Word problem exercise by cramers rule
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The document provides examples of word problems solved using Cramer's rule. The examples involve systems of linear equations with two unknowns where the goal is to determine the value of each unknown. For each problem, the document states the relevant information, sets up the corresponding system of linear equations with two unknowns, applies Cramer's rule to solve for the unknowns, and states the answer.
![WORD PROBLEM EXERCISE
BY CRAMERS RULE
Example-13:
The entry fee at Expo center is Rs.20 for children and Rs.50 for adults.
On a Sunday 2000 people attended the fair and Rs. 85000 is collected.
How many children and how many adults attended the fair at Expo
center.
Solution:
Let x be the number of children
and y be the number of adults
X + Y = 2000
20 X + 50 Y = 85000
By Cramer’s rule
D = [
1 1
20 50
] = (50)(1) – (20)(1) = 50 – 20 = 30
Dx = [
2000 1
85000 50
] = (2000)(50) – (85000)(1)
= 100000 – 85000 = 15000
Dy = [
1 2000
20 85000
] = (85000)(1) – (2000)(20)
= 85000 – 40000 = 45000
X =
Dx
𝐷
=
15000
30
= 500 (children)
Y =
Dy
𝐷
=
45000
30
= 1500 (Adults)](https://image.slidesharecdn.com/wordproblemexercisebycramersrule-200313100401/85/Word-problem-exercise-by-cramers-rule-2-320.jpg)
Word problem exercise by cramers rule
- 2. WORD PROBLEM EXERCISE BY CRAMERS RULE Example-13: The entry fee at Expo center is Rs.20 for children and Rs.50 for adults. On a Sunday 2000 people attended the fair and Rs. 85000 is collected. How many children and how many adults attended the fair at Expo center. Solution: Let x be the number of children and y be the number of adults X + Y = 2000 20 X + 50 Y = 85000 By Cramer’s rule D = [ 1 1 20 50 ] = (50)(1) – (20)(1) = 50 – 20 = 30 Dx = [ 2000 1 85000 50 ] = (2000)(50) – (85000)(1) = 100000 – 85000 = 15000 Dy = [ 1 2000 20 85000 ] = (85000)(1) – (2000)(20) = 85000 – 40000 = 45000 X = Dx 𝐷 = 15000 30 = 500 (children) Y = Dy 𝐷 = 45000 30 = 1500 (Adults)
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