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.
a11x + a12y + a13z = b1
a21x + a22y + a23z = b2
a31x + a32y + a33z = b1
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 = Δ1/Δ
y = Δ2/Δ
z = Δ3/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,
x1 = Δ1/Δ
x = -105/(-5/4)
x = 105(4/5)
x = 84
y = Δ2/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,
x1 = Δ1/Δ
x = -1.5/(-0.25)
x = 6
y = x2 = Δ2/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 = Δ1/Δ
x = (-4/30)/(-2)
x = 1/15
y = Δ2/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
Δ1 = 150(4 - 16) - 3(400 - 1000) + 2(800 - 500)
= 150(-12) - 3(-600) + 2(300)
= -1800 + 1800 + 600
Δ1 = 600
Δ2 = 2(400 - 1000) - 150(4 - 20) + 2(500 - 1000)
= 2(-600) - 150(-16) + 2(-500)
= -1200 + 2400 - 1000
Δ2 = 200
Δ3 = 2(500 - 800) - 3(500 - 1000) + 150(8 - 10)
= 2(-300) - 3(-500) + 150(-2)
= -600 + 1500 - 300
Δ3 = 600
By Cramer's rule,
x = Δ1/Δ = 600/20 = 30
y = Δ2/Δ = 200/20 = 10
z = Δ3/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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