CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 4 Marks
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CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 4 Marks
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SECTION D — Case-Based/Source-Based Integrated Questions
[4 Marks Each]
Q1. Case Study: The Seminar Hall
In a seminar, the number of participants in Hindi, English, and Mathematics are 60, 84, and 108 respectively.
2 MarksSub-question (a): Find the maximum number of participants that can be accommodated in each room if each room has the same number of participants and each room contains participants of only one subject.
Answer:
1. To find the maximum number of participants that can be accommodated in each room such that each room has the same number of participants and contains participants of only one subject, we need to find the HCF of $60$, $84$, and $108$.
2. Let us find the prime factorization of each number:
• $60 = 2^2 \times 3 \times 5$
• $84 = 2^2 \times 3 \times 7$
• $108 = 2^2 \times 3^3$
3. The HCF is the product of the lowest powers of common prime factors in all three numbers:
$\text{HCF}(60, 84, 108) = 2^2 \times 3^1 = 4 \times 3 = 12$
Result: The maximum number of participants that can be accommodated in each room is $12$.
2 MarksSub-question (b): What is the total number of rooms required?
Answer:
1. First, we find the maximum number of participants per room by calculating the HCF of $60$, $84$, and $108$, which is $12$.
2. To find the total number of rooms required, we divide the number of participants in each subject by the maximum number of participants per room ($12$) and sum them up:
• Rooms for Hindi = $\frac{60}{12} = 5$ rooms
• Rooms for English = $\frac{84}{12} = 7$ rooms
• Rooms for Mathematics = $\frac{108}{12} = 9$ rooms
3. Calculate the total number of rooms:
$\text{Total rooms} = 5 + 7 + 9 = 21$
Result: The total number of rooms required is $21$.
Q2. Case Study: The Morning Bell
Three bells toll at intervals of 9, 12, and 15 minutes respectively. If they start tolling together at 8:00 AM:
2 MarksSub-question (a): After how much time will they next toll together?
Answer:
1. To find the time after which the three bells will next toll together, we need to find the Least Common Multiple (LCM) of their individual tolling intervals: $9$, $12$, and $15$ minutes.
2. Let us find the prime factorization of each interval:
• $9 = 3^2$
• $12 = 2^2 \times 3$
• $15 = 3 \times 5$
3. The LCM is the product of the highest powers of each prime factor involved:
$\text{LCM}(9, 12, 15) = 2^2 \times 3^2 \times 5^1$
$= 4 \times 9 \times 5$
$= 180 \text{ minutes}$
Result: They will next toll together after $180$ minutes (or $3$ hours).
2 MarksSub-question (b): How many times will they toll together in the next 3 hours?
Answer:
1. First, we find the LCM of the intervals ($9$, $12$, and $15$ minutes) to determine how often the bells toll together, which is $180$ minutes ($3$ hours).
2. We are asked about the number of times they toll together in the next 3 hours (starting from 8:00 AM, up to 11:00 AM).
3. The bells toll together at the start (8:00 AM) and then next toll together after $3$ hours (at 11:00 AM).
4. Therefore, within the 3-hour duration from the starting point:
• At the exact moment they start together (0 hours / 8:00 AM): 1 time (Note: depending on whether the starting bell is counted as part of the "next 3 hours", let's check standard convention. Usually, if they start tolling together at 8:00 AM, that initial toll counts, and the next one is 3 hours later).
• Let's calculate: Total given time = $3 \text{ hours} = 180 \text{ minutes}$.
• Interval of tolling together = $180 \text{ minutes}$.
• Number of intervals in 3 hours = $\frac{180}{180} = 1$ interval.
• Including the initial toll at 8:00 AM, they toll together at the start and at the end of the 3-hour period, meaning they toll together $1 + 1 = 2$ times (at 8:00 AM and 11:00 AM).
Result: They will toll together $2$ times in the next 3 hours (at the beginning and at the end of the 3-hour duration).
Q3. Case Study: The Circular Track
Ravi and Shikha are running on a circular track. Ravi takes 12 minutes to complete one round, while Shikha takes 18 minutes.
2 MarksSub-question (a): If they start at the same point and time, after how many minutes will they meet at the starting point?
Answer:
1. To find the time after which Ravi and Shikha will meet again at the starting point, we need to find the Least Common Multiple (LCM) of the time taken by each of them to complete one round ($12$ minutes and $18$ minutes).
2. Let us find the prime factorization of each number:
• $12 = 2^2 \times 3$
• $18 = 2 \times 3^2$
3. The LCM is the product of the highest powers of each prime factor involved:
$\text{LCM}(12, 18) = 2^2 \times 3^2$
$= 4 \times 9$
$= 36 \text{ minutes}$
Result: They will meet at the starting point after $36$ minutes.
2 MarksSub-question (b): How many rounds will Ravi have completed when they meet?
Answer:
1. From the previous part, we know that Ravi and Shikha meet at the starting point after $36$ minutes, which is the LCM of their individual round times ($12$ minutes and $18$ minutes).
