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:
97 comments Page 8 of 10.
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.
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...
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
Karthik said:
1 decade ago
A = 1/4.
B&C = 1/3.
A&C = 1/2.
1ST We have to find (A+B+C) therefore (1/4+1/3) = 7/12.
This for total work.
Now we subtract (A&C) value from 7/12.
7/12-1/2=1/12 take reciprocal for that value.
B&C = 1/3.
A&C = 1/2.
1ST We have to find (A+B+C) therefore (1/4+1/3) = 7/12.
This for total work.
Now we subtract (A&C) value from 7/12.
7/12-1/2=1/12 take reciprocal for that value.
Haripriya said:
10 years ago
a = 1/4;
b+c = 1/3;
a+c = 1/2.
Therefore find the c value first:- a+c = 1/2.
1/4+c = 1/2.
c = 1/4.
Now substitute in b+c = 1/3.
b+1/4 = 1/3.
b = 1/12 hence it takes 12 hours.
b+c = 1/3;
a+c = 1/2.
Therefore find the c value first:- a+c = 1/2.
1/4+c = 1/2.
c = 1/4.
Now substitute in b+c = 1/3.
b+1/4 = 1/3.
b = 1/12 hence it takes 12 hours.
Sagar raut said:
10 years ago
A's 1 day work is 1/4.
A and B 1 day work is 1\3.
So 1/(A+B)-1\A = 1\B.
1\3 - 1\4 = 1\12.
A and B 1 day work is 1\3.
So 1/(A+B)-1\A = 1\B.
1\3 - 1\4 = 1\12.
NARESH PAL said:
10 years ago
(A + B) - B = B + C (A and B, C works equal).
1/10 - B = B + C.
1/2B = 1/10 - 1/50.
B = 25.
1/10 - B = B + C.
1/2B = 1/10 - 1/50.
B = 25.
Saurabh said:
10 years ago
Following method focuses on avoiding fractions as much as possible, because calculating with fractions takes longer amount of time than calculating with integers.
In most problems on work-time-rate, fractions can be avoided altogether, or at least the role of fractions can be minimized.
Solution:
Assume total work = LCM (4, 3, 2) units = 12 units.
A's time = 4 hr.
==> A's rate = 12 units/4 hr = 3 units/hr.....(1).
(B+C)'s time = 3 hr.
==> B's rate + C's rate = 12 units/3 hr = 4 units/hr.....(2).
(A+C)'s time = 2 hr.
==> A's rate + C's rate = 12 units/2 hr = 6 units/hr.....(3).
From (1) and (3),
C's rate = 6-3 = 3 units/hr.....(4).
From (2) and (4),
B's rate = 4-3 = 1 unit/hr.
Hence B's time = Total work/B's rate = 12/1 = 12 hours.
Note: If you want to avoid spending time calculating the LCM, you can do so by saying that total amount of work = 4x3x2 = 24 units. Further calculations will change accordingly, but final answer will still be 12 hours.
The trick is "Don't define the total work to be simply 1 unit". Whenever the total work is defined to be 1 unit, fractions are introduced in the solution almost inevitably. Instead, we keep the total amount of work problem specific, and choose the total number of units to be the multiplication (or LCM) of suitable numbers.
While there is no guarantee that the fractions will be avoided altogether, they will at least be reduced to a large extent. In many cases, even some complicated ratios can be avoided by this method.
I will try to post as many solutions as I can using this method.
In most problems on work-time-rate, fractions can be avoided altogether, or at least the role of fractions can be minimized.
Solution:
Assume total work = LCM (4, 3, 2) units = 12 units.
A's time = 4 hr.
==> A's rate = 12 units/4 hr = 3 units/hr.....(1).
(B+C)'s time = 3 hr.
==> B's rate + C's rate = 12 units/3 hr = 4 units/hr.....(2).
(A+C)'s time = 2 hr.
==> A's rate + C's rate = 12 units/2 hr = 6 units/hr.....(3).
From (1) and (3),
C's rate = 6-3 = 3 units/hr.....(4).
From (2) and (4),
B's rate = 4-3 = 1 unit/hr.
Hence B's time = Total work/B's rate = 12/1 = 12 hours.
Note: If you want to avoid spending time calculating the LCM, you can do so by saying that total amount of work = 4x3x2 = 24 units. Further calculations will change accordingly, but final answer will still be 12 hours.
The trick is "Don't define the total work to be simply 1 unit". Whenever the total work is defined to be 1 unit, fractions are introduced in the solution almost inevitably. Instead, we keep the total amount of work problem specific, and choose the total number of units to be the multiplication (or LCM) of suitable numbers.
While there is no guarantee that the fractions will be avoided altogether, they will at least be reduced to a large extent. In many cases, even some complicated ratios can be avoided by this method.
I will try to post as many solutions as I can using this method.
Suni said:
10 years ago
A's one day work = 1/4......(1).
(B+C)'s one day work = 1/3.....(2).
(A+C) 's one day work = 1/2......(3).
i.e, B+C = 1/3......(4).
(5)-(4).
A+C= 1/2......(5).
A-B = 1/2-1/3 = 1/6.
A-B = 1/6 (since A = 1/4).
1/4-1/6 = B.
B = 1/12.
Answer: 12 hrs.
(B+C)'s one day work = 1/3.....(2).
(A+C) 's one day work = 1/2......(3).
i.e, B+C = 1/3......(4).
(5)-(4).
A+C= 1/2......(5).
A-B = 1/2-1/3 = 1/6.
A-B = 1/6 (since A = 1/4).
1/4-1/6 = B.
B = 1/12.
Answer: 12 hrs.
Ajay said:
10 years ago
For these type of questions,
Try to combine a+b+c and and evaluate each type.
Try to combine a+b+c and and evaluate each type.
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