Question

a part of monthly expenses of a family on a milk is fixed for Rs 500 and the remaining varies with the ​

Answer 1

Answer:

Extra milk required = x litre

total expenditure = Rs y

cost of extra milk = Rs. 25/litre

extra expense =25×x= Rs. 25x

fixed expense = Rs. 700

Total expense = fixed expense + extra expense

⇒y=700+25x

Step-by-step explanation:

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500When x = 2, then y = 500 + 20 × 2

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500When x = 2, then y = 500 + 20 × 2= 500 + 40 = 540

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500When x = 2, then y = 500 + 20 × 2= 500 + 40 = 540When x = 3, then y = 500 + 20 × 3

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500When x = 2, then y = 500 + 20 × 2= 500 + 40 = 540When x = 3, then y = 500 + 20 × 3= 500 + 60 = 560

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500When x = 2, then y = 500 + 20 × 2= 500 + 40 = 540When x = 3, then y = 500 + 20 × 3= 500 + 60 = 560Thus, we have the following table

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500When x = 2, then y = 500 + 20 × 2= 500 + 40 = 540When x = 3, then y = 500 + 20 × 3= 500 + 60 = 560Thus, we have the following tableimage

Let the quantity of milk required extra be x L and total expenditure on milk be ₹ y.∴Required equation is y = 500 + 20x…(i)When x = 0, theny = 500 + 20 × 0 = 500When x = 2, then y = 500 + 20 × 2= 500 + 40 = 540When x = 3, then y = 500 + 20 × 3= 500 + 60 = 560Thus, we have the following tableimageNow, plot the points 4(0, 500), 6(2, 540) and C(3, 560) on a graph and join them by a line to get required garph.