2. Ravi takes $12$ minutes to complete one round.
3. To find the number of rounds Ravi has completed in $36$ minutes, we divide the total time by the time Ravi takes per round:
$\text{Number of rounds by Ravi} = \frac{36 \text{ minutes}}{12 \text{ minutes/round}} = 3 \text{ rounds}$
Result: Ravi will have completed $3$ rounds when they meet at the starting point.
Q4. Case Study: Packaging Sweets
A sweet seller has 420 kaju barfis and 130 badam barfis. He wants to stack them in such a way that each stack has the same number and they take up the least area of the tray.
2 MarksSub-question (a): What is the number of barfis that can be placed in each stack for this purpose?
Answer:
1. To find the maximum number of barfis that can be placed in each stack so that each stack has the same number and they take up the least area of the tray (meaning maximum number of barfis per stack), we need to find the HCF of $420$ and $130$.
2. Let us find the prime factorization of each number:
• $420 = 2^2 \times 3 \times 5 \times 7$
• $130 = 2 \times 5 \times 13$
3. The HCF is the product of the lowest powers of common prime factors in both numbers:
$\text{HCF}(420, 130) = 2^1 \times 5^1 = 10$
Result: The number of barfis that can be placed in each stack is $10$.
2 MarksSub-question (b): How many stacks will be formed in total?
Answer:
1. From the previous part, we know that the maximum number of barfis per stack is given by the HCF of $420$ and $130$, which is $10$.
2. To find the total number of stacks formed, we divide the number of each type of barfi by the number of barfis per stack ($10$) and sum them up:
• Stacks of kaju barfis = $\frac{420}{10} = 42$ stacks
• Stacks of badam barfis = $\frac{130}{10} = 13$ stacks
3. Calculate the total number of stacks:
$\text{Total stacks} = 42 + 13 = 55$
Result: A total of $55$ stacks will be formed.
Q5. Case Study: The Math Quiz
A quiz competition is organized with two groups, A and B. Group A has 48 members and Group B has 64 members. They need to be divided into groups of equal size for a game.
2 MarksSub-question (a): What is the maximum number of members each group can have?
Answer:
1. To find the maximum number of members each smaller group can have such that both Group A ($48$ members) and Group B ($64$ members) are divided into equal-sized sub-groups, we need to find the HCF of $48$ and $64$.
2. Let us find the prime factorization of each number:
• $48 = 2^4 \times 3$
• $64 = 2^6$
3. The HCF is the product of the lowest powers of common prime factors in both numbers:
$\text{HCF}(48, 64) = 2^4 = 16$
Result: The maximum number of members each group can have is $16$.
2 MarksSub-question (b): If the groups are formed with this size, how many total groups will participate?
Answer:
1. From the previous part, we know that the maximum size of each sub-group is given by the HCF of $48$ and $64$, which is $16$ members per group.
2. To find the total number of sub-groups formed, we divide the number of members in each original group by the size of each sub-group ($16$) and sum them up:
• Sub-groups from Group A = $\frac{48}{16} = 3$ groups
• Sub-groups from Group B = $\frac{64}{16} = 4$ groups
3. Calculate the total number of participating groups:
$\text{Total groups} = 3 + 4 = 7$
Result: A total of $7$ groups will participate.
Q6. Case Study: Irrigation System
A farmer has two rectangular fields of dimensions 120m x 80m and 150m x 100m. He wants to divide these into square plots of the same size.
2 MarksSub-question (a): Find the largest possible side length of the square plot.
Answer:
1. To find the largest possible side length of a square plot that can evenly divide the dimensions of both rectangular fields ($120\text{m} \times 80\text{m}$ and $150\text{m} \times 100\text{m}$), we need to find the HCF of all the individual side lengths: $120$, $80$, $150$, and $100$.
2. Let us find the prime factorization of each dimension:
• $120 = 2^3 \times 3 \times 5$
• $80 = 2^4 \times 5$
• $150 = 2 \times 3 \times 5^2$
• $100 = 2^2 \times 5^2$
3. The HCF is the product of the lowest powers of common prime factors across all these numbers:
$\text{HCF}(120, 80, 150, 100) = 2^1 \times 5^1 = 10$
Result: The largest possible side length of the square plot is $10\text{ m}$.
2 MarksSub-question (b): How many such plots can be made from the first field?
Answer:
1. From the previous part, we know that the largest possible side length of the square plot is given by the HCF of the dimensions, which is $10\text{ m}$.
2. The dimensions of the first field are $120\text{m} \times 80\text{m}$.
3. To find the total number of square plots that can be made from the first field, we divide the area of the first field by the area of one square plot:
• Area of the first field = $120\text{ m} \times 80\text{ m} = 9600\text{ m}^2$
• Area of one square plot = $10\text{ m} \times 10\text{ m} = 100\text{ m}^2$
• Number of plots = $\frac{9600}{100} = 96$
Alternatively, we can find the number of plots along each side and multiply them:
• Plots along the length = $\frac{120}{10} = 12$
• Plots along the breadth = $\frac{80}{10} = 8$
• Total plots = $12 \times 8 = 96$
Result: A total of $96$ square plots can be made from the first field.
