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Are you still looking for a simple procedure on how to represent the solution set of an inequation on a graph? If yes, then stay on this page. Here we are giving a detailed step by step explanation on finding Solution Set for Linear Inequations with the solved examples. So, refer to the following sections and solve the questions easily.

Graphical Representation of the Solution Set of an Inequation

We generally use a number line to represent the solution set of an inequation on a graph. Following are the steps to represent the solution set of a linear inequation on a graph.

• At first, solve the given linear inequation and find the solution set for it.
• Mark the solution set on a number line by putting dot.
• If the solution set is infinite, then put three more dots to indicate infiniteness.

Questions on Finding the Solution Set of an Inequation and their Representation

Example 1.

Solve the inequation 4x – 6 < 10, x ∈ N and represent the solution set graphically?

Solution:

Given linear inequation is 4x – 6 < 10

Add 6 to the both sides of inequation

= 4x – 6 + 6 < 10 + 6

= 4x < 16

Divide both sides of the inequation by 4

= 4x/4 < 16/4

= x < 4

So, the replacement set = {1, 2, 3, 4, . . }

Therefore, the solution set S = {1, 2, 3} or S = {x : x ∈ N, x < 4}

Let us mark the solution set graphically.

The solution set is marked on the number line by dots.

Example 2.

Solve the inequation 8x + 4 > 20, x ∈ W and represent the solution set graphically?

Solution:

Given linear inequation is 8x + 4 > 20

Subtract 4 from both sides

= 8x + 4 – 4 > 20 – 4

= 8x > 16

Divide both sides by 8

= 8x/8 > 16/8

= x > 2

Replacement set = {0, 1, 2, 3, 4, . . . }

Therefore, solution set = {3, 4, 5, . . } or S = {x : x ∈ W, x > 2}

Let us mark the solution set graphically.

The solution set is marked on the number line by dots. We put three more dots indicate the infiniteness of the solution set.

Example 3.

Solve -2 ≥ x ≥ 5, x ∈ I, and represent the solution set graphically?

Solution:

Given linear inequation is -2 ≥ x ≥ 5

This has two inequations,

-2 ≥ x and x ≥ 5

Replacement set = {. . . -3, -2, -1, 0, 1, 2, 3 . . .}

Solution set for the inequation -2 ≥ x is -2, -1, 0, 1, 2, 3, . . i.e S = {-2, -1, 0, 1, 2, 3 . . } = P

And the solution set for the inequation x ≥ 5 is 5, 6, 7, 8 . . i.e Q = {5, 6, 7, 8 . . .}

Therefore, solution set for the given inequation = P ∩ Q

= {5, 6, 7, 8, 9 . . . }

or S = {x : x ∈ I, -2 ≥ x ≥ 5}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

Example 4.

Solve 0 < 3x – 10 ≤ 12, x ∈ R and represent the solution set graphically.

Solution:

Given linear inequation is 0 < 3x – 10 ≤ 12

It has two cases.

Case I:

0 < 3x – 10

Add 10 to both sides

0 + 10 < 3x – 10 + 10

10 < 3x

Divide both sides by 3

10/3 < 3x/3

10/3 < x

Case II:

3x – 10 ≤ 12

Add 10 to both sides

3x – 10 + 10 ≤ 12 + 10

3x ≤ 22

Divide both sides by 3

3x/x ≤ 22/3

x ≤ 22/3

S ∩ S’ = {3.33 < x ≤ 7.33} x ∈ R

= {x : x ∈ R 3.33 < x ≤ 7.33}

Properties of Inequation or Inequalities page gives detailed information about six properties. Each Property is explained step by step by considering a few examples. Learn and understand the properties easily with the help of solved examples provided below. You can seek Complete Information about Linear Inequations and their definitions all on our page.

Properties of Inequation or Inequalities

The six different properties of linear inequalities or linear inequation are mentioned here. These properties describe how arithmetic operations show the effect on the linear inequations.

Property I

The inequation remains unchanged if the same quantity is added to both sides of it.

Example:

x – 3 < 4

Add 3 to the both sides

= x – 3 + 3 < 4 + 3

= x < 7

Property II:

The inequation remains unchanged if the same quantity is subtracted from the both sides of the inequation.

Example:

x + 3 < 4

Subtract 3 from both sides

x + 3 – 3 < 4 – 3

x < 1

Property III:

The inequation remains unchanged if the same positive number is multiplied to both sides of it.

Example:

x/3 > 5

multiply 3 to both sides

= x/3 x 3 > 5 x 3

= x > 15

Property IV:

The inequation changes if the same negative number is multiplied to both sides of it. Actually, it reverses.

Example:

x/5 < 6

multiply -5 to the both sides

= x/5 x (-5) < 6 x (-5)

= -x < -30

= x > 30

Property V:

The inequation remains unchanged if the same positive number is divided by both sides of the inequation.

