Quantitative Aptitude solved problem for IBPS ,SBI,MBA,RRC,SSC PO And Clerk Exam-Profit and Loss Solved Problem Pdf

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Hello Folks, Again we are back with some major info related to all of various exam like bank, ssc, railways, IAS, RPSC and any other government exam. Today we are going to talking about Profit and loss topic. All of you known that this topic is very simple, time consuming and score able part in quantitative aptitude section. More focus is given on profit and loss topic. Profit and loss is basically used in banking sector extensively as well as asked in banking exams. So today we are sharing some important information about profit and loss like basic concept, and basic logic used in profit and loss solved question. So we are advised to all of future bankers regularly visit this site for banking exam study material, banking awareness, for daily gk updates and recent banking vacancies.

profit and loss aptitude questions

Profit and loss:

Always remember important fact about Profit and Loss Topics: We are providing some important definition about profit and loss.

1.Cost Price:

  • Cost price is defined as an item is the expenditure incurred to bought and you can say to produce.

Short term of Cost Price is CP.

2.Selling Price:

  • Selling Price defined as an item is sold its called selling price.

Short term of selling price is SP.

3.Profit and Loss:

  • This is the difference between the selling price and cost price. The selling price is greater than cost price than called profit.
  • The cost price is greater than selling price than called loss.

4.Margin Price:

  • Basically, Margin price is used in percentage terms only. You can say this is the profit as percentage of Selling price (SP)

5.Marked price:

  • Marked price defined as the price of the item as displayed on the label.

6.Discount:

  • Discount term defined is the reduction given on the marked price before selling it to a customer.

7.Mark-up:

  • Mark –up term defined is basically increment on the cost price before being sold to a customer.

Always Remember some basic formulae used in Profit and Loss:

 

1.Profit/gain

 

SP-CP

 

2.Loss

 

CP-SP

 

3.Gain %

 

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{(Gain x100)/C.P}

 

 

4.Loss %

 

{(Loss x100)/S.P}

 

5.Cost Price: (CP)

 

[(100/(100+Gain)x S.P)]

 

6.Cost Price: C.P

 

[(100/(100-Gain)x S.P]

7.Selling Price: S.P [(100+Loss)/100 x S.P]
8.Selling Price: S.P [(100-Loss)/100 x S.P]
9.When a man sells two similar product one a at gain x% and other at a loss x% then a seller always gets loss given by  (common Loss and gain%/100) ^{2}=(x/10))^{2}

 

 

 

 

 

 

 

 

 

 

Quantitative Aptitude solved problem for IBPS, SBI,MBA,RRC,SSC PO and Clerk Exam -Percentage Question and Answer

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Hey friends, today we are back with new update quantitative aptitude. You know this part is most important for the entire government exam as well as private exam. And we know, all of you aware this part in past, but now we are forgot all of these thing learn in 10th standard. Yes we know, all of you know everything about these topics the main reason behind that we have to study in past time but now our mind is totally cleaned to these thing. But Aspirant Don’t stress your mind just see and analysis all of these trick and shortcut you know.

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We hope, all of you read our previous article Time, Speed and Distance; we share some tips and tricks, formulas and shortcut used in these topics. Today we are sharing some valuable information about Percentage topic. As you know that, now day’s quantitative aptitude part is compulsory for all civil service exam as well as bank exam. Percentage is most important topic in Arithmetic. It is also used in many topics like data interpretation, profit and loss, discount, Partnership and simplification etc. We are started to level low to high. We are sharing some solved and unsolved problem of percentage. And also we are given some short cut and tips and tricks for percentage problem. You know, basically bank exam is totally depend on time and accuracy. We are also giving some percentage problem in asked in previous year exam so u can easily understand about this topic and which type of question asked in exam.

PERCENTAGE:

REMEMBER IMPORTANT TIPS AND TRICKS ABOUT PERCENTAGE TOPIC:

  • Firstly you remember table 1 to 30.
  • You remember also reciprocal of 1 to 15.
  • Remember also square 1 to 30.
  • Learn cube 1 to 20
  • Square root – Up to 15
  • Cube root-Up to 15

 NOTE:All of instruction is helpful for percentage topic as well as solved simplification question in within a second.

We are covered many topics under percentage given below:

  • Basic concept percentage and of base
  • Change of base
  • Concept of successive percentage
  • Concept of Indexing

CONCEPT AND BASIC LOGIC OF PERCENTAGE:

When we are talking about percentage its means Per Cent and Per Hundred.

For example: 12% its means

                       = 12/100

 And you can say   .12

So, we can say, to convert any percentage to a number then dived it’s by Hundred.

For example. 3% means .03 and 30% = .3 and 300% =3.

Same concept applies when change to simple number to percent from its multiply by 100.

For example. 2 =200% and, .02 =20%

REMEMBER RECIPROCAL OF 1 TO 20:

1/1 1%
1/2 50%
1/3 33.33%
1/4 25%
1/5 20%
1/6 16.66%
1/7 14.22%
1/8 12.5%
1/9 11.11%
1/10 0.1%
1/11 9.09%
1/12 8.33%
1/13 7.7%
1/14 7.14%
1/15 6.66%
1/16 6.25%
1/17 5.88%
1/18 5.55%
1/19 5.26%
1/20 5%

  NOTE:This table is very useful for you.you can easily find out any fraction, ratio and percentage given problem.  

For example. 2/3, 3/4, 5/6 etc.

2/3=1/3 x 2=33.33 x 2=66.66% etc

PERCENTAGE SOLVED PROBLEM:

Example 1. 38 % of 450 +45 % of 280?

