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.
Saurabh said:
9 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.
Amit Raj Modi said:
1 decade ago
I have one another method.
A can do total work in 4 hr.
B+C can do this in 3 hr.
A+C can do this in 2 hr.
Convert all of this in unit/hr.
Take LCM of 4, 3 and 2 which is 12 Unit (total unit of Work).
A can do work 12/4=3 unit/hr.
A+C can do work in 2 hr means A done 2hr*3unit=6unit. Now remaining work is 6 unit (hint:-12unit-6unit=6unit) which has done by C. So C can do 3 unit (hint:-6unit/2hr=3unit/hr) Work in 1 hr.
Now B+C can do work in 3 hr means C done 3hr*3Unit=9 Unit. Now remaining work is 3 unit (hint:-12unit-9unit=3 unit) which has done by B. So B can do 1 unit (hint:-3unit/3hr=1unit/hr) Work in 1 hr.
B takes 12 hr (hint:-12unit*1hr) to complete whole work.
You can do this method oraly. You need only some practice on this method.
A can do total work in 4 hr.
B+C can do this in 3 hr.
A+C can do this in 2 hr.
Convert all of this in unit/hr.
Take LCM of 4, 3 and 2 which is 12 Unit (total unit of Work).
A can do work 12/4=3 unit/hr.
A+C can do work in 2 hr means A done 2hr*3unit=6unit. Now remaining work is 6 unit (hint:-12unit-6unit=6unit) which has done by C. So C can do 3 unit (hint:-6unit/2hr=3unit/hr) Work in 1 hr.
Now B+C can do work in 3 hr means C done 3hr*3Unit=9 Unit. Now remaining work is 3 unit (hint:-12unit-9unit=3 unit) which has done by B. So B can do 1 unit (hint:-3unit/3hr=1unit/hr) Work in 1 hr.
B takes 12 hr (hint:-12unit*1hr) to complete whole work.
You can do this method oraly. You need only some practice on this method.
Himanshi mishra said:
8 years ago
@Ankita.
Follow this method.
Here total work is 12 hours as lcm of 2-4-3.
A does 3 work in 1 hour ND he finishes 12 work in four hours.
In the same way, abt A and C you know A does 3 work in 1 hour thus C also does 3 work in 1 hour so total work 6 in 1 hour thus 12 work in 2 hour is completed by A and C.
And now C does 3 work in 1 hour so b does 1 work in 1 hour and thus the both do 4 work in 1 hour and finishes in 3 hours.
As a result, B takes 1 hour to do 1 work so B Will take 12 hours to do 12 work.
Follow this method.
Here total work is 12 hours as lcm of 2-4-3.
A does 3 work in 1 hour ND he finishes 12 work in four hours.
In the same way, abt A and C you know A does 3 work in 1 hour thus C also does 3 work in 1 hour so total work 6 in 1 hour thus 12 work in 2 hour is completed by A and C.
And now C does 3 work in 1 hour so b does 1 work in 1 hour and thus the both do 4 work in 1 hour and finishes in 3 hours.
As a result, B takes 1 hour to do 1 work so B Will take 12 hours to do 12 work.
Akhil said:
9 years ago
A = 4 (ie A completes the work in 4 hours)
B + C in 3.
A + C in 2.
By taking LCM 4, 3, 2 = Total work or LCM is 12hrs work.
A 1-hour work is 3/ 3*4 = 12 (ie takes 4 hours to complete total work 12 hours already given in question follow same with B + C and A + C).
B + C 1 hour work is 4/4*3 = 12.
A + C 1 hour work is 6/6*2 = 12.
A = 3
B + C = 4
A + C = 6
3A + 3C
1B + 3C
A = 3
B = 1
C = 3
So, B takes 1hr x12 = 12 hour to complete the work.
B + C in 3.
A + C in 2.
By taking LCM 4, 3, 2 = Total work or LCM is 12hrs work.
A 1-hour work is 3/ 3*4 = 12 (ie takes 4 hours to complete total work 12 hours already given in question follow same with B + C and A + C).
B + C 1 hour work is 4/4*3 = 12.
A + C 1 hour work is 6/6*2 = 12.
A = 3
B + C = 4
A + C = 6
3A + 3C
1B + 3C
A = 3
B = 1
C = 3
So, B takes 1hr x12 = 12 hour to complete the work.
Aditya said:
1 decade ago
The above methods are enough to solve this problem...but when i got this question i thought to convert all hours into days...although its a very time consuming...but i got the answer...i am just a beginner...so satisfied with it
1 day=24 hours
1/24 day=1 hour
i/6 day=4 hour....i hope you got my point...but its time consuming so don't use this method...i shared it because...i want to tell you ...what strikes on my mind.
1 day=24 hours
1/24 day=1 hour
i/6 day=4 hour....i hope you got my point...but its time consuming so don't use this method...i shared it because...i want to tell you ...what strikes on my mind.
NAVEEN CHIKATI said:
7 months ago
Hi,
A - 4.
B +C - 3.
A + C - 3.
LCM OF ALL -12
A = 12/4 = 3
B+C = 12/3 = 4
A+C = 12/2 = 6
B = ?.