Q7. Case Study: Prime Factorization in Cryptography
A security system uses two large prime numbers $p = 17$ and $q = 19$ to generate a key.
2 MarksSub-question (a): Find the value of $p \times q$.
Answer:
1. Given the values of the two prime numbers: $p = 17$ and $q = 19$.
2 MarksSub-question (b): If the system requires a number $N$ such that $N$ is the smallest number divisible by both $p$ and $q$, what is $N$?
Answer:
1. The smallest number that is divisible by both $p$ and $q$ is defined as their Least Common Multiple (LCM).
2. Given that $p = 17$ and $q = 19$ are both prime numbers, their only common factor is $1$.
3. The LCM of two distinct prime numbers is simply their product:
$\text{LCM}(p, q) = p \times q = 17 \times 19 = 323$
Result: The value of $N$ is $323$.
Q8. Case Study: Rational vs Irrational Patterns
Analyze the following sequence of numbers: $0.12, 0.12122, 0.121221222...$
2 MarksSub-question (a) (i): Identify which of these are rational.
Answer:
1. A number is rational if its decimal expansion is either terminating or non-terminating repeating (recurring).
2. Let us analyze the given numbers:
• $0.12$: This is a terminating decimal, which can be written as $\frac{12}{100} = \frac{3}{25}$. Therefore, it is a rational number.
• $0.12122$ and $0.121221222...$: These have non-terminating and non-recurring decimal expansions (as the pattern of zeros/twos keeps changing continuously).
Result: Among the given numbers, only $0.12$ is rational.
2 MarksSub-question (b) (ii): Prove that $0.121221222...$ is irrational.
Answer:
1. By definition, a real number is irrational if and only if its decimal representation is non-terminating and non-recurring (non-repeating).
2. Let the given number be $x = 0.121221222...$
3. Observing the decimal expansion of $x$:
• It goes on infinitely without stopping, meaning it is non-terminating.
• There is no repeating block of digits appearing in a fixed periodic pattern (the number of zeros or twos between the ones keeps increasing: $12$, then $122$, then $1222$, and so on), meaning it is non-recurring.
4. Since a decimal expansion that is both non-terminating and non-recurring cannot be expressed in the form of $\frac{p}{q}$ (where $p$ and $q$ are integers and $q \neq 0$), the number $0.121221222...$ is irrational.
Result: Hence, proved that $0.121221222...$ is an irrational number.
Q9. Case Study: The Library Collection
A library has 336 English books and 240 Science books. They need to be arranged in shelves where each shelf has the same number of books of a single subject.
2 MarksSub-question (a): Find the maximum number of books per shelf.
Answer:
1. To find the maximum number of books that can be placed on each shelf such that each shelf has the same number of books of a single subject, we need to find the HCF of $336$ and $240$.
2. Let us find the prime factorization of each number:
• $336 = 2^4 \times 3 \times 7$
• $240 = 2^4 \times 3 \times 5$
3. The HCF is the product of the lowest powers of common prime factors in both numbers:
$\text{HCF}(336, 240) = 2^4 \times 3^1 = 16 \times 3 = 48$
Result: The maximum number of books per shelf is $48$.
2 MarksSub-question (b): How many shelves are needed for the Science books?
Answer:
1. From the previous part, we know that the maximum number of books per shelf is given by the HCF of $336$ and $240$, which is $48$ books per shelf.
2. The total number of Science books is $240$.
3. To find the number of shelves needed for the Science books, we divide the total number of Science books by the number of books per shelf:
$\text{Number of shelves for Science books} = \frac{240}{48} = 5$ shelves
Result: A total of $5$ shelves are needed for the Science books.
Q10. Case Study: Clockwork Mechanism
Two gears in a machine have 24 and 36 teeth respectively.
2 MarksSub-question (a): After how many rotations of the smaller gear will they return to their original relative position?
Answer:
1. To find when the two gears will return to their original relative position, we need to find the Least Common Multiple (LCM) of the number of teeth on both gears ($24$ and $36$).
2. Let us find the prime factorization of each number:
• $24 = 2^3 \times 3$
• $36 = 2^2 \times 3^2$
3. The LCM is the product of the highest powers of each prime factor involved:
$\text{LCM}(24, 36) = 2^3 \times 3^2 = 8 \times 9 = 72 \text{ teeth}$
4. The smaller gear has $24$ teeth. To find the number of rotations the smaller gear must make for a total of $72$ teeth to pass, we divide the LCM by the number of teeth on the smaller gear:
$\text{Rotations of smaller gear} = \frac{72}{24} = 3 \text{ rotations}$
Result: The smaller gear will complete $3$ rotations before they return to their original relative position.
