Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 8)
                   
                                       
                                8.
A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
 
                                    Answer: Option
                                                    Explanation:
                                                
| (A + B)'s 1 day's work = | 1 | 
| 10 | 
| C's 1 day's work = | 1 | 
| 50 | 
| (A + B + C)'s 1 day's work = | ![]()  | 
    1 | + | 1 | ![]()  | 
    = | 6 | = | 3 | . .... (i) | 
| 10 | 50 | 50 | 25 | 
A's 1 day's work = (B + C)'s 1 day's work .... (ii)
| From (i) and (ii), we get: 2 x (A's 1 day's work) = | 3 | 
| 25 | 
  A's 1 day's work = | 
    3 | . | 
| 50 | 
  B's 1 day's work | 
    ![]()  | 
    1 | - | 3 | ![]()  | 
    = | 2 | = | 1 | . | 
| 10 | 50 | 50 | 25 | 
So, B alone could do the work in 25 days.
Discussion:
147 comments Page 9 of 15.
                
                        Sumasree said: 
                         
                        1 decade ago
                
                @Rohit nice your answer, but we always follow shortcuts for speed up.
                
                        Ram said: 
                         
                        1 decade ago
                
                Number of days taken by B and C to complete a work. with same number of days A alone can do it.
if C can do a work in 50 days. If B joins with C then it becomes 50 days/2 = 25 days for B and C.
A no.of days for work = B&C no.of days for work.
        
25 days = 25 days.
                if C can do a work in 50 days. If B joins with C then it becomes 50 days/2 = 25 days for B and C.
A no.of days for work = B&C no.of days for work.
25 days = 25 days.
                        VRAJ said: 
                         
                        1 decade ago
                
                A = B+C.
A+B = 10.
C = 50.
A+B+C = 1/10+1/50.
A+A = 1/10+1/50.
2A = 6/50.
A = 3/50.
A = B+C.
3/50 = B+1/50.
B = 3/50-1/50.
B = 2/50.
B = 1/25.
B CAN DONE THE WORK IN 25 DAYS.
                A+B = 10.
C = 50.
A+B+C = 1/10+1/50.
A+A = 1/10+1/50.
2A = 6/50.
A = 3/50.
A = B+C.
3/50 = B+1/50.
B = 3/50-1/50.
B = 2/50.
B = 1/25.
B CAN DONE THE WORK IN 25 DAYS.
                        Anandhan said: 
                         
                        1 decade ago
                
                a = a+b;
a+b = 1/10; together;-----------(1).
c = 1/50;
Take 1st equation,
a+b=1/10;
we know a = b+c; so substitute,
b+c+b=1/10;
2b+1/50=1/10;
2b=1/10-1/50;
2b=4/50;
b=4/100;
b=2/50;
b=25;
Thats it.
                a+b = 1/10; together;-----------(1).
c = 1/50;
Take 1st equation,
a+b=1/10;
we know a = b+c; so substitute,
b+c+b=1/10;
2b+1/50=1/10;
2b=1/10-1/50;
2b=4/50;
b=4/100;
b=2/50;
b=25;
Thats it.
                        Suresh said: 
                         
                        1 decade ago
                
                a+b = 1/10.
a = b+c eq.1.
c = 1/50.
Then a-b = 1/50 eq.2.
Solving 1 and 2 we get b = 1/25.
                a = b+c eq.1.
c = 1/50.
Then a-b = 1/50 eq.2.
Solving 1 and 2 we get b = 1/25.
                        Serina said: 
                         
                        1 decade ago
                
                How did you get 3/25, if you add 1/10 + 1/50 please explain in brief?
                
                        Sa_1 said: 
                         
                        1 decade ago
                
                @Serina.
Its just LCM nothing else. Simple method.
A + B + C = 3/25.
Now A and B + C together time are same.
A = B + C.
A + B + C = 3/25.
B + C = 3/25 - A.
So we can write
A = 3/25 - A.
2A = 3/25.
A = 3/50.
                Its just LCM nothing else. Simple method.
A + B + C = 3/25.
Now A and B + C together time are same.
A = B + C.
A + B + C = 3/25.
B + C = 3/25 - A.
So we can write
A = 3/25 - A.
2A = 3/25.
A = 3/50.
                        Shanu Verma said: 
                         
                        1 decade ago
                
                A = B+C-----(1).
A+B = 1/10------(2).
C = 1/50-----(3).
B+B+C = 1/10 (put value of A from equation 1 in equation 2).
2B+C = 1/10 (as C = 1/50).
2B = 1/10-1/50 = 4/50 = 2/25.
B = 2/50 = 1/25.
B alone could do the work in 25 Days.
                A+B = 1/10------(2).
C = 1/50-----(3).
B+B+C = 1/10 (put value of A from equation 1 in equation 2).
2B+C = 1/10 (as C = 1/50).
2B = 1/10-1/50 = 4/50 = 2/25.
B = 2/50 = 1/25.
B alone could do the work in 25 Days.
                        Alok said: 
                         
                        1 decade ago
                
                a = b+c....(i).
a+b = 1/10 (given).
a-b = c.
a-b = 1/50 (c alone can do work in 50 days putting the value of c here)....(ii).
Solving I and II.
a = 1/30.
b = 1/10-1/30.
c = 1/15.
How this one is wrong?
                a+b = 1/10 (given).
a-b = c.
a-b = 1/50 (c alone can do work in 50 days putting the value of c here)....(ii).
Solving I and II.
a = 1/30.
b = 1/10-1/30.
c = 1/15.
How this one is wrong?
                        SYAM said: 
                         
                        1 decade ago
                
                A* = B+C.
A+B = 10 days.
C = 50 days.
C is 1 day of work = 1/50.
A+B 1 day of work = 1/10.
B = 1/10-A.
B is 1 day of work = 1/10-A*.
= 1/10 - (B+C).
= 1/10 - (B+1/50).
2B = 1/10-1/50.
B = 4/100.
= 1/25.
So, 25 is the answer.
                A+B = 10 days.
C = 50 days.
C is 1 day of work = 1/50.
A+B 1 day of work = 1/10.
B = 1/10-A.
B is 1 day of work = 1/10-A*.
= 1/10 - (B+C).
= 1/10 - (B+1/50).
2B = 1/10-1/50.
B = 4/100.
= 1/25.
So, 25 is the answer.
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 A's 1 day's work =
 B's 1 day's work