CLOCKS - CALENDER

A. SYNOPSIS OF CLOCKS:

     

(I) In an hour, the hour hand covers 5 minute spaces (msp) while the minute hand covers 60 minute spaces (msp). Thus, in one hour (or) 60 minutes, the minute hand gains 55-minute spaces over the hour hand.

Hence 1 hour = 60 minutes = 55 msp (In one hour, Minute hand gains over the Hour hand)∴ 1 minute = 55/60 msp = 11/12 msp ∴ 1 minute space = 12/11 minutes

For Hour Hand For Minute Hand12 Hours = 3600∴ 1 hour = 3600/12 = 300 ∴ 60 minutes = 300 ∴ 1 minute = 300/60 = 1/20

60 minutes = 360o ∴ 1 minute = 360o/60 = 6o ∴ 1 msp = 6o

∴ The minute hand moves through 60 in each minute, where as the hour hand moves through ½0 in each minute. Thus in one minute, the minute hand gains 5 ½0 than the hour hand.

(II) The minute hand moves 12 times as fast as the hour hand.

(III) In a clock, there are four possibilities in the arrangements of hour hand and the minute hand.

They are,(a) Both the hands are coinciding with each other.(b) Both the hands are perpendicular to each other.(c) Both the hands are opposite to each other.(d) Both the hands are in gap of ‘M’ minute spaces apart.

a) Both the hands are coinciding with each other.

In between H and (H+1)'o clock, the two hands of the clock will be together is given by (60H/11) minutes past H'o clock. i.e. (5H ? 0) x 12/11 minutes past H'o clock.

In hour gap, the hour hand and the minute hand are coinciding exactly one time. But except 12 to 1'o clock. So, in 12 hours, the number of times that hour hand and the minute hand are coincide is

given by 11 times. Hence, in 24 hours, the number of times that both the hands are coinciding is given by 22

times.

b) Both the hands are perpendicular to each other.

In between H and (H+1)'o clock, the two hands of the clock will be at right angle is given by (5H ? 15) x 12/11 minutes past H'o clock

In one hour gap, except (3 to 4) and (9 to 10) the hour hand and the minute hand are perpendicular to each other in two times.

In 12 hours, both the hands are perpendicular to each other in 22 times. Hence, in 24 hours, both the hands are perpendicular to each other in 44 times.

c) Both the hands are opposite to each other.

In between H and (H+1)'o clock, the two hands of the clock will be in the same straight line but not together is given by (5H ± 30) x 12/11 minutes past H'o clock(1) When H > 6 → (5H - 30) x 12/11 minutes past H'o clock.(2) When H < 6 → (5H + 30) x 12/11 minutes past H'o clock.(3) When H = 6 → Never happens.

In hour gap, the hour hand and the minute hand are coinciding exactly one time. In between (5 to 7), exactly at 6'o clock, both the hands are opposite to each other at exactly at 6'o clock. So, in 12 hours, the number of times that hour hand and the minute hand are opposite to each other is

given by 11 times. Hence, in 24 hours, the number of times that both the hands are opposite to each other is given by 22

times.

d) Both the hands are in gap of 'M' minute spaces apart.

In between H and (H+1)'o clock, the two hands of a clock are 'M' minutes spaces apart at given by (5H + M) x 12/11 minutes past H'o clock.

 

ANGLE BETWEEN THE TWO HANDS OF A CLOCK

(1) When the minute hand is behind the hour hand, the angle between the two hands at 'M' minutes past H'o clock is given by (30( H - M/5) + M/2)0

Example : 04:10

(2) When the minute hand is ahead of the hour hand, the angle between the two hands at 'M' minutes past H'o clock is given by (30( M/5 - H) - M/2)0

Example : 02:20

  CLOCK GAINS OR LOSSES PER DAY

The minute hand of a clock overtakes the hour hand at intervals of 'M' minutes of correct time, then the clock gains or losses per day is given by (720/11 - M) (60 x 24/M) Min.

If the above value is positive then the clock will gains. If the above value is negative then the Clock will looses.

THE RELATION BETWEEN STRIKES AND MINUTE SPACES:

  To obtain one minute spaces, we required two strikes.To obtain two minute spaces, we required three strikes, and so on.

  STRIKES    MINUTE SPACES

  'N' number of strikes    (N - 1) minute spaces

  (N + 1) number of strikes    'N' minute spacesMIRROR IMAGES IN CLOCKS (LATERAL IMAGES)

To find the mirror image of any time, we can subtract the given time from 12'o clock (i.e 11 : 60)

Example: Find the mirror image of a given time 3 : 20?

Solution:                     11 : 60                   (-)3 : 20                  ___________                     08 : 40                  ____________ SOLVED EXAMPLES

 (1) If a clock strikes 12 in 33 seconds, then it will strike 6 in how many seconds?

(A) 33/2 seconds (B) 15 seconds (C) 12 Seconds (D) 22 Seconds

Explanation:      

Number of strikes Number of minute spaces Required time

12 Strikes 11 minute spaces 33 seconds

6 strikes 5 minute spaces ?

(2) At what time between 5 and 6'o clock are the hands of a clock together?

Explanation: At 5'o clock,                   Gap = 25 msp                   Required gap = 0 msp (because of both the hands are coincide)

 Displacement = Gap ± Required Gap

= 27 3/11 minutes past 5'o clock

Short-cut:

          = 27 3/11 minutes past 5'o clock(3) At what time between 4 and 5'o clock are the hands of the clocks of right angles to each other?

Explanation :

At 4'o clock,Gap = 20 mspRequired gap = 15 msp ( Hands are at right angles)

 Displacement = Gap ±Required Gap

= (20 ± 15) msp =

(20 + 15) msp (20 - 15) msp

35 x 12/11 minutes past 4'o clock 5 x 12/11 minutes past 4'o clock

38 2/11 minutes past 4'o clock 5 5/11 minutes past 4'o clock

Shortcut: (5H ± 15) x 12/11 minutes past H'o clock= (5.4 ± 15) x 12/11 minutes past 4'o clock= (20 ± 15) x 12/11 minutes past 4'o clock =  38 2/11 minutes past 4'o clock

5 5/11 minutes past 4'o clock

(4) At what time between 1 and 1.30 are the hands of the clocks of right angles to each other?

Explanation:At 1'o clock, Gap = 5 mspRequired Gap = 15 msp

 Displacement = Gap ± Required Gap

= (5 ± 15) msp = (5 + 15) msp        = (5-15) msp = 20 msp                = - 10 msp = 50 msp The given time shows that it reaching Half - an - hour. So that less msp will be our answer.So 20 x 12/11 minutes past 1'o clock= 21 9/11 minutes past 1'o clock

5) At what time between 1.30 and 2 are the hands of the clocks of right angles to each other?

Explanation : At 1'o clock, Gap = 5 mspRequired Gap = 15 msp

 Displacement = Gap ±Required Gap

= (5 ± 15) msp = (5 + 15) msp       = (5 - 15) msp = 20 msp               = - 10 msp = 50 msp

The given time shows that it completed first Half - an - hour and reaching second half - an - hour. So that more msp will be our answer. So 50 x 12/11 minutes past 1'o clock = 54 6/11 minutes past 1'o clock

(6) At what time between 8 and 9 are the hands of the clock opposite to each other?

Explanation : At 8'o clock, Gap = 40 msp Required Gap = 30 msp

 Displacement = Gap ±Required Gap

= (40 ± 30) msp = (40 + 30) msp              = (40 - 30) msp = 70 msp – 60 msp          = 10 msp = 10 mspSo required time that both the hands are opposite to each other in between 8 and 9'o clock= 10 x 12/11 minutes past 8'o clock = 10 10/11 minutes past 8'o clock

(7) In between 10 a.m. and 11 a.m., when will both the hands be 22 minutes spaces apart?

Explanation :At 10'o clock, Gap = 50 mspRequired Gap = 22 msp

 Displacement = Gap ±Required Gap

= (50 ± 22) msp= (50 + 22) msp              = (50 - 22) msp= 72 msp - 60 msp          = 28 msp = 12 msp                       = 28 x 12/11 minutes                                       Past 10'o clock= 12 x 12/11 minutes past 10'o clock= 13 1/11 minutes and 30 6/11 minutes past 10'o clock

 

(8) The hands of a clock coincide after every 64 minutes of correct time. How much is the clock fast or slow in 24 hours?

Explanation :

(9) Find the angle between the two hands of a clock at 15 minutes past 4'o clock?

Explanation : 15 minutes past 4'o clock shows that the minute hand is behind the hour hand. (04 : 15)

(10) A clock is set right at 1 p.m. If it gains one minute an hour. What is the true time when the clock indicates 6 p.m. the same day?

Explanation : 1p.m to 6 p.m. on the same day = 5 hours

SYNOPSIS OF CALENDAR

A calendar is a particular measure of time. The smallest unit of a calendar is a day, which is the average time in which earth rotates

once on its axis. We can also say that the earth completes its one full rotation in one day. Therefore One day = 24 hours.

The time in which the earth revolves around the sun is known as a Solar year and is equals to 365 days 5 hours 48 minutes 47 ½ Seconds (or) 365.2422 days approximately.

In general, the common normal year consists of 365 days. Hence, the difference between a common normal year and a solar year is 0.2422 of a day. Thus every fourth year has an increase of one full day (or) has a total of 366 days. This year is known as leap year.

A year which is not exactly divisible by 4 is called an ordinary year. An ordinary year consists of 365 days, i.e. 52 weeks + 1 odd day. Hence an ordinary year

contains one odd day. A year, which is exactly divisible by 4, is called a leap year. Thus, each one of the years

1992, 1996, 2004, and 2008 ----- is a leap year. Every 4th century is a leap year, but no other century is a leap year. Thus, each one of

400, 800, 1200, 1600, and 2000 ---- is a leap year. A leap year consists of 366 days, i.e. 52 weeks + 2 odd days. Hence a leap year consists

of two odd days. The first day of a century must be either must be either Monday (or) Tuesday (or)

Thursday (or) Saturday. The last day of a century year must be either Sunday (or) Monday (or) Wednesday (or)

Friday. The last day of a century cannot be either Tuesday, Thursday (or) Saturday. The first day of a year after a century year be either Monday (or) Tuesday (or) Thursday

(or) Saturday. 1st January 1st A.D. was Monday; therefore we must count days from Sunday.

CALCULATION OF ODD DAYS:

100 YEARS = 76 ordinary years + 24 leap years= 76 odd days + (24 x 2) odd days= 124 days = 17 weeks + 5 odd days

COUNTING OF ODD DAYS IN MONTH WISE:

SOLVED EXAMPLE:

Method – I: Find the day of the week of any given date of any year:

(1) Find the day of the week of 15th August 1947?

Explanation:

∴ 2 odd days + 3 odd days = 5 odd days = Friday.∴ 1947,August 15th was Friday.

Method – II: Find the day of the week of a previous year of the same given date

(2) If January 28th, 1975 was Tuesday. Find the day of the week of January 28th, 1974?

Explanation: Given that, January 28th, 1974 → --------January 28th, 1975 → Tuesday From Jan 28th, 1975 to Jan 28th, 1974 while coming back, February of 1974 is included, but 1974 is not a leap year. Hence required day of the week is one day behind the given day.

Hence Jan 28th, 1974 was Monday.

Method – III: Find the day of the week of a next year of the same given date:

(3) If December 25th, 1977 was Friday. Find the day of the week of December 25th, 1978?

Explanation:Given that,

December 25th, 1977 → Friday December 25th, 1978 → --------

From Dec 25th, 1977 to Dec 25th, 1978, February of 1978 is included, but 1978 is not a leap year. So the required day is one day ahead of Friday.

So December 25th, 1978 was Saturday.  

SOLVED EXAMPLE:

Method – IV: Find the dates of a given month of any year the required day will falls:

(4) On what dates of July 2004 did Monday fall?

Explanation:Let us find the day on 1st July 2004.

∴ 3 odd days + 1 odd day = 4 odd days = Thursday ∴ July 1st, 2004 was Thursday. ∴ In July, 2004 Monday will fall on the dates 5th, 12th, 19th and 26th.

Model – V: Finding the same calendar year from a given year:

If 'N' is a leap year, then same calendar year is repeated in 4 categories.

(5) In which year the same calendar of the year 2009 will serve?

Explanation: The given year 2009 is a year after a leap year.

So, the same calendar of 2009 will serve after 6 years from 2009.

So, required year = 2009 + 6 = 2015.  

TIME & WORK

If a man can do a work in 2 days, then the work done by him in one day is 1/2.Similarly, if a man can do a work in 3 days, then the work done by him in one day is 1/3.

We can now generalize the above concept and conclude that if a man can do a work in 'n' days, then the work done by him in one day is '1/n'.

The converse of this statement is also true. That is, if the work done by a man in one day is 1/n, then the number of days required by him to complete the work is 'n' days.

Example: A and B can do a work in 3 days and 6 days respectively. If both work together, in how many days can the work be completed?

Solution: Work done by A in one day = 1/3Work done by B in one day = 1/6

So, work done by both A and B together in one day = (1/3 + 1/6) = 1/2.Hence, the number of days required by both working together is 2 days.

