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Linear Equations in One Variable - Applications and Word Problems

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Understanding Linear Equations: A linear equation in one variable is an equation where the highest power of the variable (usually xx) is 1. It can be visualized as a balanced see-saw where both sides represent equal values. For example, ax+b=cax + b = c is a standard form where aa, bb, and cc are constants.

Solving Strategy: To solve word problems, first identify the unknown quantity and represent it with a variable xx. Visualize the problem as a translation process where phrases like 'is equal to' or 'gives' represent the == sign, and 'sum' or 'increased by' represent the ++ sign.

Consecutive Integers: When a problem mentions consecutive integers, imagine a number line where integers follow each other without gaps. We represent them as xx, x+1x+1, x+2x+2, etc. For consecutive even or odd integers, the gap is 2 units, so we use xx, x+2x+2, x+4x+4.

Age-Related Problems: These problems often involve comparing ages at different points in time. If the current age is xx, visualize a timeline: nn years ago, the age was (xn)(x - n), and nn years from now (hence), the age will be (x+n)(x + n). It is often helpful to organize this data in a table.

Two-Digit Numbers: A two-digit number can be visualized in its expanded form using place value. If the digit at the tens place is xx and the digit at the units place is yy, the number is represented as 10x+y10x + y. If the digits are reversed, the new number is 10y+x10y + x.

Geometry Applications: Problems often involve perimeters or angles. Visualize a rectangle where the perimeter is the total length of the boundary, calculated as 2(Length+Breadth)2(Length + Breadth). For triangles, use the angle sum property where the three interior angles add up to 180180^{\circ}.

Rational Numbers and Fractions: If the numerator of a fraction is xx and the denominator is yy, the fraction is xy\frac{x}{y}. Problems typically describe changes to these parts, such as 'if 2 is added to the numerator,' which is written as x+2y\frac{x + 2}{y}.

Speed, Distance, and Time: These problems use the relationship Distance=Speed×TimeDistance = Speed \times Time. For boat problems, visualize the boat moving with the current (downstream speed = Speedboat+SpeedstreamSpeed_{boat} + Speed_{stream}) or against the current (upstream speed = SpeedboatSpeedstreamSpeed_{boat} - Speed_{stream}).

📐Formulae

General form of a linear equation: ax+b=0ax + b = 0

Consecutive Integers: x,x+1,x+2,...x, x+1, x+2, ...

Consecutive Even/Odd Integers: x,x+2,x+4,...x, x+2, x+4, ...

Two-digit number expansion: 10×(TensDigit)+(UnitsDigit)10 \times (Tens \, Digit) + (Units \, Digit)

Perimeter of a Rectangle: P=2(l+b)P = 2(l + b)

Sum of angles in a triangle: A+B+C=180\angle A + \angle B + \angle C = 180^{\circ}

Speed Formula: Speed=DistanceTimeSpeed = \frac{Distance}{Time}

Percentage calculation: Value=Percentage100×TotalValue = \frac{Percentage}{100} \times Total

💡Examples

Problem 1:

The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 18, find the original number.

Solution:

Let the digit at the units place be xx. Since the sum of digits is 12, the digit at the tens place is (12x)(12 - x). Original Number = 10(12x)+x=12010x+x=1209x10(12 - x) + x = 120 - 10x + x = 120 - 9x. Reversed Number = 10x+(12x)=9x+1210x + (12 - x) = 9x + 12. According to the problem: ReversedNumber=OriginalNumber+18Reversed \, Number = Original \, Number + 18 9x+12=(1209x)+189x + 12 = (120 - 9x) + 18 9x+12=1389x9x + 12 = 138 - 9x 18x=12618x = 126 x=12618=7x = \frac{126}{18} = 7. Units digit = 7, Tens digit = 127=512 - 7 = 5. Original number = 57.

Explanation:

We use the place value concept to express the number. By setting up an equation based on the condition that the reversed number is 18 more than the original, we solve for the units digit and then find the full number.

Problem 2:

A father's age is 3 times the age of his son. After 12 years, his age will be twice that of his son. Find their present ages.

Solution:

Let the present age of the son be xx years. Then, the father's present age is 3x3x years. After 12 years: Son's age = (x+12)(x + 12) years Father's age = (3x+12)(3x + 12) years According to the problem: 3x+12=2(x+12)3x + 12 = 2(x + 12) 3x+12=2x+243x + 12 = 2x + 24 3x2x=24123x - 2x = 24 - 12 x=12x = 12. Son's present age = 12 years. Father's present age = 3×12=363 \times 12 = 36 years.

Explanation:

We establish the current relationship between the ages and then create expressions for their ages 12 years into the future. By applying the given condition (father being twice as old as the son), we solve for the variable xx.