Example:

5x > 20

dividing both sides by 5

= 5x/5 > 20/5

= x > 4

Property VI:

The inequation changes when the same negative number is divided by both sides. It reverses.

Example:

-3x < 12

Dividing both sides by -3

= -3x/-3 > 12/-3

= x > -4

Solved Examples on Properties of Linear Inequalities

Example 1.

Write the inequality obtained for each of the following statements.

(i) On subtracting 9 from both sides of 21 > 10.

(ii) On multiplying each side of 8 < 12 by -2.

Solution:

(i) We know that subtracting the same number from both sides of inequality does not change the inequality.

21 – 9 > 10 – 9

= 12 > 1

(ii) We know that multiplying each side of an inequality by the same negative number reverses that inequality.

= 8 x -2 < 12 x -2

= -16 < -24

= -16 > -24

Example 2.

Find the inequality obtained for the following statements.

(i) On adding 2 to both sides of 8 < 56

(ii) On dividing each side of 8 < 56 by -16.

Solution:

(i) We know that adding the same number to both sides of inequality does not change the inequality.

8 + 2 < 56 + 2

= 10 < 58

(ii) We know that dividing each side of an inequality by the same negative number reverses that.

8/-16 < 56/-16

= -1/2 > -7/2

Example 3.

Write the inequality obtained for the following statement.

On multiplying each side of 18 > 8 by 5.

Solution:

We know that multiplying each side of an inequality by the same positive number does not change the inequality.

18 x 5 > 8 x 5

= 90 > 40

On this page, we will learn about what is linear inequality and linear inequations and the steps to solve the linear inequalities problems. You will also get the properties of inequation or inequalities. Check out the representation of the solution set on the real line in the following sections. We have provided Solved Problems along with a detailed explanation so that you can better understand the concept.

What are Linear Inequation and Linear Inequalities?

Linear Inequation is a statement indicating the value of one quantity or algebraic expressions that are not exactly equal to one another.

Inequalities are nothing but the symbols enclosed between two or more algebraic expressions or quantities. The open sentence which involves <, ≤, >, ≥, and ≠ symbols are called the inequalities.

Some of the examples of Linear Inequation are listed below.

• x < 6
• y ≥ 25
• x + 3 > 40
• p ≠ 10

Linear Inequation

An inequation that contains one variable and that variable highest power is one then is known as the linear inequation in that variable. To make a linear equation into inequation, you have to replace the equal to symbol with the inequality sign. The statements of any of the forms ax + b < 0, ax + b > 0, ax + b≥ 0, ax + b ≤ 0 are the linear inequations invariable x, where a, b are real numbers and a is not equal to zero.

some of the examples of the linear inequation with variable y are included below:

• 3y + 6 ≥ 0
• 9 – y < 0
• 2y > 0
• 25 + 5y ≤ 0

Questions on Replacement Set or Domain of a Variable

Example 1.

Find the replacement set for the inequation x ≤ 9. The replacement set is a set of whole numbers?

Solution:

We know that whole numbers W = {0, 1, 2, 3 . . . }.

Replace x with some values of W. Some values of x from W satisfy the inequation and some don’t. Here, the values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 satisfy the given inequation x ≤ 9 while the other values don’t.

Thus, the set of all those values of variables that satisfy the given inequation is called the solution set of the given inequation.

Therefore, the solution set for the inequation x ≤ 9 is S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} or S = {x : x ∈ W, x ≤ 9}

Example 2.

Find the replacement set for the inequation x > 2. Let the replacement set be the set of natural numbers?

Solution:

We know that natural numbers N = {1, 2, 3, 4, 5, 6}

Replace x with some values of N. Some values of x from N satisfy the inequation and some don’t. Here, the values 3, 4, 5, . . . satisfy the given inequation x > 2 while the other values don’t.

Thus, the set of all those values of variables that satisfy the given inequation is called the solution set of the given inequation.

Therefore, the solution set for the inequation x > 2 is S = {3 4, 5, 6, . . . }

Example 3.

Find the replacement set and the solution set for the inequation x ≥ -2 when the replacement set is an integer?

Solution:

Replacement set I = {. . . -3, -2, -1, 0, 1, 2, 3, . . . }

Solution set S = {-2, -1, 0, 1, 2, . . . } or S = { x : x ∈ I, x ≥ – 2}

Example 4.

Find the solution set for the following linear inequations.