Note: If A and B can do a work in 'x' days and 'y' days respectively, then the number of days required by both working together = (xy) / (x+y).

If there are three persons A, B and C who can do a work in 'x' days, 'y' days and 'z' days respectively, then the number of days required by all three of them working together can be calculated using the formula: (xyz) / (xy + yz + zx).

Example: A and B can do a work in 5 days and 10 days respectively. A starts the work and leaves after 2 days. In how many days can B complete the remaining work?

Solution: Work done by A in 2 days = 2/5.Remaining work = 3/5.∴ The time taken by B to complete this remaining work = (3/5)(10) = 6 days.

Example: A and B can do a work in 5 days and 10 days respectively. A starts the work and B joins him after 2 days. In how many days can they complete the remaining work?

 

Solution: Work done by A in 2 days = 2/5.Remaining work = 3/5.Work done by both A and B in one day = (1/5) + (1/10) = 3/10.∴ The time taken by A and B to complete this remaining work = (3/5)(10/3) = 2 days.

Example: A, B and C can do a work in 6 days, 8 days, and 12 days respectively. In how many days can all three of them working together, complete the work?

Solution: Work done by all three of them in one day = (1/6 + 1/8 + 1/12) = 3/8.Thus, the number of days required = 8/3 = 2 2/3 days.

Example: A and B can do a work in 12 days, B and C in 30 days and C and A in 36 days. In how many days will the work be completed, if all three of them work together?

Solution: One days work of A and B = (1/12 + 1/30) = 7/60One days work of B and C = (1/30 + 1/36) = 11/180One days work of C and A = (1/36 + 1/12) = 1/9

Adding the three, we get 2(A+B+C) = (7/60) + (11/180) + (1/9)                                                 = (52 / 180) = 13/45. 

So, (A+B+C) = 13/90.Thus, the number of days required = (90/13) days.

Note: If A and B can do a work in 'x' days, B and C in 'y' days and C and A in 'z' days, then the number of days in which all three of them working together will complete the work = (2xyz) / (xy+yz+zx).

Example: A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?

Solution: One days work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = (4)(1/24 + 1/30) = 3/10.Remaining work = 7/10, which was done by A, B and C in the initial number of days.Number of days required for this initial work = 7 days.Thus, the total number of days required = 4 + 7 = 11 days.

 

 

Example:P is three times as fast as Q and working together, they can complete a work in 12 days. In how many days can Q alone complete the work?

Solution: P = 3Q.So, P + Q = 3Q + Q = 4Q.These 4 Q people can do the work in 12 days, which means Q can do the work in 48 days. Hence, P can do the work in 16 days.

Example: If 5 men and 2 boys, working together, can do four times as much work per hour as a man and a boy together. Find the ratio of the work done by a man and that of a boy for a given time.

Solution: 5M + 2B = 4(1M + 1B)Or, 5M + 2B = 4M + 4B1M = 2B.∴ The required ratio of work done by a man and a boy = 2 : 1

Example: 30 men can do a work in 40 days. When should 20 men leave the work so that the entire work is completed in 40 days after they leave the work?

Solution: 30 men can do the work in 40 days. So, the total work to be done = (30)(40) = 1200 man days.Let 20 men leave the work after 'x' days, so that the remaining work is completed in 40 days after they leave the work.∴ 40x + 20(40) = 1200. Thus, x = 10 days.

Example: A and B can do a work in 4 hours and 12 hours respectively. A starts the work at 6 AM and they work alternately for one hour each. When will the work be completed?

Solution: Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = (1/4) + (1/12) = 1/3.Thus, the total time required to complete the work = (2(3) = 6 hours.

In this topic Time & Work, there is a sub-topic: Pipes & Cisterns. Questions in this topic contain data on Pipes instead of on men. So, all the above examples are applicable to Pipes & Cisterns also.

There are two types of pipes

1. Inlet pipe = collects water (+)2. outlet pipe = empties water (-)

collecting water is considered to be a positive work

emptying water is considered to be a negative work

Example:If pipe A can fill a tank in 3 hrs and pipe B can empty the tank in 4 hrs then how long will it take to fill an empty tank when both the pipes are opened

Solution:Water collected in 1 hr

=1/A - 1/B = 1/3 - 1/4 = 1/12

= The tank will be filled in 12hr

Example:Pipe A can fill 30 gall min. pipe B can fill the tank in 6 min and Pipe C can empty the tank in 12 min. When all three pipes are opened, the tank is filled in 4 min. Find the capacity of the tank

Solution:

A can fill the tank in 6 min ∴ The capacity of the tank = 6 x 30 = 180 liters

Example:Pipe A can fill a tank in 37 ½ min and pipe B can fill the tank in 45 min. Both the pipes are

opened, find when should pipe B be turned off so that the tank may be filled in half an hour ?

Solution:

Let pipe B be closed after 'x' minthen 30 min of A + x min of B = 1

BOATS & STREAMS

Let the speed of a boat in still water be 40 kmph and let the speed of the stream / current be 10 kmph.

If the boat moves in the same direction as the flow of water in the stream, then we say that the boat is moving DOWNSTREAM and the net speed, that is, the speed of the boat downstream will be (40+10) = 50 kmph.

If the boat moves in a direction opposite to the flow of water in the stream, then we say that the boat is moving UPSTREAM and the net speed, that is, the speed of the boat Upstream will be (40-10) = 30 kmph.

Remember that always, the speed upstream will be less than the speed downstream.

We can now formulate the above concept as follows:

If the speed of a boat in still water = x and Speed of the stream = y, then Speed Downstream = (x+y)Speed Upstream = (x-y)

Similarly, if the Speed Downstream = A and the Speed Upstream = B, thenSpeed of the boat in Still water = 1/2 (Speed Downstream + Speed Upstream) = 1/2 (A + B) Speed of the stream = 1/2 (Speed Downstream - Speed Upstream) = 1/2 (A - B)

Example: The speed of a boat in still water is 60 kmph and the speed of the current is 20 kmph. Find the speed downstream and the speed upstream.

 

Solution: Speed downstream = (60+20) = 80 kmphSpeed Upstream = (60-20) = 40 kmph.

Example:The speed of a boat in Upstream is 60 kmph and the speed of the boat Downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream.

Solution: Speed of the boat in still water = 1/2 (80 + 60) = 70 kmph Speed of the stream = 1/2 (80 - 60) = 10 kmphExample: A man rows his boat 85 km downstream and 45 km upstream, taking 2 ? hours each time. Find the speed of the stream.

Solution:Speed downstream = d / t = 85 / (2 ?) = 34 kmphSpeed Upstream = d / t = 45 / (2 ? ) = 18 kmph∴ The speed of the stream = 1/2 (34-18) = 8 kmph.

Example: A man can 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?

Solution: Speed downstream = (20+6) = 26 kmph∴ Time required to cover 60 km downstream = d / S = 60 / 26 = (30/13) hours.Note: The speed upstream is always less than the speed downstream and hence, the time taken to cover a distance upstream is more than the time taken to cover the same distance downstream.

Note: If the ratio of the time upstream and the time downstream is (P:Q), then the ratio of the speed of the boat in still water to the speed of the stream = (P+Q) / (P-Q).

Example: 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.

Solution: 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 times 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.

TIME & DISTANCE

Speed is defined as the distance covered in unit time.

So, we can write Speed = (Distance) / (Time)

S = (d) / (t), where 'S' is the Speed, 'd' is the distance and 't' is the time.

In a question, speed may be given in Kilometers per Hour (Kmph). To convert it into Meters Per Second, we have to multiply with (5/18).For example, 36 kmph = (36)(5/18) = 10 mps.

Let us now look at some examples.

Example: In how much time will a car moving at 36 kmph cover a distance of 720 km?

Solution: We know S = d / t. From this, we can write 't' = d / S.

Example: What distance will be covered by a bus moving at 72 kmph in 30 seconds?

Solution: We know S = d / t. From this we can write d = (S)(t).Here, Speed is given in kmph but time is in seconds. So, we should convert kmph into mps, by multiplying with (5/18).

Thus,

Example: In how much time will a train of length 100m, moving at 36 kmph cross an electric pole?

Solution: Let us first convert speed from kmph into mps. 36 kmph = (36)(5/18) = 10 mps.

Note: If a train crosses an electric pole, telegraphic pole, a tree and a man then distance covered = length of train Speed X time = 100 meters 36 kmph X time = 100 m

∴time = 10 sec.Example:In how much time will a train of length 100m, moving at 36 kmph cross a platform of length 200m?

Note: If a train crosses a platform, a bridge or a tunnel, the distance to be covered is equal to the sum of their lengths.

Solution: Distance covered = length of train + length of platform Speed X time = (100 + 200) m 36 kmph X time = 300 m

∴ time = 30 secAverage Speed

We know that Speed = Distance / Time.So, Average Speed = Total Distance covered / Total Time taken

Example: A man goes from A to B at a speed of 20 kmph and comes back to A at a speed of 30 kmph. Find his average speed for the entire journey.

Solution: We know that Average Speed = Total Distance / Total Time.Here, distance and time are not given. Let the distance from A to B be 'd'.

So, Average speed = Total Distance / Total Time = [2d] / [(d/20) + (d/30)] = [2d] / [5d/60] = 24 kmph

Note: If we calculate the average of the two speeds: (20+30)/2 = 25 kmph; this will not be the average speed for the entire journey, it will only be the average of the two values. 

Note: If a man goes from A to B at a speed of 'x' kmph and comes back to A at a speed of 'y' kmph, then the average speed for the entire journey = (2xy) / (x+y)

 

 Note: If there are three equal distances being covered at three different speeds x, y and z, then the average speed for the entire journey can be calculated using the formula: (3xyz) / (xy + yz + zx).

Relative Speed

The speed of one body compared with the speed of another is called Relative Speed. If two bodies move in the same direction, then their relative speed is equal to the difference of their speeds and if two bodies move in opposite directions, then their relative speed is equal to the sum of their speeds.

So, if two bodies A and B are moving with speeds x and y respectively, then

If they move in the same direction, their relative speed = (x-y) and

If they move in the opposite directions their relative speed = (x+y).

Example: How much time will a train of length 200m moving at a speed of 72 kmph take to cross another train of length 300m, moving at 36 kmph in the same direction?

Solution:The distance to be covered = Sum of their lengths = 200+300 = 500 mRelative Speed = Difference of the speeds = (72 – 36) = 36 kmph = (36)(5/18) = 10 mps.

Thus,

Example: Two trains of length 100 m and 200 m are 100m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. After how much time will the trains meet?

Solution: The trains have to meet each other. So, the distance between them only has to be considered, we need not consider their lengths. As they are moving in opposite directions, relative speed is equal to the sum of their speeds.Thus, Relative speed = (54+72)(5/18) = 35 mps.Thus, the time required = d / S = 100 / 35 = (20/7) seconds.

Example: Two trains of length 100 m and 200 m are 100m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. In how much time will the trains cross each other?

Solution: The trains have to cross each other. So, the distance between them and also their lengths have to be considered. As they are moving in opposite directions, relative speed is equal to the sum of their speeds.Thus, Relative speed = (54+72)(5/18) = 35 mps.Thus, the time required = d / S = (100 + 100 + 200) / 35 = (80/7) seconds.Example: Two trains of length 100 m and 200 m are 100m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. After how much time will the trains be 100 m apart again?

Solution: The trains have to cross each other and then be at 100 m apart. So, the initial distance between them, their lengths and the final distance between them has to be considered. As they are moving in opposite directions, relative speed is equal to the sum of their speeds.

Thus, Relative speed = (54+72)(5/18) = 35 mps.Thus, the time required = d / S = (100 + 100 + 200 + 100) / 35 = (100/7) seconds.

Example: Two persons A and B start simultaneously from the same point, running in the same direction with speeds 50 mps and 150 mps, around a circular track of length 1200 meters. When will they meet for the first time, and when will they meet for the first time at the starting point?

Solution: Let us first calculate the time taken by A and B to complete one round each around the circular track.

A: t = d / S = 1200 / 50 = 24 seconds B: t = d / S = 1200 / 150 = 8 seconds

So, the time after which A and B will meet each other for the first time at the starting point = LCM of 24 and 8 = 24 seconds

Let us now calculate the time required by A and B to meet each other.Time = Distance / Speed.T = d / S = 1200 / 100 = 12 seconds.So, in the second round, they meet each other for the first time.Example: Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train?

Solution: Relative speed of the trains = (72-36)(5/18) = 10 mps.Distance covered in 27 seconds = (27)(10) = 270 meters.Therefore, the length of the faster train = 270 meters.

Note:When two trains are moving in the same direction and the faster train crosses a man in the slower train, then the distance covered is equal to the length of the faster train.

Example: Two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be clear of each other from the moment they meet?

Solution: Relative Speed = (42+30)(5/18) = 20 mps.Distance covered in passing each other = (120+280) = 400 meters.Thus, the time required = d / S = 400 / 20 = 20 seconds.

   

 

Example:The distance between two stations A and B is 208 km. Train 1 leaves A at 10 am and travels towards B at 30 kph. Train 2 leaves B at 1.20 pm and travels towards A at 24kph. When will the trains cross each other?

