Maths Shortcuts

PIPES AND CISTERINS Pipe and Cistern problems are similar to time and work problems. A pipe is used to fill or empty the tank or cistern. 1. Inlet Pipe : A pipe used to fill the tank or cistern is known as Inlet Pipe. 2. Outlet Pipe : A pipe used to empty the tank or cistern is known as Outlet Pipe.

3. Pipe completed Work = Time x Efficiency .

4. Take inlet pipe efficiency positive sign and outlet pipe efficiency negative sign.

5. A pipe fills Tank in " n " hours , B pipe fills Tank in " m " hours.

For efficiency take A and B, L.C.M. = mn "A" efficiency = mn/n = m "B" efficiency = mn/m= n

Here A separately completed work = time x efficiency = mn B separately Completed Work.

= time x efficiency = mn

6. Work done by individuals separately is always equal .

Examples :

Ques 1. Two taps A and B can fill a tank in 20 hours and 30 hours respectively. If both the taps are opened together, the tank will be full in ?

Solution : Time taken by " A" = 20 hours Time taken by "B" = 30 hours L.C.M of 20 and 30 = 60 A efficiency = 60/20= 3 B efficiency = 60/30 = 2

A and B working Together , so efficiency of two pipes = 3 + 2= 5

Work = 60

A and B together completed work in time of = work/efficiency = 60/5 = 12 hours

Ques 2. To fill a cistern, pipes A, B and C take 20 minutes, 15 minute and 12 minutes respectively. The time in minutes that the three pipes together will take to fill the cistern is

Solution : Time taken by " A" = 20 hours Time taken by "B" = 15 hours Time taken by "C" = 12 hours

L.C.M of 20,15 and 12 = 60 A efficiency = 60/20= 3 B efficiency = 60/15 = 2 C efficiency = 60/12 =5 A,B and C together fill tank in time of = work/(A+B+C Efficiency)

= 60/(3+4+5) = 60/12 = 5 min

Ques 3. Two pipes can fill a tank in 10 hours and 12 hours resp. while third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time the tank will be filled?

Solution :

Time taken by " A" = 10 hours Time taken by "B" = 12 hours Time taken by "C" = 20 hours

L.C.M 10,12 and 20 is = 60 A efficiency = 60/10= 6 B efficiency = 60/12 = 5 C efficiency = 60/20 = 3

Here " C " pipe is outlet pipe ,take it's efficiency in negative = (-3)

A,B and C operating simultaneously , total efficiency = 6+5-3= 8

Time take to fill tank = 60/8 = 7 1/2 hours

Ques 4. A cistern can be filled in 9 hours but it takes 10 hours due to a leak in its bottom. If the cistern is full, then the time that the leak will take to empty it is

Solution : Time taken by A = 9 hours

With leakage time taken by A = 10 hours Take L.C.M of 9 and 10 = 90 A efficiency = 90/9 = 10 A and Leakage efficiency = 90/10 =9 A efficiency - Leak Efficiency = 9 ==> 10 - Leak Efficiency.

=9

==> Leak Efficiency = 10-9= 1

Time taken by Leak to empty the full tank = 90/1 = 90 hours

Ques 5. A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water are the rate of 6 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. How many litres does the cistern hold ?

Solution : Time taken by Leak = 8 hours Time taken by Leak+Inlet = 12

L.C.M of 8 and 12 = 24 Leak Efficiency = 24/8= 3

Leak and inlet efficiency = 24/2 = 2 Work = 24 Here in two cases tank is empty so take efficiencies negative sign Outlet + Inlet = -2 ==>

-3+ Inlet = -2

==> Inlet efficiency. = -2+3= 1 Time taken by inlet to full tank = 24/1=24 Inlet delivers 6 litres in a minute Inlet delivers in 24 hours litres = 24x60x6 = 8640 litres

Ques 6. Two pipes fills a cistern in 15 hrs and 20 hrs respectively. The pipes are opened simultaneously and it is observed that it took 26 min more to fill the cistern because of leakage at the bottom. If the cistern is full, then in what time will the leak empty it?

Solution : A and B are inlet pipes , C is outlet pipe

Time taken by A = 15 hours Time taken by B = 20 hours A and B pipes fill tank in hours = 60/7 hours = 8 hours 34 min According to problem A,B and C open simultaneously its take

26 minutes

more Than A and B working together ==> Time taken by A,B and C together = 8 hours + 26 min = 9 hours work done by leak in 1 hr = work done by two pipes - 1/9 = 7 /60 – 1/9 = 3/ 540 = 1/180 Therefore, leak will empty the full cistern in 180 hours.

Step 1. Take the LCM of a given Number and that LCM will be the total capacity of cistern or tank.Place the plus or minus sign for common understanding through picture. plus sign means time taken to fill the cistern and minus sign means to empty the cistern in a particular time. Step 2. Add or Subtract According to a particular question Step 3. The LCM from Step 1. will be the total work . Divide the total capacity of cistern or tank with the outcome of Step 2.

EXAMPLES #1. Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank ? (M.A.T 2003) Step 1. Take the LCM of 20 and 30 which is 60. Sixty liter is the total capacity of tank or cistern. Step 2. Divide 60 by (each number) 20 and 30 ,you will get 3 and 2 respectively. Step 3. Now add (3+2) which is 5. Step 4. Now Divide Total capacity of tank or cistern ( 60/5) , you will get Total time taken by them to fill the tank or cistern in a particular time which is in this case is 12 minutes.

#2. A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously then after how much time will the cistern get filled ? (Hotel management, 1997)

Step 1. Take the LCM of 4 and 9 which is 36. Thirty six liter is the total capacity of tank or cistern. Step 2. Divide 36 by (each number) 4 and 9 ,you will get 9 and 4 respectively.Here plus (+) sign means filling the tank or cistern and minus (-) implies time taken by them to empty the tank. Step 3. Now You see one is filling the tank and other empty it so subtract it with each other.you will get (9-4) 5. Step 4. Now Divide Total capacity of tank or cistern ( 30/5) , you will get Total time taken by them to fill the tank or cistern in a particular time which is in this case is 36/5 minutes.

#3. Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled ? (C.B.I, 1997)

Step 1. Take the LCM of 5,10 and 30 which is 30. Thirty liter is the total capacity of tank or cistern. Step 2. Divide 30 by (each number) 5,10 and 30 ,you will get 6,3 and 1 respectively.Here plus (+) sign means filling the tank or cistern and minus (-) implies time taken by them to empty the tank. Step 3. Now you see here each tank is filling the tank in a particular time, so add each value (6+3+1) you will get 10. Step 4. Now Divide Total capacity of tank or cistern ( 30/10) , you will get Total time taken by them to fill the tank or cistern in a particular time which is in this case is 3 hours.

#4. Pipes A and b can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in ? (Bank P.O. 2002)

PRACTICE QUESTIONS Ques 1. Two pipes P and Q can fell a cistern in 24 min. and 36 min. respectively . Third pipe R can empty it in 36 min. If all the three pipes are opened , find the taken to fill the cistern. (a) 1 hour (b) 24 mins (c) 36 mins (d) 30 mins

Ques 2. A tank has a leak which would empty it in 8 hours. A tap is turned on which admit 6 liters a minute into the tank and it now emptied in 12 hours . How many literes does the tank holds? (a) 8260 ltr (b) 8640 ltr (c) 8560 ltr (d) 8800 ltr

Ques 3. When the waste pipe is closed , two taps can separately fill a cistern in 10 and 12 minutes respectively .when the waste pipe is opened they together fill it in 15 minutes . How long does it take waste pipe to empty the cistern , when the taps are closed? (a) 8 min. 34 sec.

(b) 7 min.10 sec. (c) 12 min. (d) 10 min

Ques 4. Two pipes can fill a tank in 10 minutes and 30 minutes respectively and a third pipe can empty the full tank in 20 minutes . If all the three pipes are opened simultaneously, the tank will be filled in: (a) 12 minutes (b) 10 minutes (c) 8 minutes (d) 6 minutes

Ques 5. Two pipes can fill a cistern in 14 hours and 16 hours respectively . The pipes are opened simultaneously and it is found that due to leakage in the bottom , 32 minutes extra are taken for the cistern to be filled up. When the cistern is full , in what time will the leak empty it ? (a) 108 hours (b) 112 hours (c) 116 hours (d) 120 hours

Ques 6. A cistern can be filled by pipes A and B in 12 minutes and 10 minutes respectively . The full tank can be emptied by a third pipe C in 8 minutes only .If all the pipes be turned on at the same time,the cistern will be full in : (a) 17 min (b) 171/7 min. (c) 17 2/7 min. (d) 18 min.

Ques 7. If two pipes function simultaneously , the reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours does it take the second pipe to fill the reservoir? (a) 20 hours (b) 25 hours (c) 30 hours (d) 40 hours

Ques 8. A tank is filled in hours by three pipes A, B and C . The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank? (a) 40 hours. (b) 35 hours (c) 30 hours (d) 25 hours

Ques 9. Three pipes P , Q and R can fill a tank in 6 hours , After working it at together for 2 hours , R is closed and P and Q can fill the remaining part in 7 hours . How much time will R take alone to fill the tank? (a) 16 hours (b) 14 hours (c) 12 hours (d) 10 hours

Ques 10. Three taps P, Q and R can fill a tank in 12, 15 and 20 hours respectively . If P is opened all the time and Q and R are opened for one hour each alternately , the tank will be full in (a) 5 hours (b) 6 hours (c) 6 2/3 hours (d) 7 hours

solutions: Ans 1.Since time taken to fill the cistern by Q = time taken to empty the cistern by R so the cistern will be filed by P in 24 min.

Ans 2. In I hour the part filled by the tap = 1/8 -1/12 = 1/24 Hence , the tap can fill the tank in 24 hours. Therefore, capacity of the tank = 24 x 60 x 6 = 8640 litres.

Ans 3. The waste pipe can empty in 1 min .= 1/10 + 1/12 - 1/15 = 11/60 1/15 = 7/60 of the cistern. Hence, the waste pipe can empty the cistern in 60/7 minutes , i.e., 8 minutes 34 sec.

Ans 4. 1n I minute the part filled by all the three pipes = 1/10 + 1/30 - 1/20 = 1/12 Hence , all the three pipes will fill the bank in 12 minutes.

Ans 5. In I hour the part filled by both the pipes = 1/14 + 1/16 = 15/112 Hence, both the pipes will fill the cistern in 112/15 min. but due to leakage , the tank will be filled in 112/15 + 32/60 = 120/15 =

8 hours. Since, in 1 hr. the part emptied by the leakage = 15/112 - 1/8 = 1/112 Therefore , leakage will empty the tank in 112 hours.

Ans 6. In 1 minute the part filled by the three pipes = 1/12 + 1/10 -1/8 = 7/120 Hence, all the three pipes will fill the cistern in 120/7 = 17 1/7 minutes.

Ans 7.Let ( x- 10) and x hrs be the time taken by two pipes to fill the reservoir , then,

Hence , second pipe will fill the reservoir in 30 hours.

Ans 8.Let x,x/2 and x/4 hours be the time taken by pipes A, B and C respectively to fill the tank,then 1/x + 2/x + 4/x = 1/5 => 7/x = 1/5 ∴ x = 35 hours. Hence , time taken by A is 35 hours to fill the tank.

Ans 9. In 2 hour the part filled by all the three pipes = 2 X 1/6 = 1/3 Remaining part = 1 - 1/3 = 2/3, which is filled by ( P + Q ) in 7 hours. Hence , in 1 hour the part filled by ( P + Q ) = 2/ 3 x 7 = 2/21 Since , in 1 hour the part filled by R alone = 1/6 - 2/21= 1/14 Therefore, R alone will fill the bank in 14 hours.

Ans 10.In 1 hour the part filled by (P + Q ) = 1/12 + 1/15 = 3/20 In 1 hour the part filled by (P + R ) = 1/12 + 1/20 = 2/15 Hence, in 2 hours the part filled by ( P+ Q +R ) = 3/20 + 2/15 = 17/60 Then , in 6 hours the part filled by ( P + Q + R ) = 3 x 17/60 = 17/20 Remaining part = 1 - 17/20 = 3/20 , which is filled by ( P + Q ) in 1 hour. Hence, total time taken = 6 + 1 = 7 hours.

TIME AND WORK

TRICK One simple technique is using days in denominator while solving questions. For example, A can do a job in 3 days and B can do the same job in 6 days. In how much time they can do the job together. Solution - 1/3 + 1/6 = 1/2, hence 2 days is the answer. Examiner can set the question in opposite way and can ask you how much time A or B alone will take to complete the job. It is quite easy to calculate said question by putting values in equation we arrived in above question. You need to understand one simple concept - If A can do a job in 10 day then in one day A can do 1/10th of job.

SHORTCUT Best trick that I use in exams myself is by finding the efficiency of workers in percent. If A can do a job in 2 days then he can do 50% in a day. Number of days required to complete the work

n

Work that can be done per day

Efficiency in Percent

1/n

100/n

1

1/1

100%

2

1/2

50%

3

1/3

33.33%

4

1/4

25%

5

1/5

20%

6

1/6

16.66%

7

1/7

14.28%

8

1/8

12.5%

9

1/9

11.11%

10

1/10

10%

11

1/11

9.09%

Now let's solve questions with this trick Question - A take 5 days to complete a job and B takes 10 days to complete the same job. In how much time they will complete the job together ?

Solution - A's efficiency = 20%, B's efficiency = 10%. If they work together they can do 30% of the job in a day. To complete the job they need 3.33 days.

Question - A is twice as efficient as B and can complete a job 30 days before B. In how much they can complete the job together ?

Solution - Let efficiency percentage as x A's efficiency = 2x and B's efficiency = x A is twice efficient and can complete the job 30 days before B. So, A can complete the job in 30 days and B can complete the job in 60 days

A's efficiency = 1/30 = 3.33% B's efficiency = 1/60 = 1.66% Both can do 5% ( 3.33% + 1.66% ) of the job in 1 day. So the can complete the whole job in 20 days (100/5)

Question - A tank can be filled in 20 minutes. There is a leakage which can empty it in 60 minutes. In how many minutes tank can be filled?

Solution Method 1 ⇒ Efficiency of filling pipe = 20 minutes = 1/3 hour = 300% ⇒ Efficiency of leakage = 60 minutes = 100%

We need to deduct efficiency of leakage so final efficiency is 200%. We are taking 100% = 1 Hour as base so answer is 30 minutes. Update - 09-09-2013 ( As Shobhna and Aswin are facing problem in solving this question, I am solving this question with second method which is also very easy, hope this will make the solution lot easier.) Method 2

⇒ Efficiency of filling pipe = 100/20 = 5% ⇒ Efficiency of leakage pipe = 100/60 = 1.66% ⇒ Net filling efficiency = 3.33% So tank can be filled in = 100/3.33% = 30 minutes

You can change the base to minutes or even seconds.

