Samacheer Kalvi Class 8 Maths Solutions Chapter 7 Information processing Exercise 7.1

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information Processing Ex 7.1

Question 1. You want to have an ice cream or a cake. There are three flavours (chocolate, strawberry and vanilla) in ice creams, and two flavours (orange and red velvet) in the cakes. In how many possible ways can you choose an ice cream or a cake?
Answer: You can choose an ice cream from 3 flavors or a cake from 2 flavors. Since you can only pick one (either an ice cream OR a cake), we add the number of choices together. So, the total number of ways to choose is \( 3 + 2 = 5 \) ways. This is an example of the Addition Principle, used when events are mutually exclusive.
In simple words: You can choose one item: either an ice cream or a cake. Add the number of ice cream choices (3) and cake choices (2) to get the total number of ways.

🎯 Exam Tip: Remember to use the Addition Principle (add the choices) when you have 'OR' situations, meaning you can choose one option from different sets, but not both at the same time.

Question 2. Shanthi has 5 chudithar sets and 4 frocks. In how many possible ways, can she wear either a chudithar or a frock?
Answer: Shanthi has 5 chudithar sets and 4 frocks. Since she wants to wear *either* a chudithar *or* a frock, we add the number of choices. Therefore, the total number of ways she can dress is \( 5 + 4 = 9 \) ways. This is another use of the Addition Principle, where two events cannot happen at the same time.
In simple words: Shanthi has 5 chudithars and 4 frocks. To find the total ways she can wear either, add the number of chudithars and frocks together.

🎯 Exam Tip: When you choose between two different types of items, like clothes here, and you can only pick one, you add the number of choices for each type.

Question 3. In a Higher Secondary School, the following groups are available in XI standard I. Science Group: (i) Physics, Chemistry, Biology and Mathematics (ii) Physics, Chemistry, Mathematics and Computer Science. (iii) Physics, Chemistry, Biology and Home Science II. Arts Group: (i) Accountancy, Commerce, Economics and Business Maths (ii) Accountancy, Commerce, Economics and Computer Science (iii) History, Geography, Economics and Commerce (ii) Home Science, Textiles and Dress Designing Theory, Textiles and Dress Designing Practical I and Textiles and Dress Designing Practical II In how many possible ways, can a student choose a group?
Answer: A student can choose a group from three categories: Science, Arts, or Vocational. There are 3 options in the Science Group, 3 options in the Arts Group, and 2 options in the Vocational Group. Since a student will choose *one* group from *any* of these categories, we add the number of options from each: \( 3 + 3 + 2 = 8 \) ways. This illustrates the Addition Principle for multiple non-overlapping choices.
In simple words: Students can pick from 3 Science groups, 3 Arts groups, or 2 Vocational groups. To find the total number of ways to pick any one group, add all these options together.

🎯 Exam Tip: Carefully identify each distinct category of choices. If a selection is made from *one* of these categories, sum up the options in each category.

Question 4. If you have 2 school bags and 3 water bottles then, in how many different ways can you choose each one of them, while going to school?
Answer: You have 2 different school bags and 3 different water bottles. To find the total number of ways to choose one of each (one bag *and* one bottle), you multiply the number of choices for each item. So, there are \( 2 \times 3 = 6 \) possible ways to select a bag and a bottle. This is known as the Multiplication Principle, used when events occur simultaneously. Each bag can be paired with any of the three bottles, forming a clear set of combinations.

In simple words: To pick one bag and one bottle, multiply the number of bag choices (2) by the number of bottle choices (3). This gives you the total ways to combine them.

🎯 Exam Tip: Use the Multiplication Principle when you need to choose one item *from each* of several different categories, as the choices are independent and happen together.