Solution:

Example:A train covers a distance of 300 km at a certain speed. If the speed of train is 15 kph faster it would take an hour less to reach. Find the original speed of train

Example:A man covers a certain distance at a certain speed. If the speed of man is 5 kph faster he could take 72 min less to reach. If the speed of man is 5 kph slower he could take 2 hrs more to reach. Find the distance traed.

CIRCULAR TRACKS

Example:A and B start at the same time from the same point and travel in the same direction around a circular track of circumference 1200 m with speeds of 50 m/min and 150 m/min respectively. (i) when will they be together for the 1 st time(ii) When will they be together for the 1 st time at the starting point?

Rule: Find the time taken by each to complete 1 round. Then the required time is their LCM. Time taken by A = 1200/50 = 24 minTime taken by B = 1200/150 = 8 minThe required time is the LCM of 8, 24 = 24 min

SIMPLE INTEREST & COMPOUND INTEREST

The money borrowed (or) lent out for a certain period is called the Principal (P).

Extra money paid for using other's money is called Interest.

There are two types of interests (i) Simple Interest (S.I)(ii) Compound Interest (C.I)

Simple Interest

If the interest on a certain sum borrowed for a certain period is reckoned uniformly is called simple Interest.Let P = Principal, R = Rate per annum, T = time in years then

We know that, Amount = Principal + Interest∴ A = P + S.I

(1) If a certain sum in 'T' years at R% per annum amounts to Rs.A, then the sum will be

Example :- What principal will amount to Rs.570 at 4% per annum in 3? years?

Explanation:- A = Rs.570; R = 4%, T = 3? yrs.

(2) The annual payment that will discharge a debt of Rs.A due in 'T' years at R% per annum

then annual payment

Example : Find the annual installment that will discharge a debt of Rs.12,900 due in 4 years at 5% per annum simple interest?

Explanation : A = Rs.12,900, T = 4 years, R% = 5% p.a

Anual payment

(3) A sum of Rs. P is divided among three such parts that amount obtained on these three parts of money after T1, T2 & T3 Years respectively at the rate of R1%, R2% and R3% per annum remains Equal.

Then the ratio of 3 parts be

Example :-A sum of Rs.1586 is divided among three such parts that amount obtained on these three parts of money after 2, 3 & 4 years respectively at the rate of 5% per annum remains equal, Find such three parts of the sum.

Explanation: - Ratio of 3 parts.

∴Total parts = 276 : 264 : 253

∴ Total parts = 276 + 264 + 253 = 793

(4) If a certain sum of money becomes 'n' times it self in 'T' years at S.I, then the rate of interest per annum is

Example:- A certain sum of money triples it self in 5 years S.I. Find rate percent per annum?

Explanation:- n = 3, T = 5 years

Example: -In what time a sum of money will double it self at a rate of S.I of 8% p.a.

Explanation :- Here n = 2, R = 8%

(5) In a certain sum of money becomes 'n' times it self in 'T' years at S.I then the time T1 in which it will become 'm' times it self

is given by

Example: A sum of money put out on simple interest doubles it self in 12? years. In how many years would it triple it self?

Explanation:- n = 2; T = 12? , m = 3

Compound Interest

Some times the borrower and the lender agree to fix up a certain unit of time, say yearly (or) Half - yearly (or) Quarterly to settle the previous account.

In such cases, the amount after first unit of time becomes the principal of second unit of time.

The amount after the second unit of time becomes the principal of 3rd unit of time. And so on.

After a certain period, the difference between the amount and the principal is called Compound Interest (C.I)

(vi) Results on Population (or) Commodity

Let 'P' be the present population,

R% be the annual increase/decrease

(c) When the rates are different for different years say R1%, R2% & R3% for Ist, IInd & IIIrd years respectively, then

Note:- For increase / Growth ___ + Decrease / depreciation ____ -

Example 1:- The population of a village increases by 5% p.a. The present population of village is 4410. Find the population of that village 2 years ago was?

Explanation:-

Example (2) :- The population of a town increases by 4% p.a. The present population of the town is 15625. Find the population of that town after 3 years?

Explanation:-

Example (3):-A machine costing Rs.5000 depreciates at the rate of 20% p.a. Find its value after 3 years.

Explanation:-

Example (4) The population of a city has been increasing at the rate of 10% for every 5 years. If the present population is 48,400 What was it 10 years ago?

Explanation:-

Example (5) The present population of a town is 1,44,000. It increases by 5% in the 1st year, decreases by 10% in the 2nd year & increases by 15% in the 3rd year respectively. Find the population after 3 years?

Explanation:-

(II) If a loan of Rs.P at R% C.I per annum is to be repaid in 'n' equal yearly installments, then each installment value 'x' is given by

Example(1) :- A sum of money is borrowed at 5% p.a. C.I. If it is paid back in two equal annual installments of Rs.8820 each, find the sum borrowed.

Explanation:- Here R = 5%, x = 8820.

Example (2) :- A sum of Rs.13,240 is borrowed at the rate of 10% p.a. C.I and paid back in 3 equal annual installments. Find the value of each installment.

Explanation:- Sum borrowed = Rs.13,240. R = 10% p.a. Each installment = x = ?

(III) Relation between S.I & C.I.

RACES & GAMES

In this topic, the data in the questions can be given in the form of four types of statements as follows:

a. In a kilometer race, A gives B a start of 100 meters.This means, when the race is for one kilometer, A starts at the initial point, where as B starts from a point which is 100 meters ahead of A.So, the distance to be covered by A = 1000mDistance to be covered by B = 900 m only.Note that in this case, if both have the same speed, B will win the race.

b. In a kilometer race, A beats B by 100 meters.This means, when the race is for one kilometer, when A and B participate, A wins the race. So, by the time A completes the race

and reaches the end point, B is 100 meters behind A, that is, B is yet to cover another 100 meters. So, the distance covered by A = 1000m.Distance covered by B = 900 m only.

c. In a kilometer race, A beats B by 10 seconds.This means, when the race is for one kilometer, when A and B participate, A wins the race. So, by the time A completes the race and reaches the end point, B is some distance behind A, that is, B is yet to cover some more distance. So, A completes the race before B by 10 seconds.So, if time taken by A to complete the race = T seconds, thenTime taken by B to complete the race = (T+10) seconds.

d. In a game of 100 points, A gives 20 points to B.This means, when the game is for 100 points, when A and B participate, A wins the game. So, A gets all the 100 points in the game while B gets 20 points less than A.So, number of points got by A = 100 pointsNumber of points received by B = (100 - 20) = 80 points

Example: In a kilometer race, A beats B by 50 meters or 10 seconds. What time does A take to complete the race?

Solution: B runs 50 meters in 10 seconds. Therefore, time taken by B to run 1000 meters = (1000)(10) / (50) = 200 seconds. Thus, time taken by A = (200 - 10) = 190 seconds.

Example: A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?

Solution: A runs 1000 meters while B runs 900 meters and C runs 800 meters. Therefore, B runs 900 meters while C runs 800 meters.So, the number of meters that C runs when B runs 1000 meters = (1000)(800) / (900) = (8000/9) meters = 888.88 meters.Thus, B can give C (1000 - 888.88) = 111.12 meters start.

Example: A can run a kilometer race in 4 ? minutes while B can run the same race in 5 minutes. How many meters start can A give B in a kilometer race, so that the race may end in a dead heat?

Solution:A can give B (5 minutes - 4 ? minutes) = 30 seconds start. Now, we have to find the distance that B can cover in these 30 seconds.The distance covered by B in 5 minutes = 1000 meters.So, distance covered in 30 seconds = (1000)(30) / (300) = 100 meters.Therefore, A can give B 100 meters start.

Example: In a race of 1000 meters, A can beat B by 100 meters, in a race of 800 meters, B can beat C by 100 meters. By how many meters will A beat C in a race of 600 meters?

Solution:When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.∴when B runs 900 meters, the distance that C runs = (900)(700) / (800) = (6300) / 8 = 787.5 meters.So, in a race of 1000 meters, A beats C by (1000 - 787.5) = 212.5 meters to C.So, in a race of 600 meters, the number of meters by which A beats C = (600)(212.5) / (1000) = 127.5 meters.

Example:In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?

Solution:A scores 60 while B scores 40 and C scores 30.∴B scores 40 while C scores 30.So, the number of points that C scores when B scores 100 = (100)(30) / (40) = 75.Hence, in a game of 100 points, B gives (100-75) = 25 points to C.

 

PARTNERSHIPS

Let us assume that two persons A and B start a business with investments Rs. 5000 and Rs. 10000 respectively. If they both invest their amount in the business for the same time duration, then at the end of that period, the ratio in which they have to share the profits = 5000 : 10000 = 1 : 2.

Note:

(a) If two persons A and B invest Rs. P and Rs. Q respectively in a business for the same amount of time, then the ratio in which they have to share the profits at the end of the year is equal to the ratio of their investments = P : Q.

(b) If two persons A and B invest the same amount for 'R' months and 'S' months respectively in a business, then the ratio in which they have to share the profits is equal to the ratio of the time periods for which they have invested = R : S

(c) If a person 'A' invests Rs. P for R months and another person 'B' invests Rs. Q for S months, then the ratio in which they have to share the profits is equal to the ratio of the products of their investments and time for which the investment is made = (PR) : (QS)

Let us now understand the above concepts through some examples:

Example: A and B start a business with Rs. 6000 and Rs. 8000 respectively. How should they share their profits at the end of one year?

Solution: The ratio of the investments made by A and B  = 6000 : 8000 = 3 : 4.As both are investing for the same time period of one year, they should share the profits in the ratio of their investments.Thus, they should share their profits in the ratio 3 : 4.

Example: A and B invest Rs. 10000 each, A invest for 8 months and B invest for all the 12 months in the year. If the total profit at the end of the year is Rs. 25000, find their shares.

Solution: As both A and B invest the same amounts, the ratio of their profits at the end of the year is equal to the ratio of the time periods for which they have invested.

Thus, the required ratio of their profits = A : B = 8 : 12 = 2 : 3.Hence, share of A in the total profit = (2/5)(25000) = Rs. 10000Similarly, share of B in the total profit = (3/5)(25000) = Rs. 15000

Example:A and B invest Rs. 8000 and Rs. 9000 in a business. After 4 months, A withdraws half of his capital and after 2 months from then, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?

Solution: The ratio in which they have to share the profits at the end of the year can be calculated as follows:

A   B  

(8000)(4) + (4000)(8) : (9000)(6) + (6000)(6)  

64000 : 90000  

64 : 90  

32 : 45   Example: A and B start a business, with A investing the total capital of Rs. 50000, on the condition that B pays A interest @ 10 % per annum on his half of the capital. A is a working partner and receives Rs. 1500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year.

Solution: Interest received by A from B = 10 % of half of Rs. 50000 = 10 % of Rs. 25000 = Rs. 2500.Amount received by A per annum for being a working partner = (1500)(12) = Rs. 18000Let 'P' be the part of the remaining profit that A receives as his share. So, total income of A = (Rs. 2500 + Rs. 18000 + Rs. P)Total income of B = only his share from the remaining profit = 'P', as A and B share the remaining profit equally.

We know that income of A = Twice the income of BSo, (2500 + 18000 + P) = 2(P)∴P = 20500.

Thus, the total profit = 2P + Rs. 18000 = 2(20500) + 18000 = Rs. 59000.

MIXTURES & ALLEGATIONS

This topic is an extension and combined application of both the topics: Ratios & Proportions and Percentage, which we have already discussed. The method of Alligations, also known as Method of Cross Difference is a very important method, due to its ability to be applied to a wide variety of concepts and topics. Let us understand its application using some examples.

Example: A mixture of 150 liters of wine and water contains 20 % water. How much more water should be added so that water becomes 25 % of the new mixture?

Solution: Number of liters of water in 125 liters of the mixture = 20 % of 150 = 1/5 of 150 = 30 liters.Let us assume that another 'P' liters of water are added to the mixture to make water 25 % of the new mixture. So, the total amount of water becomes (30 +P) and the total volume of the mixture becomes (150 + P)Thus, (30 + P) = 25 % of (150 + P).Solving, we get P = 10 liters.

Example: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and are replaced with an equal quantity of pure milk. If the process is repeated one more time, find the ratio of milk and water in the final mixture obtained.

Solution:Number of liters of milk and water in the initial mixture of 20 liters: Milk = (3/5)(20) = 12 liters; water = 8 liters.If 10 liters of the mixture are removed, amount of milk removed = 6 liters andamount of water removed = 4 liters.Thus, remaining milk = (12 - 6) = 6 litersRemaining water = (8 - 4) = 4 liters.Now, 10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.Now, the ratio of milk and water in the new mixture = 16 : 4 = 4 : 1If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = (4/5)(10) = 8 litersAmount of water removed = 2 liters.So, remaining milk = (16 - 8) = 8 litersRemaining water = (4 - 2) = 2 litersHence, the required ratio of milk and water in the final mixture obtained = 8 : 2 = 4 : 1.