2 MarksSub-question (b): What is the LCM of the number of teeth?
Answer:
1. We need to find the Least Common Multiple (LCM) of the number of teeth on the two gears, which are $24$ and $36$.
2. Let us use the prime factorization of each number:
• $24 = 2^3 \times 3$
• $36 = 2^2 \times 3^2$
3. The LCM is the product of the highest powers of each prime factor involved:
$\text{LCM}(24, 36) = 2^3 \times 3^2 = 8 \times 9 = 72$
Result: The LCM of the number of teeth is $72$.
Q11. Case Study: The School Assembly
To prepare for the annual day, a school decides to arrange students in rows. There are 72 students from Grade 9 and 96 students from Grade 10. They need to be arranged in rows such that each row has the same number of students and all students in a row belong to the same grade.
1 MarkSub-question (a): What is the maximum number of students that can be in each row?
Answer:
1. To find the maximum number of students that can be in each row so that each row has the same number of students from a single grade, we need to find the HCF of $72$ and $96$.
2. Let us find the prime factorization of each number:
• $72 = 2^3 \times 3^2$
• $96 = 2^5 \times 3$
3. The HCF is the product of the lowest powers of common prime factors in both numbers:
$\text{HCF}(72, 96) = 2^3 \times 3^1 = 8 \times 3 = 24$
Result: The maximum number of students that can be in each row is $24$.
1 MarkSub-question (b): How many rows will be formed for Grade 9 students?
Answer:
1. From the previous part, we know that the maximum number of students per row is given by the HCF of $72$ and $96$, which is $24$ students per row.
2. The total number of Grade 9 students is $72$.
3. To find the number of rows formed for Grade 9 students, we divide the total number of Grade 9 students by the number of students per row:
$\text{Number of rows for Grade 9} = \frac{72}{24} = 3$ rows
Result: A total of $3$ rows will be formed for the Grade 9 students.
1 MarkSub-question (c): How many rows will be formed for Grade 10 students?
Answer:
1. From the first part, we know that the maximum number of students per row is given by the HCF of $72$ and $96$, which is $24$ students per row.
2. The total number of Grade 10 students is $96$.
3. To find the number of rows formed for Grade 10 students, we divide the total number of Grade 10 students by the number of students per row:
$\text{Number of rows for Grade 10} = \frac{96}{24} = 4$ rows
Result: A total of $4$ rows will be formed for the Grade 10 students.
1 MarkSub-question (d): If the total number of students was 168, and they were arranged in rows of 12, how many rows would there be in total?
Answer:
1. Given the total number of students = $168$.
2. Given the number of students per row = $12$.
3. To find the total number of rows, we divide the total number of students by the number of students in each row:
$\text{Total rows} = \frac{168}{12} = 14$ rows
Result: A total of $14$ rows would be formed.
Q12. Case Study: The Running Track
Two runners, Ravi and Shikha, start running on a circular track from the same point at the same time. Ravi completes one lap in 18 minutes, while Shikha completes one lap in 24 minutes.
1 MarkSub-question (a): After how many minutes will they meet again at the starting point?
Answer:
1. To find the time when Ravi and Shikha will meet again at the starting point, we need to find the Least Common Multiple (LCM) of the time taken by each to complete one lap ($18$ minutes and $24$ minutes).
2. Let us find the prime factorization of each number:
• $18 = 2 \times 3^2$
• $24 = 2^3 \times 3$
3. The LCM is the product of the highest powers of each prime factor involved:
$\text{LCM}(18, 24) = 2^3 \times 3^2 = 8 \times 9 = 72$
Result: They will meet again at the starting point after $72$ minutes.
1 MarkSub-question (b): How many laps would Ravi have completed when they first meet at the start?
Answer:
1. From the previous part, we know that they meet again at the starting point after $72$ minutes.
2. Ravi takes $18$ minutes to complete one lap.
3. To find the number of laps completed by Ravi when they first meet at the start, we divide the total time by the time Ravi takes per lap:
$\text{Number of laps by Ravi} = \frac{72}{18} = 4$ laps
Result: Ravi would have completed $4$ laps when they first meet at the start.
1 MarkSub-question (c): How many laps would Shikha have completed when they first meet at the start?
Answer:
1. From the first part, we know that they meet again at the starting point after $72$ minutes.
2. Shikha takes $24$ minutes to complete one lap.
3. To find the number of laps completed by Shikha when they first meet at the start, we divide the total time by the time Shikha takes per lap:
$\text{Number of laps by Shikha} = \frac{72}{24} = 3$ laps
Result: Shikha would have completed $3$ laps when they first meet at the start.
1 MarkSub-question (d): If a third runner, Amit, joins and completes a lap in 12 minutes, when will all three meet at the starting point?
Answer:
1. To find when all three runners (Ravi, Shikha, and Amit) will meet again at the starting point, we need to find the Least Common Multiple (LCM) of the time taken by each to complete one lap ($18$ minutes, $24$ minutes, and $12$ minutes).
2. Let us find the prime factorization of each number:
• $18 = 2 \times 3^2$
• $24 = 2^3 \times 3$
• $12 = 2^2 \times 3$
3. The LCM is the product of the highest powers of each prime factor involved:
$\text{LCM}(18, 24, 12) = 2^3 \times 3^2 = 8 \times 9 = 72$
Result: All three runners will meet again at the starting point after $72$ minutes.
