Mr. atinga shared his cattle among his three children lordia, anita, and richmond. he gave 3/5 of cattles to anita and shared the remaining between lordia and richmond in the ratio 2: 3 respectively. in all 80 cattle were shared between lordia and richmond. how many cattle were shared among the three children how many cattle did anita receive than richmond

a). The total number of cattle shared among the three children=200

b). Anita's share=120 cattle

    Lordia's share=32 cattle

    Richmond's share=48 cattle

Step-by-step explanation:

a)

Step 1

Determine fraction shared between Lordia and Richmond as follows;

Lordia:Richmond=2:3

Let the total cattle be x

Lordia's fraction=2/(2+3)=2/5 of x=(2/5)x

Lordia's fraction=(2/5)x

Richmond's fraction=3/(2+3)=3/5 of x=(3/5) x

but;

x=80 cattle

replacing;

Total number of cattle Lordia got=(2/5)×80=32 cattle

Total number of cattle Richmond got=(3/5)×80=48 cattle

Step 2

Express the total cattle Mr. Atinga had initially as follows;

Let the initial number of cattle be y

The fraction remainder given to Lordia and Richmond=Initial number-fraction of the totat given to Anita

where;

Initial number=y

fraction of the total given to Anita=3/5 y

replacing;

The fraction remainder given to Lordia and Richmond=y-3/5 y=2/5 y

but;

2/5 y=80

y=80×5/2=

y=200

The total number of cattle shared among the three children=200

b).

Anita's share=(3/5)×200=120 cattle

Lordia's share=(2/5)×80=32 cattle

Richmond's share=(3/5)×80=48 cattle

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)