 

Example: In what ratio should two varieties of sugar at Rs. 18 per Kg and Rs. 24 Kg be mixed together to get a mixture whose cost is Rs. 20 per Kg?

Solution: 1st varietyRs. 18 / Kg

 2nd VarietyRs. 24 / Kg

 Mixture Cost:

Rs. 20 / Kg 

(24 - 20)     : (20 - 18)

= 4 : 2 = 2 : 1.

Thus, the two varieties should be mixed in the ratio 2 : 1Example: How many liters of Oil at Rs. 40 per liter should be mixed with 240 liters of a second variety of Oil at Rs. 60 per liter so as to get a mixture whose cost is Rs. 52 per liter?

Solution: Apply Alligation Method and first calculate the ratio in which they have to be mixed.

1st varietyRs. 40 / Ltr

 2nd VarietyRs. 60 / Ltr

 

 Mixture Cost:

Rs. 52 / Ltr   

(60 - 52) : (52 - 40)  

= 8 : 12 = 2 : 3.

Thus, the two varieties of oil should be mixed in the ratio 2 : 3.

So, if 240 liters of the second variety are taken, then 160 liters of the first variety should be taken.Example: In what ratio should two varieties of Tea at Rs. 60 per Kg and Rs. 120 per Kg be mixed together so that by selling the mixture at Rs. 86 per Kg, a profit of Rs. 20 % is obtained?

Solution: Let us first calculate the cost price of the mixture. The selling price of the mixture is given as Rs. 96 and the profit is given as 20%.So, Cost price = (96)(100/120) = Rs. 80.

Now, let us apply Alligation method:

1st varietyRs. 60 / Kg

 2nd VarietyRs. 120 / Kg

 

  Mixture Cost:    

Rs. 80 / Kg(120 - 80) : (80 - 60)  

= 40 : 20 = 2 : 1.

PROFIT & LOSS

Let us consider that a person purchases an article for Rs. 1000. This is called his Cost Price. He now wants to sell the article and fixes its price as Rs. 2000. This is called the Marked Price; it is not the selling price. Now, let us assume that he gives a discount of 20 % while selling. So, the discount given = 20 % of Rs. 2000 = Rs. 400. Hence, a customer who buys the article from his should actually pay Rs. 1600. This is called the Selling Price.

In this case, the Selling price is more than the Cost Price; and hence he gets a profit. Profit = Selling Price - Cost Price = Rs. 1600 - Rs. 1000 = Rs. 600.

If we want to calculate the Profit Percent, we can calculate as follows:Profit Percent = (600/1000)(100) = 60 %.

Now, let us define the terms that we have used above.

Cost Price: The price at which an article is bought is called the Cost Price (CP).

Marked Price: The price of an article mentioned on the price tag or in the price list is called the Marked Price (MP). It is also known as Written Price or Catalogue Price or List Price.

Discount: The reduction allowed on the price of an article is called the Discount or Commission.

Note: Discount is always calculated on the Marked Price only, and not on the Cost Price or the Selling Price.

Selling Price: The price at which an article is sold is called the Selling Price (SP). It is obtained after subtracting the discount, if any, from the Marked Price.

Profit & Loss: If the selling price of an article is more than its cost price, the profit is obtained, and if the selling price is less than the cost price, loss is obtained.Thus, Profit = Selling Price - Cost Price = SP - CPLoss = Cost Price - Selling Price = (CP - SP)

 

 Note: Profit percent or Loss percent is to be calculated on the Cost Price only, unless it is mentioned.

Example:A person buys an article at Rs. 500. At what price should he sell the article so as to make a profit of 20 %?

Solution:S.P = 120 % of C.P

 

=

Note: The selling price can also be calculated using the formula:SP = (CP)[(100+ P)/ 100]

Note: If loss is obtained, then the formula would be: SP = (CP)[(100- L)/ 100]

Example: By selling an article at Rs. 600, a profit of 25 % is made. Find its cost price.

Solution: SP = 125 % of C.P

Note: To calculate profit in this example observe that it should not be calculated as 25 % of the Selling Price Rs. 600; because profit is always calculated only on the cost price and not as a percentage of the selling price.

 Example: By selling an article at Rs. 800, a shopkeeper makes a profit of 25 %. At what price should he sell the article so as to make a loss of 25 %?

Solution:S.P = ____ % of C.P 800 = 125% of C.P ? = 75% of C.P

Note: The calculation in the above example can also be done as follows:

If the Profit is 25 %, for Rs. 100 of CP, the SP should be Rs. 125.Here, 125 % is given as Rs. 800So, 75 % is how much? We can calculate this: (75)(800)/125 = Rs. 480.Example: A shopkeeper buys two articles for Rs. 1000 each and then sells them, making 20 % profit on the first article and 20 % loss on the second article. Find the net profit or loss percent.

Solution: CP of each article = Rs. 1000.Profit on first article = 20 % of 1000 = Rs. 200. This is equal to the loss he makes on the second article. Thus, the net profit or loss is Zero. That is, he makes neither profit nor loss.

Example: A shopkeeper sells two articles at Rs. 1000 each, making a profit of 20 % on the first article and a loss of 20 % on the second article. Find the net profit or loss that he makes.

Solution: SP of first article = Rs. 1000Profit = 20 % CP = (SP)[100/(100+P)] = Rs. 2500/3

SP of second article = Rs. 1000Loss = 20 %CP = (SP)[100/(100-L)] = Rs. 1250

So, total SP = Rs. 2000; total CP = Rs. 6250/3As the CP is more than SP, he makes a loss.Loss = CP - SP = (6250/3) - 2000 = Rs. (250/3)

So, Loss Percent = [(250/3) / (6250/3)](100) = 4 %.

(Or)

 

Note: If two articles are sold at the same price, making P % profit on the first article and P % loss on the second article, then the net will always be a loss. This Net Loss can be directly calculated using the formula: Net Loss = (P)2 / 100.

 

   Example:

A shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 3 a rupee. Find his net profit or loss percent.

Solution: Let the total number of mangoes bought by the shopkeeper be 12. So, if he buys at the rate of 4 a rupee, his CP = Rs. 3; and if he is selling at the rate of 3 a rupee, his SP = Rs. 4So, his profit = SP - CP = Rs. 4 - Rs. 3 = Rs. 1

Example:

A dishonest dealer professes to sell his goods at Cost Price but still gets 20 % profit by using a false weight. What weight does he substitute for a kilogram?

Solution: If the cost price is Rs. 100, then to get a profit of 20 %, the selling price should be Rs. 120.

It means, if 120 kg are to be given, and then the dealer gives only 100 kg, to get a profit of 20 %. Now, we can calculate how many grams he has to give instead of one kilogram (1000 gm).

For 120 gm..............100gmSo, for 1000 gm..............?

=

PERCENTAGES

Percent means 'for every 100'.

A fraction whose denominator is 100 is called 'Percentage'.

In that fraction numerator is called 'Rate percent'.

So, when we say 40 %, it means 40 out of 100.Thus, 40 % = 40/100 = 2/5. Hence, if we want to calculate 40 % of a number, we can directly calculate 2/5 of the number.

Similarly, any percentage can be expressed as a fraction.

Example: Calculate 40 % of 625.

Solution: 40 % of a number = 2/5 of the number = 2/5 of 625 = (2/5)(625) = 250.

Let us now express some percentages as fractions:

a. 5 % = 5/100 = 1/20b. 10 % = 10/100 = 1/10c. 20 % = 20/100 = 1/5d. 25 % = 25/100 = ?e. 40 % = 40/100 = 2/5

f.    50 % = 50/100 = 1/2 g.   60 % = 60/100 = 3/5h.   75 % = 75/100 = 3/4 i.    80 % = 80/100 = 4/5j.    100 % = 100/100 = 1

 

k.   125 % = 125/100 = 5/4l.    150 % = 150/100 = 3/2m.  175 % = 175/100 = 7/4n.   200 % = 200/100 = 2

  o. = 25/800 = 1/32

p. = 25/400 = 1/16

q. = 20/300 = 1/15

r. = 25/300 = 1/12

s. = 25/200 = 1/8

t. = 50/300 = 1/6

u. = 100/300 = 1/3

v. = 75/200 = 3/8

w. = 200/300 = 2/3

x. = 175/200 = 7/8

Example: What percentage of 1600 is 40?

Solution: Let 40 be P % of 1600.So, 40 = P % of 1600

Note: P % of R = (P/100)R = PR/100R % of P = (R/100)(P) = PR/100

Example:A number, when increased by 25 % becomes 4500. Find the original number.

Solution: Let the original number be P. Increasing by 25 %, it becomes (125P)/100 .

∴ A number = 3600

Example: A number, when decreased by 30 % becomes 1050. Find the original number.

Solution: Let the original number be P.

∴ P = 1500Example: A number is first increased by 30 % and then decreased by 20 %. Find the net change in the number.

Solution: Let the original number be 100.Increasing by 30 %, it becomes 130.Now, if 130 is decreased by 20 %, it becomes 104.Thus, the net change = (104 - 100) = 4 % increase.

Note: If a number is first changed by x% and then changed by y%, then the net change in the number = [x + y + (xy/100)]. Remember that any decreasing value in this formula should be taken as 'negative' and increasing value should be taken as 'positive'.

Example: A's salary is 25 % more than B's salary. By what percent is B's salary less than A's?

Solution: If A's salary is x% more than B's salary then B's salary less than A's by

Note:

Similarly, if A's salary is x% less than B's salary, then the percentage by which B's salary is more than A's salary is

Example: If the price of sugar is increased by 25 %, find by what percentage the consumption has to be reduced to have the same expenditure.

Solution: Let the original price of sugar be Rs. 100 per kg.After increasing 25 %, the price becomes Rs. 125 per kg.Let the total expenditure be Rs. 2500.So, number of kgs of sugar obtained at the original price = 2500/100 = 25 KgNumber of Kgs of sugar obtained at the new price = 2500 / 25 = 20 Kg.Thus, decrease in the number of kgs of sugar = 25 - 20 = 5 Kg.Hence, percentage decrease in consumption = (5/25)(100) = 20 %.Note: If the price of an article is increased by P %, then the percentage by which consumption should be reduced to have the same

expenditure can be calculated by using the formula:

Note: Similarly, if the price of an article is decreased by P %, then the percentage by which consumption should be increased to have

the same expenditure can be calculated by using formula:

Note: The above two formulae can be used only if expenditure remains constant. This is because, if the expenditure is constant, price and consumption are inversely proportional to each other for most of the commodities.

Note: If expenditure also changes, then we need to use the following formula:

Percentage change in Expenditure or Revenue =

where 'P is the percentage change in Price and 'Q' is the percentage change in consumption. Remember that in this formula, as mentioned earlier, any decreasing quantity should betaken as 'negative.'Example:When the price of an article is decreased by 20 %, the consumption increased by 30 %. Find the net change in the expenditure.

Solution: Here price is decreasing by 20 % and consumption is increasing by 30 %.So, substituting x = - 20 and y = + 30 in the above formula, we get the percentage change in expenditure

= (10 - 6) % = 4%

Hence, the expenditure increases by 4 %.

Example: I spent 20 % of the amount I had on each of the four items: Rent, food, clothes and entertainment, and saved the remaining Rs. 5000. Find the total amount with me initially.

Solution: The total amount spent by me on various items = 4(20%) = 80 %. So, the remaining amount will be 20 %,20 % of income = 5000

income = Example:In an election between A and B, A got 57% of the total votes and won the election by a margin of 2100 votes. Find the total number of votes.

Solution:% of votes got by A = 57

∴ % of votes got by B = 43                           ----------                       Total = 100%

A - B = 57 - 43 = 14%

But A - B = 2100

Therefore 14% of total number of votes = 2100

Example:In an examination 31% failed and the number of students passed exceeds the number failed by 152. Find the total number of students.

Solution:%Passed = 69

%Failed = 31

Total  = 100%

P - F = 69 - 31 = 38%

But P - F = 152

38% of Total number of students = 152

Example:In an examination 65% of the boys and 85% of the girls obtained I class. If 80% on the whole obtained I class, find the number of girls students appeared in a total of 400 students.

Solution:

4 b : g = 100 : 300 = 400number of girl students = 300

Example:In an examination 52% failed in Mathematics, 42% failed in English. If 17% failed in both find the % passed in both

Solution:

Overall failures = 52 + 25 = 77%∴ % passed in both 100-77=23%

Example:In an examination 60% passed in History 50% passed in Geography. If 20% failed in both, find the% passed in both the subjects.

Solution:

Overall pass % = 100 - 20 = 80%

Therefore 60 + 50 - x = 80%

% passed in both = 110 - 80 = 30%

Example:In an examination a student who secured 30% of maximum marks failed by 90 marks. Another student who secured 50% of maximum marks got 60 marks more than necessary for passing. Find (i) maximum mark (ii) pass marks.

Solution:

RATIO & PROPORTION

The comparision between the two quantities of same kind is called the ratio between the two quantities.

Example:

If we have 20 pens and 30 pencils, we say that the ratio is: 20 : 30 = 2 : 3.A ratio is only a numerical relation, and does not contain any units.

