# A work is done by two people in 24 min. one of them can do this work a lonely in 40 min. how much time required to do the same work for the second person.

In 1 minute (A+B) can do 1/24 work. A alone can 1/40 work in 1 minuteB alone can do in 1 minute = (A+B)�s � A�s  = 1/24 � 1/40 = 1/60 workTherefore, b can do the same work in = 60 min
This question is related to TCS Interview

#### Amit Kularkar

• Sep 27th, 2005

if x is time required by person B,then time required to do a job together will be=[(40*x)/(40+x)]=24.hence on solving we get, x=60

#### gud_guy2006 Profile Answers by gud_guy2006

• Mar 7th, 2006

Plaese tell me the formula u used on solving in detail.How to know this formulaes use?ie (40*x)/(40+x)=24

#### hardeep singh

• Mar 18th, 2006

sol- p=40p*q / (p+q) = 24;40*q / (40+q) =24;=> q=60 min.

#### sriram krishnamoorthi

• Mar 30th, 2006

let A does the work in 40 mins (given)

let B does the work in x mins

amt of work done by A in 1 min = 1/40

amt of work done by B in 1 min = 1/x

amt of work done jointly by both A & B = 1/24

which implies (1/40)+(1/x) = (1/24)

on solvin the above eqn. for x we get

x =  60

#### SharnCedar

• Feb 18th, 2007

Was this a real question at an interview? Was this for Indian offshore programmers or who?  It is a funny question, it seems the interviewer is not a professional or has no IT experience. Because work done in 40 minutes by one person is done in about 3 or 4 hours by 2 people (rule of thumb from many yars experience), they must create an inteface point to share the work.  Here's the analysis:

1 person writing a module of code must write 1 API (interface to existing code ) and 1 module.

Two people writing code together must write 4 API's (programmer A writes and interface to existing code and to programmer B's code, B defines two interfaces as well) and 2 code modules.  The code modules are not necessarily "smaller" because code is often not easy to break into two pieces.  For example, how to divide up the work for writing a single loop?

So the total work is up to 5 times greater.  Any saving of time is made in the dividing of the coding of modules itself, this will only be significant on a very large piece of work, certainly something much, muvh bigger than a 40 minute piece of work.

#### ratheesh69 Profile Answers by ratheesh69

• Jun 8th, 2008

(a+b)'s 1min work=1/24
a's 1min wrk=1/40
thrfr b's 1min wrk=(a+b)-a=(1/24)-1/40=1/60

b can do a wrk in 60 mins

#### bharathfeb Profile Answers by bharathfeb

• Jul 7th, 2009

A & B can do the work in 24 mins

A can do the same work alone in 40 mins

let B can do the same work alone in 'b' mins

=> 1/40 + 1/b = 1/24

=> 1/b = 1/24 - 1/40

=> 1/b = 1/60

=> b = 60

Therefore, B can do the same work in 60 mins

#### bharathfeb Profile Answers by bharathfeb

• Jul 7th, 2009

Let A does a work in 40 mins [Given]

=> work done by A in 1 min = 1/40

Let B can do the same work in 'b' mins

=> work done by B in 1 min = 1/b

Then, work done by A & B in 1 min = 1/40 + 1/b

A & B does the same work in 24 mins [Given]

=> work done by A & B in 1 min = 1/24

So, 1/40 + 1/b = 1/24

By solving the above eq, we get b = 60

Therefore, B does the work in 60 mins

#### bharathmarri Profile Answers by bharathmarri

• Jul 8th, 2009

Let A can do a work alone in 40 mins [Given]

=> amt of work done by A in 1 min = 1/40

Let B can do the same work alone in 'b' mins

=> amt of work done by B in 1 min = 1/b

then, amt of work done by A & B in 1 min = 1/40 + 1/b ---> 1

A & B can do the same work in 24 mins [Given]

=> amt of work done by A & B in 1 min = 1/24 ---> 2

From 1 & 2,

1/40 + 1/b = 1/24

by solving the abv eq, we get b = 60

Therefore, B can do the same work in 60 mins  