In this section, you will learn how to solve word problems using Cramer's rule.

Let us consider the following system of three equations with three unknowns x, y and z.

a_{11}x + a_{12}y + a_{13}z = b_{1}

a_{21}x + a_{22}y + a_{23}z = b_{2}

a_{31}x + a_{32}y + a_{33}z = b_{1}

Now, we can write the the following determinants using the above equations.

Then, Cramer’s rule to find the values of x, y and z :

x = Δ_{1}/Δ

y = Δ_{2}/Δ

z = Δ_{3}/Δ

If Δ = 0, the system is inconsistent and it has solution.

**Problem 1 :**

In a competitive examination, one mark is awarded for every correct answer while 1/4 mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly ? (Use Cramer’s rule to solve the problem).

**Solution :**

**Total numbers of questions = 100**

**Let "x" and "y" be the number of questions answered correctly and incorrectly respectively.**

**Then, **

**x + y = 100**

**1x - (1/4) y = 80**

By Cramer's rule,

x_{1} = Δ_{1}/Δ

x = -105/(-5/4)

x = 105(4/5)

x = 84

y = Δ_{2}/Δ

y = -20/(-5/4)

y = 20(4/5)

y = 16

So, the number of questions answered correctly is 84 and wrongly is 16.

**Problem 2 :**

A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution ? (Use Cramer’s rule to solve the problem).

**Solution :**

Let "x" and "y" be the quantity of 1st and second solution.

Then,

x + y = 10 -----(1)

50% of x + 25% of y = 40% of 10

0.5x + 0.25y = 4 -----(2)

By Cramer's rule,

x_{1} = Δ_{1}/Δ

x = -1.5/(-0.25)

x = 6

y = x_{2} = Δ_{2}/Δ = -1/(-0.25)

= 4

So, the quantity of 1st solution is 6 liters and second solution is 4 liters.

**Problem 3 :**

A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself ? (Use Cramer’s rule to solve the problem).

**Solution :**

Let "x" be the number of minutes taken by sump A to fill the tank.

Let "y" be the number of minutes taken by sump B to fill the tank.

A pump fill in 1 min = 1/x

B pump fill in 1 min = 1/y

A& B together in 1 min = 1/x + 1/y

1/x + 1/y = 1/10 -----(1)

1/x - 1/y = 1/30 -----(2)

1/x = a and 1/y = b

a + b = 1/10 -----(1)

a - b = 1/30 -----(2)

By Cramer's rule,

x = Δ_{1}/Δ

x = (-4/30)/(-2)

x = 1/15

y = Δ_{2}/Δ

y = (-2/30)/(-2)

y = 1/30

A alone can fill in 15 mins

B alone can fill in 30 mins

**Problem 4 :**

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 ?

**Solution :**

Let "x", "y" and "z" be the cost of 1 dosa, 1 idly and 1 vada respectively.

2x + 3y + 2z = 150

2x + 2y + 4z = 200

5x + 4y + 2z = 250

Δ = 2(4 - 16) - 3(4 - 20) + 2(8 - 10)

= 2(-12) - 3(-16) + 2(-2)

= -24 + 48 - 4

= -28 + 48

Δ = 20

Δ_{1} = 150(4 - 16) - 3(400 - 1000) + 2(800 - 500)

= 150(-12) - 3(-600) + 2(300)

= -1800 + 1800 + 600

Δ_{1 }= 600

Δ_{2} = 2(400 - 1000) - 150(4 - 20) + 2(500 - 1000)

= 2(-600) - 150(-16) + 2(-500)

= -1200 + 2400 - 1000

Δ_{2} = 200

Δ_{3} = 2(500 - 800) - 3(500 - 1000) + 150(8 - 10)

= 2(-300) - 3(-500) + 150(-2)

= -600 + 1500 - 300

Δ_{3} = 600

By Cramer's rule,

x = Δ_{1}/Δ = 600/20 = 30

y = Δ_{2}/Δ = 200/20 = 10

z = Δ_{3}/Δ = 600/20 = 30

So, cost of 1 ildy = ₹30, cost of 1 dosa = ₹10 and cost of 1 vada = ₹30.

3x + 6y + 6z :

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

= 90 + 60 + 180

= 330

So, the amount is enough to pay the bill.

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