I am studying english and I'm not a mathematician so i need some help in solving my little brother's assignment since he is ill.
Please can anyone solve the follwing with the workings?

Indirect Variations.
The Volume of a bush to be cleared by some labourers Varies Inversly as the square of labourers to do the clearing.
When 16 hectors are to be cleared, 2 labourers are required.
Calculate:
(a) the number of labourers to clear 0.25 hectors
(b) The volume of work done by 4 men.

Please and thank you.

1. Let's assume that V=(volume of?!) area to be cleared, L=number of laborers required and k= a constant

2. Then V =k/L

3. When V= 16, L= 2 therefore VL= k that is 16 x 2=k or 32

4. When V= 0.32 then L = k/V or 32/0.25 that is L =(32 x 4 x100)/100 =128 laborers

5. When L = 4 then V =k/l or 32/4 that V =8 hectares

Hope this is helpful. Regards

1. Let's assume that :

V = volume of area to be cleared,

L = number of laborers required and

k = a constant

2. Then V =k/(L x L)

3. When V= 16, L= 2

therefore VLL= k
that is 16 x 4 = k = 64

4. When V= 0.32 then LL = k/V or 64/0.25

that is L =sqrt of (64 x 4 x100)/100 = sqrt (256) = 16 Labourers

5. When L = 4 then V =k/LL or 64/16, that is  V =4 hectares

Nice try btw.

1st post, huh?

Kudos!

4. When V = 0.25 (not 0.32)

@lakari007

The first respondent did a good job but unfortunately there was an error so I am sending u this answer

From ur question.the volume is inversley proportional to the square of the no of labourer


Let v~1\l^2

If k is a constant, then V=k/L^2

Now calculate for k, we have L=16,L=2

So cross multiplying K=V*L^2

So K=16*4=64

Question 1. For 0.25 hectares,the number of labourer will be as follows
0.25=64/L^2

So L^2=64/0.25

L^2=256
So L=16
So for 0.25 hectare,the no of labourer will be 16


Question 2. Volume of work by 4 men
V=?,L=4,K=64

So V=64/4^2
=64/16


=4 hectares

Hope this helps

Error noted. Thanks

Please am try do help my younger sister with her home work
Question: The electrical resistance of a wire varies inversely as the square of the radius if the resistance is 0.24πwhen the radius is 0.5mm and the radius when the resistance is 0.32π

Ojams1:
Please am try do help my younger sister with her home work
Question: The electrical resistance of a wire varies inversely as the square of the radius if the resistance is 0.24πwhen the radius is 0.5mm and the radius when the resistance is 0.32π

Solution

Let R=electrical resistance

Let r=radius

R~1/r²

R=K/r²

K=Rr²

When R=0.24 Π, r=0.5mm, K=?

K=(0.24Π)(0.5mm)²

K=0.06Πmm²

Here, we are required to find the radius when R=0.32Π

From R=Kr²

r=√(R/K)

r=√(0.32/0.06)

r=√(5.33333)

r=2.31mm ///