(i) x < 5 where replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}

(ii) x ≥ 7 where replacement set is { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iii) x ≠ 3 where replacement set is { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Solution:

(i) Solution Set S = {-4, -3, -2, -1, 0, 1, 2, 3, 4} S = {x : x ∈ I, -4 < x ≤ 4}

(ii) Solution set S = {7, 8, 9, 10} or S = {x : x ∈ N, 7 < x < 10}

(iii) Solution set S = {0, 1, 2, 4, 5, 6, 7, 8, 9, 10} or S = {x : x ∈ N, x ≠ 3}

Frequently Asked Questions on Linear Inequality

1. What is the difference between a linear equation and a linear inequality?

The linear equation is an equation that has one or two variables and those exponents are one. Linear inequation also has one variable whose exponent is one. Between two algebraic expressions, the = symbol is enclosed in a linear equation, linear inequality signs are enclosed in a linear inequation. The graph of inequalities is a dashed line but the equation is a solid line in any situation.

2. What is linear inequality?

Linear inequality contains any symbols of inequality. It represents the data that is not equal in graph form. It involves a linear function.

Linear inequations demonstrate the value of one quantity or algebraic expression which is not equal to another. These inequations are used to compare any two quantities. Get some solved examples on linear inequations and steps to draw graphs, the system of linear inequations in the following sections of this page.

• Linear Inequality and Linear Inequations
• Properties of Inequation or Inequalities
• Representation of the Solution Set of an Inequation
• Practice Test on Linear Inequations

What are Linear Inequations?

Linear inequalities are the expressions where any two values are compared by the inequality symbols. These are the equations that contain all mathematical symbols except equal to (=). The values can be numerical or algebraic or a combination of both.

The inequality symbols are < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to), ≠ (not equal to). The symbols < and > shows the strict inequalities and the symbols ≤ and ≥ represents the slack inequalities.

Example:

x < 3, x ≥ 5, y ≤ 8, p > 10, m ≠ 1.

Linear Inequalities Graphing

We can plot the graph for linear inequalities like an ordinary linear function. But, for a linear function, the graph represents a line and for inequalities, the graph shows the area of the coordinate plane that satisfies the inequality condition. The linear inequalities graph divides a coordinate plane into two parts. One part of the coordinate plane is called the borderline where it represents the solutions for inequality. The borderline represents the conditions <, >, ≤ and ≥.

Students who are willing to plot a graph for the linear inequations can refer to the following steps.

• Arrange the given linear inequation in such a way that, it should have one variable ‘y’ on the left-hand side of the symbol and the remaining equation on the right-hand side.
• Plot the graph for linear inequation by putting the random values of x.
• Draw a thicker and solid line for y≤ or y≥ and a dashed liner for y< or y> conditions.
• Now, draw shades as per the linear inequalities conditions.

System of Linear Inequalities

A system of linear inequalities in two variables includes at least two inequalities in the variables. By solving the linear inequality you will get an ordered pair. So basically, the solution to all linear inequalities and the graph of the linear inequality is the graph displaying all solutions of the system.

Questions on Linear Inequations

Example 1.

Solve the inequality x + 5 < 10?

Solution:

Given that,

x + 5 < 10

Move variable x to the one side of inequation.

= x < 10 – 5

= x < 5

Replacement set = {0, 1, 2, 3, 4, 5, 6, 7 . . . }

Solution set for the inequation x + 5 < 10 is 1, 2, 3, 4

Therefore, solution set s = {. . . 1, 2, 3, 4}

Let us mark the solution set graphically.

The solution set is marked on the number line by dots. We put three more dots indicate the infiniteness of the solution set.

Example 2.

Solve the inequation 3 < y ≤ 10?

Solution:

Given that,

3 < y ≤ 10

This has two inequations,

3 < y and y ≤ 10

Replacement set = {. . . -3, -2, -1, 0, 1, 2, 3, 4, , 6, 7, . . . }

Solution set for the inequation 3 < y is 4, 5, 6, 7, . . .  i.e Q = {4, 5, 6, . . . }

Solution set for the inequation y ≤ 10 is . . . . 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e P = {. . . . 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Therefore, solution set of the given inequation = P ∩ Q = {4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

Example 3.

Solve the inequality 4 ( x + 2 ) − 1 > 5 − 7 ( 4 − x )?

Solution:

Given that,

4 ( x + 2 ) − 1 > 5 − 7 ( 4 − x )

4 x + 8 − 1 > 5 − 28 + 7 x

4 x + 7 > − 23 + 7 x

− 3 x > − 30

x < 10

Replacement Set = {1, 2, 3, 4, 5 . . . }

Solution set is 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e S = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

Frequently Asked Questions on Linear Inequations

1. What is meant by linear inequations?

Linear inequations involve linear expressions in two variables by using the relational symbols like <, >, ≤, ≥, and ≠.

2. How do you solve linear inequation?

Solving linear inequations having a single variable is very easy. All you have to do is simplify both the sides of the condition and get the variable term on one side and all other terms on the other side. And them either multiply/ divide the coefficient of the variable to obtain the solution.