Q13. Case Study: The Prime Factorization
A student is given a number $N = 2^3 \times 3^2 \times 5^1$ and another number $M = 2^2 \times 3^3 \times 7^1$.
1 MarkSub-question (a): Find the HCF of $N$ and $M$.
Answer:
1. Given the numbers in their prime factorized form:
• $N = 2^3 \times 3^2 \times 5^1$
• $M = 2^2 \times 3^3 \times 7^1$
2. The HCF is the product of the lowest powers of common prime factors in both numbers:
• Common prime factors are $2$ and $3$.
• Lowest power of $2$ is $2^2$.
• Lowest power of $3$ is $3^2$.
3. Calculating the HCF:
$\text{HCF}(N, M) = 2^2 \times 3^2 = 4 \times 9 = 36$
Result: The HCF of $N$ and $M$ is $36$.
1 MarkSub-question (b): Find the LCM of $N$ and $M$.
Answer:
1. Given the numbers in their prime factorized form:
• $N = 2^3 \times 3^2 \times 5^1$
• $M = 2^2 \times 3^3 \times 7^1$
2. The LCM is the product of the highest powers of all prime factors involved in both numbers:
• Highest power of $2$ is $2^3$.
• Highest power of $3$ is $3^3$.
• Highest power of $5$ is $5^1$.
• Highest power of $7$ is $7^1$.
3. Calculating the LCM:
$\text{LCM}(N, M) = 2^3 \times 3^3 \times 5^1 \times 7^1 = 8 \times 27 \times 5 \times 7$
$\text{LCM}(N, M) = 216 \times 35 = 7560$
Result: The LCM of $N$ and $M$ is $7560$.
1 MarkSub-question (c): Verify if $\text{HCF} \times \text{LCM} = N \times M$.
Answer:
1. From the previous parts, we have:
• $\text{HCF}(N, M) = 36$
• $\text{LCM}(N, M) = 7560$
2. Calculate the Left Hand Side ($\text{HCF} \times \text{LCM}$):
$36 \times 7560 = 272,160$
3. Calculate the actual values of $N$ and $M$:
• $N = 8 \times 9 \times 5 = 360$
• $M = 4 \times 27 \times 7 = 756$
4. Calculate the Right Hand Side ($N \times M$):
$360 \times 756 = 272,160$
5. Since $\text{LHS} = \text{RHS} = 272,160$, the relation is verified.
Result: Verified that $\text{HCF} \times \text{LCM} = N \times M$.
1 MarkSub-question (d): Is $N$ a composite number? Explain why.
Answer:
1. Yes, $N$ is a composite number.
2. A composite number is a positive integer that has more than two distinct positive divisors (i.e., it can be factored into prime numbers other than just $1$ and itself).
3. The number $N$ is given in its prime factorization form as $N = 2^3 \times 3^2 \times 5^1$, which means it has factors other than $1$ and $N$ (such as $2$, $3$, and $5$). Therefore, it is a composite number.
Result: Yes, $N$ is a composite number because it has more than two prime factors and can be expressed as a product of prime numbers.
Q14. Case Study: The Rationals and Irrationals
A teacher asks students to classify numbers based on their decimal expansion and properties.
1 MarkSub-question (a): Show that $5 - \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.
Answer:
1. Let us assume to the contrary that $5 - \sqrt{3}$ is a rational number.
2. Then, there exist coprime integers $a$ and $b$ (where $b \neq 0$) such that:
$5 - \sqrt{3} = \frac{a}{b}$
3. Rearranging the equation to isolate $\sqrt{3}$:
$\sqrt{3} = 5 - \frac{a}{b}$
$\sqrt{3} = \frac{5b - a}{b}$
4. Since $a$ and $b$ are integers, $\frac{5b - a}{b}$ is a rational number. This implies that $\sqrt{3}$ is a rational number.
5. However, this contradicts the given fact that $\sqrt{3}$ is irrational. This contradiction arises from our incorrect assumption that $5 - \sqrt{3}$ is rational.
Result: Therefore, $5 - \sqrt{3}$ must be an irrational number.
1 MarkSub-question (b): Without performing long division, determine the decimal expansion of $\frac{13}{125}$.
Answer:
1. Given the rational number $\frac{13}{125}$.
2. Find the prime factorization of the denominator ($125$):
$125 = 5^3$
3. To convert it to a decimal without long division, we can make the denominator a power of $10$ by multiplying both the numerator and denominator by $2^3$:
$\frac{13}{125} = \frac{13}{5^3} = \frac{13 \times 2^3}{5^3 \times 2^3} = \frac{13 \times 8}{(5 \times 2)^3}$
4. Simplifying the numerator and denominator:
$\frac{104}{10^3} = \frac{104}{1000} = 0.104$
Result: The decimal expansion of $\frac{13}{125}$ is $0.104$.
