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:
202 comments Page 4 of 21.
Vijay said:
9 years ago
Guys. No need to get in to the equation. If 6 man and 8 boys can do a work in 10 days. So obviously 12 men's and 16 boys can complete this in 5 days. (Since the count of the mens and boys is 15 and 20 respectively. So the number of the day will be reduce more. So the only option is 4 which is below 5) so the answer is 4.
Nisha said:
1 decade ago
It can be solved more easily in this way
see in 1st ... 6 man are changing to 15 men in question ...n
8 boys r changing to 20 boys in q........so
Here we can see a change of "2.5" .. .( 6*2.5= 18 n 8*2.5=20)
as inverse relation with men n time........10 days will divded by 2.5
ans is 4days
see in 1st ... 6 man are changing to 15 men in question ...n
8 boys r changing to 20 boys in q........so
Here we can see a change of "2.5" .. .( 6*2.5= 18 n 8*2.5=20)
as inverse relation with men n time........10 days will divded by 2.5
ans is 4days
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
(169)
Shri said:
1 decade ago
2x+3y = 1/10 ----- (1).
3x+2y = 1/8 ------ (2).
We want to find: 2x+1y = ?---- (3).
Any one can solve this answer is:
10(2x+3y) = 8(3x+2y).
20x+30y = 24x+16y.
4x = 14y.
2x = 7y.
So 2x+3y = 10 ----- (1).
7y+3y = 10.
10y = 10.
2x+1y = ?---- (3).
7y+1y = 8y.
So 10yx10/8y = 25/2.
= 12 1/2 (Ans).
3x+2y = 1/8 ------ (2).
We want to find: 2x+1y = ?---- (3).
Any one can solve this answer is:
10(2x+3y) = 8(3x+2y).
20x+30y = 24x+16y.
4x = 14y.
2x = 7y.
So 2x+3y = 10 ----- (1).
7y+3y = 10.
10y = 10.
2x+1y = ?---- (3).
7y+1y = 8y.
So 10yx10/8y = 25/2.
= 12 1/2 (Ans).
Ballu... said:
1 decade ago
Lets try in other way:
Here given 6 and 8 do in 10 days. If persons halved then time doubles.
Then 3 and 4 will do work in 20 days.
By the same way,
If persons double. Then time half's. In the same way.
If 5 times the members. Then 1/5 times the days. Then 15 and 20 will do. Work in 4 days.
Here given 6 and 8 do in 10 days. If persons halved then time doubles.
Then 3 and 4 will do work in 20 days.
By the same way,
If persons double. Then time half's. In the same way.
If 5 times the members. Then 1/5 times the days. Then 15 and 20 will do. Work in 4 days.
Prem kumar said:
10 years ago
Firstly I make an equation.
6men + 8boy =1/10---------> eq1.
26men + 48boy =1/2--------> eq2.
For eq1.
6men + 8boy =1/10.
We take common.
2 (3m + 4b) =1/10.
3m + 4b = 1/20.
Then,
Multiply the common value 5.
5* (3m + 4b) = 1/20*5.
Solve the equation.
15m + 20b = 1/4.
Now answer is 4 days.
6men + 8boy =1/10---------> eq1.
26men + 48boy =1/2--------> eq2.
For eq1.
6men + 8boy =1/10.
We take common.
2 (3m + 4b) =1/10.
3m + 4b = 1/20.
Then,
Multiply the common value 5.
5* (3m + 4b) = 1/20*5.
Solve the equation.
15m + 20b = 1/4.
Now answer is 4 days.
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.
Deepak Patgar said:
1 decade ago
6 men and 8 boys.
15 men and 20 boys.
If we look at the data carefully they are in the ratio 1:2.5.
i.e. 6:15 and 8:20 are same as 1:2.5.
So the number of days required for 15 men and 20 boys will be 10/2.5=4 days.
(Note: this method can be applied only when the ratios are same).
15 men and 20 boys.
If we look at the data carefully they are in the ratio 1:2.5.
i.e. 6:15 and 8:20 are same as 1:2.5.
So the number of days required for 15 men and 20 boys will be 10/2.5=4 days.
(Note: this method can be applied only when the ratios are same).
Parmeshwar Sharma said:
5 months ago
From both conditions, equate total work:
- (6M + 8B) × 10 = (26M + 48B) × 2.
→ 60M + 80B = 52M + 96B,
→ 8M = 16B → M = 2B.
Now convert 15 men + 20 boys into boys:
→ 15 men = 30 boys → Total = 50 boys
Total work = (6M + 8B) × 10 = (12B + 8B) × 10 = 200B,
Time = 200B ÷ 50B = 4 days.
- (6M + 8B) × 10 = (26M + 48B) × 2.
→ 60M + 80B = 52M + 96B,
→ 8M = 16B → M = 2B.
Now convert 15 men + 20 boys into boys:
→ 15 men = 30 boys → Total = 50 boys
Total work = (6M + 8B) × 10 = (12B + 8B) × 10 = 200B,
Time = 200B ÷ 50B = 4 days.
(20)
Siddharth said:
2 decades ago
No need to solve equation:
we have
6x+8y=1/10 ----- (1)
26x+48y=1/2 ------(2)
we want to find: 15x+20y=?----(3)
now taking 2 common form equ(1),
we get:
2(3x+4y)=1/10
So we have, 3x+4y=1/20...ok...
now
in equ 3, take 5 common;
we get:
=5(3x+4y) (3x+4y=1/20)
=5*(1/20)
=4 (ans)
we have
6x+8y=1/10 ----- (1)
26x+48y=1/2 ------(2)
we want to find: 15x+20y=?----(3)
now taking 2 common form equ(1),
we get:
2(3x+4y)=1/10
So we have, 3x+4y=1/20...ok...
now
in equ 3, take 5 common;
we get:
=5(3x+4y) (3x+4y=1/20)
=5*(1/20)
=4 (ans)
(5)
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