5x - 2x = 3.75  x = 3.75 0.50 = 15 2   =  7.5 km

7.The ratio between the speeds of two trains is 7: 8. If the second train runs 400 kms in 4 hours, then the speed of the first train is:

Sol: Let the speeds of two trains be 7X and 8X km/hr.

Then, 8X = 400 4  = 100  X = 100 8  = 125.

Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr

8.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:

Sol: Let C's speed = x km/hr, Then, B's speed = 3x km/hr and A's speed = 6x km/hr

Ratio of speeds of A, B, C = 6x : 3x : x = 6 : 3 : 1

Ratio of times taken = 1 6 : 1 3  : 1 = 1 : 2 : 6

If C takes 6 min., then B takes 2 min.

If C takes 54 min., then B takes 2 6  x 54  min. = 18 min.

9.A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately

Sol:

Total distance travelled = (10 + 12) km/hr = 22 km/hr

Total time taken = 10 12 + 12 10  hrs = 61 30  hrs

Average speed = 22x 30 61  km/hr = 10.8 km/hr

10. 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 train to complete its journey is:

Sol:

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  x 40 = (x + 5) 60  x 50 x = 19 min.

11. An express train traveled at an average speed of 100km/hr, stopping for minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point?

Sol:

Time taken to cover 600 km = 600 100  hrs = 6 hrs

Number of stoppages = 600  - 1 = 7 75

Total time of stoppage = (3 x 7) min = 21 min.

Hence, total time taken = 6 hrs 21 min.

12.

Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

Sol:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km  = 9 54  x 60  min  =  10 min.

13.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 :

Sol: Ratio of speeds = 4 : 3 : 5

Ratio of times taken = 1 4 : 1 3 : 1 5   =  15 : 20 : 12

14. Walking ¾ of his speed, a person is 10 min late to his office. Find his usual time to cover the distance?

Sol:  Usual time = Late time / {1/ (3/4) - 1)
= 10 / (4/3 -1)
= 10 / (1/3)
= 30 minutes.

15. Running 4/3 of his usual speed, a person improves his timing by 10 minutes. Find his usual timing by 10 minutes. Find his usual time to cover the distance?

Sol:  Usual time = Improved time / {1 - (1/ (3/4)}
= 10 / { 1- (3/4) }
= 40 minutes.16. A train travelling 25 kmph leaves Delhi at 9 a.m. and another train travelling 35 kmph starts at 2 p.m. in the same direction. How many km from will they be together?

Sol: Meeting point's distance from the starting point = [25 x 35 x (2p.m. - 9 a.m)] / (35 -25)
= (25 x 35x 5) / 10
= 4375 / 10
= 437.5 km .

17. A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms?

Sol:

Distance covered in first 2 hours = (70 x 2) km = 140 km

Distance covered in next 2 hours = (80 x 2) km = 160 km

Remaining distance = 345 – (140 + 160) = 45 km.

Speed in the fifth hour = 90 km/hr

Time taken to cover 45 km = 45 90  hr = 1 2  hr

Total time taken = 2 + 2 + 1 2  = 4 1 2  hrs

18.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 the both the trips?

Sol:

Speed on return trip = 150% of 40 = 60 kmph.

Average speed = 2 x 40 x 60 40 + 60  km/hr = 4800 100  km/hr = 48 km/hr.

19.A man takes 5 hours 45 min. in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways, is:

Sol:

Let the distance be x km. Then,

(Time taken to walk x km) + (Time taken to ride x km) = 23 4  hrs

(Time taken to walk 2x km) + (Time taken to ride 2x km) = 23 2  hrs

But, time taken to ride 2x km = 15 4  hrs

Time taken to walk 2x km = 23 2 - 15 4  hrs = 31 4  hrs = 7 hrs 45 min.

20. A can complete a journey in 19 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.

Sol: Let the total distance be X km. Then,

1  x 2 21 + 1  x 2 24 = 10 x 21 + x 24 = 20

15x = 168 x 20  x = 168 x 20 15  = 224 km.

21.A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 50 meters apart, then at what speed is the train travelling?

Sol: Number of gaps between 21 telephone posts = 20

Distance traveled in 1 minute = (50 x 20) m = 1000 m = 1 km

Speed = 60 km/hr

22. Mary jogs 9 km at speed of 6 km per hour. At what speed would she need to jog during the next 1.5 hours to have an average of 9 km per hour for the entire jogging session?

Sol: Let speed of jogging be x km/hr

Total time taken = 9 6  hrs + 1.5 hrs  = 3 hrs.

Total distance covered = (9 + 1.5x) km

9 + 1.5x 3  = 9 9 + 1.5x = 27 3 2 x = 18 x = 18 x 2 3  = 12 kmph

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

Sol:

Speed = 10 x 60 12  km / hr = 50 km / hr

New speed = (50 – 5) km / hr = 45 km / hr

Time taken = 10 45  hr = 2 9  x 60  min = 13 1 3   min = 13 min 20 sec.

24.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:

Sol:

Ratio of speeds  =  3 : 4. Ratio of times taken  =  4 : 3.

Suppose A takes 4x hrs and B takes 3x hrs to reach the destination. Then,

4x - 3x  = 30 60 = 1 2   or   x = 1 2

Time taken by A = 4x hrs  = 4 x 1 2  hrs = 2 hrs

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

Sol:

Remaining distance = 3 km and Remaining time = 1 3  x 45  min = 15 min = 1 4  hour

Required speed = (3 x 4) km/hr = 12 km/hr

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