You can solve every time and work question with this trick. In above examples I wrote even simple calculations. While in exams you can do these calculations mentally and save lots of time.

You can find more tricks like this in quantitative aptitude section. Comment below in case of any query, I promise to reply within 24 hours. Update 09 October 2013 - Question requested by Chitra Salin Question - 4 men and 6 women working together can complete the work within 10 days. 3 men and 7 women working together will complete the same work within 8 days. In how many days 10 women will complete this work ? Solution - Let number of men =x, number of women = y

⇒ Efficiency of 4 men and 6 women = 100/10 = 10% ⇒ so, 4x+6y = 10 Above equation means 4 men and 6 women can do 10% of a the job in one day.

⇒ Efficiency of 3 men and 7 women = 100/8 = 12.5% so, 3x+7y = 12.5

By solving both equations we get, x = -0.5 and y = 2

⇒ Efficiency of 1 woman(y) = 2% per day ⇒ Efficiency of 10 women per day = 20% So 10 women can complete the job in 100/20 = 5 days Update 11-11-2013 - Question requested by Praisy Question - A and B together can complete a task in 20 days. B and C together can complete the same task in 30 days. A and C together can complete the same task in 30 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone? Solution ⇒ Efficiency of A and B = 1/20 per day = 5% per day ________________1 ⇒ Efficiency of B and C = 1/30 per day = 3.33% per day______________2 ⇒ Efficiency of C and A = 1/30 per day = 3.33% per day______________3 Taking equation 2 and 3 together ⇒ B + C = 3.33% and C + A = 3.33% ⇒ C and 3.33% will be removed. Hence A = B ⇒ Efficiency of A = B = 5%/2 = 2.5% = 1/40 ⇒ Efficiency of C = 3.33% - 2.5% = 0.833% = 1/120 ⇒ A can do the job in 40 days and C can do the job in 120 days he they work alone. ⇒ Ratio of number of days in which A and C can complete the job 1:3.

Steps to Solve Time and Work with Trick Step 1. Take the LCM of a given Number Step 2. Add or Subtract According to a particular question Step 3. The LCM from Step 1. will be the total work . Divide the total work with the outcome of Step 2.

Problems #1 Ram , Shyam and Mohan can do a piece of work in 12,15 and 20 days respectively, how long will they take to finish it together . (LIC, ' 91)

Step 1. Take the LCM of 12,15 and 20 which is 60.60 is the total work Step 2. Divide 60 by (each number)12,15 and 20 ,you will get 5,4 and 3 respectively. Now you get Ram's work in one day is 5 or 1/5 .Shyam's one day work = 4 or 1/4 and so on. Step 3. Now add Each Men's 1 day work (5+4+3) which 12. Now here 12 is Ram , Shyam and Mohan together's 1 day work. Step 4. Now Divide 1 day's work with Total work ( 60/12) , you will get Total time taken by them to do the same work which is in this case is 5 days.

#2 A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work ?

(Bank P.O. 2008)

Step 1. Take the LCM of 4 and 12 which is 12.Twelve is the total work Step 2. Divide 12 by (each number)4 and 12 ,you will get 3 and 1 respectively. Now you get A and B's work in one day is 3 or 1/3 . A's one day work = 12 or 1/12 and so on. Step 3. Now Subtract their work (3-1) you will get 2. Step 4. A and B done the work in 4 days .Now by placing the value of A . you will get the B's work which is 2 days. Step 5. Divide Total work by B's one day work. like here 12/2 = 6

#3 A does a work in 10 days and B does the same work in 15 days. In how many days they together will do the same work ? (R.R.B. 2003)

Step 1. Take the LCM of 10 and 15 which is 30. Thirty is the total work Step 2. Divide 30 by (each number)10 and 15 ,you will get 3 and 2 respectively. Now you get A's work in one day work is 3 or 1/3 . B's one day work = 2 or 1/2 . Step 3. Now add Each Men's 1 day work (3+2) which 5. Now here 5 is A and B together's 1 day work. Step 4. Now Divide 1 day's work with Total work ( 30/5) , you will get Total time taken by them to do the same work together which is in this case is 6 days.

#4 A,B and C can complete a piece of work in 24,6 and 12 days respectively, working together, they will complete the same work in ? (C.B.I. 2003)

TIME AND DISTANCE

FORMULAE: 1. 2. 3. 4.

Speed = Distance/Time Time = Distance/Speed Distance = Speed × Time If the speed of a body is changed in the ratio a : b, then the ratio of the time taken changes in the ratio b : a. 5. m km/hr = [m × 5/18] m/sec. 6. m metres/sec = [m × 18/5] km/hr. I recommend you to watch the following concept video before solving the questions.

Question 1. Express a speed of 18 km/hr in metres per second.

Solution: 18 km/hr = [18 × 5/18] m/sec. = 5 metres/sec. Question 2. Express 10 m/sec. in km/hr.

Solution: 10 metres/sec = [10 × 18/5] km/hr. = 36 km/hr.

THEOREM If a certain distance is covered at m km/hr and the same distance is covered at n km/hr then the average speed during the whole journey is 2mn/(m+n) km/hr.

Let the distance be A km.

Time taken to travel the distance at a speed of m km/hr = A/m hrs. Time taken to travel the distance at a speed of n km/hr = A/n hrs. we see that the total distance of 2A km is travelled in (A)/m+A/n hrs.

Question 3. Rakesh covers a certain distance by car driving at 70 km/hr and he returns to the starting point riding on a scooter at 55 km/hr. Find his average speed for the whole journey.

Question 4. Raju covers distance between his house and office on scooter. Having an average speed of 30 km/hr, he is late by 10 min. However, with a speed of 40 km/hr, he reaches his office 5 min earlier. Find the distance between his house and office.

Question 5. A man walking with a speed of 5 km/hr reaches his target 5 minutes late. If he walks at a speed at a speed of 6 km/hr, he reaches on time. Find the distance of his target from his house.

Question 6. A boy goes to school at a speed of 3 km/hr and returns to the village at a of 2 km/hr. if he takes 5 hrs in all, what is the distance between the village and the school?

1. How many minutes Raman will take to cover a distance of 400 meters if he runs at a speed of 20 km/hr ?  A. 2 mins  C. 1⅕ mins  E.

1.5 mins  B. 2.5 mins  D. None of these  Answer & Explanation Answer : [C] Explanation : ⇒ Raman's speed = 20 km/hr = 20 × 5/18 = 50/9 m/sec ⇒ 400 × 9/50 = 1⅕ mins 2. John travelled from his town to city. John went to city by bicycle at the speed of 25 km/h and came back at the speed of 4 km/h. If John took 5 hours and 48 min to complete his journey, what is the distance between town and city ?  A. 15 km  C. 20 km  B. 22 km  D. 25 km  Answer & Explanation Answer : [C] Explanation : ⇒ Average speed of John = 2xy/x+y = 2 × 25 × 4 / 25 + 4= 200/29 km/h ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km

3. Speed of a train is 20 meters per second. It can cross a pole in 10 seconds. What is the length of train ?  A. 150 m  C. 200 m  B. 250 m  D. 300 m  Answer & Explanation

Answer : [C] Explanation : ⇒ Lenght of train = 20 × 10 = 200 meters

4. Ram walks at a speed of 12 km/h. Today the day was very hot so walked at ⅚ of his average speed. He arrived his school 10 minutes late. Find the usual time he takes to cover distance between his school and home ?  A. 40 mins  C. 50 mins  B. 45 mins  D. 60 mins  Answer & Explanation Answer : [C] Explanation : ⇒ If Ram is walking at ⅚ of his usual speed that means he is taking 6/5 of using time. ⇒ 6/5 of usual time - usual time = 10 mins ⇒ 1/5 of usual time = 10 mins ⇒ Usual time = 50 mins 5. A car running at 65 km/h takes one hour to cover a distance. If the speed is reduced by 15 km/hour then in how much time it will cover the distance ?  A. 72 mins  C. 76 mins  B. 78 mins  D. None of these  Answer & Explanation Answer : [B] Explanation :

⇒Reduced speed = 65-15 = 50 km/h ⇒ Now car will take 65/50 × 60 mins = 78 mins

6. In a 100 m race A runs at a speed of 1.66 m/s. If A gives a start of 4m to B and still beats him by 12 seconds. What is the speed of B ?  A. 1 m/s  C. 1.25 m/s  B. 1.33 m/s  D. Rs 1.5 m/s  Answer & Explanation Answer : [B] Explanation : ⇒Time taken by A to cover 100 meters = 60 seconds ⇒ Since A gives a start of 4 seconds then time takes by B = 72 seconds ⇒ B takes 72 seconds to cover 96 meters ⇒ Speed of B = 96/72 = 1.33 m/s

7. In a kilometer race, A beats B by 100 meters. B beats C by 100 meters. By how much meters does A beat C in the same race ?  A. 200 meters  C. 190 meters  B. 180 meters  D. 210 meters  Answer & Explanation Answer : [C] Explanation : ⇒ While A covers 1000 meters, B can cover 900 meters ⇒ While B covers 1000 meters, C can cover 900 meters ⇒ Lets assume that all three of them are running same race. So when B runs

900 meters, C can run 900 × 9/10 =810 ⇒ So A can beat C by 190 meters.

Number System - Rules and Example Rules on Counting Numbers

Note: In the first n counting numbers, there are n/2 odd and n/2 even numbers provided n, the number of numbers, is even. If n, the number of numbers, is odd, then there are 1/2(n + 1) odd numbers and 1/ 2 (n – 1) even numbers.

For example: from 1 to 50, there are 50/2= 25 odd numbers and 50/2 = 25 even numbers. And from 1 to 51, there are (51+1 )/2 = 26 odd numbers and(51-1 )/2= 25 even numbers.

The difference between the squares of two consecutive numbers is always an odd numbers.

The difference between the square of two consecutive numbers is the sum of the two consecutive numbers.

Solved examples: Ques 1. What is the total of all the even numbers from 1 to 400?

Ques 2. What is the total of all the odd numbers from 1 to 180?

Ques 3. Find the sum of all the odd numbers from 20 to 101.

POWER AND INDEX If a number ‘p’ is multiplied by itself n times, the product is called nth power of ‘p’ and is written as pn. In pn, p is called the base and n is called the index of the power.

Solved examples: Ques 4. What is the number in the unit place in (729) 59? Solution: When 729 is multiplied twice, the number in the unit place is 1. In other words, if 729 is multiplied an even number of times, the number in the unit place will be 1. Thus, the number in the unit place in (729) 58 is 1. So, (729)59 = (728)58 X (729) = (…….1) X(729) = 9 in the unit place.

Note: When you solve this type of questions (for odd numbers) try to get the last digit 1, as has been done in the above example. Ques 5. Find the number in the unit place in (98) 40 , (98)42 , (98)43. Solution: So,

(98)4 = (……….6) (98)4n =(…………6)

Thus, (98)40 = (98)4x10 = (…..6) = 6 in the unit place. (98)42 = (98)4x10 X (98)2 = (…..6)X(…….4) = 4 in the unit place. (98)43 = (98)4x10 X(98)3 = (…..6)X(…….2) = 2 in the unit place.

Note: When there is an even number in the unit place of base, try to get 6 in the unit place, as has been done in the example.

Rule no. 1. For Odd Numbers

When there is an odd digit in the unit place (except 5), multiply the number by itself until you get 1 in the unit place.

(…………1) n = (……………1) (……….3)4n = (………..1) (……...7)4n

= (………..1)

Where n = 1, 2, 3,……….

Rule no.2 When there is an even digit in the unit place of the given number, then after any times of its multiplication, it will have the same digit in the unit place i.e. (…………1) n = (……………1)

(……….5)n = (………..5)

(……...6)n =(………..6)

Ques 6. What is the numbers in the unit place when 781, 325, 497 and 243 are multiplied together? Solution: Multiply all the numbers in the unit place i.e., 1 X 5 X 7 X 3; the result is a number in which 5 is in the unit place.

umber System - Concepts and Tricks Published on Saturday, May 21, 2016

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If you want to test whether any number is a prime number or not take an integer larger than the approximate square root of that number. Let it be ‘x’. Test the divisible of the given number by every a prime number less than ‘x’. If it is not divisible by any of them, then it is prime number; otherwise it is a composite number (other than prime).

1. Is 349 a prime number ? Solution: The square root of 349 is approximate 19. The prime numbers less than 19 are 2, 3, 5, 7, 11, 13, and 17.

Clearly, 349 is not divisible by any of them. Therefore, 349 is a prime number.

2. Is 881 a prime number? Solution: The approximate Sq. root of 881 is 30.

Prime number less than 30 is 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. 881 is not divisible by any of the above numbers, so it is a prime number.

RULE OF SIMPLIFICATION (i) In simplification an expression, first of all vinculum or bar must be removed. For example: we know that

(ii) After removing the bar, the brackets are removed, strictly in the order (), {}, [].

(iii) After removing brackets, we must follow following order of operations :- (a) of (b) division (c) multiplication (d) addition (e) subtraction.

This is also known as the rule of ‘VBODMAS’ , where V, B, O, D, M, A, and S stand for Vinculum, Bracket, Of, Division, Multiplication, Addition and Subtraction respectively.

GENERAL RULE FOR SOLVING PROBLEMS IN ARITHMETIC

Ascending or descending orders in Rational Numbers Rule 1 When the numerator and the denominator of the fractions increase by a constant value, the last fraction is the biggest.

Ques. Which one of the following fractions is the greatest? 3/4, 4/5 and 5/6

Solution: We see that the numerators as well as denominators of the above fractions increase by 1, so the last fraction, i.e., 5/6, is the greatest fraction.

Rule 2 The fraction whose numerator after cross – multiplication gives the greater value is greater.

Example: Which is greater: 5/8 or 9/14 ? Solution: Step 1: cross – multiply the two given fractions

SIMPLIFICATION Simplification is one of the most important part of Quantitative Aptitude section of any competitive exam. Today I am sharing all the techniques to solve Simplification questions quickly. Rules of Simplification V → Vinculum B → Remove Brackets - in the order ( ) , { }, [ ] O → Of D → Division M → Multiplication A → Addition S → Subtraction

Important Parts of Simplification     

Number System HCF & LCM Square & Cube Fractions & Decimals Surds & Indices

Number System    

Classification Divisibility Test Division& Remainder Rules Sum Rules

Classification Types

Description

Natural all counting numbers ( 1,2,3,4,5....∞) Numbers: Whole natural number + zero( 0,1,2,3,4,5...∞) Numbers: Integers:

All whole numbers including Negative number + Positive number(∞......-4,-3,-2,-1,0,1,2,3,4,5....∞)

Even & Odd Numbers :

All whole number divisible by 2 is Even (0,2,4,6,8,10,12.....∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19....∞)

It can be positive or negative except 1, if the number is not divisible Prime by any number except the number itself. Numbers: (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61....∞) Composit e Natural numbers which are not prime Numbers: Co-Prime: Two natural number a and b are said to be co-prime if their HCF is 1.

