How did Shankar get a partial answer to his question about the barbed wire fence around the station? What was the next part of the answer? CH The Station
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Dimension of triangle field ⇒120m,80m,50m
⇒ perimeter = 120 + 80 + 50
=250m
Area = 125(125−120)(125−80)125−50
=125×5×45×75
=25×5×5×9×5×5×15
=5×5×3×515
=37515
=375(3.88)
=1455m2
cost of fencing = 250−3
[perimeter of Remaining fence] = 247
Total cost ⇒247m×Rs20/m
=Rs4940.
as the rate are reduced by 25 %
∴ new lets is [20−20×10025]⇒20−5=15%
Rs[4950−4950×10025]
⇒4950(1−41)
4950×43
Final cost ⇒Rs3712.5
Dimension of triangle field ⇒120m,80m,50m
⇒ perimeter = 120 + 80 + 50
=250m
Area = 125(125−120)(125−80)125−50
=125×5×45×75
=25×5×5×9×5×5×15
=5×5×3×515
=37515
=375(3.88)
=1455m2
cost of fencing = 250−3
[perimeter of Remaining fence] = 247
Total cost ⇒247m×Rs20/m
=Rs4940.
as the rate are reduced by 25 %
∴ new lets is [20−20×10025]⇒20−5=15%
Rs[4950−4950×10025]
⇒4950(1−41)
4950×43
Final cost ⇒Rs3712.5
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Answer:
The old station master tried to give a vangue answer to shankar question saying that it is the lonely place hence the need of the barbed wire fence but shankar got a complete answer that time
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