Monday, 25 January 2016

Short tricks for time and work questions

Dear expert readers. Time and Work formula shortcuts Tricks Aptitude Question Answers example are presenting here. You know that Time and Work is a very easy topic and at least 2 Questions are asked from this topic in Any Exam whether it's Clerk or PO or SSC.What actually the Time and Work Topic is all about? In simple words It's how much time a person will take to Complete a Given Work. Here we are providing some cases of time and work and efficient method to solve it. Hope this would be helpful to all aspirants.




Part - I

Time and Work formula shortcuts Tricks Aptitude Question Answers pdf

Shortcut tricks are used here to solve the numerical problems on time and work. This shortcut makes the preparation of aptitude questions for exams like UPSC, SSC, CAT, MAT, GMAT, GRE, MBA competitive exams and all bank exams such as IBPS, SBI easier.

First, we need to know about the time and work basic concepts, then only we can get into solving the problems. The next thing is to know the time and work formulas, then it is easier to learn the difficult problems.

Time and work important formulas

Time and work problems formulae are given below. In the formula we have explained with men and women. For example we have taken A and B as working men or women.

Time and work problems are frequently asked in quantitative aptitude test in all exams. There are 9 rules to solve the questions faster by short cut methods and tricks.
M is person, D is days and W is work.
T is time and E is efficiency.
Rule 1: Universal Rule.
M1D1W2 = M2D2W1
M1D1T1W2 = M2D2T2W1 (When time is given )
M1D1T1E1W2 = M2D2T2E2W1 (When efficiency is added)
Rule 2 : If A can do a piece of work in n days.
In one day work done by A is 1/n.
Rule 3: A can do a work in D1 days, B can do same work in D2 days. A and B together do the same work in how many days ?
Formula= (D1*D2)/(D1+D2)
Rule 4: If A is twice as good as B, then A will take 1/2 time taken by B to complete a work.
B alone can finish in 2x days ( x is work done by A )
Rule 5: If A is thrice as good a as B, then A will take 1/3rd of the time taken by B to complete a work.
B alone can finish in 3x days
Rule 6: If A and B together can do a piece of work in x days.
B and C together takes y days and C and A together takes in z days, then same work can be done by A, B and C alone is ?
Remaining rules and formula are given in pdf notes (tutorial).

How to solve time and work problems examples

Time and work examples are given here to practice quantitative aptitude question. An Important problem with a solution for an example is derived and given.

We hope that you now understood the concept of time and work. Let us start with the different type of time and work  problems. Practice these questions with answers to get through on the concepts.

Time and work aptitude questions with solutions

Time and work solved examples are solved here in an easier way. Check out the set of four different important problems and tricks to solve those problems.


Part - II

Note -: In conventional Method work is always treated as 1

Example: So if I say that a person can complete a work in 15 days that means he will do 1/15 work in one day, It's simple maths.
Now another person does the same work in 30  days. So he will do 1/30 work in 1 day.
Now if i ask in how many days they will complete the work together. What we gonna do is Add their 1 day of work
 like 1/15 + 1/30 = (2+1)/30 = 3/30 = 1/10 
Now this 1/10 we got is actually their 1 day work, So if they do 1/10 work in one day then it's simple they will complete the whole work in 10 days.
Now that was the conventional method and I believe that you all know how to solve Questions through Conventional method.

So now lets move on to the Faster method i.e efficiency method.

In efficiency method the Work is not treated in numerical value, Like in Conventional method the work is 1 but here the work is treated as percentage.
So by common sense the work is always treated as 100%
So when i say a person completes a work in 15 days it means he will do 100/15 % work in 1 day i.e 6.66% work in 1 day
If another person does the work in 30 days that means he will do 3.33% work in 1 day.
And together they will do 6.66 + 3.33 = 9.99 or 10% work in one day So in how many days they will do the complete work, that will be 100/10 = 10 days.
Now it may sound difficult That we have to convert Each value in % but don't worry you don't have to convert each value, You just have to learn all the values till 1/30 and then it will be a cakewalk.

