Word Problem Of Linear Equations -
Many word problems can be solved easily with the help of linear equations. To solve a word problem this way, take the following steps.
Step.1) Read the problem carefully to analyze the fact given
Step.2) Denote the unknown quantity by ‘x’ or by any other variable.
Step.3) Express all other quantities mentioned in the problem in terms of the variable.
Step.4) Frame an equation by using the conditions of the problems.
Step.5) Solve the equation to find the value of the variable or the unknown, If the conditions of the given problem are satisfied by the value of the unknown, the solution is correct.
Example.1) Fours less than five times a number is 10 more than thrice the number, find the number.
Ans.) Let the number is = ‘x’
Then four less than five times a number = 5x - 4
And 10 more than thrice the number is = 3x + 10
From the given condition, the required equation would be –
5x - 4 = 3x + 10
Or, 5x – 3x = 10 + 4
Or, 2x = 14
Or, x = 7
The desired number is 7 (Ans.)
Example.2) The difference between the squares of two consecutive numbers is 121, find out the numbers.
Ans.) As per the given condition, let the number is ‘x’
So, the two consecutive number is x and x + 1
As per the given condition, we can find the desired equation is –
(x + 1)² - x² = 121
Or, x² + 2x + 1 - x² = 121
Or, 2x = 121 – 1 = 120
Or, x = 60
So, the required numbers are 60, 61 (Ans.)
Example.3) Two numbers add up to 50. One-fourth of the larger number is 20 more than one-sixth of the smaller number, find the desired numbers.
Ans.) Let the larger number be x, then the smaller number be = 50 – x
x
One-fourth of the larger number is ---------
4
50 - x
20 more than one-sixth of the smaller number = ------------ + 20 6
From the given condition, the required equation would be –
x 50 - x
--------- = ----------- + 20
4 6
x 50 – x + 120
Or, --------- = ------------------
4 6
By, cross multiplication, we find –
Or, 6x = 4 (170 – x)
Or, 6x = 680 – 4x
Or, 10x = 680
Or, x = 68
So, the required number is 68 (Ans.)
Example.4) If the same number be added to the numbers 5, 15, 20, and 25 then the resultant numbers are in proportion. Find the numbers
Ans.) Let the number be ‘x’
Then, as per the given condition proportion number would be 5 + x, 15 + x, 20 + x, and 25 + x
5 + x 20 + x
The required equation would be ------------ = -------------
15 + x 25 + x
By, cross multiplication we find –
(5 + x) (25 + x) = (20 + x) (15 + x)
Or, 125 + 25x + 5x + x² = 300 + 15x + 20x + x²
Or, 30x + 125 = 300 + 35x
Or, - 300 + 125 = 35x – 30x
Or, 5x = - 175
Or, x = - 35 (Ans.)
Example.5) If a number is subtracted from the numerator of the fraction 5/6 and thrice that number is added to the denominator, the fraction becomes 1/5. Find the number.
Ans.) Let the number be ‘x’
5 – x 1
So, as per the given condition ------------ = ---------
6 + 3x 5
Via cross multiplication -
Or, 5 (5 – x) = 6 + 3x
Or, 25 – 5x = 6 + 3x
Or, - 5x – 3x = - 25 + 6
Or, - 8x = - 19
Or, x = 19 / 8 (Ans.)
Example.6) The sum of the digits of a two-digit numbers is 10. If 5 is subtracted from the number formed interchanging the digits, the result is triple the original number. Find the original number.
Ans.) Let the digit in the units place be = ‘x’
As per the given condition, the sum of two-digit is = 10
Then the digit in the tens place = 10 – x
So, the number is = 10 (10 – x) + x
Now, as per the given condition if the number formed by the interchanging the digits = 10x + (10 - x)
From the question, as per the given condition 10x + (10 - x) – 5 = 3 {10 (10 – x) + x}
So, 10x + (10 - x) – 5 = 3 {10 (10 – x) + x}
Or, 10x + 10 – x – 5 = 30 (10 – x) + 3x
Or, 9x + 5 = 300 – 30x + 3x
Or, 9x + 27x = 300 – 5
Or, 36x = 295
Or, x = 295 / 36 = 8.19 = 8 (Ans.)
