WORD PROBLEMS ON LINEAR EQUATIONS
About "Word Problems on Linear Equations"
Word Problems on Linear Equations :
Solving word problems using linear equations is sometimes a difficult job for some students.
Actually it is not. There is a simple trick behind it.
The picture given below tells us the trick.
Word Problems on Linear Equations - Examples
Example 1 :
If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. In case, the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.
Solution :
Let "x/y" be the fraction.
Given : If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1.
So, we have
(x + 2) / (y + 1) = 1
Simplify.
(x + 2) / (y + 1) = 1
x + 2 = y + 1
x - y = - 1 ------(1)
Given : In case the numerator is decreased by 4 and the denominator by 2, the fraction becomes 1/2.
So, we have
(x - 4) / (y - 2) = 1 / 2
Simplify.
2(x - 4) = 1(y - 2)
2x - 8 = y - 2
2x - y = 6 ------(2)
Solving (1) and (2), we get
x = 7 and y = 8
Hence, the fraction is 7/8.
Example 2 :
A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750 ?
Solution :
Let "x" be the no. of adult tickets and "y' be the no. of kids tickets sold.
Given :A total of 548 tickets were sold
So, we have
x + y = 548 ------(1)
Given : Cost of each adult ticket is $10 and kid ticket is $5 and tickets were sold for a total of $3750.
So, we have
10x + 5y = 3750
2x + y = 750 ------(2)
Solving (1) & (2), we get
x = 202 and y = 346
Hence, the number of adults tickets sold is 202 and kids tickets is 346.
Example 3 :
A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297.Find the number.
Solution :
Let "x0y" be the three digit number. (As per the given information, middle digit is zero)
Given : The sum of the other digits is 9
x + y = 9 ------(1)
By interchanging the first and third digits, the number we get is
y0x
Given : The number formed by interchanging the first and third digits is more than the original number by 297
y0x - x0y = 297
(100y + x) - (100x + y) = 297
100y + x - 100x - y = 297
-99x + 99y = 297
- x + y = 3 ------(2)
Solving (1) & (2), we get
x = 3 and y = 6
So, we have
x0y = 306
Hence the three digit number is 306.
Example 4 :
A manufacturer produces 80 units of a product at a cost of $22000 and 125 units at a cost of $28750. Assuming the cost curve to be linear, find the equation of the cost curve and then use it to estimate the cost of 95 units.
Solution :
Since the cost curve is linear, its equation will be
y = Ax + B.
(Here y = Total cost, x = no. of units)
Given : The total cost of 80 units of the product is $22000.
So, we have
22000 = 80A + B
80A + B = 22000 ------(1)
Given : The total cost of 125 units of the product is $28750.
So, we have
28750 = 125A + B
125A + B = 28750 ------(2)
Solving (1) and (2), we get
A = 150 and B = 10000
Then, the equation of the cost curve is
y = 150x + 10000 ------(3)
Estimate the cost of 95 units :
Plug x = 95 in (3).
(3)------> y = 150 ⋅ 95 + 10000
y = 14250 + 10000
y = 24250
Hence, the cost of 95 units is $24250.
Example 5 :
Y is older than X by 7 years. 15 years back X's age was 3/4 of Y's age. Find the present their present ages.
Solution :
Let "x" be the present age of X and "y" be the present age of Y.
Given : Y is older than X by 7 years.
So, we have
y = x + 7 --------(1)
15 years back :
X's age = x - 15
Y's age = y - 15
Given : 15 years back X's age was 3/4 of Y's age.
So, we have
(x - 15) = 3/4 ⋅ (y - 15)
Simplify.
4(x - 15) = 3(y - 15)
4x - 60 = 3y - 45
4x = 3y + 15 ------(2)
Solving (1) and (2), we get
x = 36 and y = 43
Hence, the present ages of "x" and "y" are 36 years and 43 years.
After having gone through the stuff given above, we hope that the students would have understood "Word problems on linear equations"
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