Now we will take Some standard Cases Of time and work and you all are free to ask any problem if you have in any case.

Case 1 - A does a work in X days, B does a Work in Y days In how many days they will complete the work.
Question- A completes the work in 10 days and B completes the work in 15 days In how many days they will complete the work.
Conventional Method 
Work done by A in 1 day = 1/10
Work Done by B in 1 day = 1/15
Work done By A & B together in 1 day = 1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6
As A & B Completes 1/6 work in one day So they will complete the whole work in 6 Days.

Efficiency method.
Efficiency of A =100/10 = 10% 
Efficiency of B = 100/15 = 6.66%
Efficiency of A & B Together = 10+ 6.66 = 16.66%
So the time taken by A & B together to Complete the work will be 100/16.66 = 6 Days.

Case -2 A can do a work in X days and B can do it Y days, In how many days the work is completed if they work alternatively Started by A.
Now in these type of question the person are not actually working together, what happens in this type of question is that A works for 1 day and then on 2nd day B work and on 3rd again A work and on Fourth again B works and so on till the work is completed.
For example A can do a work in 10 days B can do it 15 days and how many days they will finish it if The work is started by A
So again work done by A in one day = 1/10
 ''     ''          ''        ''      ''  B      ''      ''      = 1/15
Now the work done by Togther will be = 1/10 + 1/15 = 1/6 [ Note now this 1/6 work is not done by them in 1 day but in 2 days Actually, See A worked for 1 day and did 1/10 work on the second day B worked and finished the 1/15 work So in total 2 days they did 1/6 work]
So in 2 days they did 1/6 work so in how many days they will complete the whole work, Simple 12 days.
Efficienecy Method
A's Efficiency = 10%
B's Efficiency = 6.66%
A + B Efficiency = 16.66%
Work done by A and B in 2 days [ remember 2 days because they are not working together but working alternatively] = 16.66%
So time taken by them to complete 100% work = 100/(16.66 = 6 [ but always remember multiply this by 2, Beacause 16.66% work is done by them in 2 days and not in 1 day.
So The answer will be 6*2 = 12 days.

Case 3: A can do a work in X days, B can do the work Y days and A leaves after working Z days.
Question- A can do a work in 10 days and B can do it in 15 days, A works for 2 days and then leaves, In how many days will the work be completed?
Now here we can se that A leaves after 2 days that means A and B only worked for 2 days and the remaining work is done by B alone.
So first we have to calculated the work done by A and B together in 2 days.
So work done be A in 1 day = 1/10
""   "         "        "  B "  "   "     = 1/15
Work done by A & B together is 1 day = 1/10 + 1/15 = 1/6
Work done by A & B together in 2 days = (1/6) * 2 = 1/3
So remaining work = 1 - 1/3 = 2/3
Now this 2/3 work has to be done by B alone.
So it takes 15 days for B to do Complete work, How much time it will be taken by B to do 2/3 work ? So it will be 15*(2/3) = 10 days
So the work will be completed in 2 + 10 days = 12 days

Efficiency method 
A's efficiency = 10% 
B's Efficiency = 6.66%
Total a+b = 16.66%
work done by A and B together in 2 days = 16.66*2 = 33.33%
Work remaining = 66.66%
time taken by B to complete 66.66% work = 66.66/6.666 = 10 days
Total time = 10+2 = 12 days