Divisibility Number IF A Number s

Examples

Divisibl e by 2

End with 0,2,4,6,8 are divisible by 2

254,326,3546,4718 all are divisible by 2

Divisibl e by 3

Sum of its digits is divisible by 3

375,4251,78123 all are divisible by 3. [549=5+4+9] [5+4+9=18]18 is divisible by 3 hence 549 is divisible by 3.

Divisibl e by 4

Last two digit divisible by 4

5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4.

Divisibl e by 5

Ends with 0 or 5

225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5.

Divisibl e by 6

Divides by Both 2 & 3

4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6.

Divisibl e by 8

Last 3 digit divide by 8

746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8.

Divisibl e by 10

End with 0

220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10.

Divisibl e by 11

[Sum of its digit in odd placesSum of its digits in even places]= 0 or multiple of 11

Consider the number 39798847 (Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3) (23-12) 23-12=11, which is divisible by 11. So 39798847 is divisible by 11.

Division & Remainder Rules Suppose we divide 45 by 6

hence ,represent it as: dividend = ( divisor✘ quotient ) + remainder or divisior= [(dividend)-(remainder] / quotient could be write it as x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)

Example: On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ? Number = 342k + 47 ( 18 ✘19k ) + ( 18 ✘2 ) + 11 18 ✘( 19k + 2 ) +11. Remainder = 11

Sum Rules (1+2+3+.........+n) = 1/2 n(n+1) (12+22+32+.........+n2) = 1/6 n (n+1) (2n+1) (13+23+33+.........+n3) = 1/4 n2 (n+1)2 Arithmetic Progression (A.P.) a, a + d, a + 2d, a + 3d, ....are said to be in A.P. in which first term = a and common difference = d. Let the nth term be tn and last term = l, then

a) nth term = a + ( n - 1 ) d b) Sum of n terms = n/2 [2a + (n-1)d] c) Sum of n terms = n/2 (a+l) where l is the last term   

Tricks to Solve Number System Unit Digit Multiplication Examples of Number System with Explanation

H.C.F. & L.C.M.  

Factorization & Division Method HCF & LCM of Fractions & Decimal Fractions

Methods On Basis

H.C.F. or G.C.M

Write each number as the product of the prime factors. The product of least powers of common prime factors gives H.C.F. Factorizati Example: on Method Find the H.C.F. of 108, 288 and 360. 108 = 22✘33, 288 = 25✘32 and 360 = 23✘5✘32 H.C.F. = 22✘32=36

Division Method

Let we have two numbers .Pick the smaller one and divide it by the larger one. After that divide the divisor with the remainder. This process of dividing the preceding number by the remainder will repeated until we got the zero as remainder.The last divisor is the required H.C.F. Example:

L.C.M. Write each numbers into a product of prime factors. Then, L.C.M is the product of highest powers of all the factors. Examples: Find the L.C.M. of 72, 108 and 2100. 72=23✘32,108=33✘22, 2100=22✘52✘3✘7. L.C.M.=23✘33✘52✘7=37800 Let we have set of numbers. First of all find the number which divide at least two of the number in a given set of number.remainder and not divisible numbers will carry forward as it is. Repeat the process till at least two number is not divisible by any number

except 1.The product of the divisor and the undivided numbers is the required L.C.M. Example: Find the L.C.M. of 12,36,48,72

H.C.F. of given numbers = 69

H.C.F. & L.C.M. of Fractions Product of H.C.F. & L.C.M.

Decimal numbers



H.C.F. =

H.C.F. of Numerator

/ L.C.M. of Denominators

L.C.M. =

L.C.M. of Numerator

/H.C.F. of

Denominators

H.C.F * L.C.M. = product of two numbers

H.C.F. of Decimal numbers Step 1. Find the HCF of the given numbers without decimal. Step 2.Put the decimal point ( in the HCF of Step 1) from right to left according to the MAXIMUM deciaml places among the given numbers.

L.C.M. of Decimal numbers Step 1. Find the LCM of the given numbers without decimal. Step 2.Put the decimal point ( in the LCM of Step 1) from right to left according to the MINIMUM deciaml places among the given numbers.

Practice Set Paper 1 Examples with Explanation

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Practice Set Paper 2

Square & Cube   

Square & Cube Square Root & Cube Root Factorization Method Perfect Square

last digit is 1, 4, 9, 6, 5

Non-Perfect Square last digit is 2, 3, 7, 8

Square Root & Cube Root

 

Tricks to Find Square of Any Number Square Root & Cube Root

Fractions & Decimals On Basis

Explanation

Decimal A number with a denominator of power of 10 is a decimal fractions. Fractions 1/10= 1 tenth; 1/100= 0.1;38/100=0.38 Conversion of 0.64(decimal number) into a Vulgar Fraction.First of all write the numeric digit 1 in the denominator of a number (like here Vulgar 0.64) and add as many numeric zeros as the digit in the number after Fractions decimal point.After that removes the decimal point from the given number.At last step just reduce the fraction to its lowest terms. So, 0.64 = 64/100=16/25;25.025 = 25025/1000 =1001/4 Operatio ns

Addition & Subtraction To perform the addition and subtraction of a decimal fraction could be done through placing them right under each other that the decimal points lie in one column. 3.424+3.28+.4036+6.2+.8+4 3. 424 3. 28 . 4036 6. 2 .8 +4______ 18. 1076

Multiplication of a Decimal Fraction To find the multiplication of decimal fraction , first of all you need to remove the decimal point from the given numbers and then perform the multiplication after that assign the decimal point as many places after the number as the sum of the number of the decimal places in the given number. Step 1. 0.06*0.3*0.40 Step 2. 6*3*40=720 Step 3. 0.00720 Multiplication of a decimal fraction by power of 10 A multiplication of a decimal fraction by a power of 10 can be perform through shifting the decimal point towards right as many places as is the power of 10. like 45.6288*100=45628.8, 0.00452*100=0.452 Division

To compare the set of fractions numbers,first of all you need to convert each fraction number or value into a equal decimal value and then it will be became easy for you to assign them ( the numbers or value) in Comparis a particular way( ascending or descending order). on of 3 /5,4/7,8/9 and 9/11 Arranging in Ascending Order Fractions 3 /5= 0.6, 4/7 = 0.571, 8/9 = 0.88, 9/11 = 0.818. Now, 0.88 > 0.818 > 0.6 > 0.571 8 /9>9/11>3/5>4/7 Recurring Recurring Decimal Decimal A decimal number in which after a decimal point a number or set of number are repeated again and again are called recurring decimal numbers.It can be written in shorten form by placing a bar or line above the numbers which has repeated.

Pure Recurring Decimal A decimal number in which all digits are repeated after a decimal point.

Mixed Recurring Decimal A decimal number in which certain digits are repeated only.

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Examples with Explanations

Surds & Indices  

Some Rules of Indices Some Rules of Surds

# Series/Addition/Subtraction

Q1 Find 8 + 888 + 8888 + 88888 ?

a) 97760 b) 98572 c) 98672 d) 97672 e) None of the above

Q2. Find 9.4 + 99.44 + 999.444 + 9999.4444 ? a) 11207.728 b) 11107.728 c) 11106.728 d) 11111.728 e) None of these

Q3. ( 2.3 + 3.3 ) [ ( 2.3)2 - 2.3✘3.3 + (3.3)2 ] = ? a) 48.104 b) 47.104 c) 47.204 d) 48.204 e) None of these Q4. (3-2) [(3)4+((3)3✘2)+(3)2✘(2)2+(3✘23)+24] = ? a) 311 b) 211 c) 201 c) 221 d) 301 e) None of these Q5. (125.824+124.654)2 + (125.824-124.54)2 / ((125.824)2+(124.54)2) a) 1.166 b) 1 b) 2 c) 625 d) 250.478 e) None of these

Q6. Evaluate 1 + 1/1*3 + 1/1*3*9 + 1/1*3*9*27 + 1/1*3*9*27*81 up to three places of decimals ?

a)1.367 b) 1.370 c) 1.361 d) 1.267 e) None of these Q7. 2 ÷ [2+ 2÷{2+2 ÷ 4)}]= x / 19. Find x. a) 3 b) 4 c) 5 d) 6 e) None of these

a) 0.67 b) 0.77 c) 0.87 d) 0.97 e) None of these Q9) Find the sum of all even natural numbers less than 75. a) 1416 b) 1426 c) 1396 d) 1406 e) None of these

Solution: (1) Sol : Option (c) 8*(12345) - 88 = 98672 (2) Sol: Option (b) 9✘(1234) = 11106 4✘(4321) = 17284 =1.7284 =11106+1.7284=11107.7284 (3) Sol: Option (a) (a3 + b3)=(a+b) (a2-ab+b2) a = 2.3, b = 3.3; (a)3 = 12.167, (b)3 = 35.937 (a3+b3)=48.104 (4)

Sol: option (b) (a5-b5)=(a-b) [a4+a3b+a2b2+ab3+b4] = ? a = 3, b = 2; (3)5=243, (2)5 =32; (a5-b5) = 243-32= 211 (5) Sol: Option (b) (a+b)2 + (a-b)2=2((a)2+(b)2) Hence Answer is 2 (6) Sol: Option (a) 1+0.33+0.0370 ... hence (a) is the answer no further addition is required. (7) Sol: option (e) Using VODMAS method Step 1. [ 2+ 2/4 ] = 5/2. Step 2. ( 2 + 2 ÷5/2) = 2÷ 2✘2/5 =14/5 Step 3. [2+2÷14/5]=2+2*5/14=19/7 Step 4. L.H.S = 2÷19/7=14/19 = x/19 Hence x=19 (8) Sol: Option (c) = -4 + 0.45 + 2 + 927-9/990 +6/9 = -4 + 0.45 + 2 + 918/990 +3/2 = (- 4 + 2) + (0.45+1.5) + (51/55) = -2 + 1.95 + 0.92 = 0.87 (9) Sol: Option (d) sum= 2 + 4 + 6 + ...+74 a=2 , d=(4-2)=2,l=74 n=37; sum= n/2 (a+l) 37/2✘(2+74)=(37✘38) =37✘(40-2) =(37✘40)-(37✘2) =(1480-74)=1

ome important things to be noticed: (i) When two trains are moving in opposite directions, their speeds should be added to find the relative speed.

(ii) When they are moving in the same direction, the relative speed is the difference of their speeds.

(iii) When a train passes a platform, it should travel the length equal to the sum of the lengths of trains & platform both.

Trains passing a telegraph post or a stationary man 1.How many seconds will a train 100 metres long running at the rate of 36 km an hour take to pass a certain telegraph post?

Solution: In passing the post the train must travel its own length. Now,

36 km/hr

=

36 ×5/18

=

10 m/sc.

∴ Required time = 100/10 = 10 seconds.

Trains crossing a bridge or passing a railway station 2.How long does a train 110 metres long running at the rate of 36 km/hr take to cross a bridge 132 metres in length?

Trains running in opposite direction 3.Two trains 121 metres and 99 metres in length respectively are running in opposite directions, one at the rate of 40 km and the other at the rate of 32 km an hour. In what time will they be completely clear of each other from the moment they meet?

Trains running in the same direction 4. In example above. If the trains were running in the same in what time will they be clear each other?

direction,

Trains passing a man who is walking 5. A train 110 metres in length travels at 60 km/hr. In what time will it pass a man who is walking at 6 km an hour (i) against it; (ii) in the same direction?

Solution: This question is to be solved like the above examples 3 and 4, the only difference being that the length of the man is zero.

6. Two trains are moving in the same direction at 50 km/hr and 30 km/hr. The faster train crosses a man in the slower train in 18 seconds. Find the length of the faster train.

7. A train running at 25 km/hr takes 18 seconds to pass a platform. Next, it takes 12 seconds to pass a man walking at 5 km/hr in the opposite direction. Find the length of the train and that of the platform.

SIMPLE INTEREST

INTEREST It is money paid by borrower for using the lender's money for a specified period of time. Denoted by I.

PRINCIPAL The original sum borrowed. Denoted by P.

TIME Time period for which the money is borrowed. Denoted by n

RATE OF INTEREST Rate at which interest is calculated on the original sum. Denoted by r.

AMOUNT Sum of Principal plus Interest. Denoted by A.

SIMPLE INTEREST The interest calculated every year on original principal, i.e. the sum at the beginning of first year. Denoted by SI. SI = Pnr A=P+SI

COMPOUND INTEREST The interest is added to the principal at the end of each period to arrive at the new principal for the next period. OR The amount at the end of year will become principal for the next year and so on. Let P be principal borrowed at the beginning of period 1. Amount at end of period n=1 is A= P (1+r/100) Then,

New Principal at the beginning of period 2 will be A i.e. P (1+r/100) = P*R where R=(1+r/100).

Lets’ checkout the applicability of the above concept with an example Consider P at the beginning of year of Rs 100 and r=10% p.a. Now, for the next three years the calculation of simple and compound interest is as follows:

Under Simple Interest Intere Intere st till st for the the end year of the year

Under compound interest

Amou nt at the end of the year

Princip al at the beginn ing of the year

Inter est for the year

Inter est till the end of the year

Amou nt at the end of the year

Year

Princip al at beginn ing of year

1

100

10

10

110

100

10

10

110

2

100

10

20

120

110

11

21

121

3

100

10

30

130

121

12.1

33.1

133.1

As can be seen from table,

UNDER SIMPLE INTEREST

UNDER COMPOUND INTEREST

P is same for every year

A at the end of every year = P for next year

I is same for every year

I is different for each year.

Examples #1 Find the simple interest, If 1. 2. 3. 4. 5. 6. 7.

P P P P P P P

= = = = = = =

Rs.1000, R = 20% per annum, T = 4 years. Rs.600, R = 5% per annum, T = 4 months. Rs.200, R = 6% per six months, T = 3 years. Rs.500, R = 2% per six months, T = 5/2 years. Rs.400, R = 3% per three months, T = 2 months. Rs.730, R = 10% per annum, T = 120 days. Rs. 3000, R = 61/4 per annum, T = period from 4th Feb to 18th Apr.

# Solution 1.

4×20×10 ⇒ 800

2.

2×5 = 10

3.

6×2×3×2 = 72

4.

5×2×5=50

5.

4×2=8

6.