TOTAL = (A) + (B+C) + (A+C),
= (3) + (4) + (6),
= 13.
TO FIND C= A + C = 6,
= 3 + c = 6,
C = 6 - 3,
C = 3.
TO FIND B = B + C = 4,
= B + 3 = 4,
B = 3 - 4,
B = 1.
B = 1/12.
B takes 12 hours.
A - 4.
B +C - 3.
A + C - 3.
LCM OF ALL -12
A = 12/4 = 3
B+C = 12/3 = 4
A+C = 12/2 = 6
B = ?.
TOTAL = (A) + (B+C) + (A+C),
= (3) + (4) + (6),
= 13.
TO FIND C= A + C = 6,
= 3 + c = 6,
C = 6 - 3,
C = 3.
TO FIND B = B + C = 4,
= B + 3 = 4,
B = 3 - 4,
B = 1.
B = 1/12.
B takes 12 hours.
(9)
Shreya said:
6 years ago
A takes 4hrs.
B+C takes 3 hrs.
A+C takes 2 hrs.
A=a.
B+C=b,
A+C=c
LCM of 4,3,2=12.
a+b+c=12units(total work done).
a=A.
So A takes 4h for 12u.
In 1 hr it does 3u.
b=B+C,
B+C take 3h for 12u.
In 1h does 4u,
c=A+C.
A+C take 2h for 12u,
in 1hr A+C do 6u.
A does 3u/h.
replacing in A+C.
3+C=6.
C=3,
C does 3u/h.
Replacing C in B+C
In 1 hr;
B+3=4.
B=1.
B does 1u/h.
1 unit of work in 1hr.
12 unit of work in 12 hrs.
B+C takes 3 hrs.
A+C takes 2 hrs.
A=a.
B+C=b,
A+C=c
LCM of 4,3,2=12.
a+b+c=12units(total work done).
a=A.
So A takes 4h for 12u.
In 1 hr it does 3u.
b=B+C,
B+C take 3h for 12u.
In 1h does 4u,
c=A+C.
A+C take 2h for 12u,
in 1hr A+C do 6u.
A does 3u/h.
replacing in A+C.
3+C=6.
C=3,
C does 3u/h.
Replacing C in B+C
In 1 hr;
B+3=4.
B=1.
B does 1u/h.
1 unit of work in 1hr.
12 unit of work in 12 hrs.
Soumyaranjan said:
5 years ago
A = 4hrs.
B+C = 3hrs.
A+C = 2hrs.
Work=LCM(4, 3,2) = 12.
The efficiency of A = 12/4 = 3.
The efficiency of BC = 12/3 = 4.
The efficiency of AC = 12/2 = 6.
The efficiency of C = efficiency of (AC-A)
= 6-3 = 3
The efficiency of B = efficiency of (BC-C)
= 4-3 = 1
T = W/E
= 12/1 = 12.
So, B alone will take 12hrs for do it.
B+C = 3hrs.
A+C = 2hrs.
Work=LCM(4, 3,2) = 12.
The efficiency of A = 12/4 = 3.
The efficiency of BC = 12/3 = 4.
The efficiency of AC = 12/2 = 6.
The efficiency of C = efficiency of (AC-A)
= 6-3 = 3
The efficiency of B = efficiency of (BC-C)
= 4-3 = 1
T = W/E
= 12/1 = 12.
So, B alone will take 12hrs for do it.
(3)
Dharam Verma said:
1 decade ago
A's 1 day work = 1/4
(B+C)'s 1 day work = 1/3
(A+C)'s 1 day work = 1/2
So, C's 1 day work = (A+C)'s 1 day work - A's 1 day work
= 1/2 - 1/4
= 1/4
Now, B's 1 day work = (B+C)'s 1 day work - C's 1 day work
= 1/3 - 1/4
= 1/12 (taking LCM)
So, B complete the work in = 12 days.
(B+C)'s 1 day work = 1/3
(A+C)'s 1 day work = 1/2
So, C's 1 day work = (A+C)'s 1 day work - A's 1 day work
= 1/2 - 1/4
= 1/4
Now, B's 1 day work = (B+C)'s 1 day work - C's 1 day work
= 1/3 - 1/4
= 1/12 (taking LCM)
So, B complete the work in = 12 days.
Akshay Ladwa said:
7 years ago
A=B+C(given) ----> (1)
one day work of A+B=1/10 --> (2)
one day work of C=1/50 ---> (3).
So for total (A+B+C)'s 1 day work, we can assume from 1--> C=A-B=1/50----> (4)
So adding equation 2 and 4 we get A=3/50.
Substituting above A in equation 2 we get B=1/25.
Hence 25 days because(If A's 1 day's work =1/n then A can finish the work in n days.)).
one day work of A+B=1/10 --> (2)
one day work of C=1/50 ---> (3).
So for total (A+B+C)'s 1 day work, we can assume from 1--> C=A-B=1/50----> (4)
So adding equation 2 and 4 we get A=3/50.
Substituting above A in equation 2 we get B=1/25.
Hence 25 days because(If A's 1 day's work =1/n then A can finish the work in n days.)).
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