G.S.Kantharao   8 years AGO

40 Men can do a work of 20mts in 15 days.
Suppose ifitis in 1 Day men needed = 600.
600 Men can do 20 mts in 1 day.
Let x Men can do 50 mts in 1 day.
x = 600*(50/20) = 1500 Men.
If the same work is shared for 3 days no.Men required =1500/3 =500.
The right option is 8.

Like? Yes (8) | No (5)  

sainath   8 years AGO

As we know the formula (M1 X D1)/H1 = (M2 X D20/H2
(40 X 15)/200 = (M2 X 3)/50
By solving M2 = 500

Like? Yes (4) | No (2)  

Devireddy Jashuva   8 years AGO

40 men - 20 m high - 15 days
? men - 50 m high - 3 days

? = 40 * (50/20) * (15/3)
= 500

Like? Yes (3) | No (1)  

savita das   8 years AGO

1 m hight wall build in 30 days by 1 man. Then 1 m night wall build in 3 days by 1*30/3.
There fore 50 m night wall build in 3 days by 1*10*50=500 men.

Like? Yes (1) | No   

vinay kumar   8 years AGO

. .
. 40 men build a 20m high wall in 15 days so for 50m high wall these men take 37.5 days.
so 1 man can build this wall in = 37.5*40 days.
now to build this wall in 3 days no. of men required=(37.5*40)/3
no.of men required=500

Like? Yes (1) | No   

ADITI GOEL   8 years AGO

40 men built 20m wall in 15 days,
40 men built wall per day= 20/15 m per day.
now,..1 man built per day=20/(40*15)m per day=1/30 m per day
therefore 1 man can built in 3 days=(1/30)*3 m=1/10 m
now,10 men built=(1/10)*10=1m
50 m wall requires=10*50=500 men

Like? Yes (1) | No   

nimisha gupta   8 years AGO

Let the number of workers be x
(40*15/20)=(x*3)/50
then x=500

Like? Yes (1) | No   

Hemanathan.R.K   8 years AGO

By Men-Days concept:
(M1*D1)/W1= (M2*D2)/W2
(40*15)/20= (x*3)/50
x=500

Like? Yes (3) | No (3)  

sai sreenivas   8 years AGO

we have the formula (m1*d1)/w1=(m2*d2)/w2.
substituting the values we get the values for m2 as 500.

Like? Yes | No   

RAVALI SIRIMALLA   8 years AGO

(men*days*hours)/work is the formula by comparing the 2 entities given we will get the answer

Like? Yes (1) | No (1)  

Sruthi Gourigari   8 years AGO

40mens------>20m
1m------>2mens
=>50m----->100mens
that's 15days
but in 3days
15/3=5
so total men's =>100×5=500

Like? Yes (4) | No (4)  

Ankit Sharma   8 years AGO

40men - 20m wall - 15 days
therefore for 50m wall we need
40*5/2=100 because 20*5/2= 50m wall
so,
100men - 50m wall - 15 days
now for 3 days we need to multiply no. of men by 5
hence 500 men will be needed for 50m wall in 3 days

Like? Yes | No   

naveenkumarmattigunta   8 years AGO

40 men can do 20 m in 15 days each can do 0.1m work in 3days so 500 need for 50 m

Like? Yes | No   

DIPTAKSHI MONDAL   8 years AGO

if in 15 days 40 men can build 20m then in 3 days men needed to complete 50m is=40*15/3*50/20=500

Like? Yes | No   

sanjeet kumar   8 years AGO

20m wall is build by 40 men in 15 days.
20m wall takes 40*20 = 600 men days.
So 50 m wall takes 1500 men days.
But to complete it in 3 days it took only 500 men.

Like? Yes | No   

kondareddy   8 years AGO

40 men build 20 m wall in 15 days.
so 40 men build 4m wall in 3 days.
in 3 days we want 50m wall.
In 3 days 4m wall build by 40 men.
so In 3 days 50 m wall build by (50/4)*40 men.
so answer is 500

Like? Yes | No   

Nisarg   8 years AGO

w1/m1*d1 = w2/m2*d2
20/40*15 = 50/x*3
60x = 3000
x=500

Like? Yes | No   

Veerabhadra rao   8 years AGO

(men1×days1)÷work1=(men2×days2)÷work2
(40×15)÷20=(3×x)÷50
x=500

Like? Yes | No (1)  

shaik jasmin   8 years AGO

according to question:
MEN M1=40 WORK W1=20 DAYS D1=15
LET MEN M2=X
WORK W2=50
DAYS D2=3
WE KNOW THAT
(M1*D1)/W1=(M2*D2)/W2
(40*15)/20=(X*3)/50
X=500
The no of men required is 500

Like? Yes | No (1)  

Sudhakarshona   8 years AGO

From the given data,
40 men can build a wall 20m high in 15 days..
Likewise, 100 men can build a wall 50m high in 15 days..
Hence, it will take 500 days to build a wall 50m high in 3 days..