The two quantities in a ratio are called the Terms; the first term is called the Antecedent and the second term is called the Consequent.Thus, in the ratio 5:6, the antecedent is 5 and the consequent is 6.

Note: A ratio remains unaltered if its numerator and denominator are multiplied or divided by the same number.Types of Ratios:

a. Duplicate and Sub-duplicate Ratio: The ratio of the squares of two numbers is called the Duplicate Ratio of the numbers.

Example:

9/16 is called the Duplicate Ratio of ?; the ratio of the square roots of two numbers is called the sub-duplicate ratio.∴ ? is called the sub-duplicate ratio of 9/16.

b. Triplicate and Sub-triplicate Ratio: The ratio of the cubes of two numbers is called the Triplicate Ratio of the numbers.

Example:

27/64 is called the Triplicate Ratio of ?; the ratio of the cube roots of two numbers is called their sub-triplicate ratio.Thus, ? is called the sub-triplicate ratio of 27/64.

c. Compound Ratio: The ratio obtained by taking the product of the antecedents to that of the consequents of two or more ratios is called the Compound Ratio.

Example:

The compound ratio of 2/3 and 4/5 = (2/3)(4/5) = 8/15

d. Inverse Ratio: The ratio obtained by interchanging the antecedent and consequent is called the inverse ratio. Ex: The inverse ratio of 5/6 is 6/5.

Comparision of two ratios: Let a/b and c/d are any two ratios, then First cross product = a X d Second cross product = b X c

(i) If ad < bc then a/b < c/d(ii) If ad > bc then a/b > c/d (iii) If ad = bc then a/b = c/d

Note: If a/b = c/d then a,b,c,d are said to be in proportionThere are two types of proportions.

If two ratios are equal, then the product of the extreme terms is equal to the product of the middle terms. 

(1) If a : b = c : d, then we can write ad = bc

Example:

4 : 5 = 16 : 20. This is called a proportion. As already discussed, in a proportion, the product of the means is equal to the product of the extremes. Thus, (5)(16) = (4)(20).

(2) If a : b = b : c, then applying the same principle, we can write (b)2 = ac. Here 'b' is called the Mean Proportional or middle term or geometric mean of 'a' and 'c' and 'c' is called the 3rd Proportional.

Combining two ratios:

Example:

 If A : B = 2 : 3, and B : C = 4 : 5, then by combining them, we get a Compound Ratio as follows:

A : B = 2 : 3B : C = 4 :5A : B : C = 2(4) : (3)(4) : (3)(5)

Thus, A : B : C = 8 : 12 : 15

 

 

 

Note: Two ratios can also be combined as follows: 

A : B = 2 : 3C : D = 4 : 5AC : BD = 8 : 15.

Example:Divide Rs. 32000 in the ratio 3 : 5.

Solution: Let the two parts be 3P and 5P.

3P + 5P = 32000; 8P = 32000 and hence P = 4000.Thus, the two parts are: 3P = 12000, 5P = 20000.

We can also solve as follows: (3/8)(32000) = Rs. 12000(5/8)(32000) = Rs. 20000

Example: Divide Rs. 118000 among three persons A, B and C such that the ratio of the shares of A and B is 3 : 4 and that of B and C is 5 : 6.

Solution: Let us calculate the compound ratio A : B : C.

A : B = 3 : 4B : C = 5 : 6A : B : C = 3(5) : (4)(5) : (4)(6)

So, A : B : C = 15 : 20 : 24

Now, we can divide Rs. 118000 in this ratio.

Share of A = (15/59)(118000) = Rs. 30000Share of B = (20/59)(118000) = Rs. 40000Share of C = (24/59)(118000) = Rs. 48000

Example: The incomes of two persons A and B are in the ratio 3:4. If each saves Rs. 100 per month, the ratio of their expenditures is Rs. 1:2. Find their incomes.

Solution: Let the incomes of A and B be 3P and 4P.If each saves Rs. 100 per month, then their expenditures = Income - Saving = (3P-100) and (4P-100).The ratio of their expenditures is given as 1 : 2. Therefore, (3P-100) : (4P-100) = 1 : 2 Solving, we get P = 50. Substitute this value of P in 3P and 4P. Thus, their incomes are: Rs. 150 and Rs. 200.

Example: Find the fraction which has the same ratio to 2/13 that 5/34 has to 7/48.

Solution: Let the required fraction be P.(P) : (2/13) = (5/34) : (7/48)As the product of the means is equal to the product of the extremes,(P)(7/48) = (2/13)(5/34). Solving, we get P = 240/1547.

Example: A bag contains certain number of 50 paise coins, 20 paise coins and 10 paise coins in the ratio 2 : 3 : 4. If the total value of all the coins in the bag is Rs. 400, find the number of coins of each kind.

Solution:

 

20x = 400 X 10 x = 200

∴ Number of coins of 50 paise = 2x = 2 X 200 = 400 Number of coins of 20 paise = 3x = 3 X 200 = 600Number of coins of 10 paise = 4x = 4 X 200 = 800

 Example: 125 liters of a mixture of milk and water contains milk and water in the ratio 3 : 2. How much water should now be added so that the ratio of milk and water becomes 3: 4?

Solution: Amount of milk present in the mixture = (3/5)(125) = 75 liters. Therefore, water = 50 liters.75 : (50 + P) = 3 : 4, where P is the number of liters of water added so that the ratio of milk to water becomes 3 : 4.150 + 3P = 300; P = 50.Hence, 50 liters of water are to be added for the ratio to become 3 : 4.

Example: The food in a Camp lasts for 30 men for 40 days. If ten more men join, how many days will the food last?

Solution: 30 men can consume the food in 40 days.So, one man can consume the food in (30)(40) days = 1200 days.If 10 more men join, the total number of men = 40.The number of days the food will last = (1200)/40 = 30 days.

 

 Example:If 12 men can reap 120 acres of land in 36 days, how many acres of land can 54 men reap in 54 days?

Solution:

12 men 120 acres 36 days  54 men        ? 54 days  

M1 D1W2 = M2 D2 W1

12 X 36 X W2 = 54 X 54 X 120

Example: 10 camels cost as much as 24 horses, 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants. If the cost of 10 elephants is Rs. 170000, find the cost of a camel.

Solution: Let the cost of one camel = Rs. P

10 camels = 24 horses16 horses = 4 oxen6 oxen = 4 elephants10 elephants = Rs. 170000

Note: If a series of quantities are connected with one another like in the above example, we can apply the above method of calculation, which is called the Chain Rule.

Example:A contractor undertakes to complete the construction of a tunnel 720 meters long in 240 days and employs 60 men for the purpose. After 120 days, he finds that only 240 meters of the tunnel is made. How many more men should be employ in order to complete the work in time?

Solution:In 120 days, only 240 meters of the tunnel is constructed by 60 men.Therefore, in the remaining 120 days, 480 meters of the tunnel can be constructed by 120 men.Hence, additional number of men required = 120 - 60 = 60 men.

AVERAGES

 

 

Example:Find the average of the first 20 natural numbers.

Solution: Average of the first 'n' natural numbers = (n+1)/2Substituting n = 20, we get (20+1)/2 = 10.5

Example: The average of 13 numbers is 60. Average of the first 7 of them is 57 and that of the last 7 is 61. Find the 7th number.

 

Solution: Sum of all the 13 numbers = (13)(60) = 780.Sum of the first 7 of them = 7(57) = 399Sum of the last 7 of them = 7(61) = 427∴ 7 th number = sum of first 7 + sum of last 7 - sum of 13 numbers                     = 399 + 427 - 780 = 46

Example:A batsman makes a score of 64 runs in the 16th inning and thus increased his average by 3. Find his average after the 16th inning.

Solution: Let the average after the 16th inning be P.So, the average after the 15th inning will be (P-3).Hence, we can write: 15(P-3) + 64 = 16P.Solving, we get P = 19, which is the average after the 16th inning.

 Example: The average marks of a class of 30 students is 40 and that of another class of 50 students is 60. Find the average marks for all the students.

Solution: Sum of the marks for the class of 30 students = (30)(40) = 1200Sum of the marks for the class of 50 students = (50)(60) = 3000Sum of the marks for all the 80 students = 1200 + 3000 = 4200.Average marks of all the students = 4200 / 80 = 52.5

Example: The average salary of a person for the months of January, February, March, and April is Rs. 8000 and that of the months February, March, April, and May is Rs. 8500. If his salary for the month of May is Rs. 6500, find his salary for the month of January.

Solution: Sum of the salaries of the person for the months of January, February, March, and April = 4(8000) = 32000...........(I)Sum of the salaries of the person for the months of February, March, April and May = 4(8500) = 34000...........(II)

Subtracting II from I, we get: Jan - May = -2000As we know that his salary for May is Rs. 6500, the salary for January = Rs. (- 2000 + 6500) = Rs. 450

Example: The average age of a family of 8 members is 24 years. If the age of the youngest member is 2 years, find the average age of the family at the birth of the youngest member.

Solution:

Sum of the ages of the family = 8(24) = 192.At the birth of the youngest member (age 2 years), the age of each of the other 7 members of the family was 2 years less. So, the sum of the ages for the remaining 7 members of the family 2 years ago= 192 – 8(2) = 192 – 16 = 176. So, the required average = 176/7 = 25.14

Example:The average age of a group of 10 persons was increased by 3 years when one person, whose age was 42 years was replaced by a new person. Find the age of the new person.

Solution: Age of new person = Age of removed person + total increase = 42 years + 10 (3) = 72 years

Example: The average of 10 numbers is 23. If each number is increased by 4, what will be new average?

Solution: Sum of the 10 numbers = 230.If each number is increased by 4, the total increase = 4(10) = 40.The new sum = 230 + 40 = 270.Thus, the new average = 270/10 = 27.

Note: If the average of some numbers is P and if each number is increased by 'n', then the new average will be (P + n)

Example:The average of 35 numbers is 25. If each number is multiplied by 5, find the new average.

Solution: Sum of the 35 numbers = 35(25) = 875.If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5. Thus, the new average is 25(5) = 125.

 

Example: The average of the marks of 12 students in a class is 36. If the marks of each student are doubled, find the new average.

Solution: The sum of the marks for the 12 students = (12)(36) = 432.If the marks of each student are doubled, the sum also will be doubled. Thus, the new sum will be 432(2) = 864.So, the new average = 864/12 = 72.

Or, the same question can also be done by directly doubling the average = 2(36) = 72.NUMBERS & AGES

a. Any single digit number can be represented by using a variable X, Y or Z.b. Any three consecutive integers can be represented as (X-1), X & X+1c. Any five consecutive integers can be represented as (X-2), (X-1), X, (X+1), (X+2)d. Any 2-digit number can be represented as (10X+Y)e. Any 3-digit number can be represented as (100X+10Y+Z)f. Any even number can be represented as 2Xg. Consecutive even numbers can be represented as (2X), (2X+2), (2X+4), (2X+6),......h. Any odd number can be represented as (2X+1) or (2X-1)i. Consecutive odd numbers can be represented as (2X+1), (2X+3), (2X+5),.......or (2X-1), (2X-3), (2X-

5),.........Example: The sum of three consecutive integers is 102. Find the least of them.

Solution:Any three consecutive numbers can be taken as (P-1), P and (P+1).So, (P-1)+P+(P+1) = 102.3P=102; P = 34.So, the least of them is: (P-1) = 34-1 = 33.

Example: The sum of five consecutive integers is 115. Find the largest of them.

Solution: Any five consecutive integers can be taken as (P-2), (P-1), P, (P+1), (P+2).So, (P-2),(P-1),P,(P+1) & (P+2) = 115.5P = 115; P = 23. Thus, the largest of them is: (P+2) = 25.

Example: A 2-digit number is such that the sum of the digits is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number.

Solution: Any two digit number can be written as (10P+Q), where P is the digit in the tens place and Q is the digit in the units place.

  From the given data, we can write: P+Q = 10..........(I)

(10Q+P)-(10P+Q) = 549(Q - P) = 54;(Q - P) = 6......(II)

From II, and I we can solve for P and Q. P = 2 and Q = 8.Thus, the required number is = 10P + Q = 10 (2) + 8 = 28.

Example:

The sum of three consecutive even numbers is 42. Find the middle of the three numbers.

Solution:Any three consecutive even numbers can be written as (2P-2), 2P and (2P+2).

So, (2P-2)+2P+(2P+2) = 42.6P = 42; P = 7. Thus, the middle number is: 2P = 14.

Example: A fraction is such that the numerator is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes ¾. Find the original fraction.

Solution:Let the denominator be P, then the numerator will be (P-8).

    So, the original fraction = .    

According to the given condition

4P - 20 = 3P - 9∴ P = 11hence the original fraction =

= 3/11

 

  Example: The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence.

Solution: A + B = 60; A = 2B.Solving, we get A = 40, B = 20.

∴ The sum of their ages 5 years hence = (40 + 5) + (20 + 5) = 80 years

Example: The ratio of the present ages of P and Q is 3:4. 5 years ago, the ratio of their ages was 5:7. Find their present ages.

Solution:As the ratio of their present ages is 3:4, let their present ages be 3X and 4X.

