Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 6)
6.
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
Answer: Option
Explanation:
Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.
| Then, 6x + 8y = | 1 | and 26x + 48y = | 1 | . |
| 10 | 2 |
| Solving these two equations, we get : x = | 1 | and y = | 1 | . |
| 100 | 200 |
| (15 men + 20 boy)'s 1 day's work = |  |
15 | + | 20 |  |
= | 1 | . |
| 100 | 200 | 4 |
15 men and 20 boys can do the work in 4 days.
Discussion:
201 comments Page 19 of 21.
Waseem said:
3 years ago
Thanks for explaining this @Zoya.
(4)
Devil said:
3 years ago
Thanks for the good explanation @Parkavi.
(1)
Sajid said:
3 years ago
6M+8B *10 = 26M + 48B*2.
4M = 8B.
Ratio M&B = 2/1.
Total work = 2(6M+8B) * 10.
200.
Acc to question.
15M+20B=? ,
2(15M+20B),
50.
TIME = 200/50 = 4 Days.
4M = 8B.
Ratio M&B = 2/1.
Total work = 2(6M+8B) * 10.
200.
Acc to question.
15M+20B=? ,
2(15M+20B),
50.
TIME = 200/50 = 4 Days.
(12)
HYACINTH S BABU said:
3 years ago
1M = 2BOY.
(15B+20B) = (6M+8B) * 10,
(15M+10M) * DAYS = (6M+4M)*10.
= 4 days is the right answer.
(15B+20B) = (6M+8B) * 10,
(15M+10M) * DAYS = (6M+4M)*10.
= 4 days is the right answer.
(15)
Sujan gautam said:
3 years ago
15/6=2.5, 20/8=2.5.
Then , 10(6m and 8b) /2.5=4.
Then , 10(6m and 8b) /2.5=4.
(19)
Yamuna said:
2 years ago
6m + 8b = 1/10.
2(3m + 4b) =1/10.
3m + 4b = 1/20.
5(3m + 4b) = 5(1/20).
15m + 20b = 1/4.
2(3m + 4b) =1/10.
3m + 4b = 1/20.
5(3m + 4b) = 5(1/20).
15m + 20b = 1/4.
(60)
Arun said:
2 years ago
From given data,
6M+8B ---> 10days.
26M+48B ---> 2days.
15M+20B ---> t days.
By using Product approach
WF1*T1=WF2*T2.
(6M+8B)*10=(26M+48B)*2,
4M=8B,
M/B=2/1,
M=2;B=1.
Now substitute We get;
20 --->10 days.
100 --->2 days.
50 ---> t days.
100*2=50*t(Product approach)
t=4 i.e.,4 days.
6M+8B ---> 10days.
26M+48B ---> 2days.
15M+20B ---> t days.
By using Product approach
WF1*T1=WF2*T2.
(6M+8B)*10=(26M+48B)*2,
4M=8B,
M/B=2/1,
M=2;B=1.
Now substitute We get;
20 --->10 days.
100 --->2 days.
50 ---> t days.
100*2=50*t(Product approach)
t=4 i.e.,4 days.
(41)
Kumar said:
2 years ago
6m + 8b = 1/10.
2(3m + 4b) =1/10.
3m + 4b = 1/20.
5(3m + 4b) = 5(1/20).
15m + 20b = 1/4.
2(3m + 4b) =1/10.
3m + 4b = 1/20.
5(3m + 4b) = 5(1/20).
15m + 20b = 1/4.
(130)
YASEEN said:
2 years ago
Here in 1st case, there are 6 men and 8 boys = [6+8=14] and their work done is 10.
So, 10*14 = 140[total work done by men and boys,
Similarly,
In 2nd case 26+48 = 74,
74*2 = 140.
The Total work done by 15 men and 20 boys is 15+20=35 and let their work done be x.
Now 35 * x
.
From the above work done take anyone.
140 = 35*x,
x = 4.
So, 10*14 = 140[total work done by men and boys,
Similarly,
In 2nd case 26+48 = 74,
74*2 = 140.
The Total work done by 15 men and 20 boys is 15+20=35 and let their work done be x.
Now 35 * x
.
From the above work done take anyone.
140 = 35*x,
x = 4.
(71)
YASEEN said:
2 years ago
Here in 1st case there are 6 men and 8 boys = [6+8=14] and their work done is 10 so 10*14=140[total workdone by men and boys]
Similarly in 2nd case 26+48=74 , 74*2=140
Total work done by 15 men and 20 boys is 15+20=35 and let their work done be x.
Now 35*x
from the above work done take anyone
140=35*x
x=4
Similarly in 2nd case 26+48=74 , 74*2=140
Total work done by 15 men and 20 boys is 15+20=35 and let their work done be x.
Now 35*x
from the above work done take anyone
140=35*x
x=4
(164)
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