Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 29)
29.
A and B can do a job together in 7 days. A is 1
times as efficient as B. The same job can be done by A alone in :
times as efficient as B. The same job can be done by A alone in :Answer: Option
Explanation:
| (A's 1 day's work) : (B's 1 day's work) = | 7 | : 1 = 7 : 4. |
| 4 |
Let A's and B's 1 day's work be 7x and 4x respectively.
| Then, 7x + 4x = | 1 |  11x = |
1 | x = |
1 | . |
| 7 | 7 | 77 |
 A's 1 day's work = |
 |
1 | x 7 |  |
= | 1 | . |
| 77 | 11 |
Discussion:
53 comments Page 5 of 6.
Girish said:
1 decade ago
Why we should multiply by 7 at the end ?
Pradeep Oram said:
1 decade ago
Given: A+B = 1/7......(1).
And A = 7/4B (efficient than B).
Put in(1). We get.
B=4/77(work of B).
Then,
Work of A = (1/7)-(4/77) = 7/77 = 1/11 (from eq.(1)).
So, A finish job in 11 days.
And A = 7/4B (efficient than B).
Put in(1). We get.
B=4/77(work of B).
Then,
Work of A = (1/7)-(4/77) = 7/77 = 1/11 (from eq.(1)).
So, A finish job in 11 days.
Torres Dungs said:
1 decade ago
I think this would be an easy approach:
A=7/4B (A is as efficient as 1*3/4 times as B).
=> On cross multiplying we get, 4A=7B-----------> 1.
=> From question, A+B=1/7 (A,B is one days work so equating to 1/7).
Therefore again cross multiply,
We get, 7A+7B = 1.
=>from 1, we have 4A=7B.
So substitute 4A in 7B.
We get,
7A+4A=1 ==>11A=1===> A=1/11(which is A's one day's work).
Hence,
Total days A need==> 11 days.
Hope you'd have undersTood :).
A=7/4B (A is as efficient as 1*3/4 times as B).
=> On cross multiplying we get, 4A=7B-----------> 1.
=> From question, A+B=1/7 (A,B is one days work so equating to 1/7).
Therefore again cross multiply,
We get, 7A+7B = 1.
=>from 1, we have 4A=7B.
So substitute 4A in 7B.
We get,
7A+4A=1 ==>11A=1===> A=1/11(which is A's one day's work).
Hence,
Total days A need==> 11 days.
Hope you'd have undersTood :).
Heena said:
1 decade ago
@nidhi
if A=(7/4)B
then Eq2 should be
7/4(B)+B=1/7
if A=(7/4)B
then Eq2 should be
7/4(B)+B=1/7
Jimmy John said:
1 decade ago
A+B=1/7
7A+7B=1-----1
A=7/4B
4A=7B-----2
sub 2 in 1
7A+4A=1 (Since 7B=4A)
therefore 11A=1
A=1/11
therefore A can finish in 11 days
7A+7B=1-----1
A=7/4B
4A=7B-----2
sub 2 in 1
7A+4A=1 (Since 7B=4A)
therefore 11A=1
A=1/11
therefore A can finish in 11 days
(1)
Varun said:
1 decade ago
Ratio type problem solve by 'x' value.
Nidhi said:
1 decade ago
A and B collectively can do the work in 7 days.
So their one day work is 1/7.
A+B=1/7-- (1).
A is 7/4 times efficient then B which means.
A= (7/4) B.
Or.
B= (4/7) A - (2).
Put the value of (2) in (1).
(4/7) *B+B=1/7.
B=1/11 THIS IS HIS ONE DAY WORK.
ALONE B CAN FINISH IN 11 DAY.
So their one day work is 1/7.
A+B=1/7-- (1).
A is 7/4 times efficient then B which means.
A= (7/4) B.
Or.
B= (4/7) A - (2).
Put the value of (2) in (1).
(4/7) *B+B=1/7.
B=1/11 THIS IS HIS ONE DAY WORK.
ALONE B CAN FINISH IN 11 DAY.
UMESH PRASAD said:
1 decade ago
A and B collectively can do the work in 7 days
so their one day work is 1/7
A+B=1/7-------(1)
A is 7/4 times efficient then B which means
A=(7/4)B ----------
or
B=(4/7)A ------------(2)
put the value of (2)in (1)
(4/7)*B+B=1/7
B=1/11 THIS IS HIS ONE DAY WORK
ALONE B CAN FINISH IN 11 DAY
so their one day work is 1/7
A+B=1/7-------(1)
A is 7/4 times efficient then B which means
A=(7/4)B ----------
or
B=(4/7)A ------------(2)
put the value of (2)in (1)
(4/7)*B+B=1/7
B=1/11 THIS IS HIS ONE DAY WORK
ALONE B CAN FINISH IN 11 DAY
Pankaj said:
1 decade ago
How Total Working Days=7*4=28 ?
M.vignesh said:
1 decade ago
A=7/4
B=1
A:B=7/4:1
So A:b=7:4
Total Working Days=7*4=28
a=28/7=4
b=28/4=7
a=a+b
a=4+7=11
A alone=11
B=1
A:B=7/4:1
So A:b=7:4
Total Working Days=7*4=28
a=28/7=4
b=28/4=7
a=a+b
a=4+7=11
A alone=11
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