# 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 :

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 7*x* and 4*x* 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:

52 comments Page 1 of 6.
Menya Alex said:
2 years ago

Let A+B = 7 days.

Then (A+B)'s one day work = 1/7,

Let X represent B,

=> 7/4X + X = 1/7,

=>7X + 4X = 4/7,

=>11X = 4/7,

=>X = 4/77,

But A = 7/4X.

=> A = (7/4*4/77),

=> A = 1/11,

Giving us 11 days.

Then (A+B)'s one day work = 1/7,

Let X represent B,

=> 7/4X + X = 1/7,

=>7X + 4X = 4/7,

=>11X = 4/7,

=>X = 4/77,

But A = 7/4X.

=> A = (7/4*4/77),

=> A = 1/11,

Giving us 11 days.

(8)

Lingaraj Behera said:
2 years ago

(A+B) 1day work is 1/7 --------------> (1)

Let 1-day work of A is 1/A, such as 1-day work of B is 1/B

Then 1/A + 1/B = 1A + 1/A * 7/4.

1/A + 4/7A = 11/7A--------------(2)

11/7A = 1/7,

7A = 77,

A = 11.

Let 1-day work of A is 1/A, such as 1-day work of B is 1/B

Then 1/A + 1/B = 1A + 1/A * 7/4.

1/A + 4/7A = 11/7A--------------(2)

11/7A = 1/7,

7A = 77,

A = 11.

(4)

Sourav Kalal said:
3 years ago

Here is my simple approach

------------------------------------

A=7 (A's 1 day work)

B=4 (B's 1 day work)

------------------------------------

A+B = 11 (A+B's 1 day work).

A+B can finish work in 7 days,

So, Total work = 11*7=77 unit.

A can do 7 units of work in 1 day.

for 77 unit he will take 77/7= 11 days.

(7 ----> 1,

77 ---> ?) = 77/7.

------------------------------------

A=7 (A's 1 day work)

B=4 (B's 1 day work)

------------------------------------

A+B = 11 (A+B's 1 day work).

A+B can finish work in 7 days,

So, Total work = 11*7=77 unit.

A can do 7 units of work in 1 day.

for 77 unit he will take 77/7= 11 days.

(7 ----> 1,

77 ---> ?) = 77/7.

(10)

Nesha said:
6 years ago

Thank you @Jimmy John.

(1)

Faisal said:
6 years ago

(A+B)'s 1 day = 1/7

And given that,

B=(4/7)A.

(A+ (4/7)A) 1 day = 1/7

(11/7) A 1 day = 1/7

11 A 1 day= 1

A's 1 day= 1/11.

Therefore, A needs 11 days alone to complete the required work.

And given that,

B=(4/7)A.

(A+ (4/7)A) 1 day = 1/7

(11/7) A 1 day = 1/7

11 A 1 day= 1

A's 1 day= 1/11.

Therefore, A needs 11 days alone to complete the required work.

(2)

Hellani said:
7 years ago

Can you please explain me? Please.

(1)

Prabhu said:
7 years ago

Nice explanation @Jimmy John.

Padma said:
7 years ago

Efficiency of A:B = 7/4k : k.

No of days taken by A: B = 4k/7k = 4/7 (no of days taken = ratio of the reciprocals of the efficiency).

Let the work done in one day = 1/s.

then , work done by A in one day = 4/s.

work done by B in one day = 7/s.

work done by A+B = 1/7.

4/s + 7/s = 1/7.

11/s = 1/7.

s= 77.

Work done by A in one day = 4/s.

= 4/77.

the efficiency of A = 7/4.

Hence, work done by A in one day = 4/77 * 7/4.

= 1/11

Thus, A can do the job in 11 days.

No of days taken by A: B = 4k/7k = 4/7 (no of days taken = ratio of the reciprocals of the efficiency).

Let the work done in one day = 1/s.

then , work done by A in one day = 4/s.

work done by B in one day = 7/s.

work done by A+B = 1/7.

4/s + 7/s = 1/7.

11/s = 1/7.

s= 77.

Work done by A in one day = 4/s.

= 4/77.

the efficiency of A = 7/4.

Hence, work done by A in one day = 4/77 * 7/4.

= 1/11

Thus, A can do the job in 11 days.

Padma said:
7 years ago

Why have they taken efficiency as such without converting it to number of days?

Swapna said:
7 years ago

7+4=11......28 ,7....? 11/7*28 = 44 ans.

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