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 17 of 21.
Vineeth TC said:
6 years ago
To apply in equation we need to convert days in to work they did. Let's assume they make 20 chairs (LCM).
6m + 8b =2 -->(1) (total 20 chairs;per day 2 chairs).
26m+48b=10 -->(2).
6m + 8b =2 -->(1) (total 20 chairs;per day 2 chairs).
26m+48b=10 -->(2).
Ritik Jain said:
6 years ago
Given : 6 men and 8 boys can do a piece of work in 10.
To find: 15 men and 20 boys in doing the same type of work will be:
So here we can see that the number of men and boys are increased by a factor of 2.5
Therefore we decrease the number of days by factor 2.5.
So, the answer is 10/2.5 = 4.
To find: 15 men and 20 boys in doing the same type of work will be:
So here we can see that the number of men and boys are increased by a factor of 2.5
Therefore we decrease the number of days by factor 2.5.
So, the answer is 10/2.5 = 4.
Pravin said:
6 years ago
6x+8y = 1/10,
2(3x+4y) = 1/10,
3x+4y = 1/20,
Both side multiple with 5,
15x+20y = 1/4.
So 1/4 work completes = 1 day.
total work = 4days.
2(3x+4y) = 1/10,
3x+4y = 1/20,
Both side multiple with 5,
15x+20y = 1/4.
So 1/4 work completes = 1 day.
total work = 4days.
Rupak said:
6 years ago
Good, thanks @Ritik.
Kermiki Bang said:
6 years ago
Apply the value of x = 1/100 in Eqn 1, we get,
6/100+8y = 1/10.
Take LCM you will get the answer y = 1/200.
6/100+8y = 1/10.
Take LCM you will get the answer y = 1/200.
Anshul said:
6 years ago
Manpower 6M+8B 26M+48B
Rate 1 5
Time 10 2
Work 10 10
From this use M1*R1*T1/W = M2*R2*T2/W2.
We will get 30M+40B = 26M+48B.
From this equation, we can see that if 4 men are decreased then 8 boys are required to do the same work at the same time.
Now, we need to find the days required for 15M+20B.
15M+20B is nothing but 25 men. (because of 1 man = 2 boys).
Now if we find the rate at which 15M+20B works T can be easily calculated.
To find the rate,
Take 6M+8B works at rate of 1. i.e 10 men work at rate of 1.
We need to find rate at which 25 men will work.
therefore,10 men at rate of 1,20 men at 2 and hence 25 at 2.5.
Manpower 6M+8B 26M+48B 15M+20B
Rate 1 5 2.5
Time 10 2 4
Work 10 10 4
you can do this mentally pretty fast if practiced.
Rate 1 5
Time 10 2
Work 10 10
From this use M1*R1*T1/W = M2*R2*T2/W2.
We will get 30M+40B = 26M+48B.
From this equation, we can see that if 4 men are decreased then 8 boys are required to do the same work at the same time.
Now, we need to find the days required for 15M+20B.
15M+20B is nothing but 25 men. (because of 1 man = 2 boys).
Now if we find the rate at which 15M+20B works T can be easily calculated.
To find the rate,
Take 6M+8B works at rate of 1. i.e 10 men work at rate of 1.
We need to find rate at which 25 men will work.
therefore,10 men at rate of 1,20 men at 2 and hence 25 at 2.5.
Manpower 6M+8B 26M+48B 15M+20B
Rate 1 5 2.5
Time 10 2 4
Work 10 10 4
you can do this mentally pretty fast if practiced.
Devendra said:
6 years ago
Given that.
6M+8B = 1/10 (1'd work). ----------------[1].
26M+49B = 1/2 (1'd work). ----------------[2].
Sub eq (2) - eq (1).
20M+40B = 1/2-1/10=4/10.
2M+4B = 4/100 = 1/25.
Total time 25.
6M+8B = 1/10 (1'd work). ----------------[1].
26M+49B = 1/2 (1'd work). ----------------[2].
Sub eq (2) - eq (1).
20M+40B = 1/2-1/10=4/10.
2M+4B = 4/100 = 1/25.
Total time 25.
(1)
Bantu Venkatesh said:
6 years ago
(6M+8B)10=(26M+48B)2.
8M=16B.
M:B
2:1.
12B + 8B-----10days
30B + 20B------?
=> 20/50 *10=4days.
8M=16B.
M:B
2:1.
12B + 8B-----10days
30B + 20B------?
=> 20/50 *10=4days.
(1)
Ravikiran said:
5 years ago
If 6 men and 8 boys take 10days to complete a work.
Which means 3 men and 4 boys take 20days to complete a work.
Then 3*5 (15 men) and 4*5(20 boys) take (20/5) which is 4 days.
Which means 3 men and 4 boys take 20days to complete a work.
Then 3*5 (15 men) and 4*5(20 boys) take (20/5) which is 4 days.
(2)
Balu said:
5 years ago
If X is the overall work then,
X=10(6m+8b)=60m+80b-----1st eqn.
X=2(26m+48b)=52m+96b-----2nd eqn.
Equating the both eqns 1 and 2
We get m:b=1:2 it means 1man equal to 2 boys so 15m+20b=30b+20b=50b.
From eqn 1,12b+8b=20b can do in 10days.
Then cm to 20b=10.
50b=y.
Y=4 days.
X=10(6m+8b)=60m+80b-----1st eqn.
X=2(26m+48b)=52m+96b-----2nd eqn.
Equating the both eqns 1 and 2
We get m:b=1:2 it means 1man equal to 2 boys so 15m+20b=30b+20b=50b.
From eqn 1,12b+8b=20b can do in 10days.
Then cm to 20b=10.
50b=y.
Y=4 days.
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