Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 7)
7.
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?
Answer: Option
Explanation:
| A's 1 hour's work = | 1 | ; |
| 4 |
| (B + C)'s 1 hour's work = | 1 | ; |
| 3 |
| (A + C)'s 1 hour's work = | 1 | . |
| 2 |
| (A + B + C)'s 1 hour's work = |  |
1 | + | 1 |  |
= | 7 | . |
| 4 | 3 | 12 |
| B's 1 hour's work = | |
7 | - | 1 | |
= | 1 | . |
| 12 | 2 | 12 |
B alone will take 12 hours to do the work.
Discussion:
95 comments Page 1 of 10.
Rucha said:
1 decade ago
How the answer for a+b+c = 7/12?
Varseyny said:
1 decade ago
A=1/4 (given)
(B+C)=1/3 (given)
(A+B+C)=(1/4+1/3) then take L.C.M
3+4/12=7/12
FIND FOR B
A+B+C=7/12
B=7/12-(A+C) given(A+C=1/2)
B=7/12-(1/2)=1/12
ANS:12
(B+C)=1/3 (given)
(A+B+C)=(1/4+1/3) then take L.C.M
3+4/12=7/12
FIND FOR B
A+B+C=7/12
B=7/12-(A+C) given(A+C=1/2)
B=7/12-(1/2)=1/12
ANS:12
Nagarjun said:
1 decade ago
A's 1 hour work is 1/4
(A+C)'s 1 hour work is 1/2
(B+C)'s 1 hour work is 1/3
so, A+C=1/2 ->X and B+C=1/3 ->Y
subtract X-Y
we will get A-B=1/6
now, substitute A's value in d equation u vl get B's value as 1/12.
which is 1 day,s work of B...
hence B alone require 12 days...
I HOPE IT VL B HELPFULL...
(A+C)'s 1 hour work is 1/2
(B+C)'s 1 hour work is 1/3
so, A+C=1/2 ->X and B+C=1/3 ->Y
subtract X-Y
we will get A-B=1/6
now, substitute A's value in d equation u vl get B's value as 1/12.
which is 1 day,s work of B...
hence B alone require 12 days...
I HOPE IT VL B HELPFULL...
Nagendramurthy said:
1 decade ago
Hi friends,
(A+C-A)=C
(1/2-1/4)=1/4
so C's 1 hour work will be 1/4.
Similarly
(B+C-C)=B
(1/3-1/4)=1/12
Therefore B's 1 hour work is 1/12.
(A+C-A)=C
(1/2-1/4)=1/4
so C's 1 hour work will be 1/4.
Similarly
(B+C-C)=B
(1/3-1/4)=1/12
Therefore B's 1 hour work is 1/12.
Ashish said:
1 decade ago
Good question. Reasoning is difficult for this one.
Hindu said:
1 decade ago
Take the given data as 2 equations as follows
A=1\4 ,B+C=1\3--eq1 and A+C=1\2---eq2
put A=1\4 in eq2
1\4+C=1\2
C=1\2-1\4 we get
C=1\4
put C=1\4 in eq1
B+1\4=1\3
B=1\3-1\4 we get
B=1\12
Therefore B alone take 12hours to do a work
A=1\4 ,B+C=1\3--eq1 and A+C=1\2---eq2
put A=1\4 in eq2
1\4+C=1\2
C=1\2-1\4 we get
C=1\4
put C=1\4 in eq1
B+1\4=1\3
B=1\3-1\4 we get
B=1\12
Therefore B alone take 12hours to do a work
HARRY said:
1 decade ago
C=1/2-1/4=>1/4(C'S 1 DAY WORK)
B=1/3-1/4=>1/12(B'S 1 DAY WORK)
SO 12 DAYS TO COMPLETE THE WORK
B=1/3-1/4=>1/12(B'S 1 DAY WORK)
SO 12 DAYS TO COMPLETE THE WORK
Jayanth babu said:
1 decade ago
Hi friends, in above problem
A=1/4----->(1)
B+C=1/3----->(2)
A+C=1/2------>(3)
now solving (2)
B+C=1/3
c=B-1/3--------->(4)
substitute (1) and (4) in (3)
then..
(1/4)+B-(1/3)=1/2
by solving this equation
B=1/12
Thus, B alone take 12hrs.
A=1/4----->(1)
B+C=1/3----->(2)
A+C=1/2------>(3)
now solving (2)
B+C=1/3
c=B-1/3--------->(4)
substitute (1) and (4) in (3)
then..
(1/4)+B-(1/3)=1/2
by solving this equation
B=1/12
Thus, B alone take 12hrs.
Pavan@9966606261 said:
1 decade ago
Hi friends
b+c=1/3;
a+c=1/2;
a=1/4;
c=1/2-1/4=1/4;
b=1/3-1/4=1/12;
time taken for B alone to complete is 12 hours
b+c=1/3;
a+c=1/2;
a=1/4;
c=1/2-1/4=1/4;
b=1/3-1/4=1/12;
time taken for B alone to complete is 12 hours
Piyus said:
1 decade ago
(b+c)of 1 hour-(a+c) of 1 hour = 1/3-1/2
=> b(1hr)-a(1hr)=(-1/6)
=> b(1hr)=1/4-1/6
=>b's 1hr work= 1/12
=>b ll take 12 hr to complete the work alone
=> b(1hr)-a(1hr)=(-1/6)
=> b(1hr)=1/4-1/6
=>b's 1hr work= 1/12
=>b ll take 12 hr to complete the work alone
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