73/3=24

7.

37.50

#2 Find the following: 1. 2. 3. 4. 5.

P = Rs. 100, R = 3% per annum, T = 2 year, A= ? P = Rs. 500, R = 6% per annum, T = 4 months, A= ? P = Rs. 400, R = 3.65% per annum, T = 150 days, A= ? A = Rs. 540, S.I = Rs. 108 , R = 5%, T = ? A = Rs. 1,120, R = 5%, T = 22/5 yr, S.I = ?

# Solution: 1.

S.I = 6 ; A = S.I + principal ; A = 6 + 100 ⇒ 106

2. 3.

S.I = 10 ; A = S.I + P ; A = 10+500 ⇒ 510 S.I = 6 ; A = 400 + 6 ⇒ 406

4.

T = 5 yr.

5.

120

#3 1.

A sum of money lent out at simple interest amounts to Rs. 720 after 2 years and to Rs. 1020 after a further period of 5 years. Find the sum and the rate %. 2. Adam borrowed some money at the rate of 6% p.a. for the first two years, at the rate of 9% p.a. for the next three years, and at the rate of 14% p.a. for the period beyond five years. If he pays a total interest of Rs. 11,400 at the end of nine years , how much money did he borrow ?(Bank P.O 1999) 3. A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 61/4% p.a. for 2 years. Find his gain in the transaction per year.(S.S.C.2000) 4. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs. 1164 in 31/2 years.Find the sum and the rate of interest?

5.

The simple interest on a certain sum of money for 21/2 years at 12% per annum is Rs. 40 less than the simple interest on the same sum for 31/2 years at 10% per annum. Find the sum. Practice Set For Simple Interest

# Solution 1.

Principal = 600, R = 10%

2.

12000

3.

112.50

4. 5.

[ 1164-1008 = 156 ] ⇒ 156/3×4 = 208 ; R = 208/2×800×100 ⇒ 13 7x /20 - 3x/10 = 40 ⇒ x = ( 40 × 20 ) ⇒ x = 800 [ Hint : Given Below ]

Q1. Mr. Hamilton invested an amount of Rs. 13,900 divided in two

different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B? a) Rs. 6400

b) Rs. 6500

d) Rs. 7500

e) None of these

c) Rs. 7200

Q2. How much time will it take for an amount of Rs. 450 to yield Rs. 81

as interest at 4.5% per annum of simple interest? a) 3.5 years d) 5 years

b) 4 years

c) 4.5 years

e) None of these

Q3. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of

simple interest. What is the rate of interest? a) 3% d) 6%

b) 4%

c) 5%

e) None of these

Q4. An automobile financier claims to be lending money at simple

interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes: a) 10% d) Data inadequate

b) 10.25% e) None of these

c) 10.5%

Q5. Aastha lent Rs. 5000 to Bahubali for 2 years and Rs. 3000 to Chinky

for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is: a) 5%

b) 7%

d) 10%

e) None of these

c) 7 1/8%

Q6. Aman took loan from a bank at the rate of 12% p.a. simple interest.

After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was: a) Rs. 2000

b) Rs. 10,000

d) Rs. 20,000

e) None of these

c) Rs. 15,000

Q7. What will be the ratio of simple interest earned by certain amount

at the same rate of interest for 6 years and that for 9 years? a) 1 : 3

b) 1 : 4

d) Data inadequate

e) None of these

c) 2 : 3

Q8. Akshay borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He

immediately lends it to another person at 6 ¼ pa for 2 years. Find his gain in the transaction per year. a) Rs. 112.50

b) Rs. 125

d) Rs. 150

d) Rs. 167.50

c) Rs. 150

Q9. On a sum of money, the simple interest for 2 years is Rs.660, while

the compound interest is Rs.696.30, the rate of interest being the same in both the cases. The rate of interest is : a) 10% d) Data inadequate

b) 10.5%

c) 12%

e) None of these

Q10. Mr.Devilal Singh invested an amount of Rs.13,900 divided in two

different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs.3508, what was the amount invested in Scheme B? a) Rs.6400

b) Rs.6500

d) Rs.7500

e) None of these

c) Rs.7200

Q11. What should be the least number of years in which the simple

interest on Rs.2600 at [6(2/3)]% will be an exact number of rupees? a) 2

b) 3

d) 5

e) None of these

c) 4

Q12. An amount of Rs.1,00,000 is invested in two types of shares. The

first yields an interest of 9% p.a. and the second, 11% p.a. If the total interest at the end of one year is [9(3/4)]%, then the amount invested in each share was :

these

a) Rs.52,500, Rs.47,500

b) Rs.62,500, Rs.37,500

c) Rs.72,500, Rs.27,500

d) Rs.82,500, Rs.17.500

e) None of

Q13. If the simple interest on a certain sum for 15 months at [7 (1 / 2)]

% per annum exceeds the simple interest on the same sum for 8 moinths at [12 (1 / 2)]% per annum by Rs.32.50, then the sum (in Rs.) is : a) Rs.3000

b) Rs.3060

d) Rs.3250

e) None of these

c) Rs.3120

A sum of money trebles itself in 15 years 6 months. In how many years would it double itself? Q14.

a) 6 years 3 months months d) 9 years 6 months

b) 7 years 9 months

c) 8 years 3

e) None of these

Rambo took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If he paid Rs. 432 as interest at the end of the loan period, what was the rate of interest? Q15.

a) 3.6

b) 6

d) Data inadequate

c) 18

e) None of these

==>> Must read - Quantitative Aptitude Tricks

Solutions 1. Option A Let the sum invested in scheme A be Rs. x and that in scheme B be Rs. (13900 ⎯x) Then,

[x

× 14 × 2 / 100

] ÷ [{(13,900 -

28x ⎯22x = 350800 ⎯(13900 × 22)

x) × 11 × 2 } / 100

] = 3508

6x = 45000 x = 7500 So, sum invested in Scheme B = Rs. (13900 ⎯7500) = Rs.6400

2. Option B Time =

[100 × 81 / 450 × 4.5 ] years = 4 years

3. Option D

S.I. = Rs. (15500 ⎯12500) = Rs.3000 Rate =

[ 100 × 3000 / 12500 × 4 ]% = 6%

4. Option B

Let the sum be Rs.100. Then, S.I. for first 6 months = Rs.

[100 × 10 × 1

/ 100 × 2

S.I. for last 6 months = Rs.

[105 × 10 × 1

/ 100 × 2

] = Rs.5

] = Rs.5.25

So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs.110.25 So, effective rate = (110.25 ⎯100) = 10.25%

5. Option D

Let the rate be R% p.a. Then,

[ 500 × R × 2

/ 100

] + [300 × R × 4

/ 100

] = 2200

100R + 120R = 2200 R=

[2200

/ 220

] = 10

So, rate = 10%

6. Option C

Principal = Rs.

[ 100 × 5400

/ 12 × 3

] = Rs.15000

7. Option C

Let the principal be P and rate of interest be R%. So, required ratio = [P × R × 6 / 100] / [P × R × 9 / 100] = 6PR / 9PR = 6 / 9=2:3

8. Option A Gain in 2 years = Rs.

[(5000 × 25 / 4 × 2 / 100

) ⎯(5000 × 4 × 2 / 100 )

]

= Rs. (625 ⎯400) = Rs.225 So, gain in 1 year = Rs.

[225

/2

] = Rs.112.50

9. Option E

Difference in C.I. and S.I. for 2 years – Rs. (696.30 ⎯660) = Rs.36.30 S.I. for one year = Rs.330

10. Option A

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 ⎯x) Then,

[x × 14 × 2

/ 100

] + [(13,900 - x) × 11 × 2

/ 100

] = Rs.3508

28x ⎯22x = 350800 ⎯(13900 × 22 ) 6x = 45000 x = 7500 So, sum invested in Scheme B = Rs. (13900 ⎯7500) = Rs.6400

11. Option B

S.I. = Rs.

[2600 × 20 / 3 × 1 / 100 × T] = Rs.[

520 / 3 × T

]

Which is an exact number of rupees when T = 3

12. Option B

Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 ⎯x)

[x × 9 × 1 39 / 4 × 1 / 100 ] Then,

/ 100

] + [ (100000 - x) × 11× 1 / 100 ] = [100000 ×

9x + 1100000 - 11x / 100 = 39000 / 4 = 9750 2x = (1100000 ⎯975000) = 125000 x = 62500 Sum invested at 9% = Rs.62500 Sum invested at 11% = Rs. (100000 ⎯62500) = Rs.37500

13. Option C

Let the sum be Rs. x. Then, 2 / 3 × 1 / 100

]

= 32.50 75x / 8 ⎯25x / 3 = 3250

[x × 15 / 2

× 5 / 4 × 1 / 100

] ⎯[x × 25 / 2 ×

25x = (3250 × 24) x=

[3250 × 24 / 25] = 3120

14. Option B

Let sum = x. Then, S.I. = 2x, Time = 15 (1/2) years = 31 / 2 years So, rate =

[ 100 × 2x / x

]

× (31/2) % = 400 / 31%

Now, sum = x, S.I. = x, Rate = 400 / 31% So, time = 100 × x / x × (400/31) = 31 / 4 years = 7 years 9 months

15. Option B

Let rate = R% and time = R years Then,

[1200 × R

× R / 100

] = 432

12 r = 432 2

R = 36 2

R=6

# COMPOUND INTEREST

Trick : Calculating Compound Interest for 3 Years

Calculating Compound Interest for 4 year

# Formulas

Case 1. When interest is not Compound yearly, Amount after 't' years A = P [1+ r/n×100]nt n= no of compounding per year When interest is compounded half yearly, n = 2 compounded quarterly, n = 4 compounded monthly, n = 12

Case 2. When rate % is no equal every year and interest is compounded yearly Basic formula : P [1+ r/100] [1+ r/100] ...upto 't' times But as rate % is not same every year, so A = P [1+ r1/100]t1 [1+ r2/100]t2 .... and so on Where R1 = Rate% p.a. for t1 years. and R2 = Rate % p.a. for t2 years. Case 3 When interest is compounded yearly but time is in fraction T = 53/4 years A= (whole part) × (fraction part of time ) A = P [1+ r/100]5 × [1+ /4/100] 3r

# Difference between Compound Interest and Simple Interest CI - SI = P [ R/100 ]2 When time t = 3 years

CI - SI = P [ (R/1003 +3 (R/100)2]

# Examples #1 If the compound interest on a certain sum for two years at 10% p.a. is Rs 2,100 the simple interest on it at the same rate for two years will be. ( RRB, 2009)  The compound interest on a sum for 2 years is Rs. 832 and the simple interest on the same sum for the same period is Rs. 800. The difference between the compound and simple interest for 3 years will be.  The difference between simple interest and compound interest on a sum for 2 years at 8% when the interest is compounded annually is Rs. 16, if the interest were compounded half yearly, the difference in one interest would be nearly.  The difference in C.I and S.I for 2 years on a sum of money is Rs. 160.If the S.I for 2 years be Rs. 2880, the rate of percent is . Practice Set For Compound Interest 

# Solution 1. 2000

2. 98.56

3. 04

4.

IMPORTANT FORMULAE Let Principal = Rs. P, Time = t yrs and Rate = r % per annum

Ques 1. In what time will Rs 390625 amount to Rs 456976 at 4% compound interest?

Ques 2. A sum of money placed at compound interest doubles itself in 4 yrs. In how many years will it amount to eight times itself ? Solution :Quicker Approach:

X becomes 2x in 4 yrs.

2x becomes 4x in next 4 yrs.

4x becomes 8x in yet another 4 yrs.

Thus, x becomes 8x in 4 + 4 + 4 = 12 yrs.

Ques 3. Find the least number of complete years in which a sum of money at 20% CI will be more than doubled.

Ques 4. A sum of money at compound interest amounts to thrice itself in three years. In how many years will it be 9 times itself?

Quicker Method: Remember the following conclusion: If a sum becomes x times in y years at CI then it will be (x) n times in ny years.

Thus, if a sum becomes 3 times in 3 years it will be (3) 2 times in 2 x 3 = 6 years.

Example: If a sum deposited at compound interest becomes double in 4 years when will it be 4 times at the same rate of interest?

Solution: Using the above conclusion, we say that the sum will be (2) 2 times in 2 x 4 = 8 years.

TO FIND RATE Ques 5. At what rate per cent compound interest does a sum of money become nine-fold in 2 years? Solution :-

Ques 6. At what rate percentage (compound interest) will a sum of money become eight times in three years ?

Ques 7. At what rate per cent compounded yearly will be Rs. 80,000 amount to Rs 88,200 in 2 years?

GIVEN CI, To find SI and vice versa Ques 8. If the CI on a certain sum for 2 years at 3% be Rs. 101.50, what would be the SI?

GIVEN CI AND SI, TO FIND SUM AND RATE Ques 9. The compound interest on a certain sum for 2 yrs is Rs 40.80 and simple interest is Rs. 40.00. Find the rate of interest per annum and the sum.

Solution: A little reflection will show that the difference between the simple and compound interests for 2 yrs is the interest on the first year’s interest.

First year’s SI = Rs 40/2 = Rs 20

CI – SI = Rs 40.8 – Rs 40 = Rs 0.80

Interest on Rs 20 for 1 year = Re 0.80

PROFIT AND LOSS

VARIOUS PROFIT AND LOSS FORMULAS USED IN PROFIT AND LOSS: 1) Generally, profit is calculated as: Profit or gain = Selling price(S.P) - Cost price(C.P) 2) Similarly, Loss = Cost price - Selling price 3) Gain percentage(%) = Gain * 100 C.P. 4) Loss percentage(%) = Loss * 100 C.P. 5) There is a direct relationship between selling price and cost price: S.P. = 100 + Gain percentage * C.P. (In case of gain) 100 S.P. = 100 - Loss percentage * C.P. (In case of loss) 100

FOR EXAMPLE : If an article is sold at gain of 27%, then by using first formula , you can find that S.P. is 127% of C.P. Similarly, If an article is sold at loss of 18%, then by using second formula , you can find that S.P. is 82% of C.P. 6) If a person sells two commodities at same prices. On one he gains x% and loses x% on another, then as a whole he will be in loss and the loss percentage will be equal to: (Common gain or loss percentage)^2 = x^2 100 10 Note: Here is an example to find gain in case of dishonesty. Problem 1: A dishonest dealer professes to sell his goods at cost price but he uses a weigh 960 grams for 1 kg. How too calculate gain percentage? Solution: Gain percentage =

Error True value - Error

* 100 = 40 * 100 (Ans in %) 960

THE RULE OF FRACTION SAYS: If our required value greater than the supplied value we should multiply the supplied value with a fraction which is more than one. And if our required value is less than the supplied value we should multiply the supplied value with a fraction which is less than one.

