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 2 of 10.
Elayabharathi said:
1 decade ago
No need of fraction, general understanding.
take lcm for 4,3,2 = 12 parts.
A =3p/d, BC=4p/d AC=6p/d.
so C alone do in (AC-A) 3p/d,
B(BC-C)=1p/d. So for 12 parts it ll take 12 DAYS for B do complete the same work.
question can be twisted like what is the ratio for A:B:C?
what will be the potential for cC?
ABOVE STEPS HAVE ALL THE ANSWERS YOU NEED!
take lcm for 4,3,2 = 12 parts.
A =3p/d, BC=4p/d AC=6p/d.
so C alone do in (AC-A) 3p/d,
B(BC-C)=1p/d. So for 12 parts it ll take 12 DAYS for B do complete the same work.
question can be twisted like what is the ratio for A:B:C?
what will be the potential for cC?
ABOVE STEPS HAVE ALL THE ANSWERS YOU NEED!
Asha said:
1 decade ago
There is one more method for this question, a shortcut formula. A theorem says if A and B together can do a piece of work in x days, and A alone can do it in y days, then B alone can do the work in xy/y-x days.
Therefore for AC pair C's work hrs = 4*2/4-2= 4 hrs
now for BC pair B's work hrs = 4*3/4-3 = 12 hrs :)
Therefore for AC pair C's work hrs = 4*2/4-2= 4 hrs
now for BC pair B's work hrs = 4*3/4-3 = 12 hrs :)
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...
Dhananjay said:
2 years ago
A 1 hours work = 1/4 ----> (i)
B+C 1 hors work = 1/3 ----> (ii)
A+C 1 hour work = 1/2 ----> (iii).
By subtracting from (iii) to (ii) we get A-B = 1/6.
Now in place of A put the value of A = 1/4.
1/4-B = 1/6,
1/4-1/6 = B.
1/12 IS B'S 1 Hour's work.
B will take 12 hours.
B+C 1 hors work = 1/3 ----> (ii)
A+C 1 hour work = 1/2 ----> (iii).
By subtracting from (iii) to (ii) we get A-B = 1/6.
Now in place of A put the value of A = 1/4.
1/4-B = 1/6,
1/4-1/6 = B.
1/12 IS B'S 1 Hour's work.
B will take 12 hours.
(7)
Akshay Sen said:
6 years ago
Work done by A in one hr = 1/4.
As A+C = 2hr.
so, C's working for 1 hr will be = 1/2 -1/4 (A's working).
C's working for 1hr = 1/4.
Now As B+C = 3hrs.
so, we can find the working of B in 1 hr.
therefore working of B in 1 hr = 1/3 - 1/4.
B's = 1/12.
Hence 12hrs taken by B.
As A+C = 2hr.
so, C's working for 1 hr will be = 1/2 -1/4 (A's working).
C's working for 1hr = 1/4.
Now As B+C = 3hrs.
so, we can find the working of B in 1 hr.
therefore working of B in 1 hr = 1/3 - 1/4.
B's = 1/12.
Hence 12hrs taken by B.
Vinod Anand said:
1 decade ago
If A = 4 -(1)
B+C = 3 -(2)
A+C = 2 -(3)
From Eq.(1) and (2)
4C = 2-1, C = 4 (4' is coming from LCM and 4/4 = 1 and 4/2 = 2,
So 2-1 = 1 and 4/1 = 4).
Put the value of C in eq.(2).
12 B = 4-3.
Than B = 12 hours, Option -C.
B+C = 3 -(2)
A+C = 2 -(3)
From Eq.(1) and (2)
4C = 2-1, C = 4 (4' is coming from LCM and 4/4 = 1 and 4/2 = 2,
So 2-1 = 1 and 4/1 = 4).
Put the value of C in eq.(2).
12 B = 4-3.
Than B = 12 hours, Option -C.
Aman Jadhao said:
4 years ago
Hi;
a=4h
B+c=3h
A+c=2h
B=?.
A b+c. A+c
4. 3 2
Take LCM of
Lcm=12
After LCM value is
3. 4. 6.
Adding the value we get
A + c = 2
C = 1.
Then solve
B+c =4,
B=3.
Simply multiple with LCM values of b&c
3 * 4 = 12 hours.
a=4h
B+c=3h
A+c=2h
B=?.
A b+c. A+c
4. 3 2
Take LCM of
Lcm=12
After LCM value is
3. 4. 6.
Adding the value we get
A + c = 2
C = 1.
Then solve
B+c =4,
B=3.
Simply multiple with LCM values of b&c
3 * 4 = 12 hours.
(3)
Pallabi Sethi said:
4 years ago
A 1 hours work=1/4 ----> (i)
B+C 1 hors work=1/3 ----> (ii)
A+C 1 hour work=1/2 ----> (iii).
By subtracting from (iii) to (ii) we get A-B = 1/6.
Now in place of A put the value of A = 1/4.
1/4-B = 1/6,
1/4-1/6 = B.
1/12 IS B'S 1 Hour's work.
B will take 12 hours.
B+C 1 hors work=1/3 ----> (ii)
A+C 1 hour work=1/2 ----> (iii).
By subtracting from (iii) to (ii) we get A-B = 1/6.
Now in place of A put the value of A = 1/4.
1/4-B = 1/6,
1/4-1/6 = B.
1/12 IS B'S 1 Hour's work.
B will take 12 hours.
Suni said:
9 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.
Sandeep choudhary said:
8 years ago
A = 4.
B + C = 3.
A + C = 2.
Take LCM = 12.
ONE DAY WORK OF A = 12/4 = 3.
Similarly, B + C = 12/3 = 4------>1.
A+C = 12/2 = 6------> 2.
TOTAL ONE DAY WORK = 13.
PUT VALUE OF A in Eqn 2 and got the value of B then put in eqn 1 and got value of c=12.
B + C = 3.
A + C = 2.
Take LCM = 12.
ONE DAY WORK OF A = 12/4 = 3.
Similarly, B + C = 12/3 = 4------>1.
A+C = 12/2 = 6------> 2.
TOTAL ONE DAY WORK = 13.
PUT VALUE OF A in Eqn 2 and got the value of B then put in eqn 1 and got value of c=12.
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