3. What are the examples of linear inequations?

Some of the examples of linear inequality are x < 8, y > 15, z ≠ 7 + 9, p + 1 ≤ 2, 3 ≥ (z + 15).

Practice Test on Framing the Formula let you know about the different problems available on the Framing the Formula. Most of the below questions appear on the exam which helps you to get good marks in the exam. Practice all the questions available below and have a perfect grip on Framing the Formula Problems.

All the concepts such as formulas, framing a formula, find the value of the variable using a change of subject of formula or an equation, change the subject of the formula, and method of substitution are included in the below article. Learn Tricks to solve Formula and Framing the Formula problems easily.

1. Write the formulas for the following statements.

(a) Area of the square is equal to the square of its side (b)
(b) Area B of the rhombus is equal to half the product of its diagonals (s₁, s₂).
(c) Perimeter (p) of a parallelogram is thrice the sum of its adjacent sides.
(d) Area of four walls (D) of a room is the product of two times the sum of length (l), breadth (b) and height (h).
(e) Profit (P) is calculated by taking the difference of cost price (C) and selling price (S).

Solution:

(a) Given that Area of the square is equal to the square of its side (b)
Area of the square = A
A = b²
(b) Area B of the rhombus is equal to half the product of its diagonals (s₁, s₂).
B = s₁ × s₂
(c) Perimeter (p) of a parallelogram is thrice the sum of its adjacent sides.
sum of its adjacent sides = x
p = 3(x)
(d) Area of four walls (D) of a room is the product of two times the sum of length (l), breadth (b) and height (h).
D = 2 (l + b + h)
(e) Profit (P) is calculated by taking the difference of cost price (C) and selling price (S).
P = S – C

2. Mention the formulas for the following statements.

(a) A side of a chessboard (D) is 3.14 times the side of a board (e).
(b) The difference between the two different variables is 36.
(c) The sum (X) of all the interior angles of a regular polygon of m sides is 2 less than m times 180°.
(d) Subtracting 2/5 from a number and multiplying this difference by 4 gives two times the same number.
(e) Fourteen years from now Sam’s age will be three times her present age.

Solution:

(a) Given that A side of a chessboard (D) is 3.14 times the side of a board (e).
D = 3.14e
(b) The difference between the two different variables is 36.
Two different variables are a and b
a – b = 36
(c) The sum (X) of all the interior angles of a regular polygon of m sides is 2 less than m times 180°.
X = (m – 2) × 180°
(d) Subtracting 2/5 from a number and multiplying this difference by 4 gives two times the same number.
Let the number is X.
X = 4(X – 2/5)
(e) Fourteen years from now Sam’s age will be three times her present age.
Sam’s age is C
C + 14 = 3C

3. Change the subject of the formula and find the value of the new subject.

(a) X = 2YZ make the subject Z. X = 20, Y = 5
(b) l² = r² + h², l is the height of the cone, h is the height and r is the radius. Make the subject h when l = 12 and r = 6.
(c) M = 1/2 × b × h where M is the area of a triangle with base b and height h. Make h the subject, find h when M = 50, b = 24.
(d) D = d × q + r where D is the dividend, q is the quotient, d the divisor, and r is the remainder. Make the subject r, when D = x² – x, d = x – 2, q = x + 3.
(e) M × N = c × d, Make N as the subject where M = 2, c = 4, d = 10.

Solution:
(a) X = 2YZ make the subject Z. X = 20, Y = 5
Given that X = 2YZ
Divide both sides with 2Y
X/2Y = 2YZ/2Y
X/2Y = Z
Therefore, Z = X/2Y
Substitute the given values X = 20, Y = 5
Z = 20/2(5) = 20/10 = 2
Z = 2.

The final answer is Z = 2.

(b) l² = r² + h², l is the height of the cone, h is the height and r is the radius. Make the subject h when l = 12 and r = 6.
Given that l² = r² + h², l is the height of the cone, h is the height and r is the radius.
l² = r² + h²
Subtract r² on both sides
l² – r² = r² – r² + h²
l² – r² = h²
h = √(l² – r²)
Substitute the given values l = 12 and r = 6
h = √((12)² – (6)²) = √144 – 36 = √108

The final answer is h = √108

(c) M = 1/2 × b × h where M is the area of a triangle with base b and height h. Make h the subject, find h when M = 50, b = 24.
Given that M = 1/2 × b × h where M is the area of a triangle with base b and height h. Make h the subject, find h when M = 50, b = 24.
M = 1/2 × b × h
Divide 1/2 × b on both sides.
M/(1/2 × b) = (1/2 × b × h)/(1/2 × b)
M/(1/2 × b) = h
Therefore, h = M/(1/2 × b)
Substitute the given values M = 50, b = 24.
h = 50/(1/2 × 24)
h = 50/12