So, 5 years ago, as the ratio of their ages was 5:7, we can write

solving we get X = 10 The present ages are 3X = 30 and 4X = 40

Example: A man said to his son, "I was two-thirds of your present age when you were born." If the present age of the man is 48 years, find the present age of the son.

Solution: Let the present age of the son be P, which means he was born P years ago. At the time the son was born, the age of the man was: (48-P).So, according to the statement made by the man, his age when the son was born should be equal to 2/3 of P.

Therefore, (48-P) = (2/3)P. Solving, we get P = 28.8 years; which is the present age of the son.

LCM & HCF

LCM stands for Least Common Multiple and HCF stands for Highest Common Factor. HCF is also known as Greatest Common Divisor (GCD) or Greatest Common Measure (GCM).

Least Common Multiple (L.C.M)

Let us consider the two numbers, 2 and 3.

The multiples of 2 are: 2,4,6,8,10,12,14,16,18,20,.......The multiples of 3 are: 3,6,9,12,15,18,21,..........

Now, if we compare the multiples of 2 and 3, we find that the common multiples for them are: 6, 12, 18, etc. Among these common multiples, 6 is the least number. Thus, 6 is called the Least Common Multiple of 2 and 3.

Thus, LCM of two numbers is the least number, which is exactly divisible by both of them.

Method of finding LCM of two given numbers:

LCM can be calculated in two methods: Factorization Method and Division Method.

Let us consider the numbers 8 and 12.

Factorization Method of finding LCM:

The multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64 72,.........The multiples of 12 are: 12, 24, 36, 48, 60, 72,...........

Thus, the common Multiples are: 24, 48, 72, etc, of which 24 is the least and hence is the LCM of 8 and 12.

Division method of finding LCM:

In the division method, remember that we should

a. always divide with a prime numberb. continue dividing until only prime numbers remainc. continue dividing until there is no repetition of any number 

 

LCM will be equal to the product all the divisors and the final remaining numbers.

Let us now use this method for 8 and 12:

L.C.M of 8 & 12 = 2 X 2 X 2 X 3 = 24. 

Highest Common Factor (HCF)

Let us consider the numbers 6 and 8

Factors for 6 are: 1, 2, 3 and 6Factors for 8 are: 1, 2, 4, and 8

The common factors for 6 and 8 are: 1 and 2 and 2 is the highest between them. Thus, 2 is called the Highest Common Factor of 6 and 8.Thus, HCF of two given numbers is the highest number that exactly divides the given two numbers.

 Methods of finding HCF of two given numbers:

HCF of two given numbers can be calculated by using either the Division Method or the Factorization Method.Let us consider the numbers 8 and 12 again.

Factorization Method of finding HCF:

The factors of 8 are: 1, 2, 4 and 8.The factors of 12 are: 1, 2, 3, 4, 6, and 12.

The common Factors are: 1, 2, and 4. As 4 is the highest among them it is the HCF of 8 and 12.

Division method of finding HCF:

Divide the greater number with the smaller number and continue dividing the divisor with the remainder obtained until 0 is obtained. The corresponding divisor when 0 is obtained will be the HCF of the two numbers.

Thus, HCF of 8 and 12 is 4.

Example:Find the least number, which is exactly divisible by 2, 3 and 4.

Solution: We need to calculate the LCM of 2, 3 and 4, which is 12.

Note: Remember that whenever we find the term 'divisible by' or 'divided by' in a question, its an indication that we need to calculate LCM. Similarly, if we find the term 'divides' or 'while dividing', we need to calculate the HCF.

Example: Find the greatest 4-digit number, which is exactly divisible by 3, 4 and 5.

Solution: We know that the greatest 4-digit number is 9999.LCM of 3, 4 and 5 is 60.Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 - 39 = 9960.

 

 Example:Find the least 4 digit number which when divided by 3, 4 or 5 leaves a remainder 2 in each case.

Solution: We know that the least 4-digit number is 1000. LCM of 3, 4 and 5 is 60.Dividing 1000 by 60, we get the remainder 40. Thus, the least 4-digit number which is exactly divisible by 3, 4 & 5 is 1000 + (60 - 40) = 1020. Now, add the remainder 2 that's required. Thus, the answer is 1022.

Example: Find the greatest number, which exactly divides 35, 91 and 840.

Solution:The greatest number which exactly divides 35, 91 and 840 is the HCF of the three numbers. So, calculating HCF we get the answer as 7.

Note: Observe that the term 'divides' is used in the question and hence we have to calculate the HCF.

Example:Find the greatest number, which while dividing 19, 83 or 67 gives a remainder 3 in each case.

 

Solution: First, subtract the remainder 3 from each of the given numbers: (19 - 3) = 16; (67 - 3) = 64; & (83 - 3) = 80. Now, find the HCF of the results 16, 64 and 80. We get '16'.Thus, '16' is the greatest number, which while dividing 19, 67 or 83 gives a remainder 3 in each case.

Example: Find the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each case.

Solution:If the remainder is given, then we could have subtracted it from each value and then calculated HCF like in the previous example. But, now, we only know that the remainder is same in each case, but we do not know its value. So, first let us calculate the differences, taking two numbers at a time as follows:(215 - 47) = 168; (365 - 215) = 150; (365 - 47) = 318; Now, calculate the HCF of the results: 168, 150 and 318.We get '6', which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder '5' in each case.

 

 

Example: A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the room.

Solution:Let us take both the length and width of the room in centimeters.Length = 6 meters and 24 centimeters = 624 cmWidth = 4 meters and 32 centimeters = 432 cm.As we want the least number of square tiles required, it means the length of each square tile should be as maximum as possible. Further, the length of each square tile should be a factor of both the length and width of the room.Hence, the length of each square tile will be equal to the HCF of the length and width of the room= HCF of 624 and 432 = 24.

 

 

  Example: A drink vendor has 80 liters of Maaza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required?

Solution: If we want to pack the drinks in the least number of cans possible, then each Can should contain the maximum numbers of liters possible. As each can contains the same number liters of a drink, the number of liters in each Can is a common factor for 80, 144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.So, the number of liters in each can= HCF of 80, 144 and 368 = 16 liters.Now, number of cans of Maaza = 80/16 = 5Number of cans of Pepsi = 144/16 = 9Number of cans of Sprite = 368/16 = 23.Thus, the total number of cans required = 5+9+23 = 37 cans.

NUMBER SYSTEM

This is a very important topic in Arithmetic as it deals with all the basic properties of numbers. Questions from this topic can be of very high difficulty level, and unless you are thorough with fundamental concepts and properties of numbers, it will be difficult to answer them.

Let us first discuss the basic properties of numbers.

a. Natural Numbers: 1, 2, 3, 4, 5,.......∞, the set of natural numbers is denoted by 'N'.b. Whole Numbers: 0, 1, 2, 3, 4, 5,.......∞, the set of whole numbers is denoted by 'W'.

Note: Natural numbers including Zero form the set of whole numbers.

c. Integers: All positive and negative numbers including zero are together called integers; the set of integers is denoted by I.

I = {.......,-3, -2, -1, 0, 1, 2, 3,.........}

Note: Decimals are not a part of integers and only complete numbers are included.

 

d. Rational Numbers: Any number in the form 'p/q', where 'p' and 'q' are both integers and 'q' is not equal to Zero is called a Rational number. the set of rational numbers is represented by Q.

Ex: 1/3, 5/8. 23/6, etc.

Note: All integers are rational numbers. Zero also can be considered to be a rational number because zero can be written as (0/2 or 0/100).

Note: Decimals are of four types as follows:

i. Recurring-terminating decimals: Ex: 0.333, 0.125, etcii.Recurring-non-terminating decimals. Ex: 0.3333333......∞ , 1.37373737........∞ iii.Non-recurring terminating decimals. Ex: 2.367, 0.45879, etc.iv.Non-recurring non-terminating decimals. Ex: 0.23987643..........∞

Note: All decimals except non-recurring non-terminating decimals can be converted into the form 'p/q' and hence are rational numbers.

Conversion of Decimals into fractions:

i. Terminating decimals: Example:0.333. This can be written as 333/10001.9054. This can be written as 19054/10000

ii. Non-terminating decimals: As already mentioned earlier, only recurring-non-terminating decimals can be converted into rational numbers, non-recurring non-terminating decimals cannot be converted into rational numbers.

Example: Convert '0.3333...........' into a rational number.

Solution:Let 0.3333.......... be equal to X. So, X = 0.3333..............- Equation I

Multiply both sides with '10' because only one digit '3' is repeated.  We get 10X = 3.3333..............- Equation IISubtract the Equation (I) from Equation (II).

10 x = 3.333............∞ x = 0.333............∞__________________________

9 x = 3∴ x = 3/9 = 1/3 _Thus we have converted decimal 0.333..........∞ = 0.3

Note: If two digits are repeated in the decimal, we need to multiply with 100 and if three digits are repeated, we need to multiply with 1000.

Example: Convert '3.71717171.........' into a rational number.

Solution: Let 3.71717171...........be equal to XSo, X = 3.71717171............Equation (I)

Now Multiply with 100 since two digits (both 7 and 1) are being repeated.

We get 100X = 371.717171............Equation (II)Now let us subtract Equation (I) from Equation (II).

100X = 371.7171.............∞ X = 3.717171.............∞ ___________________________

99X = 368

X = 368/99

Thus, the given decimal is converted into a rational number.

Example: Which of the following is not a rational number?

a) 2.367 b) 0.98989898   c) 0.34670123........ d) 3.142857142857142857  e) None of these

Solution: The decimal given in option 'd' is a non-recurring non-terminating decimal and hence is not a rational number.Thus, the answer is (d).

Now, let us continue with our discussion of the properties of numbers.

e. Irrational numbers: Numbers which are nsot rational are said to be irrational numbers.

Example: √2, √3, ∏, etc.

Note: All non-recurring non-terminating decimals are irrational numbers.

Note: We know that √2 = 1.414; and 1.414 is a rational number, since it can be written as 1414/1000. So, we might get a doubt as to whether √2 is also a rational number.

 The actual value of √2 is not 1.414, it is only an approximate value. We can say that is only an approximate value because, if we square 1.414, we don't get back 2. And, if we observe the actual decimal value of √2, we find that it is a non-recurring non-terminating decimal, and hence, is not a rational number, but is an irrational number.

Similarly, √3 is also an irrational number.

Note: We know that ∏ = 3.14; which can be written as 314/100, which is a rational number. So, we might get a doubt whether ∏ is a rational number. But we need to understand that 3.14 is only the

approximate value of ∏, and is not the exact value.Even if we take ∏ = 22/7, let us divide 22 with 7 and see what decimal we get.

We observe that the digits '142857' are repeated again and again as we go on dividing 22 with 7. So, the decimal we get is a recurring-non-terminating decimal, and is hence a rational number.

But again, 22/7 is only an approximate value of ∏ and is not the exact value; if we calculate the actual value of ∏, we find that it is actually a non-recurring non-terminating decimal, and hence is not a rational number. Thus, ∏ is also an irrational number.

f. Real Numbers: All the numbers we have discussed till now, together are called as Real numbers; the set of real numbers is denoted by R

g. Imaginary numbers: Any number, which is in the form of i2 = -1 means i = √-1 is called an imaginary number

Note: The square root of any negative number is an imaginary number. Imaginary numbers may also be classified as Complex numbers.

h. Even Numbers: Any number, which is exactly divisible by 2 is said to be an even number. Any number, it consists in its unit's place any one of the digit from 0, 2, 4, 6, or 8 is called an even number

Example:2, 4, 6, 8, 10, 24, 268, etc

Note: Zero is also divisible by 2, so we might get a doubt whether it is also even number. Please remember that we use the two terms even and odd with respect to natural numbers only.

Note: The smallest even number is 2.

i. Odd Numbers: Any number, which is not exactly divisible by 2 is an odd number. Any number, it consists in its unit's place any one of the digit from 1, 3, 5, 7, or 9 is called an odd number

Example:5, 13, 17, 29, 81, etc.

Note: The smallest odd number is 1.

j. Prime Numbers: A number, which does not have any other factor except 1 and itself is said to be a prime number.

Example: 2, 3, 5, 7 13, 71, etc

Note: 2 is the smallest prime number.Note: 2 is the only even number which is also a prime number. No other even number is a prime number.

k. Composite Numbers: A number, which has at least one factor other than 1 and itself is said to be a composite number.

Example: 4, 6, 8, 9, 10, 15, 33, 140, etc.

Note: 1 is neither a prime number nor a composite number.

l. Relatively Prime Numbers: Two numbers, which do not have any other common factor except 1 are said to be relatively prime.

Example: 8 and 9; 16 and 25

Let us take 8 and 9.

Factors for 8 are: 1, 2, 4, and 8Factors for 9 are: 1, 3, and 9

So, the common factor for 8 and 9 is '1' only, there is no other common factor for them. So, 8 and 9 are said to be relatively prime numbers.

Similarly we can conclude for 16 and 25 also.

Note: 5, 6 and 7 are three relatively prime numbers.

