Question

A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is ₹ 150. The cost of the two dosai, two idlies and four vadais is ₹ 200. The cost of five dosai, four idlies and two vadais is ₹ 250. The family has ₹ 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?

Answer 1

Let the Cost of dosai, Idlies and vadais be x, y, z

2x + 3y + 2z = 150

2x + 2y + 4z = 200

5x + 4y + 2z = 250

Δ = `|(2, 3, 2),(2, 2, 4),(5, 4, 2)|`

= 2(4 – 16) – 3(4 – 20) + 2(8 – 10)

= 2(– 12) – 3(– 16) + 2(– 2)

= – 24 + 48 – 4

= 20 ≠ 0

Δ_x = `|(150, 3, 2),(200, 2, 4),(250, 4, 2)|`

= 150(4 – 16) – 3(400 – 1000) + 2(800 – 500)

= 150(– 12) – 3(– 600) + 2(300)

= – 1800 + 1800 + 600

= 600

Δ_y = `|(2, 150, 2),(2, 200, 4),(5, 250, 2)|`

= 2(400 – 1000) – 150(4 – 20) + 2(500 – 1000)

= 2(– 600) – 150(– 16) + 2(– 500)

= – 1200 + 2400 – 1000

= 200

Δ_z = `|(2, 3, 150),(2, 2, 200),(5, 4, 250)|`

= 2(500 – 800) – 3(500 – 1000) + 150(8 – 10)

= 2(– 300) – 3(– 500) + 150(– 2)

= – 600 + 1500 – 300

= 600

x = `Delta_x/Delta = 600/20` = 30

y = `Delta_y/Delta = 200/20` = 10

z = `Delta_z/Delta = 600/20` = 30

x = Rs 30, y = Rs 10, z = Rs 30

There are 3 dosai, 6 idlies and 6 vadais

= 3x + 6y + 6z

= 3(30) + 6(10) + 6(30)

= 90 + 60 + 180

= ₹ 330

They can eat within the amount.