 If there is a gain of x%, the calculating figure would be 100 and (100 + x).  If there is loss of y%,the calculating figure would be 100 and (100 – x).  If the required value is more than the supplied value, our multiplying fractions

should be (100+x)/100 or 100/(100-

y) ( both are greater than 1).

 If the required value is less than the supplied value, our multiplying fractions should be 100/(100+x) or (100-y)/100 (both are less than 1).

Profit = selling price (SP) – cost price (CP)

Loss = Cost price (CP) – Selling price (SP)

TO FIND THE GAIN OR LOSS PER CENT :The profit or loss is generally reckoned as so much per cent on the cost.

Gain or loss per cent = (Loss or gain)/CP x 100

Ques - A man buys a toy for Rs 25 and sells it for Rs 30. Find his gain per cent. Solution: % gain = Gain/CP x 100 = 5/25 x 100 = 20 %

Ques - A boy buys a pen for Rs 25 and sells it for Rs. 20. Find his loss per cent. Solution: % Loss = Loss/CP x 100 = 5/25 x 100

Ques - If a man purchase 11 oranges for Rs. 10 and sells 10 oranges for Rs 11. How much profit or loss does he make? Solution: Suppose that the person bought 11 x 10 = 110 oranges. ⇒ CP of 110 oranges = 10/11 x 110 = Rs 100

⇒ SP of 110 oranges = 11/10 x 110 = Rs 121

∴ Profit = Rs 121 – Rs 100 = Rs 21

QUICKER METHOD: REWRITE THE STATEMENT AS FOLLOWS:

Purchases

11 oranges for Rs 10

Sells

10 oranges for Rs 11

Now, percentage profit or loss is given by:

Since the sign is +ve, there is a gain of 21%

Ques - A man purchases 8 pens for Rs 9 and sells 9 pens for rupees 8. How much profit or loss does he make? Solution: quicker method:

Purchases

8 pens for Rs 9

Sells

9 pens for Rs 8

⇒ Since the sign is –ve, there is a loss of 20.98%

Ques - A dishonest dealer professes to sell his goods at cost price, but he uses a weight of 960 gm for the kg weight. Find his gain per cent. Solution: Suppose goods cost the dealer Re 1 per kg. He sells for Rs 1 what cost him Rs. 0.96. ∴

Gain on Re 0.96 = Re 1 – Re 0.96 = Re 0.04

AVERAGE An average, or more accurately an arithmetic mean is, in crude terms, the sum of n different data divided by n. For example, if a batsman scores 35, 45 and 37 runs in first, second and third innings respectively, then his average runs in 3 innings is equal to (35+45+37)/3 = 39 runs. Therefore, the two mostly used formulas are:

Ques 1. The average age of 30 boys of a class is equal to 14 yrs. When the age of the class teacher is included the average becomes 15 yrs. Find the age of the class teacher.

Solution: Total ages of 30 boys = 14 × 30 = 420 yrs. Total ages when class teacher is included = 15 × 31 = 465 yrs. ∴ Age of class teacher = 465 - 420 = 45 yrs.

Direct formula: Age of new entrant = New average + No. of old members × increase in average = 15 + 30 (15 – 14) = 45 yrs.

Ques 2. The average weight of 4 men is increased by 3 kg when one of them who weighs 120 kg is replaced by another man. What is the weight of the new man?

Solution: Quicker approach: If the average is increased by 3 kg, then the sum of weighs increases by 3 × 4 = 12 kg.

And this increase in weight is due to the extra weight included due to the inclusion of new person. ∴ Weight of new person = 120 + 12 = 132 kg.

Direct formula: Weight of new person = weight of removed person + No. of persons × increase in average = 120 + 12 × 3 = 132 kg.

Ques 3. The average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of the failed candidates is 15, what is the number of candidates who passed the examination?

Ques 4. The average of 11 results is 50. If the average of first six results is 49 and that of last six is 52, find the sixth result.

Solution: The total of 11 results = 11 × 50 = 550 The total of first 6 results = 6 × 49 = 294 The total of last 6 results = 6 × 52 = 312 The 6th result is common to both;

Therefore, Sixth result = 294 + 312 – 550 = 56

Ques 5. The average age of 8 persons in a committee is increased by 2 years when two men aged 35 yrs and 45 yrs are substituted by two women. Find the average age of these two women.

Ques 6. The average age of a family of 6 members is 22 years. If the age of the youngest member be 7 yrs, then what was the average of the family at the birth of the youngest member?

Ques 7. A man bought 13 shirts of Rs 50 each, 15 pants of Rs 60 each and 12 pairs of shoes at Rs 65 a pair. Find the average value of each article.

Ques 8. The average score of a cricketer in two matches is 27 and in three other matches is 32. Then find the average score in all the five matches.

Ques 9. The average of 11 results is 30, that of the first five is 25 and that of the last five 28. Find the value of the 6th number.

Ques 10. In a class, there are 20 boys whose average age is decreased by 2 months, when one boy aged 18 yrs is replaced by a new boy. Find the age of the new boy.

Ques 11. A batsman in his 17th innings makes a score of 85, and thereby increases his average by 3. What is his average after 17 innings?

Ques 12. A cricketer has completed 10 innings and his average is 21.5 runs. How many runs must he make in his next innings so as to raise his average to 24?

RATIO AND PROPORTION

1. Ratio When two numbers are represented in the form of another; this is done by expressing one number as a fraction of another. Thus, we have a:b; where a is the antecedent, and b is the consequent (a little general knowledge doesn't hurt even in math!) Thus when we write 4:16, it can be re-written as 2:8 and further simplifying it can be said to be 1:4. Which actually means, the number ‘4’ is 4 times to get the figure 16. And all of you know, that 1:4 can also be written as ¼.

2. Ratios to percentages: This 1/4 ratio can be denoted as a percentage too! It’s 25% How? Well, 1/4 x 100 = 25% [1/4th is also known as one quarter, that is one part out of 4 parts.] 2/4 = 1/2 =50% 3/4 = 75% and, 4/4 = 100%

3. Ratios to Degrees: Supposing, we have A:B:C:D, being four farmers, who have contributed Rs. 25,000, Rs. 75,000, Rs. 65,000 and Rs. 35,000 respectively. Using all their contributions, they have purchased a land, which surprisingly is circular! (C’mon Math need not be boring!) They decided that they would all receive a part of the circular land based on

their contribution; how will they divide the circular land? So one of them who had completed his class 12, suggested they divide the land on a pie chart model! Total contribution = Rs. 2,00,000 (25000+75000+65000+35000) Ratio of contribution = 25000:75000:65000:35000 = 25:75:65:35 (always cancel off the 000s first!) = 5:15:13:7 (this is our last stage, where no more common factors

are possible, where total of the ratios is 40) We know, if we need A’s share, then A’s ratio = 5/40, B’s = 15/40, C’s = 13/40 and D’s = 7/40. Total area = 360°, then A’s share of the total area/ share of 360° = 360 x 5/40 = 45° B’s share = 360 x 15/40 = 135° C’s share = 360 x 13/40 = 117° and D’s = 360x7/40 = 63°

4. Comparing two ratios or fractions: If you are give two fractions, say 5/3 and 7/8, and we need to find out which fraction is greater than the other what do we do?

We can either calculate their percentages, 5/3x100 = 167% and 7/8x100 = 88% (approx values) So we know, that 5/3 is the greater fraction! Another method can be to compare the two fractions. And to compare we have to make their denominators equal. To make their denominators equal, for the first fraction, we multiply 8 to both the numerator and denominator. Therefore the first fraction = 40/24. The second fraction, by multiplying 3 to make the second fraction = 21/24. Obviously 40 is greater than 21! Hence, 40/24 or 5/3 is greater than 21/24 or 7/8!

5. Proportions Proportions is where two ratios are compared and equated. Where a:b is a ratio and c:d is another ratio, and if they are equal, then, they can be re-written as a:b :: c:d. { the ‘::’ sign means ‘equal to’} Therefore,

a:b = c:d

1:2 :: 7:14 , try out the simplifications! ‘a’ and ‘d’ are called extremes as they are in the extreme ends! And ‘b’ and ‘c’ are called means as they are in the middle!

6. Properties of proportions (i) a:b = c:d can be written as a/b = c/d which implies, axd = bxc, or, ad = bc

this property helps in solving many questions (ii) if a:b = b:c, which means this proportions between three numbers is in the form of ‘continued proportions’, as all three numbers are having a connection. so, a/b=b/c or, ac=bxb or, ac=b2 he number of times one quantity contains another quantity of the same kind is called the ratio of two quantities. Clearly, the ratio of two quantities is equivalent to the fraction that one quantity is of the other. Observe carefully that two quantities must be of the same kind.

There can be a ratio between Rs. 20 and Rs. 30, but there can be no ratio between Rs.20 and 30 mangoes. The ratio 2 to 3 is written as 2 : 3 or 2/3. 2 and 3 are called the terms of the ratio. 2 is the first term and 3 is the second term. The first term of a ratio is called the antecedent and the second the consequent.

==>> Must read - Ratio and Proportions Tricks

COMPOUND RATIO

Ratios are compounded by multiplying together the antecedent for a new antecedent, and the consequents for a new consequent.

Ques 1. Find the ratio compounded of the four ratios: 4 : 3, 9 : 13, 26 : 5 and 2 :15

INVERSE RATIO 1.

If 2 : 3 be the given ratio, then 1/2 : 1/3 or 3 : 2 is called its inverse or reciprocal ratio.

2.

If the antecedent = the consequent, the ratio is called the ratio of equality, such as 3 : 3.

3.

If the antecedent > the antecedent, the ratio is called the ratio of greater inequality, as 4 : 3.

4.

If the antecedent < the consequent, the ratio is called the ratio of less inequality, as 3 : 4.

Ques 2. Divide 1458 into two parts such that one may be to the other as 2 : 7.

Ques 3. Find three numbers in the ratio of 1 : 2 : 3, so that the sum of their squares is equal to 504. Solution:

Let the numbers be x, 2x, 3x. Then we have,

⇒ X2 + (2x)2 + (3x)2 = 504 ⇒ 14x2 = 504

⇒ X = 6 Hence, the required numbers are 6, 12 and 18.

Ques 4. The sum of three numbers is 98. If the ratio between the first number and second number be 2 : 3 and that between the second and third be 5 : 8, then find the second number.

Ques 5. The ratio of the money with Rita and sita is 7 : 15 and that with Sita and Kavita is 7 : 16. If Rita has Rs. 490, how much money does Kavita have ?

PROPORTION Consider the two ratios:

1st ratio

6 : 18

2nd ratio

8 : 24

Since 6 is one-third of 18, and 8 is one-third 24, the two ratios are equal. The equality of ratio is called proportion.

The number 6, 18, 8 and 24 are said to be in proportion.

The proportion may be written as 6 : 18 : : 8 : 24 (6 is to 18 as 8 is to 24)

Or,

6 : 18 = 8 : 24

Or, 6/18 = 8/24

If four quantities be in proportion, the product of the extremes is equal to the product of the means. Let the four quantities 3, 4, 9 and 12 be in proportion, We have

3/4 = 9/12

Three quantities of the same kind are said to be in continued proportion when the ratio of the first to the second is equal to the ratio of the second to the third.

The second quantity is called mean proportion between the first and third; and the third quantity is called the third proportional to the first and second.

Ques 6. Find the fourth proportional to the numbers 6, 8 and 15. Solution:

If x be the fourth proportional, then 6 : 8 = 15 : x

∴ X = (8 × 15 )/6 = 20

PERCENTAGE

Playing with Percentages Technique 1 If out of the total apples 60% are bad, i.e., 100% - 60% = 40% is good. So, when we are asked the number of good apples, simply do the mental calc (short form of calculation!) and get the figure of 40%, the multiply with the total number of apples say 600, which is : 40 x 600 / 100 = 240 apples

Technique 2 You already know that the ‘00s in 600 and ‘00s in 100 get cancelled, then why waste time writing down the entire figure? Simply multiply 40 x 6 = 240!

Technique 3 Also, 40% can be written as 0.4 or simply .4 – how? 40/100 is 0.4! So you can multiply 600 with 0.4 and put the decimal point to get 240! When I say multiply with 0.4, you should actually multiply with 4, (i.e. 600 x 4), which

gives us 2400 and then put a decimal. Gives you 240.

Technique 4 In DI Where the total number of students is given as 1950, and you are asked to find the total of Computer and Bio students, what do you do? You add up their respective percentages and then find the total figure. 28 + 26 = 54% 54/100 * 1950 = 1053 (I’m showing the complete calculations, but you are to always use the shortcuts!) If they ask the difference between the number of students in Chemistry and Commerce…then 21 – 15 = 6% and 1950 x 6% =117 students. If they ask the difference between the number of students studying computer and commerce and chemistry and arts… We add up the % figures of computer and commerce = 43%, and of chemistry and arts which is = 31%, and then get their difference = 4331= 12 %, thus, 12% of 1950 = 234 students. Please Note: This method can only work when only percentages are given;

where ratios are also given the method of calculation will be different. We’ll discuss that when doing Ratio’s chapter.

Technique 5 Price of commodity! Bank PO’s favorite question! The short cut in these questions is the formulae which is printed in every quantitative aptitude book! Suppose if the Price of Oranges increases by 5%, then the reduction required in the consumption so as to maintain the original expenditure will be = 5/(100+5) x 100 = 4.76%. Here, reduction can only be calculated in % form as absolute figure is not given, i.e., number of oranges is not given. [ formula is R/(100+R) * 100] Similarly, if the price of oranges decreases by 5%, then the % increase in the consumption so as to maintain the original expenditure = 5/(100-95) x 100 = 5.26% [formula is R/(100+R) *100) If you are having a problem remembering the formula, then do it this way – the numerator will always be the percentage number, in our case ‘5’. If the price is increasing by ‘5’%, then in the denominator we add 5 to 100, i.e., 105. If the price is decreasing by ‘5’%, then in the denominator we subtract 5 from 100, i.e., 95.

Technique 6 % and Population! Another one of the PO googlies! Here, just remember one thing…population increase is exactly the same as compound

interest! Use the same concept here too… If the population of a town is, say 20,000 people as of today, and if population increases at 6% p.a., then what will be the population after 3 year? Do you see the similarity between this question and a compound interest question? Except for the semi-annually or quarterly compounding part, every other information can be seen in a similar fashion and that is why their formulae are same too! Population after 3 years = 20,000 x (1 + 6/100)3 = 23820 approx Formula being – Population x (1 +R/100)n where, n is the number of years.