The final answer is h = 50/12

(d) D = (d × q) + r where D is the dividend, q is the quotient, d the divisor, and r is the remainder. Make the subject r, when D = x² – x, d = x – 2, q = x + 3.
Given that D = d × q + r where D is the dividend, q is the quotient, d the divisor, and r is the remainder. Make the subject r, when D = x² – x, d = x – 2, q = x + 3.
D = (d × q) + r
Subtract (d × q) on both sides
D – (d × q) = (d × q) – (d × q) + r
D – (d × q) = r
Therefore, r = D – (d × q)
Substitute the given values D = x² – x, d = x – 2, q = x + 3.
r = (x² – x) – ((x – 2)(x + 3))
r = (x² – x) – (x² + 3x – 2x – 6)
r = x² – x – x² – x + 6
r = -2x + 6

The final answer is r = -2x + 6

(e) M × N = c × d, Make N as the subject where M = 2, c = 4, d = 10.
Given that M × N = c × d
Divide M on both sides
MN/M = cd/M
N = cd/M
Substitute the given values M = 2, c = 4, d = 10
N = (4 × 10)/2
N = 40/2
N = 20

The final answer is N = 20.

3. If the base of the triangle is 3/2 times its height, then find the area of the triangle.

Solution:
Given that the base of the triangle is 3/2 times its height, then find the area of the triangle.
The base of the triangle = b
Height of the triangle = h
The area of the triangle = A
A = 1/2 × b × h
Base of the triangle is 3/2 times its height
b = 3/2 × h
A = 1/2 × 3/2 × h × h
A = 3/4 × h²

4. If ‘S’ is equal to the 3/4 of the r, then find r.

Solution:
Given that If ‘S’ is equal to the 3/4 of the r, then find r.
S = 3/4 r
Divide 3/4 on both sides.
S/(3/4) = (3/4)/(3/4) .  r
4S/3 = r

The final answer is r = 4S/3

5. If x, y, z are in continued proportion, then find the value of z.

Solution:
Given that If x, y, z are in continued proportion, then find the value of z.
x/y = y/z
x . z = y . y
xz = y²
y² = xz
y = √xz

The final answer is y = √xz

6. If y workers can build a wall in 24 days, in how many days will 16 workers take to build a same wall.

Solution: Given that y workers can build a wall in 24 days
So, find the time taken to build a wall by one worker.
y workers = 24 days
1 worker = 24y days
Now, the time taken by 16 workers to build a wall.
1 worker = 24y days
16 worker = 24y/16 = 3/4y
Therefore, The time taken to build a wall by 16 workers is 3/4y.

The final answer is 3/4y.

7. A shirt is marked $ y and the shopkeeper allows 100 rupees off as a discount on it. What is its selling price?

Solution:
Given that a shirt is marked $ y and the shopkeeper allows $100 off as a discount on it.
Subtract $100 from $y to get the selling price.
$y – $100

The selling price is $y – $100.

8. A weighs 3 kg more than B and B weighs 6 kg less than C. If the weights of A, B, C is n, find the weights of A, B, C separately.

Solution:
Given that A weighs 3 kg more than B
A = 3 + B
B weighs 6 kg less than C
B = C – 6; C = B + 6
A + B + C = n
3 + B + B + B + 6 = n
3 + 3B + 6 = n
3B + 9 = n
3B = n – 9
B = (n – 9)/3
A = 3 + B
A = 3 + (n – 9)/3
C = B + 6
C = (n – 9)/3 + 6

The values of A, B, and C is A = 3 + (n – 9)/3, B = (n – 9)/3, and C = (n – 9)/3 + 6.

9. Half of a herd of deer are grazing in the field and 2/3 of the remaining are playing nearby. The rest 8 are drinking water from the pond. Find the number of deer in the herd?

Solution:
Let the total no.of deer = y
Half of a herd = y/2
2/3 of remaining half herd = (y/2)(2/3) = y/3
remaining deer = 8
From the given information, y = (y/2) + (y/3) + 8
y = (3y + 2y)/6 + 8
y = (5y + 48)/6
Multiply 6 on both sides
6y = (5y + 48)/6 × 6
6y = 5y + 48
Move 5y to the left side
6y – 5y = 48
y = 48.

The total number of deers is 48.

10. Arun is twice as old as Sam six years ago his age was four times Shriya’s age. Find their present ages.

Solution:

Let Arun age is x and Sam age is y
Arun age is twice as old as Sam
x = 2y if it is five years ago
x – 5 = 4(y – 5)
x – 5 = 4y – 20
x – 4y + 15 = 0
Substitute x = 2y in x – 4y + 15 = 0
2y – 4y + 15 = 0
-2y + 15 = 0
2y – 15 = 0
2y = 15
y = 15/2 = 7.5
x = 2y = 2 (15/2) = 15

Arun’s age is 15 and Sam’s age is 7.5

11. A car travels 12 km at the speed of x km/hr. Find the time taken by the car to reach the destination.

Solution:
Given that a car travels 12 km at the speed of x km/hr.
We know that Speed = Distance/Time
Time = Distance/Speed
Time = 12km/x km/hr.
The time = 12 hrs.