Note: Relatively prime numbers does not mean that the numbers are prime numbers, the numbers can be composite numbers also, as we have seen in the case of 8 and 9 and also 16 and 25.But if we compare both of them their common factor should be 1 only; then they can be called as

relatively prime numbers.

Face Value and Place Value of a digit:

Let us consider the number 54321. This is a 5-digit number.

We read it is fifty four thousand three hundred and twenty one.

This is because the digit '5' is in 'ten-thousands' place, 4 is in thousands place, 3 is in hundreds place, 2 is in tens place and 1 is in ones place.

Face Value is also called as Absolute Value of a digit, The unchanged value of a digit is known as its absolute value.If the value of the digit is not absolute, it is called Local value of the digit. Ex: In the number '37489', the absolute value of '4' is '4'; the absolute value of '8' is '8'.The local value of '4' is '400'; the local value of '8' is '80'.

Note: In any number, the value of unit's digit is equal to its absolute value.

Example: Find the difference between the Place value and face value of the digit '5' in the number '45678.

Solution:Place value of '5' in 45678 is 5000; the face value is 5.Therefore, the required value = 5000 - 5 = 4995.

Note:'0' is called insignificant digit, where as the other digits from 1 to 9 are called significant digits.

The value of the digit itself is called its Face value and the value obtained by multiplying the face value of a digit with the value of the place in which it is present is called its Place value.

So, in the number 54321, the face value of 4 is 4 and that of 5 is 5.

The place value of 5 is: 5(10000) = 50000The place value of 4 is: 4(1000) = 4000The place value of 3 is: 3(100) = 300The place value of 2 is: 2(10) = 20The place value of 1 is: 1(1) = 1

By adding place values of all the digits, we get the total value of the number as 54321.

Note:i.  Single-digit numbers are: 1 to 9: 9 numbersii.  2-digit numbers are: 10 to 99: 90 numbersiii. 3-digit numbers are: 100 to 999: 900 numbersiv. 4-digit numbers are: 1000 to 9999: 9000 numbersv. 5-digit numbers are: 10000 to 99999: 90000 numbers

Based on the above concept, questions can be asked in different ways. Let us see some examples:

Example: How many times is the digit 8 used in writing all numbers from 600 to 900?

Solution: All these numbers from 600 to 900 are 3-digit numbers. So, each number will have three places: Unit's place, Ten's place and Hundred's place. 

First let us consider numbers from 600 to 699. 

All these numbers are starting with the digit '6' and so, the digit '8' is not used at all in the hundreds place.  When we consider numbers: 680, 681, 682, 683, 684, 685, 686, 687, 688, and 689, in all these numbers, the digit '8' is used in the tens' place TEN times.

When we consider the numbers: 608, 618, 628, 638, 648, 658, 668, 678, 688 and 698, in all these numbers, the digit '8' is used in the unit's place TEN times.

Thus, when we write all numbers from 600 to 699, the digit '8' is totally used 20 times (0 times in the hundred's place, 10 times in the ten's place and 10 times in the unit's place)

Similarly, we can calculate the number of times '8' is used in writing numbers from 700 to 799 and then 800 to 899.

This data can be represented as follows:

600 - 699 700 - 799 800 - 899

H T U H T U H T U0 10 10 0 10 10 100 10 10 20 times 20 times 120 times

Thus, the total number of times the digit '8' is used in writing numbers from 600 to 900 is: (20 + 20 + 120) = 160 times

Example:How many numbers from 600 to 900 either begin with or end with the digit '8'?

Solution:We know have to calculate the number of numbers either beginning with 8 or ending with 8. So, we need not consider the ten's place, we have to consider only the hundred's place and the unit's place.

So, applying the same logic as we used in the previous question, the required data can be represented as follows:

600-699           700-799              800-899 

H     T     U         H     T     U             H     T     U

 

0         0         10               0         0         10                 100         0         10

10 numbers         10 numbers           110 numbers

But, the TEN numbers shown in the last, i.e., 808, 818, 828, 838,..........., 898 are already included in the 100 numbers we showed in the hundred's place. Hence, they need not be counted again.

Thus, the number of numbers from 600 to 900 which either begin with or end with the digit '8' are: 10 + 10 + 100 = 120 numbers, and not 130 numbers.

 

 

Example: How many times should the keyboard of a computer be pressed to type all numbers from 1 to 500?

Solution:We need to type all numbers from 1 to 500.

If we want to type the number '8' we need to press the key having the digit '8'on it, which means, we have to press the keyboard once. Similarly, if we want to type the number '53', we need to first press the key with the digit '5' on it and then we have to press the key with the digit '3' on it, which means, we need to press the keyboard twice.

Thus, for every single digit number, we need to press the keyboard once, for every 2-digit number; we need to press twice, for every 3-digit number, thrice, and so on. This data can be represented as follows:

Single digit numbers: 1 to 9: 9 numbers: 9(1) = 9 times 2-digit numbers: 10 to 99: 90 numbers: 90(2) = 180 times 3-digit numbers: 100 to 500: 401 numbers: 401(3) = 1203 times TOTAL:   1392 times 

Thus, we have to press the keyboard of a computer 1392 times to type all numbers from 1 to 500.Note: If we want to calculate the number of numbers from 10 to 99, it is equal to (99-10) + 1 = 89 + 1 = 90 numbers.We should add '1' because the starting number 10 is also included.Similarly, the number of numbers from 100 to 900 = (900-100) + 1 = 800 + 1 = 801 numbers.

Example:In a canteen, tokens are issued in terms of currency for different food items. Each token has only one digit (any digit from 0 to 9) written on it. If you were asked to form all numbers from 400 to 1200, how many tokens would you require?

Solution:To form a single-digit number, only one token is required, to form a 2-digit number, two tokens will be required and so on.

So, the total number of tokens required to form all numbers from 400 to 1200 can be calculated as follows:

3-digit numbers: 400 to 999: 600 numbers: 600(3) = 1800 times4-digit numbers: 1000 to 1200: 201 numbers: 201(4) = 804 timesTOTAL: 2604 times  Some formulae based on natural numbers:

 

Example:In a pond, the number of leaves of a lotus plant keep increasing by one every day. i.e., 1 leaf on the first day, 2 leaves on the second day, 3 leaves on the fourth day, and so on. Find the total number of leaves by the end of 250 days.

Solution: The total number of leaves = 1+2+3+.........+250 = n(n+1)/2 = 250(251)/2 = 125(251) = 31375.Example:Find the sum of all the odd numbers starting from 1 and ending up to the greatest 4-digit number.

Solution: Greatest 4-digit number is 9999. We need to find the sum of all odd numbers from 1 to 9999, which are totally 9999 in number. As we know, 9999 is an odd number and hence there are 1/2 (9999+1) = 5000 odd numbers. ∴The sum of these 5000 odd numbers = (5000)2 = 25000000.

Note: Let us consider the first 'n' natural numbers. If 'n' is even, then there will be (n/2) odd numbers and n/2 even

 

numbers.If 'n' is odd, then there will be (n+1)/2 odd numbers and (n-1)/2 even numbers.

Example: Find the sum of the first 60 multiples of 48.

Solution: The required sum = 48(1)+48(2)+48(3)+48(4).......+48(60)                        = 48(1+2+3+.......+60)                         = 48(60 x 61/2)= 48 x 30 x 61 = 87840

Note: Sum of the first 'P' multiples of 'Q' = Q(sum of the first 'P' natural numbers) = Q[P(P+1)/2] = QP(P+1)/2 Note: The sum of two consecutive numbers is always an odd number and the difference of two consecutive numbers is always 1. Hence, the difference between the squares of two consecutive numbers is always an odd number. Thus, if P and Q are two consecutive numbers, then (P2 - Q2) = (P+Q)(P-Q) = (P+Q)(1) = (P+Q).

Note: The difference of the squares of two consecutive numbers is always equal to the sum of the two consecutive numbers.

Note: In general, the number of 'n' digit numbers is 9(10)(n-1).

Relation between Dividend, Divisor, Quotient and Remainder:

Let us divide the number 35 with 4:

4) 35 (8 32 3

From the above division, we can write: 35 = (4)(8) + 3

Here, 35 is called the Dividend, 4 is called the Divisor, 8 is the Quotient and 3 is the Remainder.

Hence, Dividend = (Divisor)(Quotient) + RemainderNote: Dividend = Divisor (Quotient) + Remainder Divisor = (Dividend - Remainder) / Quotient Quotient = (Dividend - Remainder) / Divisor.

Note: When the second divisor is a factor of the first divisor, the second remainder is obtained by dividing the first remainder by the second divisor.

Note: The remainder obtained by dividing a given number by the method of successive division is called Complete Remainder.Example: Find the number nearest to 16438, which is exactly divisible by 4 as well as 7.

Solution: Any number, which is exactly divisible by both 4 and 7, is also divisible by 28. ∴Divide 16438 by 28.

The remainder is 2, which is less than half the divisor. ∴ the number nearest to 16438 which is exactly divisible by both 4 and 7 is (16438 - 2) = 16436.

Note: To find the nearest number to P, which is exactly divisible by Q, find the remainder obtained by dividing P with Q; let the remainder be 'R'.

Now, if R is less than half the divisor Q, then the nearest number is P - R and if R is more than half the divisor Q, then the nearest number is P + (Q-R). Observe that if R is equal to half the divisor Q, then the question cannot be asked. Note: If two numbers, when divided by a third number give the same remainders, then the difference between the two numbers is also divisible by the third number.

Example:A number when divided by 36 gives a remainder 17. Find the remainder when the same number is divided by 9.

Solution:Let the number be N.Thus, we can write N = (36)(Q) + 17, where Q is the quotient obtained when the number N is divided by 36.

Now, if we divide the same number N with 9, it means we are dividing the value (36Q + 17) with 9. As 36 is a multiple of 9, we get a remainder '0' when 36Q is divided by 9 and hence, 17 divided by 9 gives us the remainder '8', which is the required answer.Example:A number when divided by 5 gives a remainder 3. Find the remainder when the square of the number is divided by 5.

Solution: Let the number be N.

Thus, we can write it as: N = 5Q + 3, where Q is the quotient obtained when the number N is divided by 5.

So, the square of the number = (N)? = (5Q + 3)? = 25Q? + 30Q + 9.

If we divide this value by 5, the remainder would be the same as remainder obtained by dividing 9 with 5, as the first two terms in the expression are multiples of 5 and hence give remainder '0' when divided with 5.

So, 9, when divided with 5, gives the remainder 4, which is the required answer.Example:What least number should be subtracted from 1236 to get a number exactly divisible by 7:

Solution:

Let us first divide 1236 with 7 and find the remainder.

7) 1236 (176 7 53 49 46 42 4

As the remainder is 4, this is the least number that should be subtracted from 1236 to get a number exactly divisible by 7.

Observe that 1236-4 = 1232 is exactly divisible by 7.Example:Find the least number that should be added to 1236 to get a number exactly divisible by 7.

Solution:Let us divide 1236 with 7 and find the remainder.

        7) 1236 (176             7             53             49               46               42                  4

Thus, we need to add divisor-remainder = 7 - 4 = 3, to 1236 to get a number exactly divisible by 7.

Observe that 1236 + 3 = 1239 is exactly divisible by 7.

Note: If you want to find out the least number that should be added, add (divisor - Remainder) and if you want to find the least number that should be subtracted, subtract the Remainder.

 

 Example:Find the greatest 5-digit number that is exactly divisible by 40.

Solution:We know that the greatest 5-digit number is 99999.Let us first divide this by 40 and find the remainder.

40) 99999 (2499 80 199 160 399 360 399

360 39

As the remainder is 39, subtract this from 99999. We get 99960, which is the greatest 5-digit number exactly divisible by 40, and hence is the required answer.Example:Find the smallest 4-digit number which is exactly divisible by 30.

Solution:We know that the smallest 4-digit number is 1000. Let us divide this with 30 and find the remainder.

30) 1000 (33 90 100 90 10

Since the remainder is 10, if we subtract this from 1000, we get 990, which becomes a 3-digit number. But, we require a 4-digit number. So, let us add (divisor-remainder) to 1000. We get 1000 + (30 - 10) = 1000 + 20 = 1020. This is the smallest 4-digit number, which is exactly divisible by 30.Successive Division:

Consider the following sequence of divisions:

5) 144 (28 3) 28 (9 7) 9 (1 140 27 7 4 1 2

First, we divided 144 with 5, getting quotient as 28 and remainder 4. Next we divided 28 (the quotient in the first case) with 3 and got the remainder 1. Then we divided 9 (the quotient) in the second case with 7 and got the remainder 2.

This method of division, where the quotient in each case is taken as the dividend in the next case is called successive division or continuous division.

Based on this concept, the question can be asked as follows:Example: When a number is successively / continuously divided by 5, 3 and 7, it gives remainders 4, 1 and 2 respectively. Find the remainder when the same number is divided by 105.

Solution: The divisors are given as 5, 3 and 7; the remainders are given as 4, 1 and 2 respectively.

Using the concept of successive division, let us assume that Q1, Q2 and Q3 are quotients in each case respectively.

The values of Q1, Q2 and Q3 are not given in the question, but as it is successive division, we can conclude that the Quotient in each case becomes the dividend in the next case.