1 MarkSub-question (c): Is $0.121221222...$ a rational or irrational number? Give a reason.
Answer:
1. The given number has a decimal expansion of $0.121221222...$
2. Analyzing the pattern, the decimal expansion is **non-terminating** and **non-repeating** (or non-recurring), as the number of zeros or the pattern of digits between the ones changes continuously (e.g., one $2$, then two $2$s, then three $2$s).
3. By definition, any number with a non-terminating and non-recurring decimal expansion is an irrational number.
Result: It is an irrational number because its decimal expansion is neither terminating nor repeating.
1 MarkSub-question (d): Find the smallest number which is divisible by both 306 and 657.
Answer:
1. The smallest number which is divisible by both given numbers is their Least Common Multiple (LCM).
2. We are given two numbers: $a = 306$ and $b = 657$. We also know their HCF (commonly given or calculated as $\text{HCF}(306, 657) = 9$).
3. Using the relationship between HCF, LCM, and the two numbers:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
Result: The smallest number divisible by both $306$ and $657$ is $22,338$.
Q15. Case Study: Gift Distribution
A shopkeeper has 120 liters of oil and 180 liters of ghee. He wants to fill them into tins of equal capacity so that each tin is completely filled.
1 MarkSub-question (a): What is the maximum capacity of each tin?
Answer:
1. To find the maximum capacity of each tin that can measure both quantities of oil ($120$ liters) and ghee ($180$ liters) completely, we need to find the Highest Common Factor (HCF) of $120$ and $180$.
2. Let us find the prime factorization of each number:
• $120 = 2^3 \times 3 \times 5$
• $180 = 2^2 \times 3^2 \times 5$
3. The HCF is the product of the lowest powers of common prime factors involved:
$\text{HCF}(120, 180) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$
Result: The maximum capacity of each tin is $60$ liters.
1 MarkSub-question (b): How many tins of oil will be required?
Answer:
1. From the previous part, the maximum capacity of each tin is $60$ liters.
2. The total quantity of oil is $120$ liters.
3. To find the number of tins required for oil, we divide the total quantity of oil by the capacity of each tin:
$\text{Number of oil tins} = \frac{120}{60} = 2$
Result: $2$ tins of oil will be required.
1 MarkSub-question (c): How many tins of ghee will be required?
Answer:
1. From the first part, the maximum capacity of each tin is $60$ liters.
2. The total quantity of ghee is $180$ liters.
3. To find the number of tins required for ghee, we divide the total quantity of ghee by the capacity of each tin:
$\text{Number of ghee tins} = \frac{180}{60} = 3$
Result: $3$ tins of ghee will be required.
1 MarkSub-question (d): If the shopkeeper decides to use 10-liter tins, how many total tins will he need for both oil and ghee?
Answer:
1. The total quantity of oil is $120$ liters, and the total quantity of ghee is $180$ liters.
2. The capacity of each tin being used is $10$ liters.
3. Calculate the number of tins required for oil:
$\text{Number of oil tins} = \frac{120}{10} = 12$
4. Calculate the number of tins required for ghee:
$\text{Number of ghee tins} = \frac{180}{10} = 18$
5. Calculate the total number of tins required:
$\text{Total tins} = 12 + 18 = 30$
Result: The shopkeeper will need a total of $30$ tins for both oil and ghee.
Q16. Case Study: The Marathon Route
A marathon organizer wants to place markers at equal intervals along a 400m track and a 600m path.
1 MarkSub-question (a): What is the greatest distance (in meters) between markers so that they are placed at equal intervals on both paths?
Answer:
1. To find the greatest distance between markers that can evenly divide both paths ($400$ meters and $600$ meters), we need to find the Highest Common Factor (HCF) of $400$ and $600$.
2. Let us find the prime factorization of each number:
• $400 = 2^4 \times 5^2$
• $600 = 2^3 \times 3 \times 5^2$
3. The HCF is the product of the lowest powers of common prime factors involved:
$\text{HCF}(400, 600) = 2^3 \times 5^2 = 8 \times 25 = 200$
Result: The greatest distance between markers is $200$ meters.
1 MarkSub-question (b): How many markers are needed for the 400m track?
Answer:
1. From the previous part, the greatest distance between markers is $200$ meters.
2. The track is a circular or closed track of length $400$ meters (or alternatively, if considered as segments, but for a circular track, the number of intervals equals the number of markers).
3. To find the number of markers required for the $400$m track, we divide the total length of the track by the distance between each marker:
$\text{Number of markers} = \frac{400}{200} = 2$
Result: $2$ markers are needed for the $400$m track.
1 MarkSub-question (c): How many markers are needed for the 600m path?
Answer:
1. From the first part, the greatest distance between markers is $200$ meters.
2. The total length of the path is $600$ meters.
3. Assuming a linear path where markers are placed at the start and end (or depending on the path type, dividing the length by the interval gives the number of intervals/markers):
$\text{Number of markers} = \frac{600}{200} = 3$
Result: $3$ markers are needed for the $600$m path.
1 MarkSub-question (d): If the organizer decides to place markers every 50 meters, will all markers align at the same positions?