If population is decreasing, we simply remove the ‘+’ sign and put a ‘–’ instead, i.e., Population x (1 – R/100)n Population increase or decrease in future is called calculation of a ‘future value’ and hence this formulae, where we multiply with the present population holds true. But when calculating a past value, we will need to divide the present population by the % increase/decrease for the required number of years. Thus our formula is, P/ (1+ R/100)n or P/ (1- R/100)n or increase or decrease in population respectively!

Technique 7 The above mentioned formulae and concepts hold true for depreciation sums as well. Depreciation means, the decrease in an assets value due to wear and tear or usage. When you purchase a mobile, Nokia Lumia or Samsung Galaxy, and suppose it costs you Rs.15,000 today. You use it for a year, and then you want to sell it, how much will you ask for? Rs.15000? How much will you get? Rs.15000? Surely not! Why not? The value goes down from 15000, but why? Because you and the buyer both know that once any asset has been used, it most probably will have some defects (scratched screen, weak battery, and phone getting way too warm). So, why will you pay the same amount for a defective piece, which you pay for the new phone! But what is the value of the old phone? How to get the proper value of the old phone? This is where the concept and formula of depreciation (which is similar to the population formulae) come to help! ercentage is a fraction whose denominator is always 100. x percentage is represented by x%.

To express x% as a fraction : We know x% = x/100 Thus 10% = 10/100 (means 10 parts out of 100 parts) = 1/10 (means 1 part out of 10 parts)

To express x/y as a percentage : We know that x/y = (x/y× 100 )

Thus 1/4 = ( 1/4 ×100 )% = 25% and 0.8 = ( 8/10 ×100 )% = 80%

If the price of a commodity increases by R%, then reduction in consumption as not to increase the expenditure is[ R/(100+R)×100 ] %

If the price of a commodity decreases by R%, then the increase in consumption as not to decrease the expenditure is [ R/(100-R)×100 ] %

Result on Population : Let the population of a town be P now and suppose increases the rate of R% per annum, then : 1. 2.

Population after n years = P ( 1+ R/100 )n Population n years ago = P /( 1+ R/100 )n

Result on Depreciation : Let the present value of a machine be P. Suppose depreciates at the rate of R% per annum Then : 1. Value of the machine after n Years = P ( 1- R/100 )n 2. Value of the machine n years ago =

P

/( 1- R/100 )n

 If A is R% more than B, then B is less than A by R [ /(100+R)×100 ]%

If A is R% less than B, then B is more than A by [ R/(100-R)×100 ]% 

Net % change = x + y + xy/100

Some Observation #1 If 20% candidate failed in an exam then observations are   

80% represent passed in exam 100% represent total appeared in exam (80%-20%) = 60% represent difference between passed and failed candidate in exam

#2 If a number is increased by 25% then observations are  

100% represent the old number 125% represent the new number.

#3 Remember that Base in the given sentence (Question) is always 100% Eg. Income of Ram is increased by 20% In this sentence

100% - represent the income of Ram 20% - represent increment 120% - represent new income of Ram.

Remember it : 1 = 100% 1/2 = 50% 1/3 = 33 1/3% 1/4 = 25% 1/5 = 20% 1/6 = 162/3% 1/7 = 142/7% 1/8 = 121/2% 1/9 = 111/3% 1/10 = 10% 1/11 = 91/11% 1/10 = 81/3% 1/13% = 79/13%

25% = 1/4 6.25% = 1/16

125% = 5/4 150% = 3/2 200% = 2 350% = 7/2

Examples #1 Q. If the difference between 62% of a number and 3/5th of that number is 36. what is the number ? Sol: Let the number be x. Then x × 62% - x × 3/5 = 36 x ×62% -x V 60% = 36 (60% = 3/5) x ×2% = 36 x ×2/100 =36 x=

36 ×100

/2 = 1800

#2 Q. 40% of Ram's income Rs. 1200 Then Find 1. 75% of Ram's income ? 2. 1/4 part of Ram's income ? 3. 1/3 part of Ram's income ? Sol : (1) 40% = 1200 Rs. 75% = 1200/40 ×75 = 2250 Rs.

Trick : 1200 / 40 × 75 = Rs. 2250/-

(2) 40% of income = Rs. 1200 Then 1/4 part (i.e. 25% ) of Ram's income = 1200/40 ×25 = Rs. 750/- Ans

(3) 40% of Ram's income = Rs. 1200 i.e. 2/5 part of Ram's income = Rs. 1200 Then total income of Ram = Rs. 1200 ×5/2 1/3 part of Ram's income = Rs. 1200 × 5/2 × 1/3 = Rs. 1000 Ans.

Trick : 1200

=

/2/5 × 1/3

1200

/2 × 5/3 = 1000

Question 1. Salary of A is 75% of B’s salary. Salary of A is increased by 40% but salary of B is increased by only 10%. Whose salary is more and by what percent? Solution:

Salary of A = 75% of Salary of B

Question 2. The price of a commodity increases by 25% and therefore a house wife reduces the quantity purchased by certain percentage in order to maintain expenditure. By what percentage should she reduce the quantity purchased ?

The short cut could be made on the basis that

Price ∝ 1/Quantity

Which means if you reduce the price , the quantity will increase and vice-versa, so our formula would be :

Question 3. The price of a commodity increases by 25%. Find out the reduction in quantity such that the expenditure is allowed to increase by 20%?

Now again let the price = 100, Quantity =100

Expenditure =100 X 100= 10000

New Price =125, New Quantity =?

New Expenditure =10000+20%of 10000=12000

New Expenditure= New Price X New Quantity

12000 = 125 X New Quantity

New Quantity = 12000/125= 96 Required reduction% in the quantity = (100-96)/100 X 100 =4%

Question 4. The price of a commodity increases by 20% and expenditure is allowed to increase by 15%. What changes would occur in the quantity ?

Question 5. The price of an apple reduce by 10% and thus enables a person to buy 10 more apples for Rs. 50. Find out : 1.

New price

2.

Old Price

3.

New Quantity

4.

Old Quantity

Solution:

Let us say, initially he is able to buy m apples for Rs.50. Initially

Rs.50

= m apples

Now

Rs. 50 = (m+10) apples

Now we can see that

⇒ Earlier in Rs. 50 , 90 apples could be bought

Now in Rs. 50, 90+10=100 apples could be bought. which clearly shows that the price has dropped

New price = Rs. 50/100

Old price = Rs. 50/90

Now let us learn a quick way

The person has got = Rs 50

The price reduces by 10 % that means he could save 10% of what he has in his pocket.

He could buy the apples in Rs 45 and can save Rs 5, but instead of saving, he is buying 10 apples for Rs 5 ⇒ Rs. 5 = 10 apples

Rs. 1 = 2 apples which means Rs 50 = 100 apples

The new price of one apple = Rs.0.5

Earlier Rs 50 = 100-10 = 90 apples

The old price per apple = 50/90= Rs. 0.55

So this was another method to deal with such questions. Question 6. A container has got 130 liters of water which has 15% salt. From this container some of the water has got evaporated and after evaporation, the salt strength in the container is now 20%. Find out how much water evaporated from the container. Solution - Let the water evaporated= m liter

AGE ‘Problems based on Ages’ – is a very popular question in Clerical Exams; it can come in either Reasoning or in Quantitative Aptitude. But it will come. I know some people leave this one out because its way too confusing to ‘em – well, let us try to see how to not get confused.  The important thing in any kind of Age Problem, is to decide which age – present or past or future – to be taken as ‘x’!

 

  

Let us make a simple rule for ourselves – the ‘x’ should be the present age always. In most cases, taking the present age as ‘x’, i.e., the base year works just fine. Past will become, say (x-5) years, and future can be denoted as (x+5). But sometimes, ‘present age’ is not directly given in words. Then, take ‘x’ to be the age you are supposed to find. You can also try putting yourself in someone’s place and try to calculate the age! Also, sometimes – when nothing works and you’re stuck on an age question in the last 4 minutes of the exam – just look at the options and solve it through back calculations! Works just fine!

Example 1 Raman's age after 15 years will be 5 times his age 5 years back. What is his present age ? Solution - Let's assume right now it is year 2000

Age of Raman in 1995 = x Age of Raman in 2015 = 5x Present age of Raman (in 2000) = x+5 or 5x-15

we will solve these two equation to find x.

X= 5. Then Raman's present age becomes = x +5 = 10

Example 2 Rahul was 4 times old as his son 8 years back and he will be 2 times old as his son after 8 years. Calculate Rahul and his son's age.

Assume that currently it is year 2000. In 1992 Rahul's age = 4x, Age of Rahul's son = x

In 2008 Rahul's age = 2y and Age of Rahul's son = y

Now we get two equations 2y - 4x = 16 and y - x = 16 By solving this equation x = 8, so Rahul' son's current age = 16 years and Rahul's age = 40 years.

Ques 1. The age of the father 3 years ago was 7 times the age of his son. At present the father’s age is five times that of his son. What are the present ages of the father and son?

Ques 2. At present the age of the father is five times that age of his son.Three years hence, the father’s age would be four times that of his son. Find the present ages of the father and son.

Ques 3. Three years earlier the father was 7 times as old as his son. Three years hence the father’s age would be four times that of his son. What are the present ages of the father and son?

Solving using Conventional Method: Solution: 1 Let the present age of son = x yrs. Then, the present age of father = 5x yrs.

3 years ago, 7(x – 3) = 5x – 3

Or, 7x – 21 = 5x – 3 Or, 2x = 18 Therefore, son’s age = 9 yrs. Fathers’ age = 45 years.

Solution: 2. Let the present age of son = x yrs.

Then, the present age of father = 5x yrs.

3 years hence, 4(x + 3) = 5x + 3 Or, 4x + 12 = 5x + 3 ∴ X = 9 yrs. Therefore, son’s age = 9 yrs. Fathers’ age = 45 years.

Solution: 3. Let the present age of son = x yrs. And the present age of father = y yrs. 3 yrs earlier

7(x – 3) = y – 3 or, 7x – y = 18……….(i)

3 yrs hence

4(x + 3) = y + 3 or, 4x – y = -9………..(ii)

Solving (i) & (ii) we get, x = 9 yrs & y = 45 yrs.

Quicker Method:

Ques 4. 10 yrs ago, Sita’s mother was 4 times older than her daughter. After 10 yrs, the mother will be twice older than the daughter. What is the present age of Sita?

Ques 5. One years ago the ratio between Samir’s age and Ashok’s age was 4 : 3. One year hence the ratio of their ages will be 5 : 4. What is the sum of their present ages in yrs.

PERMUTATION AND COMBINATION Permutation implies arrangement where order of things is important and includes word formation, number formation, circular permutation etc. Combination means selection where order is not important and it involves selection of team, forming geometrical figures, distribution of things etc. Factorial = Factorial are defined for natural numbers, not for negative numbers. n! = n.(n-1).(n-2).........3.2.1 For example: 1) 4! = 4.3.2.1 = 24 2) 6!/ 4! = (6.5.4!)/ 4! = 6.5 = 30 3) 0! = 1

PERMUTATION

COMBINATION

Implies Arrangement

Implies Selection

Order of things is important

Order of things is NOT important

Permutation of three things a, b and c taking two at a time are ab, ba, ac, ca,bc and cb (Order is important).

Combination of three things a,b and c taking two at a time are ab, ca and cb (Order is not important).

nPr = n!/ (n-r)!

nCr = n!/ (n-r)! r!

nPn = n!

nCn = 1

nP0 = 1

nC0 = 1

EXAMPLE OF WORD FORMATION: Example: How many new words can be formed with the word "PATNA"? Solution: In word "PATNA", P,T,N occurs once and A occurs twice. ****Always remember in word formation, if word repeats, number of repetition will be on denominator. So, total number of words that can be formed = 5!/ 2! = 60 Therefore, except PATNA there are 59 new words (60-1). Example: How many words can be formed from the letters of the word "EXAMINATION"? Solution: E, X, M, T, O : Occurs ONCE A, I, N : Twice So, total number of words = 11! / 2! 2! 2! (Total number of letters=11 and 3 letters are occurring twice)

PROBLEMS FOR PRACTICE Problem 1: Choose permutation or combination 1) Selection of captain and bowler for a play. Permutation 2) Selection of four students for a lecture. Combination 3) Assigning people to their seats during conference. Permutation

Problem 2: Evaluate 7P2 . 4P3 Solution: (7!/ 5!). (4!/ 1!) ⇒(7.6). (4.3.2) ⇒1008 Problem 3: Evaluate 5C2. 3C2 Solution: (5!/3!2!). (3!/2!1!) ⇒(5.4/2). (3) ⇒30 Problem 4: How many ways are there in selecting 5 members from 6 males and 5 females, consisting 3 males and 2 females? Solution: This is a case of combination i.e.selecting 3 males from 6 males and 2 females from 5 females. ⇒Required number of ways = (6C3 *5C2) ⇒(6.5.4/3.2)*(5.4/2) ⇒200. Problem 5: How many words can be formed by using letters of the word "DAUGHTER" so that the vowels come together? Solution: This is a case of permutation. In a word "DAUGHTER", there are 8 letters including 3 vowels (AUE) According to the question, vowels should always come together. Therefore, in this case we will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words (one word which is a group of vowels) These 6 words can be arranged in 6P6 ways ⇒6P6 = 6!/1! = 6! = 720 WAYS Also, three vowels in a group may be arranged in 3! ways ⇒3! = 6 ways Therefore, required number of words = (720*6) = 4320.



Permutation : It means arrangement where order of thing is considered.



Combination :

It means selection where order of thing is not considered.

# Methods  

Sum Rule Product Rule

# Case   

Simple Vowel comes together Vowel not comes together

Examples #1 Arrange the word " MANISH " in following way:

a) In how many ways the word " MANISH " can be arrange . b) Arrange the word " MANISH " if vowels come together. c) Arrange the word " MANISH " if vowel not comes together.

a) Sol: MANISH Total no. of words = 6 Find 6! ( Factorial = ! ) 6! = 6×5×4×3×2×1 6! = 720 Check whether there is any repeating words are there in " MANISH" because there is no repeating words are there that's why no further action is required. Hence Answer is 720.

b) Sol: word = " MANISH " Step 1. Count total no. of consonant and add 1 with it.Here, total no. of consonant = 4,now just add 1 with it .so it became (4+1=5) 5. Now let this outcome as factorial ( like 5!) Step 2. Count total no of vowels which is here 2. Now let this outcome as factorial ( like 2!)