The time taken by the car to reach the destination is 12 hrs.

12. Ram had $192 with him. He purchased x kg potatoes for $40 a kg and y kg tomatoes for $45 a kg and z kg onions at $43 a kg. Find the money left with him?

Solution:
Given that Ram had $192 with him.
He purchased x kg potatoes for $40 a kg and y kg tomatoes for $45 a kg and z kg onions at $43 a kg.
Total cost of potatoes = 40x
The total cost of tomatoes = 45y
Total cost of onions = 43z
Total amount spent = 40x + 45y + 43z.
Money left with Ram is $192 – (40x + 45y + 43z)

Total Money left with Ram is $192 – (40x + 45y + 43z).

Learn about the concept of Two Trains Passing in Opposite Direction completely by referring to the entire article. Know How to Calculate Speed Time and Distance when Two Trains Run in Opposite Direction. Refer to the Formulas and Solved Examples on Two Trains Passes in Opposite Direction and get a good grip on it. Detailed Solutions provided for each and every problem makes it easy for you to understand the entire concept.

How to find Relative Speed while Two Trains Running in Opposite Direction?

When Two Trains Passes through a Moving Object having a certain length in the Opposite Direction

Let us assume the Length of the faster train is l meters and the length of the slower train is m meters

Speed of faster train = x km/hr

Speed of slower train = y km/hr

Relative Speed = (x+y) km/hr

Time taken by faster train to cross the slower train = (l+m) m/(x+y) km/hr

Using this Simple Formula you can calculate the measures easily when they run on parallel tracks in the opposite direction.

Solved Problems on Two Trains Running on Parallel Tracks in the Opposite Direction

1. Two trains of length 130 m and 100 m respectively are running at the speed of 52 km/hr and 40 km/hr on parallel tracks in opposite directions. In what time will they cross each other?

Solution:

Speed of faster train = 52 km/hr

Speed of slower train = 40 km/hr

Relative Speed of Trains = (52 km/hr – 40 km/hr)

= 12 km/hr

= 12*5/18

= 3.33 m/sec

Length of first train = 130 m

Length of Second Train = 100 m

Time taken by the two trains to cross each other = sum of the length of trains/relative speed of trains

= (130+100) m/12 km/hr

= 230 m/3.33 m/sec

= 69.06 sec

Therefore, Two Trains Crosses each other in 69.06 sec

2. Two trains 170 m and 145 m long are running on parallel tracks in the opposite directions with a speed of 50 km/hr and 40 km/hr. How long will it take to cross each other?

Solution:

Speed of faster train = 50 km/hr

Speed of slower train = 40 km/hr

Relative Speed of Trains = (50 km/hr +40 km/hr)

= 110 km/hr

= 110*5/18

= 30.5 m/sec

Length of first train = 170 m

Length of second train = 145 m

Time taken by two trains to cross each other = Sum of Length of Trains/Relative Speed of Trains

= (170+145) m/30.5 m/sec

= 315 m/30.5 m/sec

= 10.3 sec

3. Two trains travel in opposite directions at 50 km/hr and 30 km/hr respectively. A man sitting in the slower train passes the faster train in 12 s. The length of the faster train is?

Solution:

Speed of faster train = 50 km/hr

Speed of second train = 30 km/hr

Time taken to cross each other = 12 sec

Relative Speed of Trains = (50 Km/hr +30 Km/hr)

= 80 km/hr

Relative Speed of Trains in m/sec = 80*5/18

= 22.22 m/sec

Length of faster train = 22.22 m/sec * 12 sec

= 266.6 m

Therefore, the length of the faster train is 266.6 m

Get acquainted with the Concept of Two Trains Passes in the Same Direction better by going through the entire article. Refer to the Solved Problems on Two Trains running in the Same Direction along with Solutions for better understanding. Check out the Formulas for Speed Time and Distance while Two Trains Passes in the Same Direction. We have provided a detailed procedure on how to find the when Two Trains Passes a moving object in the Same Direction.

Two Trains Passes a Moving Object in the Same Direction

When two trains passes a moving object in the same direction.

Consider Length of the faster train be l meters and length of the slower train be m meters

The speed of the faster train be x km/hr

The speed of the slower train be y km/hr

Relative Speed = (x-y) km/hr

Time taken by the faster train to cross the slower train = (l+m) m/(x-y) km/hr

Solved Problems on Two Trains Running on Parallel Tracks in the Same Direction

1. Two trains 110 m and 150 m long are running on parallel tracks in the same direction with a speed of 60 km/hr and 45 km/hr. How long will it take to clear off each other from the moment they meet?