Case 4
A can do a piece of Work in 10 days and B can do it in 15 days, In how many days will the work be completed if B leaves 2 days before the completion on work.
Now in this question B leaves before 2 days
Together they can do the work in what = 1/10 + 1/15 = 1/6
That means 6 Days.
That means Together they could have completed the work in 6 days but B works only till 4th day and The remaining work will be done by A alone
So they worked together for 4 days in all So work done by them in 4 days = (1/6)*4 = 4/6 = 2/3
remaining work = 1/3 
Now this 1/3 work will be done by A alone 
Now A can do the complete work in 10 days, So the time taken by him to do 1/3 work = 10 * (1/3) = 10/3 days or 3.33 days
So total time taken = 4+ 3.33 days = 7.33 days
Efficiency method
A's efficiency = 10%
B's efficiency = 6.66%
Total = 16.66%
Work will be completed in 6 days
Work done in 4 days = 66.66%
remaining = 33.33%
A will complete the remaining in = 33.33/10 = 3.33 
Total = 4+3.33 = 7.33

Case 5: A can do a Work in X days B can Do it in Y days, In how many days The work will get completed if B leaves 2 days before the actual completion of work.

Question: A can do a work in 10 days B can do it in 15 days, In how many days The work will get completed if B leaves 2 days before the Actual Completion of Work.what is the difference between this Actual completion of work and Completion of Work?
See in last example the work was supposed to get completed in 6 days, So we just Solved the question taking into consideration that B leaves 2 days before the completion of work i.e B worked for 4 days and the rest 2 days work was don by A alone and Completes that work in 3.33 days Total 7.33 days.
So if i ask In this question If B left 2 days before the actual completion then it means B should have left on 5.33 days Got it ?
Now back to the question.
Now just imagine that the work gets completed in x days.
So A would work for x days[ A works for the whole time ]
And B would work for x-2 days[ because B left 2 days before the actual completion of work]
So now According to Question
x/10 + (x-2)/15 = 1 [ Beacuse work is always one ]
(3x+2x-4)/30 = 1
5x -4 = 30
5x = 34
x = 6.8 days.
So the work will be completed in 6.8 Days.
It can also be asked That after how many days B left, So the answer would be Simple 6.8 - 2 = 4.8 days

Efficiency Method
A's Efficiency = 10%
B's Efficiency = 6.66%
Let the no. of days be x 
so Accordring to question
10x + 6.66(x-2) = 100[ Work is always 100% in efficiency method ]
10x + 6.66x - 13.33 = 100
16.66x = 113.33
x = 113.33/16.66 = 6.8 
Answer = 6.8 days

Download IBPS Bank Exam Free Tips, Tricks, Quiz and notes pdf

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  • Bank Jobs Quantitative Aptitude - Study Notes : PROBLEMS ON TRAINS
  • Bank Jobs Quantitative Aptitude - Study Notes : TIME AND DISTANCE
  • Bank Jobs Quantitative Aptitude - Study Notes : TIME AND WORK
  • Bank Jobs Quantitative Aptitude - Study Notes : PARTNERSHIP
  • Bank Jobs Quantitative Aptitude - Study Notes : RATIO AND PROPORTION
  • Bank Jobs Quantitative Aptitude - Study Notes : PROFIT AND LOSS
  • Bank Jobs Quantitative Aptitude - Study Notes : PERCENTAGE
  • Bank Jobs Quantitative Aptitude - Study Notes : SURDS AND INDICES
  • Bank Jobs Quantitative Aptitude - Study Notes : AVERAGE
  • Bank Jobs Quantitative Aptitude - Study Notes : SQUARE ROOTS AND CUBE ROOTS
  • SIMPLIFICATION
  • Bank Jobs Quantitative Aptitude - Study Notes : DECIMAL FRACTIONS
  • Bank Jobs Quantitative Aptitude - Study Notes : H.C.F. AND L.C.M.
  • Quantitative Shortcuts:Ratio and Fractions

  • Short tricks for time and work questions in Hindi

    (1).दूरी, समय और गति के बीच संबंध:
    Distance = speed x Time

    (2).किसी वस्तु की गति को  किमी प्रति घंटा से  मीटर प्रति सेकंड में परिवर्तित करने के लिए गति को गुणा, 1000 / 3600 = 5 / 18 