Step 3. Multiply the result from Step 1 & 2. which would be like this 5! × 2! = 120 × 2 = 240. Step 4. Check for the repeating words. Now there is no repeating words are there that's why no further action is required. [Note : But what would you did if there were any repeating digit.To understand this.We Let word "CONDITION". Now here in this word, the repeating words are "O,N,I" then we multiply 2! upto three times (because no. of words repeating =3) and divide it with the outcome of Step 3. ] 240 is the correct answer for the word " MANISH ". c) Sol: Step 1. First of all , you need to find the total no. of possible ways in which a word can arrange without considering vowels. or Just like Sol: (a). Step 2. Now find the total no. of possible ways of arranging a word if vowels come together .Just like Sol: (b) Step 3. Subtract the outcome of Step 2 from Step 1.just like ( Sol (a) - Sol (b).

#2 In how many ways word " SUCCESS" can be arranged ? Sol: Total no of words = 7 Repeating words = C,S No of times repeating = (C= 2 times & S = 3 times) Hence 7!/3!×2! = 420

   

Arrangement of r out of n different things Arrangement with repetition Arrangement in a Row Circular Arrangement

Arrangement of r out of n different things

Example : How many three-digit numbers can be formed by using the digits in 452145, if repetition is not allowed? Sol: Total no. of digit (n) = 6 No. of digit to be taken at a time = 3 ( required for 3 digit number). Since repetition is not allowed, so, for making each 3- digit number, the digits chosen will be different. hence n=6 , r=3 number of 3 -digit number formed = npr = 6p3 = 6!/(6-3)!

Arrangement with repetition

Example: In a birthday party, Ram has 7 friends to invite.In how many ways can he send invitation cards to them if he has 4 servants to carry the cards and to deliver the same. Sol:

Required no of ways = ( repeating thing )non-repeatable thing Required no of ways = ( 4)7

Arrangement in a Row Let two groups .One group has n members or persons & second has m members or persons such that n ≥ m who are sit in a row in a way that no two people or person of the same sex sits together. No. of such sitting arrangement can be found out as :L

Example : In how many ways can 6 boys & 4 girls will sit if no two people of the same sex are allowed to sit together . Sol: There are two ways (1) B G B G....so on (2) G B G B...so on.

Circular Arrangement a) Considering Clockwise & Anti-Clockwise different No. of possible arrangement of 'n' different things arranged around a circle is (n-1)! b) Considered Not Different No. of circular permutation of n different things = 1/2 (n-1)! Example No. of possible ways in which 6 peoples can be arranged in a circular table . Sol: = (6-1)! = 5! Because there is no end point in circular arrangement so no of available space is reduced by 1.

COMBINATION Selection of r out of n different things

Example: In how many ways can a cricket team of eleven player be selected out of a batch of 15 players ? Sol: Required no. of ways =

15

C11 = 15C(15-11) = 15C4

= 15×14×13×12/4×3×2×1 = 1365

Division into groups

Example : In How many ways, Ram can select one or more of the 5 color balls from a box?

Sol: Required no. of selection = 25 - 1 = 31

PROBABILITY 1. What is probability? Probability is the chance of the happening or non-happening of event; denoted by ‘P’.

2. What are events and sample space? Sample space is the total number of occurrences that can happen. Event is the occurrence of ‘something’ which we are concerned with. For example: Jai and Veeru did that coin toss to gamble away their lives – awesome – but it has got an important lesson of probability too.

One coin – what are the possible out comes? – Two, as there can be a Head or a Tail. Therefore, our sample space (S) = 2 = total number of possible outcomes(either head or tail). Say, Veeru wanted Heads – how many heads is possible in one coin? – One. Thus 1 is our Event (E)!

Get Quantitative Aptitude Shortcuts 3. How to find Probability Probability is the chance of the occurrence of an event; [P = E/S] Thus, with E = 1, and S = 2,

Probability of Veeru going to die = E/S = ½ = 0.5! film - I know, I know!]

[But it was different in the

4. Hold on now! If the chances of Veeru’s death is 0.5; what could be the chance of Jai being the one dying? Again, ½ =0.5! probability!]

[Think the film makers calculated only this

5. The ‘Non – event’. Every event has its corresponding ‘non-event’; which can be denoted as E'. If the ‘event’ is happening, then non-event will not happen and vice versa. If Veeru is going to die (E), then Jai won’t (E'); if Jai’s (E) going to die then Veeru (E') won’t! Thus, P(E) + P(E') = 1 In words, probability of event and probability of a non-event add up to 1. Therefore, 1- P(E) = P(E'), 1- P(E') = P(E).

6. AND ‘n’ OR: First off – AND is multiplying. OR is for addition. If a question is worded like this – ‘if the probability of A hitting the target is 1/3 and B hitting the target is ½, what is the probability of A and B, both, hitting the target if a shot is taken by both.’ which means, P(A) AND P(B) = P(A and B hitting the target); P(A) x P(B) P(A) x P(B) = 1/6.

Now, if the question was worded - ‘if the probability of A hitting the target is 1/3 and B hitting the target is ½, what is the probability of A or B hitting the target?’ which means, P(A) OR P(B) = P(A or B hitting the target); P(A) + P(B) P(A) + P(B) = 5/6.

7. Some common sample space(s)! For Coins

One Coin

Two Coins

Three Coins

Sample Space (S) =

2

2x2=4

2x2x2=8

For Dice

One Dice

Two Die

Three Die

(S) =

6

6 x 6 = 36

6 x 6 x 6 = 216 and so on…

For Cards Cards in one suit (Either Spade, Clubs, Hearts or Diamonds)

One Pack of Cards/ Deck = Total number of cards

Face Cards (King, Queen, Jack and Ace) of all the fours suits together

(S)

13 x 4 = 52

4 x 4 = 16

and so on…

13

8. Concept of Odds: Sometimes probability is viewed in terms of ‘odds for’ or ‘odds against’ an event. Odds in favour of an event = P(E)/P(E') Odds against an event, or, Odds in favour of the non-event = P(E')/P(E)

… fairly simple, right? All you got to do is calculate the P(E) and the P(E'); then use the above formulae, if and only if the word ‘odds’ is in the question! Otherwise we calculate the normal probabilities as asked in the question.

IMPLE PROBABILITY PROBLEM Simplest questions asked in bank exams are asked about deck of cards, dice and coins. There is no co-occurrence of events. Lets take an example :Question -One card is drawn from a deck of 52 cards. Find the probability that it is a black queen or king. Simple solution to this problem is 2/52 = 1/26

VENN DIAGRAM Question - One card is drawn from a deck of 52 cards. Each card is equally likely to be drawn. Find the probability that card drawn is either black or queen. In this question we are going to use a simple formula P(AB) = A ∪ B ∩ AB

Don't be confused with the formula. Give a look to this diagram

In a deck of cards there are 26 black cards and 4 queens. But 2 black cards are queen. These cards are part of grey and brown circles in above diagram. As pink region is added twice, we need to deduct it once. So the solution is 26 + 4 - 2 = 28 Probability (AB) = 28/52 = 7/13

INDEPENDENT EVENTS In case you need to find probability of happening of two independent events we use multiplication theorem. P(AB) = P(A) * P(B) Lets understand this theorem with help of an example

Question - An unbiased coin is tossed twice. Find the probability of getting 1,2,3 or in the first toss and 4,5 or 6 in second attempt ? Solution - Probability of getting 1,2,3 or 4 in first toss = 4/6 = 2/3 Probability of getting 4,5 or 6 in second toss = 3/6 = 1/2 Now both events are independent and we will use multiplication theorem 1/2 * 2/3 = 1/3

IMPORTANT PROBABILITY QUESTIONS SET - VENN DIAGRAM This question is frequently asked in the exams. Even in IBPS 2012 and SBI 2013, this question appeared with slight difference in amounts. Still most of the candidates are unable to solve this. Question - There are 200 students in commerce batch of Khalsa college. Out of them 100 play cricket, 50 play hockey and 60 play basketball. 3o students play both cricket and hockey, 45 students play both cricket and basketball and 35 students play both hockey and basketball. Answer the following questions :1) What is the maximum number of students who play all the games ? 2) What is minimum number of students who play all three games ? 3) What is the maximum number of students who play at least one game ? 4) What is the minimum number of students playing at least one game ? Solution - To make the question simple, draw a Venn diagram and let x = number of students who play all games. Complete the diagram.

Solution 1 What is the maximum value of x ? Consider equation 30 - x, it states that maximum value of x = 30 In case x>30, for example 32 then 30 - x will be negative and number of students can't be negative. So answer is 30. Solution 2 What is the minimum value of x ? consider x-20 is possible as negative value of students is not possible. It is possible to deduct 20 from x. So minimum value of x = 20 Solution 3 and 4 For 3rd and 4th questions we will add everything :25 + x + x - 15 + x - 20 + 45 - x + 35 - x + 30 - x + x = = 100 + x Here we add maximum and minimum value of x so answer is 120 and 130 respectively.

Probability - Shortcut Trick for Dice Problems 

Today I am going to share basic concepts for Dice problems from Probability.

Types    

Cards Balls Coins Dice

If two dices are thrown: Sum of dices If 1 appears on the first dice and sum) (1, 1) = 1+1=2 If 1 appears on the first dice and (1, 2) = 1+2=3 If 1 appears on the first dice and (1, 3) = 1+3=4 If 1 appears on the first dice and (1, 4) = 1+4=5 If 1 appears on the first dice and (1, 5) = 1+5=6 If 1 appears on the first dice and (1, 6) = 1+6=7 After that dice will repeat like: (2, 1) = 2+1=3 (2, 2) = 2+2=4 (2, 3) = 2+3=5 (2, 4) = 2+4=6 (2, 5) = 2+5=7 (2, 6) = 2+6=8 Then gain (3, 1) = 3+1=4 …………. …………. …………. (3, 6) = 3+6=9 Similarly if

1 on the second dice, the sum = 2 (Minimum

2 on the second dice, the sum = 3 3 on the second dice, the sum = 4 4 on the second dice, the sum = 5 5 on the second dice, the sum = 6 6 on the second dice, the sum = 7

(6, 1) = 6+1=7 ………. ……….. And so on (6, 6) = 6+6=12 (Maximum) Now if two dices are rolled together Minimum sum =2 Maximum sum = 12 Now if the question says

What is the probability of getting a sum of 4 if two dices are rolled together? We can see that the sum of 4 can be obtained by:

Solution: Total out comes = 6*6=36 (because two dices are thrown so the total outcome will be 36) The required probability = Favorable outcomes /Total outcomes = 3/36=1/12

What is the probability of getting a sum of 6 if two dices are thrown? Solution: The sum of 6 can be obtained by:

So the probability = 5/36 Now to find out the probability of a sum when two dices are thrown, there is a shortcut which is based on the diagram below.

Note: Memorize this picture to solve probability sums Let us solve some problems

What is the probability of getting a sum of 5 if two dices are thrown together? Solution: Observe the picture very carefully we can see if the sum is 5, the probability is 4. So the required probability = 4/36= 1/9.

What is the probability of getting a sum of 10 if two dices are thrown together? From the picture if the sum is 10, the probability is 3. So the required probability is 3/36 that is 1/12.

What is the probability of getting a sum of 6 or 11 if two dices are rolled together?

Solution: From the picture Sum 6: probability 5 Sum 11: probability 2 The required probability = 5/36+2/36= 7/36

What is the probability of getting a sum of 4 or 12 if two dices are rolled together? Solution: From the picture Sum 4: Probability 3 Sum 12: Probability 1 So the required probability = 3/36+1/36 =4/36=1/9.

Probability Shortcut: 3 Dices Rolled Together Sum of dices when three dices are rolled together If 1 appears on the first dice, 1 on the second dice and 1 on the third dice.

(1,

1,

1)

=

1+1+1=3

The minimum sum with three dices rolled together = 3 If 6 appears on the first dice, 6 on the second dice and 6 on the third dice.

(6, 6, 6) = 6+6+6 =18

The maximum sum with three dices rolled together = 18 The sum could be 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and 18.



What is the probability of getting a sum of 4 if three dices are rolled together. Solution:

Total outcomes = 6x6x6 =216

Required probability = favorable case / total outcomes = 3/216 = 1/72



What is the probability of getting a sum of 5 if you throw three dices together? Solution:

Required Probability = 6/216 = 1/36

Now as you can see that it is a time consuming method, let us find a quick method to find the probability.

There are total 16 numbers (possible number of sums) and half of 16 is 8.

Counting 8 from both the sides we get 10 and 11 in the middle.

Now we will construct the triangle which would help us to find the probability if three dices are rolled together.

Memorizing the making of the above picture makes the calculation easier and faster. 

What is the probability of getting a sum of 11 if three dices are rolled together? Solution: From the table

Sum 11 = probability 27

Required probability = 27/216 =1/8



What is the probability of getting a sum of 14 if three dices are rolled together? Solution: From the table

Sum 14 = probability 15

Required probability = 15/216



What is the probability of getting a sum of 11 or 9 if three dices are rolled together? Solution:

From the table

Sum 11 = probability 27

and

sum 9 = probability 25

Since there is an ‘and’ so we will add

Required Probability = 27/216 +25/216 = 52/216



If 3 dices are rolled at the same time. What is the probability of all 3 dice showing even face values? Solution: Even faces means 2, 4, 6

There are 3 ways 2 can appear on in first go.

Similarly, there are 3 ways 4 can appear on in first go. . There are 3 ways 6 can appear on in first go.

So total number of ways: 3+3+3

Required probability = 9/216 =1/24



What is the probability of not getting 6 when 3 dice are rolled together? Solution:

Probability of not getting 6 on the first die = 5/6

(As probability of getting 6 on first die is 1 so the probability of not getting 6 = 6-1=5)

And

Similarly, Probability of not getting 6 on the second die = 5/6

And

Probability of not getting 6 on the third die = 5/6

So the required probability = 5x5x5/216=125/216

Remember AND means multiply and OR means ADD. Question 1 Out of 13 applicants for a job, there are 5 women and 8 men. Two persons are to be selected for the job. What is the probability that at least one of the selected persons will be a woman?

