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:
196 comments Page 1 of 20.
Vinutha said:
1 year 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.
(202)
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
(155)
Chahat Mishra said:
7 months ago
Very Easy to Understand.
6m + 8b = 1/10,
2(3m + 4b) = 1/10,
3m + 4b = 1/20,
5(3m + 4b) = 5(1/20),
15m + 20b = 1/4.
We are getting the solution without the involvement of 26 men and 48 boys.
6m + 8b = 1/10,
2(3m + 4b) = 1/10,
3m + 4b = 1/20,
5(3m + 4b) = 5(1/20),
15m + 20b = 1/4.
We are getting the solution without the involvement of 26 men and 48 boys.
(146)
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.
(128)
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.
(67)
Nav Raj Bhatta said:
4 years ago
Find the relation between first, second to third equation.
6M+8B=10
26M+48B=2
15M+20B=?.
Now taking first equation;
6M + 8B = 10.
60M + 80B = 1.
30M + 40B = 2.
15M + 20B = 4.
6M+8B=10
26M+48B=2
15M+20B=?.
Now taking first equation;
6M + 8B = 10.
60M + 80B = 1.
30M + 40B = 2.
15M + 20B = 4.
(65)
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)
Sanjay Chauhan said:
8 months ago
6 men = 1/10.
1 men = 1/10 * 1/6 = 1/60.
8 boys = 1/10.
1 boy = 1/10 * 1/8 = 1/80.
Now,
15 men = 15/60
20 boys = 20/80
15/60 + 20/80 = 4.
Ans - 4days.
1 men = 1/10 * 1/6 = 1/60.
8 boys = 1/10.
1 boy = 1/10 * 1/8 = 1/80.
Now,
15 men = 15/60
20 boys = 20/80
15/60 + 20/80 = 4.
Ans - 4days.
(59)
Lung said:
12 months ago
(6M + 8B)× 10 = (26M+48B)×2
30M+40B = 26M+48B
M=2, B=1 {their efficiency}
Now, total work = 6×2+8×10 = 200
therefore work done by 15M + 20B = 200/15×2+20 = 4.
30M+40B = 26M+48B
M=2, B=1 {their efficiency}
Now, total work = 6×2+8×10 = 200
therefore work done by 15M + 20B = 200/15×2+20 = 4.
(40)
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.
(39)
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