Question

20 men can dig 40 holes in 60 days. if so, 10 men can dig 20holes in how many days?

Answer 1

Given:

First class of Labor:

M₁=20

T₁=60 days

W₁=40 holes

Second Class of Labors:

M₂=10

W₂=20 holes

T₂=?

To Find:

Find the value of T₂

Solution:

As we know that,

M₁×T₁/W₁     =     M₂×T₂/W₂

20×60/40=  10×T₂/20

We get, T₂= 60

Hence 10 men can dig 20 holes in 60 days.

Answer 2

According to the question;

1st situation

Men (M₁) =20

Holes (H₁ ) =40

Days (D₁)=60

2nd situation

Men(M₂) =10

Holes (H₂)=20

Days (D₂)=?

now, as we know;

$\frac{M_{1} * D_{1} }{H_{1} } =\frac{M_{2} * D_{2} }{H_{2} }$

M₁ × D₁ × H₂ = M₂ × D₂ × H₁

20 × 60 × 20 = 10× D₂ × 40

$D_{2} = \frac{20 * 60 * 20}{10 * 40}$

D₂ = 60

If 20 men can dig 40 holes in 60 days. if so, 10 men can dig 20holes in 60 days.

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