Solution:

Speed of faster train = 60 km/hr

Speed of slower train = 45 km/hr

Length of first train = 110 m

Length of second train = 150 m

Relative Speed = (60 km/hr – 45 km/hr)

= 15 km/hr

= 15*5/18 m/sec

= 4.16 m/sec

Time taken by train to clear off each other = Sum of Lengths of both the Trains/Relative Speed

= (110+150)m /4.16 m/sec

= 260 m/4.16 m/sec

= 62.5 sec

2. The two trains are running on parallel tracks in the same direction at 80 km/hr and 55 km/hr respectively. The faster train passes a man 20 seconds faster than the slower train. Find the length of the faster train?

Solution:

Relative Speed of Trains = (80 km/hr – 55 km/hr)

= 25 km/hr

Relative Speed of Trains in m/sec = 25*5/18

= 6.94 m/sec

Length of faster train = Relative Speed * Time Taken by Train to Pass

= 6.94 m/sec * 20 sec

= 138.8 m

3. Two trains are moving in the same direction at 70 km/hr and 40 km/hr. The faster train crossed a man in the slower train in 30 seconds. Find the length of the faster train?

Solution:

Speed of Faster Train = 70 km/hr

Speed of Slower Train = 40 km/hr

Relative Speed = (70 km/hr – 40 km/hr)

= 30 km/hr

Relative Speed in m/sec = 30 *5/18

= 8.33 m/sec

Time taken to cross = 30 sec

Length of faster train = Speed * Time

= 8.33 m/sec * 30 sec

= 250 m

Looking for help on the concept Train Passes through a Bridge then you can find all of it here. Refer to Solved Examples on How to find the Speed Time and Distance when a Train Crosses a Bridge. We have provided solutions for all the Problems covered on the concept of Train Passing through a Bridge. or Tunnel or any Stationary Object.

In this case, the Length of the Train and Length of the Bridge are added to get the Distance. Practice the Train Passes through a Bridge Problems with Answers and get the elaborate explanation provided to understand the concept better.

How to Solve Train Passing through a Bridge or Tunnel Problems?

Let us consider the Length of the Train = x meters

Length of the Stationary Object = y meters

The Speed of the Train is z km/hr

Time Taken by the Train to Cross the Bridge = (Length of Train + Length of Bridge)/Speed of Train

= (x+y)m /z km/hr

To change between km/hr to m/sec multiply with 5/18

Solved Examples on Train Crossing a Stationary Object having Some Length

1. A train 250 m long crosses a bridge which is 100 m long in 50 seconds. What is the speed of the train?

Solution:

Length of the Train = 250 m

Length of the Bridge = 100 m

Speed of the Train = (Length of the Train+Length of Stationary Object)/Time

= (250 m+100 m)/50

= 350 m/50 sec

= 7 m/sec

2. A train 260 m long is running at a speed of 40 km/hr. What time will it take to cross an 80 m long tunnel?

Solution:

Length of the Train = 260 m

Speed of the Train = 40 km/hr

= 40 *5/18 m/sec

= 11.11 m/sec

Length of the Tunnel = 80 m

Time = Distance/Speed

= 260 m/11.11 m/sec

= 23.40 sec

Train takes 23.40 sec to travel the tunnel.

3. Find the time taken by a 180 m long train passes through a bridge which is 75 m long, running at a speed of 54 km/hr?

Solution:

Length of the Train = 180 m

Length of the Bridge = 75 m

Speed of the Train = 54 km/hr

Speed of train in m/sec = 54*5/18

= 15

Time = Distance/Speed

= 225 m/15 m/sec

= 9 sec

Therefore, train takes 9 sec to cross the bridge.

4. A 120 m long train is running at a speed of 45 km/hr. If it takes 20 seconds to cross a platform, find the length of the platform?

Solution:

Speed of the Train = 45 km/hr

Speed of Train in m/sec = 45*5/18

= 12.5 m/sec

Time taken to cross the platform = 20 sec

Length of the Train = 120 m

Consider the Length of Platform = x

Speed = (Length of Train +Length of Platform)/Time

12.5 m/sec = (120+x)/20

12.5*20 = 120+x

250 = 120+x

x = 250 – 120

= 130 m

Therefore, Length of Platform is 130 m.

Practice the Questions available below when Train Passes through a Pole and get full-fledged knowledge about the concept. Know How to Calculate Speed Time and Distance when a train crosses a stationary object in the coming modules. Train Passes a Pole, Man or Tree Problems with Solutions along with step by step explanation makes it easy for you to understand the concept. Refer to them and learn how to solve the Train Passing through a Pole Problems easily.