Thus, we can say that Q1 = 28, Q2 = 9. But, we do not know the value of Q3 from the question; so, we cannot calculate the original number.

But, observe that the three divisors 5,3 and 7 are Relatively Prime and the final divisor 105 is their LCM. (Least Common Multiple).In this case, to find the required remainder, we can directly use the formula: (d1 d2 r3 + d1 r2 + r1), where d1, d2 and d3 are the three divisors and r1, r2 and  r3 are the three remainders respectively.

From the given data, d1 = 5, d2 = 3, d3 = 7 and r1 = 4, r2 = 1 and  r3 = 2.

Thus, the required answer = (d1 d2 r3 + d1 r2 + r1)

  = (5)(3)(2) + (5)(1) + 4 

  = 30 + 5 + 4 

  = 39

Note: If there are only two divisors d1, d2 and two remainders r1 and r2, then we need to use the formula: (d1 r2 + r1).

Example: A number when successively divided by 5 and 2 gives remainders 3 and 1 respectively. Find the remainder when the same number is divided by 10.

Solution: Here d1 = 5, d2 = 2; r1 = 3 and r2 = 1Apply the formula discussed above.The required remainder = (d1 r2 + r1) = (5)(1) + 3 = 8.Note: If two numbers, when divided by a third number give remainders r1 and r2 and if the sum of the two numbers when divided by the same divisor (third number) gives remainder r3, then the divisor can be calculated using the formula: (r1 + r2 - r3)

Example: wo numbers, when divided by a third number give remainders 6 and 5 respectively. When the sum of the two numbers is divided by the same divisor, the remainder is 3. Find the divisor.

Solution:Here r1 = 6, r2 = 5 and r3 = 3.

Applying the formula mentioned above, the divisor = (r1 + r2- r3)

= (6 + 5 -3) = 8.Factor and Multiple of a Number:

If we consider the numbers 2 and 8, 2 exactly divides 8, or we can say that 8 is exactly divisible by 2. Then, 2 is called the factor of 8 and 8 is called the multiple of 2.

The multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, 16, etcThe factors of 8 are: 1, 2, 4, 8.

So, out of two numbers P and Q, if P is exactly divisible by Q, then P is said to be a Multiple of Q and Q is said to be the Factor of P.

Factors of 30 are: 1, 2, 3, 5, 6, 10, 15, and 30.Multiples of 30 are: 30, 60, 90, 120, 150, 180, 210, etc.Perfect Square:

The square of a natural number is called to be a perfect square.

For example, let us consider the number 144, which can be written as (12)?. Thus, 144 is said to be a perfect square. Similarly, 1296 is also a perfect square as it can be written as (36)?.

Note: Numbers like 72, 140, 250 are not perfect squares, as they cannot be written as squares of any other numbers.

Let us now find the factors of 144. They are: 1, 2, 3, 4, ..etc.

144 can be written as: (9)(16) = (3)2 (2)4 (3 square and 2 to the power of 4)

Observe that both the base numbers (2 and 3) are prime numbers and the powers (2 and 4) are even numbers. As both the powers are even numbers, 144 is a perfect square.Note: If a number can be written in the form [(P)m][(Q)n], where 'P' and 'Q' are prime numbers, then if 'm' and 'n' are both even numbers, then the number is a perfect square.

900 can be written as [(2)2][(3)2][(5)2]. As all the powers are even numbers, 900 is a perfect square.

Note: A number, which is a perfect square, cannot end with 2, 3, 7 or 8.

Example: Find the least number with which 180 should be multiplied to get a perfect square.

Solution:Let us write the given number 180 in the form [(P)m][(Q)n], where P and Q are prime numbers.

180 = [(2)2][(3)2][(5)1]

The powers of 2 and 3 are both even but the power of 5 is 1, which is not an even number, not odd. But, for a perfect square, all the powers should be even.Hence, the power of 5 also should be made even. So, if we multiply with another 5, the power of 5 will become (5)2; 2 is even.

Hence, 180 should be multiplied with 5 to get a perfect square.Observe that (180)(5) = 900, a perfect square.  Example:

Find the least number, which should be added to 43481 to get a perfect square.

Solution:We know that 43481 is not a perfect square. 

(208)? = 43264 and (209)? = 43681. 

Thus, by trial and error, we can find that 43264 is a perfect square which is just less than 43481 and 43681 is the perfect square greater than 43481. So, the least number that should be added to 43481 to get a perfect square is (43681 - 43481) = 200.

Factors of a Number:

The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.Thus, there a totally 12 factors for 72.

As 72 is a smaller number, we could list out all factors and count them. But, if a bigger number is given, it would be difficult to list out all factors and count them. 

So, we can use the following method to do so.

If a number can be written in the form [(P)m][(Q)n], where P and Q are prime numbers, then

a.The total number of factors = (m + 1)(n + 1) b.The number of prime factors = (m + n)

c.The sum of all factors =

We know that 72 can be written as: (8)(9) = (2)?(3)?  So, applying the above formulae, we get:

a. Total number of factors = (m + 1)(n + 1) = (3 + 1)(2 + 1) = (4)(3) = 12b. Number of prime factors = (m + n) = (3 + 2) = 5c. Sum of all factors =

                              

Example: Find the total number of factors for 1001.

Solution: Let us write 1001 in the form [(P)m][(Q)n], where P and Q are prime numbers.

 

1001 = [(7)1][(11)1] [(13)1]So, the total number of factors = (1 + 1)(1 + 1)(1 + 1) = 8 Observe that the number of prime factors would be (1 + 1 + 1) = 3

 To find the number in the unit place of (P)Q, where P and Q are positive numbers:

In (P)Q, P is called the base and Q is called the Power or Index.

Case 1: When P is a number ending in 0, 1, 5 or 6, the number in the units place of (P)Q will also be 0, 1, 5 and 6 respectively, for all possible values of Q.

Example: The digit in the units place of (3050)28 is 0The digit in the units place of (6941)35 is 1The digit in the units place of (9025)76 is 5The digit in the units place of (5676)432 is 6

Case 2: When P is a number ending either in 4 or 9:We know that (4)1 = 4, (4)2 = 16, (4)3 = 64, (4)4 = 256, and so on, which implies that when the power is an odd number, the units place digit is 4 and when the power is even number, the units place digit is 6.

Example: (564)238 contains 6 in the units place because the power 238 is even and (724)431 contains 4 in the units place because the power 431 is odd.Similarly, (9)n contains 9 in the units place when n is odd and 1 in the units place when n is even.So, the digit in the units place of (389)23 is 9 and the digit in the units place of (649)34 is 1.Case 3: When P is a number ending in 2, 3, 7 or 8:Divide the power Q by 4 and replace the power Q by the remainder 1, 2, 3 or 0.

Example:(2343)74: 2343 ends in 3. So, divide 74 by 4, we get the remainder 2. Now, replacing the power 74 by 2, we get (2343)2, which ends in 9.

Note: The same method can be applied for any number ending in 2, 3, 7 or 8.Rules of Divisibility: These rules help us to know whether a given number is divisible by a certain divisor, without performing the actual division.

For 2: Any number which has either 0 or an even number (2, 4, 6 or 8) in its units place is divisible by 2. Or, we can say that if the digit in the unit's place of a number is divisible by 2, then the number is also divisible by 2.Ex: 32, 74, 98, 240, 546, etc.

For 3: If the sum of all the digits in a number is divisible by 3, then the number is divisible by 3. Ex: 4569: 4+5+6+9 = 24, which is divisible by 3 and hence 4569 is also divisible by 3.

For 4: If the last two digits of a number is divisible by 4, then the number is divisible by 4. Ex: 632: the last two digits are 32, which is a number divisible by 4.

For 5: If the units place digit is either 0 or 5, then the number is divisible by 5.

Ex: 655, 84530.

For 6: If a number is divisible by both 2 and 3, then it is also divisible by 6. Or, we can say that any even number, which is divisible by 3, is also divisible by 6. Ex: 636Note: observe that 2 and 3 are both relatively prime numbers; only such set of numbers can be used while checking the divisibility of a number.

For 7: Actual division is applicable

For 8: If the last three digits of a number form a number divisible by 8, then the number is divisible by 8.Ex: 23456 is divisible by 8 because the last three digits are: 456, which is divisible by 8.

For 9: If the sum of all the digits in a number is divisible by 9, then the number is also divisible by 9. Ex: 67347.

For 10: If a number ends in 0, then it is divisible by 10. Ex: 3450

For 11: If the sum of the digits in the odd places of a number is equal to the sum of the digits in the even places, or if the two sums differ by a number divisible by 11, then the given number is divisible by 11.

Ex:14641: Sum of digits in odd places = 1+6+1 = 8, sum of digits in even places = 4+4 = 8. Both the sums are equal; so the number is divisible by 11.

Ex:759: Sum of odd places = 7+9 = 16, sum of even places = 5. The difference of the sums = 16 - 5 = 11, which is divisible by 11. Hence, 759 is divisible by 11.For 12: Any number, which is divisible by both 3 and 4, is also divisible by 12.

Ex: 744 is divisible by both 3 and 4 and hence is divisible by 12 also.

Observe that 3 and 4 are also relatively prime. Also note that though 2 and 6 are also factors for 12, they cannot be used to check the divisibility by 12 as they are not relatively prime.

For 14: Any number, which is divisible by both 2 and 7, is also divisible by 14.

For 15: Any number, which is divisible by both 3 and 5, is also divisible by 15.

For 16: If the last 4 digits of a number form a number divisible by 16, then the number is also divisible by 16. Ex: 238000: 8000 is divisible by 16, and hence we can conclude that 238000 is also divisible by 16.

For 18: Any number, which is divisible by both 2 and 9, is also divisible by 18. Ex: 720.

Observe here that though 3 and 6 are also factors for 18, they cannot be used to check the divisibility for 18, as they are not relatively prime.

For 20: Any number, which is divisible by both 4 and 5, is also divisible by 20.

For 21: Any number, which is divisible by both 3 and 7, is also divisible by 21.For 22: Any number, which is divisible by both 2 and 11, is also divisible by 22.

For 24: Any number, which is divisible by both 3 and 8, is also divisible by 24.

Note that (2, 12) and (4,6) cannot be used to check the divisibility by 24, as they are not relatively prime pairs.

For 25: If the last two digits of a number form a number divisible by 25, then the number is also divisible by 25. Or, if a number ends with either "00" or 25 or 50 or 75, then it is divisible by 25.

For 26: Any number which is exactly divisible by both 2 and 13 is also divisible by 26.

For 28: Any number which is divisible by 4 and 7 is also divisible by 28.

For 30: Any number which is divisible by both 2 and 15 is also divisible by 30. Note that we can also use (3, 10) and (5,6) to check the divisibility by 30, as they are also relatively prime pairsExample: Find the least value of P so that the number 60748P is exactly divisible by 2.

Solution: We know that any number that is divisible by 2 should end with a digit divisible by 2. Hence, the least possible value for P is 0.

Example: Find the least value of Q so that the number 45376Q is exactly divisible by 3.

Solution: For a number to be divisible by 3, we know that the sum of all the digits in it should be divisible by 3.4+5+3+7+6+Q = 25+Q.For this to be divisible by 3, the least value possible for Q is 2.Fractions:

A fraction is part of a quantity. The upper part of a fraction is called the Numerator and the lower part is called the Denominator.

Example: In two-sevenths (2/7), 2 is the numerator and 7 is the denominator.

A fraction in which the numerator is less than the denominator is called a Proper fraction or Vulgar fraction or Common Fraction.

Example: 2/7, 5/6, 3/11, etc.

A fraction in which the numerator is more than the denominator is called an Improper Fraction.

Example: 7/2, 6/5, 11/3, etc.

An improper fraction can be converted into a Mixed Fraction.

Example: 7/2 can be written as 3 ?. This is called a Mixed Fraction.

A fraction whose denominator is a multiple of 10 is called a Decimal Fraction.

Example: 3/10, 43/100, etc.

A fraction is always expressed in its lowest terms. This is done by canceling all common factors for the numerator and the denominator.

Example: 44/60 can be written as 11/15.Comparison of Fractions:

Fractions can be compared in any of the following three methods:

a. Decimal Method:

Example: let us take the fractions 2/3, 4/5, 5/6

By dividing, we find that 2/3 = 0.666....., 4/5 = 0.8, 5/6 = 0.83Thus, we can say 2/3 < 4/5 < 5/6.

b. LCM method:

Example: Let us take the same fractions: 2/3, 4/5, 5/6LCM for the denominators is 30.Thus, taking the LCM and converting numerator into number of times the denominator divides the LCM, we get: 20, 24, 25. Thus, we can again conclude that 2/3 < 4/5 < 5/6. c. Cross Multiplication Method:

Example: Let us again take the same fractions: 2/3, 4/5, 5/6

First if we cross multiply 2/3 and 4/5 , we get (2)(5) = 10 and (4)(3) = 12. As 10<12, we can conclude that 2/3 < 4/5. Similarly, we can compare 4/5 and 5/6 and also 5/6 and 2/3,which will help us to finally conclude that 2/3 < 4/5 < 5/6.