So, the digit in the units place = 295 / 36 = 8.19 = 8 (Approximately in round figure)
Digit in the tens place = 10 – 8 = 2
Hence, the original number = 10 (10 – x) + x = 10 (10 – 8) + 8 = 10 X 2 + 8 = 28 (Ans.)
Example.7) In a two-digit number, the digit in the tens place exceeds the digit in the units place by 5. If 3 more than six times the sum of the digits is subtracted from the number, the digits are reserved. Find the original number.
Ans.) let the digit in the unit place be = ‘x’
Then the digit in the tens place = x + 5
So, the number = 10(x + 5) + x
And, the sum of digits = x + (x + 5) = 2x + 5
After reversing the digit, the new number = 10x + (x + 5)
As per the given condition, we can get the required equation is –
{10(x + 5) + x} – { 3 + 6(2x + 5)} = 10x + (x + 5)
Or, 10x + 50 + x - 3 - 12x - 30 = 10x + x + 5
Or, 17 - x = 11x + 5
Or, 11x + x = 17 - 5
Or, 12x = 12
Or, x = 1
So, the original number is 2x + 5 = 2.1 + 5 = 7 (Ans.)
Example.8) The present age of a man is twice that of his son. 5 years hence their ages will be in the ratio 3 : 2. Find the son’s present age
Ans.) Let the age of the son is ‘x’, so as per the given condition the age of the person would be 2x
After 5 years the age of the son and his father’s age will be (x + 5) and (2x + 5) respectively
So, as per the given condition –
2x + 5 3
--------------- = ----------
x + 5 2
by, cross multiplication –
2 (2x + 5) = 3 (x + 5)
Or, 4x + 10 = 3x + 15
Or, 4x – 3x = 15 – 10
Or, x = 5
The present age of the son is 5 years (Ans.)
Example.9) A car covers the distance between two cities in 6 hours. A van covers the same distance in 5.5 hours by traveling 4 km/h faster than the car. What is the distance between the two cities? find the speeds of the car and the van
Ans.) Let the distance between two cities is ‘x’ km
As we know, the formula for the speed of any moving object is (here the speed of car)
Distance x
Speed = -------------- = ----------- km/h
Time 6
x
Speed of car = ---------- km/h
6
x 10x 2x
Speed of van is = -------- km/h = --------- km/h = --------- km/h
5.5 55 11
Now, as per the given condition, the desired equation would be –
2x x
----------- = ---------- + 4
11 6
2x (x + 24)
Or, ----------- = ---------------
11 6
Or, 12x = 11 (x + 24)
Or, 12x – 11x = 24 X 11
Or, x = 264
x 264
Now, the speed of the car is = --------- = ---------- = 44 km/h
6 6
2x 2 x 264
And, the speed of Van is = -------- = ----------
11 11
= 2 X 24 = 48 km/h (Ans.)
Example.10) The length of a rectangular field is 15 m more than its width. If the length is decreased by 10 meters and the width is decreased by 20 m, the area decreased by 250 m². Find the length and the width of the field.
Ans.) Let the width of the field is = ‘x’ m
Then the length as per given condition is = (x + 15) m
As per formula, the area of field is = x (x + 15) m²
If the length of the field is decreased by 10 m then the length would be = (x + 15) – 10 = (x + 5) m
The width of the field when it has been decreased by 20 m then the desired width would be = (x – 20) m
The new area of same field would be = (x + 5) (x – 20) m²
As per the given condition, the new area will be decreased by 250 m²
So, the equation would be –
x (x + 15) – {(x + 5) (x – 20)} = 250
or, x² + 15x – (x² + 5x – 20x – 100) = 250
or, x² + 15x – x² + 15x + 100 = 250
or, 30x = 250 – 100 = 150
or, x = 150 / 30 = 5
so, the width of the field is 5m
and the length of the field is = (x + 15) m = (5 + 15) m = 20 m (Ans.)