Question 5. Roll numbers are created with a letter followed by 3 digits in it, from the letters A, B, C, D and E and any 3 digits from 0 to 9. In how many possible ways can the roll numbers be generated? (except A000, B000, C000, D000 and E000)
Answer: A roll number consists of one letter followed by three digits. There are 5 choices for the letter (A, B, C, D, E). For each of the three digit places, there are 10 choices (0 to 9). So, the total number of unique three-digit numbers is \( 10 \times 10 \times 10 = 1000 \). To find the total possible roll numbers, we multiply the number of letter choices by the number of three-digit number choices: \( 5 \times 1000 = 5000 \) ways. However, the problem states that roll numbers like A000, B000, C000, D000, and E000 are not allowed. There are 5 such forbidden roll numbers. So, the final count of allowed roll numbers is \( 5000 - 5 = 4995 \) ways. This combines the Multiplication Principle with subtraction for exclusions.
In simple words: First, pick one of 5 letters. Then, pick 3 digits from 0-9 for the next three spots. This gives \( 5 \times 10 \times 10 \times 10 = 5000 \) possible roll numbers. But, we must remove 5 specific ones (like A000, B000), leaving \( 5000 - 5 = 4995 \) ways.

🎯 Exam Tip: When calculating combinations or permutations, always consider any exclusions mentioned in the problem and subtract them from the total possible ways at the end.

Question 6. A safety locker in a jewel shop requires a 4 digit unique code. The code has the digits from 0 to 9. How many unique codes are possible?
Answer: A 4-digit unique code means there are four positions for digits. For each of these four positions, there are 10 possible digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) that can be chosen. Since each digit choice is independent, we use the Multiplication Principle. So, the total number of unique 4-digit codes possible is \( 10 \times 10 \times 10 \times 10 = 10^4 = 10,000 \) ways. This shows how quickly the number of possibilities grows with more choices, demonstrating the power of independent selections.
In simple words: For a 4-digit code, there are 10 choices (0-9) for each of the four spots. Multiply 10 by itself four times to get the total number of codes.

🎯 Exam Tip: When digits can be repeated and there are no restrictions on position, the number of possibilities for each position is simply the total number of available digits.

Question 7. An examination paper has 3 sections, each with five questions and students are instructed to answer one question from each section. In how many different ways can the questions be answered?
Answer: The exam paper has 3 sections, and each section contains 5 questions. Students must answer one question *from each* section. This means they have 5 choices for the first section, 5 choices for the second section, and 5 choices for the third section. Using the Multiplication Principle, the total number of ways to answer the questions is \( 5 \times 5 \times 5 = 125 \) ways. This allows for a wide range of unique answer combinations, highlighting the many paths a student could take.
In simple words: There are 3 sections, and you pick one question from each section. Since each section has 5 questions, you multiply \( 5 \times 5 \times 5 \) to find all the possible ways to answer.

🎯 Exam Tip: When a selection requires choosing items from multiple independent categories (like questions from different sections), multiply the number of choices available in each category.

Question 8. The given spinner is spun twice and the two numbers got are used to form a 2 digit number. How many different 2 digits numbers are possible?
Answer: The spinner has 5 numbers (1, 2, 3, 4, 5). When spun twice, the first spin gives the tens digit and the second spin gives the ones digit. So, for the first digit, there are 5 choices, and for the second digit, there are also 5 choices. Using the Multiplication Principle, this gives \( 5 \times 5 = 25 \) possible two-digit numbers. However, the problem asks for 'different' numbers, and the source specifies removing repetitions where both digits are the same (like 11, 22, 33, 44, 55). There are 5 such repeating numbers. So, removing these, we get \( 25 - 5 = 20 \) different two-digit numbers. This approach filters out specific cases from the total.

In simple words: The spinner has 5 numbers. For a two-digit number, the first spin gives the tens digit (5 choices) and the second spin gives the ones digit (5 choices). This makes \( 5 \times 5 = 25 \) possible numbers. If we don't count numbers where both digits are the same (like 11, 22, etc.), then we subtract those 5, leaving 20 different numbers.