Answer:
1. Let us check if $50$ meters is a common divisor (or factor) of both track lengths ($400$ meters and $600$ meters).
2. For the $400$m track:
$\frac{400}{50} = 8$ (markers can be placed completely without remainder)
3. For the $600$m path:
$\frac{600}{50} = 12$ (markers can be placed completely without remainder)
4. Since $50$ divides both $400$ and $600$ evenly, every marker placed at $50$-meter intervals will coincide with the multiples of $50$ on both tracks/paths.
Result: Yes, all markers will align at the same positions because $50$ is a common factor of both $400$ and $600$.
Q17. Case Study: Prime Number Patterns
Given a number $X = p^2q^3$ and $Y = p^3q$, where $p$ and $q$ are prime numbers.
1 MarkSub-question (a): What is the HCF of $X$ and $Y$ in terms of $p$ and $q$?
Answer:
1. Given the numbers in prime factorization form: $X = p^2q^3$ and $Y = p^3q$.
2. The Highest Common Factor (HCF) is the product of the smallest powers of each common prime factor involved in the numbers ($p$ and $q$).
3. For $p$, the smallest power between $p^2$ and $p^3$ is $p^2$.
4. For $q$, the smallest power between $q^3$ and $q^1$ is $q^1$ (or $q$).
5. Combining these, $\text{HCF}(X, Y) = p^2q$.
Result: The HCF of $X$ and $Y$ is $p^2q$.
1 MarkSub-question (b): What is the LCM of $X$ and $Y$ in terms of $p$ and $q$?
Answer:
1. Given the numbers in prime factorization form: $X = p^2q^3$ and $Y = p^3q$.
2. The Least Common Multiple (LCM) is the product of the greatest powers of each prime factor involved in the numbers ($p$ and $q$).
3. For $p$, the greatest power between $p^2$ and $p^3$ is $p^3$.
4. For $q$, the greatest power between $q^3$ and $q^1$ is $q^3$.
5. Combining these, $\text{LCM}(X, Y) = p^3q^3$.
Result: The LCM of $X$ and $Y$ is $p^3q^3$.
1 MarkSub-question (c): If $p=2$ and $q=3$, what is the numerical value of $X$?
Answer:
1. Given the expression for $X$: $X = p^2q^3$
2. Substitute the given values $p = 2$ and $q = 3$ into the expression:
$X = (2)^2 \times (3)^3$
3. Calculate the powers:
• $2^2 = 4$
• $3^3 = 27$
4. Multiply the results:
$X = 4 \times 27 = 108$
Result: The numerical value of $X$ is $108$.
1 MarkSub-question (d): For these values, does the product $HCF \times LCM$ equal the product $X \times Y$?
Answer:
1. We know the standard property for any two positive integers: $\text{HCF}(X, Y) \times \text{LCM}(X, Y) = X \times Y$.
2. Let us verify this in terms of $p$ and $q$:
• $\text{HCF} \times \text{LCM} = (p^2q) \times (p^3q^3) = p^{2+3}q^{1+3} = p^5q^4$
• $X \times Y = (p^2q^3) \times (p^3q) = p^{2+3}q^{3+1} = p^5q^4$
3. Since both products result in the exact same expression ($p^5q^4$), they are equal.
Result: Yes, the product of $\text{HCF} \times \text{LCM}$ always equals the product $X \times Y$.
Q18. Case Study: Decimal Expansions
Students are analyzing the fraction $\frac{17}{80}$.
1 MarkSub-question (a): Is the decimal expansion terminating or non-terminating?
Answer:
1. Consider the given fraction $\frac{17}{80}$, which is in its simplest form.
2. Find the prime factorization of the denominator ($80$):
$80 = 2^4 \times 5^1$
3. Since the prime factorization of the denominator is of the form $2^n 5^m$ (where $n = 4$ and $m = 1$), the rational number has a terminating decimal expansion.
Result: The decimal expansion is terminating.
1 MarkSub-question (b): Write the prime factorization of the denominator.
Answer:
1. The denominator of the given fraction $\frac{17}{80}$ is $80$.
2. Let us find the prime factorization of $80$ by successive division:
• $80 \div 2 = 40$
• $40 \div 2 = 20$
• $20 \div 2 = 10$
• $10 \div 2 = 5$
• $5 \div 5 = 1$
3. Combining all the prime factors:
$80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^4 \times 5^1$
Result: The prime factorization of the denominator is $2^4 \times 5^1$.
1 MarkSub-question (c): After how many decimal places will the expansion terminate?
Answer:
1. From the prime factorization of the denominator $80 = 2^4 \times 5^1$, the highest exponent between the prime factors $2$ and $5$ is $4$.
2. The number of decimal places after which the decimal expansion of a rational number terminates is equal to the maximum of the exponents of $2$ and $5$ in its denominator's prime factorization.
3. Here, the maximum exponent is $\max(4, 1) = 4$.
Result: The decimal expansion will terminate after $4$ decimal places.
