Aptitude - Time and Work - Discussion

Let the time taken by A be x days,
then time taken by B will be 3x days.
Given that,
time taken by A to finish the work = time taken by B -60
i.e, x=3x-60
=> x=30.
Therefore, time taken by A to finish the work is 30 days.
so B takes 90 days.
Now ,work done by A & B in 1 day=1/30 + 1/90
=4/90
=2/45
Time taken by A & B to complete the work=45/2 days
=22 1/2 days

#include<stdio.h>
void main()
{
float a,b;
int c,d;
printf("enter ratio of A : B\n");
scanf("%f",&a);
printf("enter no of days lesser than B by A to complete work\n");
scanf("%f",&b);
c=ab/(a-1);
d=c-b;
printf("no of days to complete work by A and B is %d",(c*d)/(c+d));
}

If difference of time 2 days B takes 3 days

Similarly why don't we take difference of time 60 days B takes 60*3 is 180 days. Please explain to me ?

Friends, don't make it complex: Here we go.

Ratio of time taken by A & B = 1:3

And A takes 60 days less then B

Suppose, days taken by B = x then days taken by A = (x - 60)

Thus,

A : B = (x - 60) : x
1 : 3 = (x - 60) : x
1/3 = (x - 60) / x
X = 3 (x - 60)
X = 3x - 180
180 = 3x - x
180 = 2x
X = 180 / 2 = 90

Hence,
Days taken by B = x = 90 days
Days taken by A = (x - 60) = 90 - 60 = 30 days

1 day work of A and B = 1/30 + 1/90
= 4/90
= 2/45

So,
Total days taken by A & B to complete the work = 45/2 = 22.5 days

Formula-->nx- (n+1) /nx.
4x-4x/3=60 so x=45/2.

So by the explanation above we have to take in the ratio 1:3.

A is twice as good a workmen as B and together they can finish a piece of work in 18days. In how many days will A alone finish the work?

Can we solve the above problem using the formula k*x/k^2-1.

Let A takes x days to do the work alone
Therefore days taken by b is (x+60).

Now A is 3 times faster than B
So no.of days taken by A will be 3 times less than B.
So
3*x=(x+60)
2x=60
x=30 days //work done by A
Work done by B=30+60=90 days.

Now we can find work done together
1/30 +1/90 to get the answer.

We have a direct formula for these kind of questions .. i.e.,

Work done together is given by = (m*d)/(m^2 - 1),

Where d is no of days taken by B and m is no of time A is faster than B,
Therefore = (3*60)/(3^2-1) = 22.5.

@All.

Please refer to Sheetal's response as given below. Its the best and easiest way to solve such questions.

PS:Thanks sheetal.

"A is K times as good a worker as B & takes X days less than B to finish the work. Then the amount of time required by A & B working together is K * X / K'S SQUARE - 1.

SO IN QUESTION K IS 3. X IS 60. SO PUTTING THE VALUES IN ABOVE MENTIONED FORMULA.

K * X / K'S SQUARE - 1
=3 * 60 / 9 - 1
=3 * 60 / 8
=180 / 8
=22.5 DAYS.

AS MENTIONED OPTION B.