🎯 Exam Tip: Understand if 'different' means distinct combinations, or if identical numbers formed by different sequences (e.g., first 1 then 2 is different from first 2 then 1) should be counted once. The problem implies removing exact repeats like 11 or 22 from the set of possible outcomes.

Question 9. Ramya wants to paint a pattern in her living room wall with minimum budget. Help her to colour the pattern with 2 colours but make sure that no two adjacent regions have the same colour. The pattern is shown in the picture.
Answer: To color the pattern with a minimum budget using only 2 colors, Ramya needs to make sure that any two regions sharing a border (adjacent regions) are painted with different colors. The provided solution shows an example of such a coloring, where alternating colors are used for adjacent sections. This is a common technique in graph theory, ensuring distinctness between linked areas and is efficient for simple patterns.

In simple words: Ramya can use two colors. The trick is to make sure that any two parts of the pattern that touch each other have different colors. This way, the colors will alternate and the pattern will be clear.

🎯 Exam Tip: For coloring problems, start by picking a color for one region. Then, for its neighbors, pick the other color. Continue this process, ensuring no two touching regions have the same color. If you run out of colors, you need more than two.

Question 10. Colour the regions in the maps with few colours as possible but make sure that no two adjacent countries are of the same colour.
Answer: The goal is to color map regions so that no two adjacent (touching) regions share the same color, using as few colors as possible. This is a classic map coloring problem. For the first map (checkerboard-like), two colors are sufficient, as regions can alternate colors. For the second map, at least three colors are required because there are regions that all touch each other in a chain or around a central region. For instance, imagine a central country surrounded by others; it must be a different color from all its neighbors, which might then have to be different from each other. The Four Color Theorem states that any map can be colored with at most four colors.

In simple words: When coloring maps, make sure any two parts that touch each other have different colors. Use the smallest number of colors possible. Some simple maps only need two colors, while others, like one with a central country, might need three or more.

🎯 Exam Tip: For map coloring, start with a complex region (one touching many others). Assign it a color, then assign different colors to its neighbors. Work systematically through the map, minimizing new colors.

Objective Type Questions

Question 11. In a class there are 26 boys and 15 girls. The teacher wants to select a boy or a girl to represent a quiz competition. In how many ways can the teacher make this selection?
(a) 41
(b) 26
(c) 15
(d) 390
Answer: (a) 41
In simple words: To pick one student, either a boy or a girl, just add the number of boys (26) and the number of girls (15) together.

🎯 Exam Tip: The keyword 'or' in selection problems usually means you should add the number of choices from each distinct group.

Question 12. How many outcomes can you get when you toss three coins once?
(a) 6
(b) 8
(c) 3
(d) 2
Answer: (b) 8
In simple words: Each coin has 2 possible results (Heads or Tails). If you toss three coins, multiply 2 by itself three times to find the total number of different results.

🎯 Exam Tip: For multiple independent events, like coin tosses, the total number of outcomes is found by multiplying the number of outcomes for each individual event.

Question 13. In how many ways can you answer 3 multiple choice questions, with the choices A,B,C and D?
(a) 4
(b) 3
(c) 12
(d) 64
Answer: (d) 64
In simple words: Each of the 3 questions has 4 possible answers. To find the total ways to answer all of them, multiply 4 by itself three times.

🎯 Exam Tip: For multiple-choice questions, if each question has the same number of options, raise the number of options to the power of the number of questions.

Question 14. How many 2 digit numbers contain the number 7?
(a) 10
(b) 18
(c) 19
(d) 20
Answer: (b) 18
In simple words: For 2-digit numbers, count all numbers that start with 7 (like 70, 71, etc.). Then, count numbers that end with 7 (like 17, 27, etc.) but don't start with 7. Add these two counts together.

🎯 Exam Tip: For counting problems with specific digit conditions, break it down by place value (tens digit, units digit) and be careful to avoid counting any number more than once.