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Algebra - Using Expressions in Practical Situations

Grade 6ICSE

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

🔑Concepts

Algebraic variables like x,y,z,x, y, z, or nn are used to represent unknown quantities in practical situations. Visually, think of a variable as an empty container or a 'placeholder box' where different values can be placed depending on the context.

Translating English phrases into mathematical expressions is essential for solving real-world problems. For instance, '5 more than xx' is written as x+5x + 5, while '7 less than yy' is written as y7y - 7. Visually, 'more than' represents extending a line segment, while 'less than' represents cutting a piece off.

Geometric perimeters can be expressed algebraically. For a square with side length ss, the perimeter is the total boundary length, calculated as s+s+s+ss + s + s + s, which simplifies to 4s4s. Visually, this is represented by four equal segments forming a closed loop.

Algebraic rules can be derived from patterns, such as matchstick shapes. If one 'L' shape requires 2 matchsticks, then nn such shapes require 2n2n matchsticks. Visually, you can imagine a repeating sequence of shapes where each addition adds a fixed number of units.

Age-related problems use variables to relate different people's ages. If a person's current age is aa, their age 55 years ago was a5a - 5 and their age 1010 years from now will be a+10a + 10. Visually, this can be plotted on a horizontal timeline with the 'present' in the center.

Practical cost calculations involve multiplying quantity by unit price. If the cost of one notebook is rr, the cost of 1212 notebooks is 12r12r. Visually, this represents a grid or an array where the total area corresponds to the total price.

Distance, speed, and time relationships are expressed algebraically. If a car travels at a speed of vv km/h for tt hours, the distance dd is given by v×tv \times t. Visually, this can be seen as a vector or an arrow representing movement over a specific duration.

📐Formulae

Perimeter of a square: P=4sP = 4s (where ss is the side)

Perimeter of a rectangle: P=2(l+b)P = 2(l + b) (where ll is length and bb is breadth)

Perimeter of an equilateral triangle: P=3sP = 3s (where ss is the side)

Total Cost: C=n×pC = n \times p (where nn is number of items and pp is price per item)

Distance: d=s×td = s \times t (where ss is speed and tt is time)

Age nn years from now: Current Age+n\text{Current Age} + n

Age nn years ago: Current Agen\text{Current Age} - n

💡Examples

Problem 1:

Leela is Radha's younger sister. Leela is 44 years younger than Radha. Can you write Leela's age in terms of Radha's age? Let Radha's age be xx years.

Solution:

  1. Let Radha's age be xx years.
  2. Leela is 44 years younger than Radha.
  3. To find Leela's age, we subtract 44 from Radha's age.
  4. Therefore, Leela's age = (x4)(x - 4) years.

Explanation:

In this problem, we identify the independent variable (Radha's age) and apply the operation 'younger than' which corresponds to subtraction.

Problem 2:

A bus travels at vv km per hour. It is going from Daspur to Beespur. After the bus has travelled 55 hours, Beespur is still 2020 km away. What is the distance from Daspur to Beespur? Express it using vv.

Solution:

  1. Speed of the bus = vv km/h
  2. Time travelled = 55 hours
  3. Distance travelled in 55 hours = Speed×Time=v×5=5v\text{Speed} \times \text{Time} = v \times 5 = 5v km
  4. Remaining distance to Beespur = 2020 km
  5. Total distance = Distance travelled+Remaining distance=(5v+20)\text{Distance travelled} + \text{Remaining distance} = (5v + 20) km

Explanation:

We first use the distance formula (d=s×td = s \times t) to find out how far the bus has gone, then add the remaining distance to find the total length of the journey.