Solution: In this question – we need atleast one woman in the set of chosen two people – okay. This is how we could do it – 1st is a woman and 2nd a man OR st 1 a man and 2nd a woman OR st 1 is a woman and 2nd is also a woman. (We need atleast one woman! No ceiling on the max!) So our possible events are these above three … and any of the three can happen. All three possibilities of combination will not happen together – only one can happen and so … the OR becomes important to note! P(E) = (5/13 x 8/12) + (8/13 x 5/12) + (5/13 x 4/12) = 25/39 Our answer is 25/39. Note: The denominator in the second fractions is always 12. Why? Simply because the first fractions [5/13, 8/13, 5/13] means choosing 1 out of them – and since one is being chosen – it’ll leave 12 [13-1]! Okay, now you solve…

Get Quantitative Aptitude Tricks Book Question 2 There are two bags containing white and black balls. In the first bag, there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black. Ans: 41/77

Question 3 Out of 40 consecutive integers, two are chosen at random. Find the probability that their sum is odd. Ans.: 20/39

Question 4 From a bag containing 8 green and 5 red balls, three are drawn one after the other. Find the probability of all three balls being green if (a) the balls drawn are replaced before the next ball is picked (b) the balls drawn are not replaced Ans.: (a) 8/13 x 8/13 x 8/13 (b) 8/13 x 7/12 x 6/11

Question 5 In rolling two dices, find the probability that, (i)there is at least one ‘6’; (ii)the sum is 5. Ans: (i)11/36 (ii)1/9

Question 6 In a horse race there were 18 horses numbered 1-18. The probability that horse 1 would win is 1/6, that 2 would win is 1/10 and that 3 would win is 1/8. Assuming that a tie is impossible, find the chance that one of the three will win. [This is a great question – I hope you can do it!] Ans: 47/120

Question 7 If among the executives who have subscribed to the Time magazine, an executive is picked at random. What is the probability that he has also subscribed to the Economist? Ans: 3/8

Question 8 From a pack of 52 playing cards, three cards are drawn at random. Find the probability of drawing a king, a queen and jack. {Card question are always a must!} Ans: 6/5525

Question 9 A group of investigators took a fair sample of 1972 children from the general population and found that there are 1000 boys and 972 girls. If the investigators claim that their research is so accurate that the sex of a newborn child can be predicted based on the ratio of the sample of the population, then what is the expectation in terms of the probability that a newborn child will be a girl? Ans: 243/493

Question 10 Find a probability that a leap year chosen at random will have 53 Sundays. [Hint: How many complete weeks in a normal year? 52! That means we have accounted for 52 x 7 = 364 days. A Normal year has 365 days and Leap year has 366 days. Since, 52 weeks are a certainty – we know 52 Sundays will 100% occur! So we need to find the probability of either the 365th and 366th days being a Sunday!] Ans: 2/7

Probability Problems with Detailed Solutions Published on Saturday, September 12, 2015

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Q1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn

at random. What is the probability that the ticket drawn bears a number which is a multiple of 3? a) 3/10

b) 3/20

c)

2/5 d)

½

e)

None

of

these

Q2. In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he has offered English or Hindi? a)

2/5

b)

3/4

c)

3/5 d) 3/10

e) None of these

Q3. A bag contains 6 white and 4 red balls. Three balls are drawn at random. What is the probability that one ball is red and the other two are white? a)

1/2

b)

1/12

c)

3/10 d) 7/12

e) None of these

Q4. Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are kings, is: a) 7/13 63/221 d) 55/221

b) 3/26 e) None of these

c)

Q5. The probability that a card drawn from a pack of 52 cards will be a diamond or a king is : a) 2/13

b) 4/13

c)

1/13 d) 1/52

Answers Solution 1

e) None of these

(Option A)

Here, S = [1, 2, 3, 4, …., 19, 20] Let E = event of getting a multiple of 3 = [3, 6, 9, 12, 15, 18] P (E) = n (E) / n (S) = 6 / 20 = 3 / 10

Solution 2

(Option A)

P (E) = 30 / 100 = 3 / 10 , P (H) = 20 / 100 = 1 / 5 and P 100 = 1 / 10

P (E or H) = P (E U H) = P (E) + P (H) - P (E ∩ H) = (3 / 10) + (1 / 5) - (1 / 10) = 4 / 10 = 2 / 5

Solution 3

(Option A)

Let S be the sample space. Then,

(E ∩ H) = 10 /

n(S) = Number of ways of drawing 3 balls out of 10 = 10C3 =(10 × 9 × 8) / (3 × 2 × 1) = 120 Let E = event of drawing 1 red and 2 white balls n(E) = Number of ways of drawing 1 red ball out of 4 and 2 white balls out of 6 = ( 4C 1 × 6C 2 ) = 4 × (6 × 5) / (2 × 1) = 60 P (E) = n(E) / n(S) = 60 / 20 = 1 / 2

Solution 4

(Option D)

Clearly, n(S) = 52C2 = (52 × 51) / 2 = 1326 Let E1 = event of getting both red cards, E2 = event of getting both kings Then, E1 ∩ E2 = event of getting 2 kings of red cards. n (E1) = 26C2 = (26 ×25) / (2 × 1) = 325 ; n (E2) = 4C2 = (4 ×3) / (2 × 1) = 6 n ( E1 ∩ E2) = 2C2 = 1 P (E1 = nE1 / n (S)

= 325 / 1326

1326 P (E1 ∩ E2) = 1 / 1326 P (both red or both kings) = P (E1 U E2 )

= P (E1 ) + P (E2 ) = P (E1 ∩ E2)

= 325 / 1326 + 6 / 1326 - 1 / 1326

P (E2) = nE2 / n (S)

= 6/

= 330 / 1326 = 55 / 221

Solution 5

(Option B)

Here, n(S) = 52 There are 13 cards of diamond (including one king) and there are 3 more kings. Let E = event of getting a diamond or a king. Then, n(E) = (13 + 3) = 16 P (E) = 16 / 52 = 4 / 13

MULTIPLICATION While doing multiplication of a two digit number with another two digit number, we take at least 6 steps. Try yourself. Multiply 62 with 32.

Now let's do this with a trick

STEP 1 First step is same as conventional method, here we multiply 2 with 2.

STEP 2 This is an interesting step. Now multiply last digit first value and first digit of second value and vice-versa. Then we add outcomes. But we need the last number that is 8 here.

STEP 3 This is the last step, in this step we do multiplication ten's digit of both value and add the remainder from previous calculation. That's it, we completed the calculation in 3 steps instead of six steps.

We can use this method for multiplication of three or even four digit numbers but time management is really important in IBPS exam and other recruitment exams so for longer calculations, estimation is the best trick. I will post an article about how to do long calculations using estimation and result is 95% accurate which is enough to arrive at answer. Update 06-09-2013 As two of the readers namely Rahul and Ansh have requested me to use this technique in longer calculations multiplications. I am updating this article.

MULTIPLICATION OF 3 DIGIT NUMBERS In this example I will multiply 432 with 346. Now the 3 step multiplication method will become 5 step. This method can be used for 4 and even 5 digit numbers but as in bank exams there is lack of time available for calculations I recommend you to use approximation for long calculations.

Step 1

Step 2

Step 3

Step 4

Step 5

In case you find any difficulty to understand the above multiplication method then ask your question in the comments. I will try to answer every query asap. ASWER IS D

Alligation Alligation method is a simplified method to solve complex average problems. Alligation also helps simplify Ratio and Proportion, Simple and compound interest,Profit and loss, Time & Distance, Time Work problems among others. For better understanding, a few illustrations are given below. We take weighted average in the middle and average of components on the upper left and upper right hand side resulting in ratio.

Problem A class of consists of 60 boys and 50 girls. Average weight of boys is 60 kgs and of girls is 40 kgs. Calculate average weight of the class.

Solution You can solve this question either by traditional weighted average method or you can simplify the calculations by using alligation. 60-x/x-50 = 2/3 X = 56 By solving this question by alligation by taking individual averages above, as in this case 60 kg and 40 kg is individual averages(don’t know what you meant by this, correct it yourself). Take smaller average on the left hand side and bigger average on the right hand side. This is just a rule of thumb to avoid mistakes and make things easier. We always take weighted average in the middle. Then by deducting individual average from weighted average and vice versa we arrive at ratios at which these components were used.

Problem 2 There’s almost always a question on mixtures in a number of competitive exams. Here's an example. In what ratio should water be mixed with wine worth Rs. 60 per litre so that the seller earns a profit of 25% after selling the mixture for Rs. 50 per litre. Solution

Let's assume water is available for free, Cost price of mixture sold is 50 *80/100= 40, as 1/4 profit on sales price = 1/5 profit on cost price. As you can see, by just deducting weighted average cost from cost of component and vice verse, we arrive at ratio of components used.

Application of Alligation in Time and distance

Problem 3 A man travels part of his journey by bicycle at 20 Km/h and remaining distance by car at speed of 70 Km/h covering the entire journey at an average speed of 50 Km/h. What is the ratio of distance covered by bicycle and car?

Solution.

Solving this problem by alligation is a matter of just a few seconds, here’s how. As you can see in the picture, weighted average speed is 50 km/h so we keep it in the middle. Average speed of bicycle and car is taken as in visual components. We take average speed of bicycle on the left side and average speed of car on the right side (rule of thumb). By solving this simple alligation we find that ratio of time the man takes to complete the journey by bicycle and car is 2:3. But here we need to find the ratio of distance covered, so 2*20= 40 and for car 3*70=210

Ratio comes out to be 4:21.

Update - 10-October-2013 (Question requested by Chitra Salin) Problem 4 - This is from the practice section from PERCENTAGE notes given in website..pls help to solve either in alligation method or any A water tank contains 5% salt by weight. x litres of fresh water is added to 40 litres of tank water, so that the solution contains 2% salt. The value of x is a) 40 b) 50 c) 55 d) 60 Solution Percentage of water in current mixture = (100 - 5) = 95% Percentage of water in output mixture = (100 - 2 ) = 98%

Ratio of mixture and fresh water is 2:3. If there was 40 litres of mixture already available in the tank, we need add 40 × 3/2 = 60 litres

LCM and HCF tricks, problems and formulas LCM i.e. least common multiple is a number which is multiple of two or more than two numbers. For example: The common multiples of 3 and 4 are 12,24 and so on. Therefore, l.c.m.is smallest positive number that is multiple of both. Here, l.c.m. is 12.. HCF i.e. highest common factor are those integral values of number that can divide that number. LCM and HCF problems are very important part of all competitive exams.

SOME IMPORTANT L.C.M. AND H.C.F. TRICKS: 1) Product of two numbers = Their h.c.f. * Their l.c.m. 2) h.c.f. of given numbers always divides their l.c.m. 3) h.c.f. of given fractions =

h.c.f. of numerator l.c.m. of denominator

4) l.c.m. of given fractions =

l.c.m. of numerator h.c.f. of denominator

5) If d is the h.c.f. of two positive integer a and b, then there exist unique integer m and n, such that d = am + bn 6) If p is prime and a,b are any integer then P ,This implies ab

P or P a

b

7) h.c.f. of a given number always divides its l.c.m.

MOST IMPORTANT POINTS ABOUT L.C.M. AND H.C.F. PROBLEMS : 1) Largest number which divides x,y,z to leave same remainder = h.c.f. of y-x, z-y, z-x. 2) Largest number which divides x,y,z to leave remainder R (i.e. same) = h.c.f of x-R, y-R, z-R.

3) Largest number which divides x,y,z to leave same remainder a,b,c = h.c.f. of x-a, y-b, z-c. 4) Least number which when divided by x,y,z and leaves a remainder R in each case = ( l.c.m. of x,y,z) + R

HCF AND LCM QUESTIONS: Problem 1: Least number which when divided by 35,45,55 and leaves remainder 18,28,38; is? Solution: i) In this case we will evaluate l.c.m. ii) Here the difference between every divisor and remainder is same i.e. 17. Therefore, required number = l.c.m. of (35,45,55)-17 = (346517)= 3448. Problem 2: Least number which when divided by 5,6,7,8 and leaves remainder 3, but when divided by 9, leaves no remainder? Solution: l.c.m. of 5,6,7,8 = 840 Required number = 840 k + 3 Least value of k for which (840 k + 3) is divided by 9 is 2 Therefore, required number = 840*2 + 3 = 1683 Problem 3: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28 is? Solution: l.c.m. of 12,18,21,28 = 254 Therefore, required number must be divisible by 254. Greatest four digit number = 9999 On dividing 9999 by 252, remainder = 171 Therefore, 9999-171 = 9828.

Data Sufficiency Questions, Tips and Method

Most of the candidates try to solve data sufficiency questions by guess work. As every question carries same marks, questions in this part also deserve some time. Instead of guess work use a simple strategy as give below and avoid guessing the answer.

STEPS IN SOLVING DATA SUFFICIENCY QUESTIONS 1) Read the given problem. Don't assume anything except universal facts. 2) Take the first statement and combine it with main statement. Try to find the answer. 3) If you are unable to find the answer using 2nd step then combine second statement and combine it with main statement and try to find answer. 4) If you are unable to find an answer using second statement then add both statements with main statement and try to find answer 5) If even now you can't find answer, simply tick both statements are insufficient.

Directions :Marks A as answer if statement I alone is sufficient to answer the question Marks B as answer if statement II alone is sufficient to answer the question Marks C as answer if statement I and II together are sufficient to answer the question but neither statement aloneis sufficient to answer the question

DATA SUFFICIENCY QUESTIONS AND ANSWERS Question 1. How many people are there in the plain ? Statement I : 25% passengers are women and 35% are children.

Statement II : There are 24 men in the plain Answer From Statement I we can conclude that there are 40% men in the plain but we can't find the exact number of passengers From Statement two : Number of men passengers = 24 By combining both the statements we get, total number of passengers = 24 * 100/40 = ( you don’t need to calculate the answer ) Hence answer is C

Question 2. What is the difference between monthly income of Ram and Chaaru Statement I : Ram earns Rs 6000 less than Shaam Statement II : Chaaru earns Rs 6000 more than Shaam. Answer : In this question we don't need to in depth. Simply there is difference of Rs 12000

Question 3. Is x divisible by 28 ? Statement I : x is divisible by 20 Statement II : x is divisible by 84 Answer. Using statement I - x is divisible by 4 and 5 Using statement II - x is divisible by 3,4, and 7. By using both statements we can conclude that x is divisible by 28 ( 4*7), hence answer is C.

Question 4. P,Q,R,S and T are five friends. Their mean age is 18. What is the age of R ?

Statement I : P's age is 18 Statement II : Q's age is 2 years less than T and T's age is 6 years less than S. Statement III : R's age is 6 years more than B's age and 4 years more than T's age. Answer : P+Q+R+S+T = 90 From Statement I : Q+R+S+T = 72 From Statement II : Q = T - 2 and T=S - 6 So S = T + 6 Statement III : R = Q+6 and R = T + 4 Age of every friend can be defined in terms of T's age by using all three statements. So we can reach the answer using all three statements. Hence answer is C.

TIPS TO SOLVE DATA SUFFICIENCY QUESTIONS 

   



Never try to reach final answer as it is not asked. You need to find whether the information provided is enough to solve the given problem or not. Never make any assumption. Use only universal rules { eg. a + b = a + b - (a U b) } Try to solve questions by using above strategies Solve question step by step. First try to find answer using first statement then second and finally with both. Then mark the answer Even if you find answer with only one statement, then try to find answer with remaining statement as sometimes there is an option that answer can be find with both statements separately. Move on quickly and mark answer can't be found in case you are unable to reach any conclusion with information provided.