Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 4)
4.
A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
Answer: Option
Explanation:
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
| If difference of time is 60 days, B takes |  |
3 | x 60 |  |
= 90 days. |
| 2 |
So, A takes 30 days to do the work.
| A's 1 day's work = | 1 |
| 30 |
| B's 1 day's work = | 1 |
| 90 |
| (A + B)'s 1 day's work = | |
1 | + | 1 | |
= | 4 | = | 2 |
| 30 | 90 | 90 | 45 |
 A and B together can do the work in |
45 | = 22 | 1 | days. |
| 2 | 2 |
Discussion:
281 comments Page 7 of 29.
SANJEEV said:
1 decade ago
#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));
}
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));
}
SscASPIRANT said:
9 years ago
Hi guys
Here is a trick to solve it.
If B = 1, A = 3,
Therefore efficiency is A : B = 3 : 1.
Now, 60 days less means the difference of their efficiency.
2 ---> 60
1 ---> 30
Now you got 30.
A = 3 * 30 = 90,
B = 1 * 30 = 30,
LCM = 90.
Which is total work and total days = total work/total efficiency.
T.D = 90/4 = 45/2 = 22 (1/2)days.
Here is a trick to solve it.
If B = 1, A = 3,
Therefore efficiency is A : B = 3 : 1.
Now, 60 days less means the difference of their efficiency.
2 ---> 60
1 ---> 30
Now you got 30.
A = 3 * 30 = 90,
B = 1 * 30 = 30,
LCM = 90.
Which is total work and total days = total work/total efficiency.
T.D = 90/4 = 45/2 = 22 (1/2)days.
Chawdharya said:
6 years ago
A : B
TIME 1/3 : 1.
DAYS X-60 : X.
X-60/X = 1/3,
X= 90,
X-60 = 30.
A work efficiency = 1/90,
B work efficiency = 1/30,
A+B work efficiency = 1/90+1/30,
= 2/45,
work efficiency = work/time.
work is constant,so time = 45/2.
= 22.5.
TIME 1/3 : 1.
DAYS X-60 : X.
X-60/X = 1/3,
X= 90,
X-60 = 30.
A work efficiency = 1/90,
B work efficiency = 1/30,
A+B work efficiency = 1/90+1/30,
= 2/45,
work efficiency = work/time.
work is constant,so time = 45/2.
= 22.5.
Deepti said:
1 decade ago
It is not so big and easy to understand.
let work done by B in 1 day = 1/x
so work done by A in 1 day = 1/(x-60)
1/(x-60) = 3/x
so x = 3x - 180
x = 90
work done by A and B in 1 day is 30 and 90 respectively
work done by both together in 1 day = 1/30 + 1/90 = 2/45.
total work done = 45/2
let work done by B in 1 day = 1/x
so work done by A in 1 day = 1/(x-60)
1/(x-60) = 3/x
so x = 3x - 180
x = 90
work done by A and B in 1 day is 30 and 90 respectively
work done by both together in 1 day = 1/30 + 1/90 = 2/45.
total work done = 45/2
Prashant said:
1 decade ago
Suppose B takes x days to finish the work and then according to question A will take (60-x) days to work. in question the ratio of time taken is 1:3 therefore
1/3= (60-x)/x
x=45
So A can complete the work in (60-x)=(60-45)=15 days
So (A+B) one day's work
=(1/15 + 1/45)
=4/45
So (A+B) complete the work in 45/4 = 11 1/4 days.
1/3= (60-x)/x
x=45
So A can complete the work in (60-x)=(60-45)=15 days
So (A+B) one day's work
=(1/15 + 1/45)
=4/45
So (A+B) complete the work in 45/4 = 11 1/4 days.
Pramod patel said:
8 years ago
I think the best and universal approach is this:
A can do the work in X days, B takes X+60 days for same work,
But (X+60)/X=3/1 as time taken by A and B is in 1:3 ratio,
Solving we get X=30,
A's one day work is 1/30,
B's one day work is 1/90,
combining their one day work we get 1/30+1/90=2/45,
total time taken by them is 45/2.
A can do the work in X days, B takes X+60 days for same work,
But (X+60)/X=3/1 as time taken by A and B is in 1:3 ratio,
Solving we get X=30,
A's one day work is 1/30,
B's one day work is 1/90,
combining their one day work we get 1/30+1/90=2/45,
total time taken by them is 45/2.
Mohit Agnihotri said:
1 decade ago
Work done ratio (A:B) = 3:1.
And let x be no. of days taken by B.
So, A:B = x-60:x {Given- A finish a job in 60 days less than B}.
Now, (x-60)/x = 1/3.
3x-180 = x => x = 90.
It means A takes 30 days and B takes 90days.
One day work of A and B is 1/30+1/90 = 4/90.
Thus, 90/4 = 22.5 days required to complete the work.
And let x be no. of days taken by B.
So, A:B = x-60:x {Given- A finish a job in 60 days less than B}.
Now, (x-60)/x = 1/3.
3x-180 = x => x = 90.
It means A takes 30 days and B takes 90days.
One day work of A and B is 1/30+1/90 = 4/90.
Thus, 90/4 = 22.5 days required to complete the work.
Rohan said:
6 years ago
A and B
Work 3:1
Time 1:3
i.e. 30:90 ......multiply by 30.
And also they have given that the difference between the time taken to complete the work is 60.
Hence when A and B time to work together,
1/30+ 1/90 = 2/45,
Defer for A and B to complete the entire work they require 45/2 which is equivalent to 22 and A half-day.
Work 3:1
Time 1:3
i.e. 30:90 ......multiply by 30.
And also they have given that the difference between the time taken to complete the work is 60.
Hence when A and B time to work together,
1/30+ 1/90 = 2/45,
Defer for A and B to complete the entire work they require 45/2 which is equivalent to 22 and A half-day.
Raj said:
1 decade ago
Easy lets assume A works total days = X.
So A one day work = 1/X.
B is working 3 times less than A.
So Total days of B is = 3X.
So B Works one day = 1/3X.
Different between total days of B and A is 60.
So 3X-X = 60.
2X = 60.
X = 30.
Now normal maths. Replace X with numbers. You will get answer.
So 1/30-1/90 = 22.5.
So A one day work = 1/X.
B is working 3 times less than A.
So Total days of B is = 3X.
So B Works one day = 1/3X.
Different between total days of B and A is 60.
So 3X-X = 60.
2X = 60.
X = 30.
Now normal maths. Replace X with numbers. You will get answer.
So 1/30-1/90 = 22.5.
@verma ji said:
9 years ago
Hello Friends: this shortcut formula is better to solve these type question in less time.
Alternate Method: Since 'A' is 'm' times as efficient as 'B' & takes 'D' days less than 'B', then the time required to complete the job together is given by,
T = m * D/(m^2 " 1)
T = 3 * 60 / [(3)^2 - 1] = 180 / 8 = 22 1/2 days.
Alternate Method: Since 'A' is 'm' times as efficient as 'B' & takes 'D' days less than 'B', then the time required to complete the job together is given by,
T = m * D/(m^2 " 1)
T = 3 * 60 / [(3)^2 - 1] = 180 / 8 = 22 1/2 days.
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