    (3).किसी वस्तु की गति को मीटर प्रति सेकंड से  किमी प्रति घंटा में परिवर्तित करने के लिए गति को गुणा = 3600 / 1000 = 18/ 5

    (4).यदि A और B का गति अनुपात a:b है तो निश्चित दूरी को कवर करने के लिए समय का अनुपात  = 1/a : 1/b = b : a होगा|


    (5).यदि एक व्यक्ति x किमी प्रति घंटा गति के साथ निश्चित दूरी तय करता है और y किमी प्रति घंटा गति से लौटता है तो पूरी यात्रा में उसकी औसत गति है, औसत गति = 2xy/(x+y)KMPH

    (6).यदि एक निश्चित दूरी को 3 भिन्न गतियों किमी प्रति घंटा, y किमी प्रति घंटा और z किमी प्रति घंटा के साथ साथ तय की जाती है तो पूरी यात्रा की औसत गति है, औसत गति  3xyz/(xy+yz+zx)KMPH

    (7).यदि  दो भिन्न दूरियां क्रमशः  किमी प्रति घंटा, y किमी प्रति घंटा गति के साथ तय की जाती है लेकिन आवश्यक वही समय है तो पूरी यात्रा में औसत गति है. औसत गति  = (x+y)/2 KMPH

    (8).यदि 2 ट्रेन भिन्न स्थानों से एक ही समय पर चलती हैं मान लीजिये A और B क्रमशः एक दूसरे की ओर बढ़ती हैं और पार करने के बाद वह क्रमशः B और A बिंदु पर पहुँचने के लिए यदि वे a और b सेकंड लेती हैं तो
    (A’s speed) : (B’s speed) = Öb : Öa
     ट्रेन सम्बंधित प्रश्नों पर आधारित सूत्र
    सम्बंधित गति 
    (9)यदि दो ट्रेन किमी प्रति घंटा, y किमी प्रति घंटा गति के साथ एक ही दिशा में चल रही हैं जहां  x>y है तो उस संबंध में उनकी सम्बंधित गति  :(x-y) KMPH इस प्रकार होगी| 

    (10)यदि दो ट्रेन किमी प्रति घंटा, y किमी प्रति घंटा गति के साथ विपरीत दिशा में चल रहीं है तो इस संबंध में उनकी गति  (x+y) KMPH होगी| 

    आइए अब समय और दूरी पर क्विज हल करते हैं  
    1.अपनी सामान्य गति के 7/8 गति से चलने पर एक व्यक्ति गंतव्य तक पहुँचने में अपने सामान्य समय से 16 मिनट देर से पहुँचता है|  गंतव्य तक पहुँचने के लिए उसके द्वारा लिया गया समान्य समय ज्ञात कीजिये?  
    (a) 1 घंटा, 44 मिनट  
    (b) 1 घंटा, 52 मिनट 
    (c) 1 घंटा, 36 मिनट 
    (d) 1 घंटा, 40 मिनट

    2.एक व्यक्ति ट्रेन से ऑफिस जाता है| वह ट्रेन पकड़ने के लिए अपने घर के नज़दीक रेलवे स्टेशन तक पैदल चलता है| एक वह 4 किमी/घं चलता है और 5 मिनट से अपनी ट्रेन खो देता है| अगले दिन वह 6 किमी/घं चलता है और ट्रेन के आने से 7 मिनट पहले स्टेशन पहुँच जाता है| उसके घर और स्टेशन के बीच की दूरी ज्ञात कीजिये| 
    (a) 2.4 किमी 
    (b) 1.8 किमी  
    (c) 3.6 किमी  
    (d) 3 किमी

    3. अशोक 225 किमी की दूरी कुछ इस प्रकार तय करता है कि पहले 45 किमी/घं में 15 किमी, अगली 60 किमी/घं में 120 किमी और शेष यात्रा 90 किमी/घं में तय करता है| उसकी 225 किमी की यात्रा की औसत गति ज्ञात कीजिये| 
    (a) 65 किमी/घं   
    (b) 67.5 किमी/घं  
    (c) 70 किमी/घं  
    (d) 73.5 किमी/घं 

