Book cover for Computer Science and Information Technology

Computer Science and Information Technology

Trishna Knowledge Systems

ISBN #9789352868469

2018 Edition

1,832 Questions

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27,660 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

Chapter 1 introduces the fundamental aspects of simple equations, focusing on the techniques required to isolate variables using inverse operations. It demonstrates solving one-step and two-step equations and emphasizes checking solutions by substituting back into the original equation. Mastery of these basic concepts is crucial for progressing to more complex algebraic problems.

Learning Objectives

1

Understand the concept of a simple equation and the role of variables, coefficients, and constants.

2

Learn to isolate the variable using inverse operations.

3

Solve one-step and two-step equations efficiently.

4

Verify solutions by substituting back into the original equation.

Key Concepts

CONCEPT

DEFINITION

Equation

A mathematical statement asserting that two expressions are equal.

Variable

A symbol, often represented by a letter, that stands for an unknown value.

Coefficient

The numerical factor multiplied by the variable in an equation.

Constant

A fixed numerical value in an equation that does not change.

Inverse Operation

An operation that reverses the effect of another operation, such as subtraction being the inverse of addition.

Example Problems

Example 1

Select the correct alternative from the given choices. $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are successive even natural numbers in ascending order. Five times $\mathrm{R}$ is eight more than seven times $\mathrm{P}$. Find $\mathrm{Q}$ (A) 6 (B) 8 (C) 12 (D) 14

Example 2

Select the correct alternative from the given choices. Divide $1 \mathrm{~kg}$ weight into two parts such that the sum of the parts is $5 / 4^{\text {th }}$ the difference. (A) $550 \mathrm{gm}, 450 \mathrm{gm}$ (B) $200 \mathrm{gm}, 800 \mathrm{gm}$ (C) $900 \mathrm{gm}, 100 \mathrm{gm}$ (D) $400 \mathrm{gm}, 600 \mathrm{gm}$

Example 3

Select the correct alternative from the given choices. A is greater than $\mathrm{B}$ by $1 / 3^{\text {nd }}$ the sum of $\mathrm{A}$ and $\mathrm{B}$. If $\mathrm{B}$ is increased by 40 , it becomes greater than twice $\mathrm{A}$ by $10 .$ Find $\mathrm{A}, \mathrm{B}$. (A) 30,20 (B) 60,30 (C) 20,10 (D) 20,40

Example 4

Select the correct alternative from the given choices. Ajay was asked to find $(2 / 9)^{\text {th }}$ of a number. He instead multiplied the number by $(9 / 2)$ and obtained an answer which was 4235 more than the correct answer. Find the number (A) 900 (B) 945 (C) 990 (D) 810

Example 5

Select the correct alternative from the given choices. An amount of $? 5,600$ is divided among $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. The sum of the shares of $\mathrm{B}$ and $\mathrm{C}$ is equal to thrice the share of $\mathrm{A}$. The sum of the shares of $\mathrm{A}$ and $\mathrm{C}$ is equal to ninefifths the share of $\mathrm{B}$. What is the share of $\mathrm{C}$ ? (A) $? 1,400$ (B) $? 2,400$ (C) $? 2,200$ (D) $? 2,000$
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Step-by-Step Explanations

QUESTION

How do we solve the equation 2x + 3 = 7?

STEP-BY-STEP ANSWER:

Step 1: Subtract 3 from both sides of the equation to isolate the term containing x. This gives: 2x + 3 - 3 = 7 - 3, simplifying to 2x = 4.
Step 2: Divide both sides by 2 to solve for x: 2x/2 = 4/2, resulting in x = 2.
Final Answer: x = 2.

Solving the equation 2x + 3 = 7

QUESTION

How do we solve the equation x - 5 = 10?

STEP-BY-STEP ANSWER:

Step 1: Add 5 to both sides to get x by itself: x - 5 + 5 = 10 + 5, which simplifies to x = 15.
Final Answer: x = 15.

Solving the equation x - 5 = 10

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Common Mistakes

  • Not performing the same operation on both sides of the equation, which breaks the balance.
  • Confusing addition with subtraction or vice versa during inverse operations.
  • Forgetting to divide by the coefficient after isolating the variable term.
  • Neglecting to check the solution by substituting it back into the original equation.