1 MarkSub-question (d): Perform the division to find the exact decimal value.
Answer:
1. We need to find the decimal value of $\frac{17}{80}$.
2. We can make the denominator a power of $10$ by multiplying the numerator and the denominator by $5^3$ (since $80 = 2^4 \times 5^1$, we need three more $5$s to make $2^4 \times 5^4 = (2 \times 5)^4 = 10^4$):
$\frac{17}{2^4 \times 5^1} = \frac{17 \times 5^3}{(2^4 \times 5^1) \times 5^3} = \frac{17 \times 125}{2^4 \times 5^4}$
3. Calculate the numerator:
$17 \times 125 = 2125$
5. Divide the numerator by the denominator:
$\frac{2125}{10000} = 0.2125$
Result: The exact decimal value is $0.2125$.
Q19. Case Study: Proofs and Logic
Consider the statement: "The product of a non-zero rational number and an irrational number is always irrational."
1 MarkSub-question (a): Is the statement true or false?
Answer:
1. Let a non-zero rational number be $r$ (where $r \neq 0$) and an irrational number be $s$.
2. According to number theory, the product of any non-zero rational number and an irrational number is always an irrational number.
Result: The statement is true.
1 MarkSub-question (b): Give an example to support your answer.
Answer:
1. Let us choose a non-zero rational number, such as $3$.
2. Let us choose an irrational number, such as $\sqrt{2}$.
3. Find their product:
$3 \times \sqrt{2} = 3\sqrt{2}$
4. Since $3\sqrt{2}$ cannot be expressed as a ratio of two integers and has a non-terminating, non-recurring decimal expansion, it is an irrational number.
Result: An example is $3$ (rational) $\times \sqrt{2}$ (irrational) $= 3\sqrt{2}$ (irrational).
1 MarkSub-question (c): Is $7 \times \sqrt{5}$ rational or irrational?
Answer:
1. Here, $7$ is a non-zero rational number and $\sqrt{5}$ is an irrational number.
2. By the property that the product of a non-zero rational number and an irrational number is always irrational, their product $7\sqrt{5}$ must be irrational.
Result: $7 \times \sqrt{5}$ is irrational.
1 MarkSub-question (d): Is $\sqrt{2} + \sqrt{3}$ rational or irrational?
Answer:
1. Let us assume to the contrary that $\sqrt{2} + \sqrt{3}$ is a rational number, say $r$.
3. Rearranging the terms to isolate the radical:
$2\sqrt{6} = r^2 - 5$
$\sqrt{6} = \frac{r^2 - 5}{2}$
4. Since $r$ is rational, $\frac{r^2 - 5}{2}$ must be rational. However, $\sqrt{6}$ is a known irrational number, which leads to a contradiction.
Result: $\sqrt{2} + \sqrt{3}$ is irrational.
Q20. Case Study: Clockwork
Three electronic bells toll at intervals of 9, 12, and 15 minutes respectively. They all toll together at 10:00 AM.
1 MarkSub-question (a): Find the LCM of 9, 12, and 15.
Answer:
1. Find the prime factorization of each number:
• $9 = 3^2$
• $12 = 2^2 \times 3$
• $15 = 3 \times 5$
2. The Least Common Multiple (LCM) is the product of the highest powers of each prime factor involved ($2$, $3$, and $5$):
• Highest power of $2$ is $2^2$
• Highest power of $3$ is $3^2$
• Highest power of $5$ is $5^1$
1 MarkSub-question (b): After how many minutes will the bells toll together again?
Answer:
1. The bells will toll together again after a time interval equal to the Least Common Multiple (LCM) of their individual tolling intervals ($9$, $12$, and $15$ minutes).
2. From the previous calculation, $\text{LCM}(9, 12, 15) = 180$ minutes.
Result: The bells will toll together again after $180$ minutes (or $3$ hours).
1 MarkSub-question (c): At what time will they next toll together?
Answer:
1. The bells toll together next after $180$ minutes, which is equivalent to $3$ hours ($180 \div 60$).
2. They last tolled together at $10:00\text{ AM}$.
3. Adding $3$ hours to $10:00\text{ AM}$ gives $1:00\text{ PM}$.
Result: They will next toll together at $1:00\text{ PM}$.
1 MarkSub-question (d): How many times will they toll together between 10:00 AM and 1:00 PM?
Answer:
1. The time period from $10:00\text{ AM}$ to $1:00\text{ PM}$ is $3$ hours ($180$ minutes).
2. The bells toll together every $180$ minutes.
3. Counting the occurrences:
• At the start ($10:00\text{ AM}$), they toll together.
• After $180$ minutes ($1:00\text{ PM}$), they toll together again.
4. Therefore, they toll together $1$ time strictly *between* $10:00\text{ AM}$ and $1:00\text{ PM}$ (or $2$ times if including the starting or ending boundary, depending on context; usually, in these word problems, it refers to the exact next occurrence at the end of the interval, which is $1$). Let's clarify: counting the next event at $1:00\text{ PM}$, it happens $1$ time after the initial stroke.
Result: They will toll together $1$ time within the duration (at $1:00\text{ PM}$).
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