Like? Yes | No (1)  

Dinesh Sharma   8 years AGO

apply formula: (M1*R1*T1 )W1= (M2*R2*T2)/W2

M = Men/Female; R = Rate; T = Time; W = Work

Therefore,
(40*1*15)/20 = (x*1*3)/50
i.e. x = 500 men required

Like? Yes (1) | No (2)  

MD TAHERA sultana   8 years AGO

m1d1/w1=m2d2/w2
40 men -m1,w1-20,d1-15days
xmen-m2, w2- 50mts height wall
d2 -3 days
(40* 15)/20=(x*3)/50
500

Like? Yes | No (1)  

Rahul Ranjan   8 years AGO

Man1*day1/work1=man2*day2/work2

40*15/20= x*3/50
On solving
X=500 Man's

Like? Yes | No (1)  

Narasimha   8 years AGO

40 Men 20m
then 50m is 100..
after men and days are inversely proportional so 15/3=5..
500

Like? Yes | No (1)  

JAYA TRIPATHI   8 years AGO

m1d1/w1 = m2d2/w2
40*15/20 = m2*3/50
15*2 *50/3 = m2
=> m2 =10*50 = 500
hence the no of men required is 500. option (8).

Like? Yes (1) | No (2)  

Nwobi Chibuzor    8 years AGO

The man power of 40men in 15days to build a wall is 40*60= 600. While the man power of Xmen in 3days to build a similar wall is 3*X= 3X.
So 600 = 20m high wall and 3X = 50m high wall. 600*50 / 20*3X ; 30000/60X gives X=500.
Therefore 500men can build 50m high wall in 3days

Like? Yes (1) | No (2)  

KK   8 years AGO

20m in 15 days by 40 men
so,20m in 3 days will require 200 men
and hence 50m in 3 days will require 500 men.

Like? Yes | No (1)  

Anamika singh   8 years AGO

for the extra height=40×50/20=100
for shorter time=100×15/3=500. option no.(8)👍

Like? Yes | No (1)  

MANZAR ALI   8 years AGO

500

Like? Yes | No (1)  

Nishu Kumari   8 years AGO

40 men 20m high 15 days
? Men 500 high 3 days

So m1d1/w1=m2d2/w2
Where m1= men for first work, d1 = days for first work nd w1= first work. And m2= men for second work, d2= days for 2nd work w2= second work

40*15/20=?*3/50

=500.

Like? Yes | No (1)  

Sabarisan   8 years AGO

40 men finished the construction of 20 m in 15 days... 200 men can finish the wall of 20 m in 3 days... simmilarly, 500 men can finish the 50 m wall in 3 days

Like? Yes | No (1)  

Yahia boudah   8 years AGO

(40x5x2.5)= 500
200 men can build 20m in 3 days.
(15/5=3).... (40x5=200)
500 men can build 50m in three days.
(200x2)+(200/2)=500
similarly (20x2)+(20/2)= 50.

Like? Yes | No (1)  

sai krishna   8 years AGO

work done in 20 days is 600units=20m
1m=30units
n*3=30*50=1500
n=500

Like? Yes | No (1)  

Nidhi Mishra   8 years AGO

m1*d1/w1=m2*d2/w2
here m1=40,m2=?,d1=15,d2=3,w1=20,w2=50.
substituting these values in the above equation,we get m2=500

Like? Yes | No (1)  

Nandini JB   8 years AGO

m1*d1/w1 = m2*d2/w2
m1 =m2*d2*w1/d1*w2
m1 = 40*15*50/20*3
m1 =600*50/60
m1=10*50=500

Like? Yes | No (1)  

Satishbabu   8 years AGO

Number of men M1=40,
Number of Days D1=15,
Work done W1=20 m
=> W2=50, D2=3, M2=?
=> M1D1/W1=M2D2/W2
=>M2=500

Like? Yes (1) | No (2)  

kanhaiya jee   8 years AGO

40 men 20m high 15 days
1 man 20 m high 600 days
now 600 days 1 man
so 3 days 200 men
and again 20 m high 200 men
so 50 m high 500 men

Like? Yes | No (1)  

Veera Vignesh   8 years AGO

40*15/20.......(1)
x*3/50.............(2)
1=2
x=500

Like? Yes | No (1)  

prakash sharma   8 years AGO

(M1*D1)/W1=(M2*D2)/W2
where M1=40
D1=15
W1=20
M2=?
D2=3
W2=50
so 40*15/20=M2*3/50
=>M2=500(Option 8)

Like? Yes | No (2)  

Lakshmipathi   8 years AGO

m1d1/w1=m2d2/w2

40*15/20=m2*3/50
500

Like? Yes | No (2)  

Priyanka devi E   8 years AGO

15*40*50=3*20*x;
x=30000/60;
x=500;

option 8 is correct

Like? Yes | No (2)