What quantity of Urea is required for top dressing of two acre of wheat field at a dose of 60 kg N//ha ?
Dose of nitrogen: 60 kg N/ha Field area: 2 acres 2 acres = 2 * 0.4047 hectares = 0.8094 hectares Total nitrogen requirement = Dose of nitrogen * Field area = 60 kg N/ha * 0.8094 hectares = 48.564 kg N Quantity of urea required = Total nitrogen requirement / Nitrogen content in urea = 48.564 kg N / 0.46 ≈ 105.558 kg
Quantity I: The price of rice is decreased by 30%, by how much % the consumption is increase so that the expenditure will decreased by 10%?
Quant...
I. 4x2+ 25x + 36 =0
II. 2y2+ 5y + 3 = 0
I. 2(x+2)+ 2(-x)=5
II. (1/(y+1)+ 1/(y+5))=(1/(y+2)+ 1/(y+4))
I. 56x² - 99x + 40 = 0
II. 8y² - 30y + 25 = 0
I. 8x² - 74x + 165 = 0
II. 15y² - 38y + 24 = 0
I. 2 x ² + x – 1 = 0
II. 2 y ² - 3 y + 1 = 0
...I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0
I. 144x² - 163x - 65 = 0
II. 91y² - 128y -48 = 0
I. 5x + 2y = 31
II. 3x + 7y = 36
I. 5x + y = 37
II. 4y+ x = 15