How to Solve Train Passes through a Pole Problems?

When a Train Passes through a Pole or Tree the formulas to find the Time Speed and Distance are given below

Suppose the Length of the Train = x mts

Speed of the Train = y km/hr

Time taken by the Train to Cross a Pole or Stationary Object = Length of the Train/Speed of the Train

= x meters/y km/hr

To Change km/hr to m/sec simply multiply with 5/18.

Solved Examples to Calculate When Train Passes through a Pole

1. A train 200 m long is running at a uniform speed of 75 km/hr. How much time will it take to cross a pole?

Solution:

Length of the Train = 200 m

Speed of the Train = 75 km/hr

To convert it to m/sec multiply with 5/18

= 75*5/18

= 20.83 m/sec

Time taken by the train to cross the pole = Length of the Train/Speed of the Train

= 200 m/20.83 m/sec

= 9.6 sec

Therefore, Train takes 9.6 sec to cross the pole.

2. Find the time taken by a train 350 m long, running at a speed of 60 km/hr in crossing the pole?

Solution:

Length of the Train = 350 m

Speed of the Train = 60 km/hr

Time taken by the Train to cross the pole = Length of the Train/Speed of the Train

= 350m/60 km/hr

To change from km/hr to m/sec multiply with 5/18 i.e. 60*5/18 = 16.66 m/sec

= 350 m/16.66m/sec

= 21.008 sec

Thus, the train takes 21.008 sec to cross the pole.

3. A train is running at a speed of 54 km/hr. It crosses a tower in 7 seconds. Find the Length of the train?

Solution:

Speed of the Train = 54 km/hr

Time taken by train to cross tower = 7 seconds

Length of the Train = Speed * Time

= 54*(5/18)*7

= 105 m

Therefore, the Length of the Train is 105 m

4. A train is running at a speed of 140 km/hr. if it crosses a pole in just 8 seconds, what is the length of the train?

Solution:

Speed of the Train = 140 km/hr

Time taken to cross the pole = 8 seconds

Length of the Train = Speed * Time

= 140 km/hr*8

= 140*(5/18)*8

= 311.11 m

Therefore, the length of the train is 311.11 m

Are you looking for help on the concept Train Passes a Moving Object in the Opposite Direction? Then you have reached the right place. Learn the Formulas for Speed Time and Distance in the case of a train crossing a moving body in the opposite direction. Get Solved Examples for finding the Train Crossing a Moving Object in Opposite Direction long with detailed solutions. Practice the Problems available and get a good hold of the concept.

How to calculate Time Speed and Distance for Train Crossing a Moving Object in Opposite Direction?

Let us assume the length of the train = l m

Speed of the train =  x km/hr

Speed of the object = y km/hr

Relative Speed = (x+y) km/hr

Time taken by train to cross the moving object = Distance/Speed

= l m/(x+y) km/hr

Simply rearrange the formula to obtain the other measures if few are known.

Solved Problems on Train Passes a Moving Object in the Opposite Direction

1. A train 200 m long is running at a speed of 50 km/hr. In what time will it pass a man who is running at the speed of 4 km/hr in the opposite direction in which the train is moving?

Solution:

Length of the Train = 200 m

Speed of the Train = 50 km/hr

Speed of Man = 4 km/hr

Relative Speed = (50+4) Km/hr

= 54 km/hr

Time taken by train to cross a man = Distance/Relative Speed

= 200 m/54 Km/hr

= 200 m/ (54*5/18) m/sec

= 200 m/15 m/sec

= 13.33 sec

Therefore, the train takes 13.33 sec to cross the man.

2. Two trains 125 meters and 170 meters long are running in the opposite direction with speeds of 60 km/hr and 45 km/hr. In how much time they will cross each other?

Solution:

Total Distance Covered = 125+170

= 295 m

Speed of first train = 60 kmph

Speed of second train = 45 kmph

Relative Speed = 60+45

= 105 kmph

Relative Speed in m/sec = 105*5/18

= 29.1 m/sec

Time taken by trains to cross each other = Distance/Speed

=295m/29.1 m/sec

= 10.1 sec

Therefore, it takes 10.1 sec for both the trains to cross each other.

3. Two trains running in opposite directions cross a man standing on the platform in 30 seconds and 21 seconds respectively and they cross each other in 25 seconds. The ratio of their speeds is?

Solution:

Let us consider the speed of two trains be x m/sec and y m/sec

Length of the first train = 30x m(since distance = Speed*Time)

Length of the second train = 21y m

From the given data

Time taken by both trains to cross each other = 25 sec

(30x+21y)/(x+y) = 25

30x+21y = 25x+25y

30x-25x = 25y-21y

5x=4y

x/y = 4/5

Therefore, the Ratio of Speeds is 4:5