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
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 :
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
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.
E
XAMPLES
#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)
P
RACTICE
Q
UESTIONS
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?
(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.
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
T
RICK
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.
S
HORTCUT
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
Work that can be done per day
Efficiency in Percent
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.)
⇒ 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
⇒ 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 NumberStep 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 ?
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. Speed = Distance/Time 2. Time = Distance/Speed 3. Distance = Speed × Time
4. 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.
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 : [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]
⇒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
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 p
n. In p
n, p is called the base and n is called the index of the
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)
58is 1.
So, (729)
59= (728)
58X (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: (98)
4= (……….6)
So, (98)
4n=(…………6)
Thus, (98)
40= (98)
4x10= (…..6) = 6 in the unit place.
(98)
42= (98)
4x10X (98)
2= (…..6)X(…….4) = 4 in the unit place.
(98)
43= (98)
4x10X(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.
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)
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 Share on Facebook Share on Twitter Share on Google+
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
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 1When 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?
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 ?
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
Numbers: all counting numbers ( 1,2,3,4,5....∞) Whole
Numbers: natural number + zero( 0,1,2,3,4,5...∞)
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....∞)
Prime Numbers:
It can be positive or negative except 1, if the number is not divisible by any number except the number itself.
(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61....∞)
Composit e Numbers:
Natural numbers which are not prime
Co-Prime: Two natural number a and b are said to be co-prime if their HCF is 1.
Divisibility
NumberDivisibl 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 places-Sum 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 6hence ,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) = 1/4n2 (n+1)2Arithmetic 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.
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 L.C.M.
Factorizati on Method
Write each number as the product of the prime factors. The product of least powers of common prime factors gives H.C.F.
Example:
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
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 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:
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
H.C.F. of given numbers = 69
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. & L.C.M. of Fractions 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 Product of H.C.F. & L.C.M.
H.C.F * L.C.M. = product of two numbers
Decimal 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 2
Square & Cube
Square & Cube Square Root & Cube Root Factorization Method
Perfect Square Non-Perfect Square
Tricks to Find Square of Any Number Square Root & Cube Root
Fractions & Decimals
OnBasis Explanation
Decimal Fractions
A number with a denominator of power of 10 is a decimal fractions. 1/
10= 1 tenth; 1/100= 0.1;38/100=0.38
Vulgar Fractions
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 0.64) and add as many numeric zeros as the digit in the number after 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 Comparis on of Fractions
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 a particular way( ascending or descending order).
3/5,4/7,8/9 and 9/11 Arranging in Ascending Order 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 Decimal Recurring 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.
Examples with Explanations
Surds & Indices
Some Rules of Indices Some Rules of Surds
# Series/Addition/Subtraction
Q1 Find 8 + 888 + 8888 + 88888 ?
Q2. Find 9.4 + 99.44 + 999.444 + 9999.4444 ?
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
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 + <sup>927-9</sup>/<sub>990</sub> +6/9 = -4 + 0.45 + 2 + <sup>918</sup>/990 +3/2 = (- 4 + 2) + (0.45+1.5) + (<sup>51</sup>/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 direction, in what time will they be clear each other?
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.
S
IMPLE
I
NTEREST
INTEREST
It is money paid by borrower for using the lender's money for a specified period of time.
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)
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 Under compound interest
Year Princip al at beginn ing of year Intere st for the year Intere st till the end of the year 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 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
Examples
#1
Find the simple interest, If
1. P = Rs.1000, R = 20% per annum, T = 4 years. 2. P = Rs.600, R = 5% per annum, T = 4 months. 3. P = Rs.200, R = 6% per six months, T = 3 years. 4. P = Rs.500, R = 2% per six months, T = 5/
2 years.
5. P = Rs.400, R = 3% per three months, T = 2 months. 6. P = Rs.730, R = 10% per annum, T = 120 days.
7. P = 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
4. 5×2×5=50
5. 4×2=8
7. 37.50
#2
Find the following:
1. P = Rs. 100, R = 3% per annum, T = 2 year, A= ? 2. P = Rs. 500, R = 6% per annum, T = 4 months, A= ? 3. P = Rs. 400, R = 3.65% per annum, T = 150 days, A= ? 4. A = Rs. 540, S.I = Rs. 108 , R = 5%, T = ?
5. A = Rs. 1,120, R = 5%, T = 22/
5 yr, S.I = ?
# Solution:
2. S.I = 10 ; A = S.I + P ; A = 10+500 ⇒ 510 3. S.I = 6 ; A = 400 + 6 ⇒ 406
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
4. [ 1164-1008 = 156 ] ⇒ 156/
3×4 = 208 ; R = 208/2×800×100 ⇒ 13 5. 7x/
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 c) Rs. 7200 d) Rs. 7500 e) None of these
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 b) 4 years c) 4.5 years d) 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% b) 4% c) 5% d) 6% 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% b) 10.25% c) 10.5% d) Data inadequate e) None of these
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% c) 7 1/8% d) 10% e) None of these
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 c) Rs. 15,000 d) Rs. 20,000 e) None of these
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 c) 2 : 3 d) Data inadequate e) None of these
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 c) Rs. 150 d) Rs. 150 d) Rs. 167.50
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% b) 10.5% c) 12% d) Data inadequate 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 c) Rs.7200 d) Rs.7500 e) None of these
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 c) 4 d) 5 e) None of these
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 :
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 these
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 c) Rs.3120 d) Rs.3250 e) None of these
Q14. A sum of money trebles itself in 15 years 6 months. In how many years would it double itself?
a) 6 years 3 months b) 7 years 9 months c) 8 years 3 months
d) 9 years 6 months e) None of these
Q15. 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?
a) 3.6 b) 6 c) 18 d) Data inadequate 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 - x) × 11 × 2 } / 100]
= 3508 28x ⎯ 22x = 350800 ⎯ (13900 × 22)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 years3. 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]
= Rs.5 S.I. for last 6 months = Rs.[
105 × 10 × 1 / 100 × 2]
= Rs.5.25 So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs.110.255. 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 CLet 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
= Rs. (625 ⎯ 400) = Rs.225
So, gain in 1 year = Rs.
[
225 / 2]
= Rs.112.509. 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
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) Then,
[
x × 9 × 1 / 100]
+[
(100000 - x) × 11× 1 / 100]
=[
100000 × 39 / 4 × 1 / 100]
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 CLet the sum be Rs. x. Then,
[
x × 15 / 2 × 5 / 4 × 1 / 100]
⎯[
x × 25 / 2 × 2 / 3 × 1 / 100]
= 32.50
25x = (3250 × 24)
x =
[
3250 × 24 / 25]
= 312014. 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 r2 = 432R2 = 36
R = 6
Trick :
Calculating Compound Interest for 3 Years
# 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+ 3r/4/100]
# Difference between Compound Interest and Simple Interest
CI - SI = P [ R/100 ]2CI - 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. 20004.
IMPORTANT FORMULAE
Ques 1.
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:
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.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
PROFIT AND LOSS
V
ARIOUS
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