    4. एक व्यक्ति P से Q a किमी/घं की औसत गति से और Q से R b किमी/घं की औसत गति से और R से S c किमी/घं की औसत गति से जाता है| यदि  PQ = QR = RS तो P से S तक पहुँचने की  औसत गति क्या होगी?
    (a) (a + b + c)/ 3 
    (b) 3abc/(ab + bc + ca) 
    (c) 3abc/(a + b + c) 
    (d) 3(ab + bc + ca)/(a + b + c)

    5. कार  P, शहर X से शहर Y की ओर जाने के लिए चलती है| कार  Q, Y से X की ओर जाने के लिए चलती है|  दोनों कारें एक साथ शुरू होती हैं और एक समान गति से यात्रा करने के बाद अपने गंतव्य तक पहुँचती हैं|  XY = 200 किमी| दोनों कारें  2 घंटे बाद मिलती हैं| यदि  P और Q एक ही दिशा में चलती हैं तो दोनों कारें 4 घंटे में मिलती हैं| P की गति ज्ञात कीजिये|  
    (a) 60 किमी/घं  
    (b) 85 किमी/घं  
    (c) 75 किमी/घं 
    (d) 80 किमी/घं

    6. ट्रेन  P अपने से दुगनी लम्बी ट्रेन Q को ओवरटेक करती है और ट्रेन P की आधी गति से 36 सेकंड में यात्रा करती है| ट्रेन P, ट्रेन R को विपरीत दिशा में अपनी गति को दुगना कर 8 सेकंड में पार कर जाती है| यदि ट्रेन P की गति 72 किमी/घं है तो ट्रेन R की लम्बाई ज्ञात कीजिये| 
    (a) 330 मी  
    (b) 360 मी  
    (c) 390 मी  
    (d) 420 मी

    7. एक 480 मीटर लम्बी ट्रेन 72 किमी/घं से यात्रा करती है|  वह ट्रेन की दिशा में चल रहे साइकिल यात्री को पार करने में 32 सेकंड लेती है| साइकिल यात्री की गति ज्ञात कीजिये|  
    (a) 12 किमी/घं  
    (b) 15 किमी/घं  
    (c) 18 किमी/घं 
    (d) 9 किमी/घं

    8. एक 180 मी लम्बी ट्रेन 120 मीटर लम्बे प्लेटफ़ॉर्म को पार करने में 20 सेकंड लेती है और अन्य ट्रेन उसी गति से एक इलेक्ट्रिक पोल को पार करने में 10 सेकंड लेती है| विपरीत दिशा में यात्रा करने पर वह एक दूसरे को कितने  समय में पार कर लेंगे?
    (a) 11 सेकंड  
    (b). 13 सेकंड  
    (c) 12 सेकंड  
    (d) 14 सेकंड

    9. एक वृत्ताकार रास्ते पर, A और B जब विपरीत दिशा में चलते हैं तो वह उसी दिशा में लिए गए समय से 1/4 समय लेते हैं|  उनकी गति का अनुपात ज्ञात कीजिये?  
    (a) 5: 3
    (b) 6 : 5 
    (c) 4 : 3 
    (d) 3 : 2

    10. तीन व्यक्ति एक ही बिंदु से शुरू करते हैं और 2 किमी वृत्ताकार रास्ते के आसपास 4 किमी/घं, 6 किमी/घं  और 8 किमी/घं की यात्रा करते हैं तो वह तीनों आरंभिक बिंदु पर कितने समय बाद मिलेंगे?  
    (a) 1/2 घं  
    (b) 1घं 
    (c) 1.5 घं  
    (d) 2 घं

    उत्तर

    1. (b)
    2. (a)
    3. (b)
    4. (b)
    5. (c)
    6. (b)
    7. (c)
    8. (a)
    9. (a)
    10. (b)

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