Solution: (38/100) X 450 + (45/100) X 280

               =161 +126

                =287

 TO BE CONTINUE ……………….

 

 

 

 

 

 

 

 

{QUANTITATIVE APTITUDE} TIME,SPEED AND DISTANCE SOLVED PROBLEM IN PDF

Hello bankersherald reader, Today we are going to talk about Quantitative aptitude part. All of you know Quantitative aptitude part is very important role of different type of competitive exam like banking sector exam, railways, ssc and other competitive exam.So guys today we are going to discussing about Time, Speed and distance question and basic formulas used in this section. So Bankers herald team are providing many solved and unsolved problem of Time , Speed and Distance.

TIME, SPEED AND DISTANCE

The major formulae used in TSD:

1.Distance (D)=Speed(S)x Time(T)

2.Speed (S)=Distance(D)/Time(T)

3.Time (T)=Distance (D)/Speed(S)

 Important Point Always Remember:

  1. When distance is given in kilometer (Km), then speed should be used in Km/Hr and time in hours.
  2. When distance is given in meter, then speed should be used in m/sec and time in second
  3. Point number third , if speed is given Km/Hr, then in order to convert it , in m/sec, it is directly multiplied by 5/18

For example, 36Km/hr=36×5/18=10m/sec

4.If speed is given in m/sec, then in order to convert it, in Km/hr its directly multiplied by 18/5

For example, 20 m/sec=20×18/5=36Km/hr

Example 1.Ram moves from A to B travelling a distance of 10 km in 4hr. What is ram’s Speed?

Answer- Distance (D)=10km

                  Total time =4hr

                  Speed= Distance (D)/Time (T)=10Km/4Hr=2.5Km/hr

Example 2.If ram moving at the speed of 2.5 Km/hr travels from Delhi to Agra in 1 day and total time take by ram 10 hr. Find the distance between Delhi to Agra.

Answer-Firstly Check unite

                Then Apply  Distance(D)=Speed x time

                                                         =2.5Km/hr x (1 day+10hr)

                                                         =2.5Km/hr x 34hr

                                                         =85Km

Relationship between Time, Speed and Distance:

1.You can see, speed is directly proportional to distance. So if twice the distance is traveled in same time, speed has become twice.

2.speed is inversely Proportional to time. So if speed x times, time taken to cover same distance will become 1/x times.

3.Distance is directly proportional to time taken , for example time taken at same speed becomes 1/x, it means distance also become 1/x th .

Note-All of point, try to avoid the use of equations. It will help you increase your speed tremendously.

Some  solved question are in below

Example3.Ramesh sees Uma standing at a distance of 200 m from his position .He increase his speed by 50% and hence takes 20s now to reach her.

(a). If he travels at the  original speed , how much time will be taken ?

(b).What was his original speed (in Km/hr)

(c).What is his new speed?

Explanation:

Part(a):Travelling at 1505of his original Speed(3/2)his original speed. Ramesh should take 2/3of the original time, 20s, thus his original time is 30 s.

Part(b):Speed=Distance/time

                          =200/30 m/s=20/3 x 18/5=24 Km/hr

Part(c):His new speed is 50 more =24 x 1.5

                                                             =36 Km/hr.

Relative Speed:

Case 1.If two persons are moving in opposite direction at speed V & v respectively.

The Relative Speed  is given by  Vr=V+v

Case2.It two persons are moving in same direction at speed V & v respectively.

The Relative Speed is given by Vr=|V-v|

Important Point Remember:

  • A man walking at S  km/hr reaches his home t1 minutes late . if we walk V km/hr , he reaches there t2 minutes earlier  then u always remember this formula

Distance(d) = {(S x V/V-S)X (t1+t2)/60 Km}

Ex[4].A man walking at 2 km/hr , reaches his office 6 minutes late. if he walks 3 km/hr , he reaches there 6 minutes earlier . How far is the office from his house?

Answer:  Distance(D) =(3 X 2/3-2) X (6+6)/60

                                        =6 X (12/60)=6/5 Km

Ex[5].Two car cover the same distance at 15 km/hr and 16 km/hr respectively . Find the distance travelled by each , if one take 16 min longer then other. 

Answer: Suppose required distance x km.

                 x/15-x/16 =16/60

            solve the above equation

             we get , 16 x- 15 x  =64

          the required distance =64km

Ex.6 Ram Covers a distance of 12 km in 45 minutes. if he covers half the distance in 2/3rd of the time then , what speed should be maintain to cover the remaining distance in the remaining time ?

Answer:

Total distance =12 km

  half distance = 6 km

 Total time  =   45mint

 Time taken cover the half distance =45 x 2/3

                                                                  =30 minutes

  Time remaining =45-30 =15 minutes

Ex[7].A person cover a distance of 120 km at 60 Km/hr and next  150 km at 50 Km/hr . What is hi average speed for his whole journey of 270 km ?

Answer: Total time =120/60+150/50

                                    =5Hr

               Total distance =270

               Average speed=270/5

                                          =54 km/hr

Ex[8].A and B are two different cities.Ram travels on car from A to B at a speed of 15 km/hr and returns back at the rate of 10 Km/hr.Find his average speed for the whole journey?

Answer: you can use formula of average speed:

                   Average speed=2 u v/u+v

                  you get

                                   2 x 15×10/25

                             =12 km/hr

Ex[9].Ram travels 2/15 of his  total  journey by car .and 9/20 by cycle and the renaming 10 km on foot . What ihis total journey?

Answer-Suppose total journey is x km

                                  then, 2x/15+9x/20+10 =x

                                  solved above equation we get x=24

                                   total journey is 24

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