Quantitative Aptitude: Time and Work Set 3

 Time and Work Problems with tricks for SBI PO, IBPS PO, NIACL, NICl, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams

1. A and B can do a piece of work in 25 days. B and C can do the same piece of work in 30 days. C and A can do the same work in 37.5 days. In how many days will A alone do the same work?
   A) 100 days
   B) 60 days
   C) 50 days
   D) 75 days
   E) None of these
   
   View Answer

   Option B
   Solution :
   A+B =25 ————6 (1 days work) [total=150]
   B+C =30————-5 (1 days work)
   C+A =37.5———–4(1 days work)
   Total = LCM of 25,30,37.5
   1 day work=total work/days
   adding the three equation
   2(A+B+C) = 15
   A+B+C 1 day work = 7.5
   A+B+C – (B+C) =7.5 – 5 = 2.5
   A =150/2.5 = 60 days
   
2. A and B started a work together and after 7 days A left the work and B did the remaining work in 56 days. Then find in how many days A alone can do the same work if A and B together can do the work in 21 days.
   A) 30 days
   B) 27 days
   C) 28 days
   D) 29 days
   E) None of these
   
   View Answer

   Option C
   Solution :
   A+B=21 days
   Let total work = 21
   A+B 1 day work = 1
   7 day work = 7
   remaining work = 21-7 = 14
   B did 14 work in 56 days
   1 work in 4 days
   21 work in 84 days
   A+B = 21
   B = 84 hence A will be 28
3. 12 men can do a work in X days and 18 women can do the same work in (X+2) days. If the ratio of 18 men and 36 women work in same time is 9:10 then find the value of X
   A) 10 days
   B) 12 days
   C) 18 days
   D) 15 days
   E) None of these
   
   View Answer

   Option A
   Solution :
   18 men work/36 women work = 9/10
   To get 1 men work/1 women work divide 9 by 18 and 10 by 36
   1 men work/1 women work = (9/18)/(10/36)
   = 18 : 10
   Now multiply by 12*18 and 18*10 (no. of men and women as given in ques 12 men and 18 women)
   12*18 : 18*10 = 6 : 5 (This is the ratio of work)
   Hence ratio of days = 5 : 6
   difference => 1 = 2
   hence 5 = 10 days
4. Ram and Shyam can complete an assignment of data entry in 6 days. If Ram is 60% of Shyam’s speed and the total key depression in the assignment are 96000, What is Ram’s speed in key depression per hour if they work for 10 hour daily?
   A) 300
   B) 600
   C) 400
   D) 1000
   E) None of these
   
   View Answer

   Option B
   Solution :
   Ram : Shyam = 3:5
   multiply the ratio with 60 (day*hour) i.e.6*10
   Ram : Shyam = 180 : 300
   Total = 480 = 96000
   1 = 200
   3 = 600
   
5. If 32 men and 24 women can do a work in 2 days and 13 men and 18 women can do the same work in 4 days then find in how many days will 11 men do the work?
   A) 16 days
   B) 8 days
   C) 4 days
   D) 12 days
   E) None of these
   
   View Answer

   Option B
   Solution :
   32 M +24 W = 2
   => 64M+48 W = 1 day —— (i)
   13 M + 18 W = 4 day
   => 52 M + 72 W = 1 day—–(ii)
   using (i) and (ii)
   1 M = 2W
   put it in eq 1
   1 M = 88
   =>11 men = 8 days
6. 2 men or 4 women or 6 boys can do a work in 22 days. Find in how many days will 1 men, 1 women and 1 boy together can do the work?
   A) 22 days
   B) 12 days
   C) 24 days
   D) Cannot be determined
   E) None of these
   
   View Answer

   Option C
   Solution :
   2M = 22=——- 1 (1 day work—total work =22)
   4W = 22———1
   6B = 22———-1
   2M = 1 work => 1 M = 1/2 work
   1 W = 1/4 work
   1B = 1/6 work
   1 M + 1W +1B = 1/2+1/4+1/6 = 11/12 work
   days required = 22/(11/12) = 24 days
   
7. A is 50% more efficient than B. If A can do a work in 30 days, then find in how many will C do the work if C is 20% efficient of A and B together?
   A) 15 days
   B) 30 days
   C) 20 days
   D) 10 days
   E) None of these
   
   View Answer

   Option D
   Solution : B= 50% more efficient than A
   hence A:B=150:100=3:2=30:20 (in terms of day)
   A+B = 5
   C = 20% = 1/5
   C = 1
   1 = 10
   A : B : C = 3 : 2 : 1
   C = 1 = 10 days
8. A contract is given to 40 person to do a work in 45 days. After 25 days it is found that only half work is done, then find how many more persons are required to complete the work on time?
   A) 10
   B) 15
   C) 20
   D) 8
   E) None of these
   
   View Answer

   Option A
   Solution :
   (40*25)/(1/2) = (x+40)*20 /(1/2)
   x=10
9. A pipe can fill a cistern in 140 mins and B pipe can fill it in 105 mins. They are opened together but due to a leakage they take 10 min more to fill the tank. Find in how many minutes will leakage empty the tank?
   A) 240 mins
   B) 360 mins
   C) 420 mins
   D) 210 mins
   E) None of these
   
   View Answer

   Option C
   Solution :
   A=140————3 (total=420)
   B=105————4
   A+B=420/7=60 mins
   A+B+leakage=70 mins
   420/70=6 (1 mins work)
   7-6=1
   leakage=420/1=420 mins
   
10. In a tank theer are 3 inlet pipes A,B and C. A can fill the tank 1.5 times faster than B. B can fill the tank in half time taken by C and tank C is 3 times less efficient as A.There is one outlet pipe D which can empty the tank in 10 mins and the time taken by D is half of the time taken by A. Find in how much time will A and B together fill the tank?
    A) 12 mins
    B) 15 mins
    C) 10 mins
    D) 20 mins
    E) None of these
    
    View Answer

    Option A
    Solution :
    —————A:B
    Work——–3:2
    Days———2:3
    B is half of C, hence C=6
    A:B:C=2:3:6 = 20:30:60
    time =D:A=1:2=10:20
    A=20
    B=30
    A+B=12 mins