Case Based Questions: Real Numbers | Mathematics (Maths) Class 10 PDF Download

Q1: Read the source below and answer the questions that follow:

i. Co-prime numbers are those numbers which do not have any common factor other than 1. Is this statement true?
ii. Find the sum of the powers of all different prime factors of the numbers 10, 16 and 20. 
iii. If all of them started together, then what time will they start preparing a new card together? 
iv. What is the common time to make one card? 

Ans:
i. True 
ii. By prime factorisation, 10=2¹× 5¹ 16=2x2x2x2=24 20 2x2x5=22x5¹ .. Required sum = sum of the power of 2 + sum of the power of 5 = (1 + 4 + 2) + (1+1)=7+2=9 
iii. The required number of minutes after which they start preparing a new card together is the LCM of 10, 16 and 20 min. Now, 10=2x5 16=2x2x2x2 20=2x2x5=22x5 .. LCM (10, 16, 20) = 24 x 51 = 16 x 5 = 80 min So, they will start preparing a new card together after 80 min i.e., 1 h 20 min.
iv. The common time to make one card = HCF of (10, 16, 20) = 2 min 

Q2: Read the source below and answer the questions that follow:

Ans:
i. True 
ii. By prime factorisation, 240=2x2x2x2x3x5=24 x 31 x 5¹ 90=2×3×3×5=2¹×32x5¹ 120=2×2×3×2×5=23×3¹×5¹ :. Required sum = sum of the power of 2+ sum of the power of 3+ sum of the power of 5=1+1+1=3. 
iii. Minimum required distance to reach the juice shop = LCM (240, 90, 120) 240 = 2x2x2x2x3x5=24x3x5 90 = 2×3×3×5=2x32x5 and 120 = 2×2×2×3×5=23x3x5 Now, LCM=24x32x5=16x9 × 5=720 Hence, required minimum distance is 720 cm. 
iv. The number of common steps covered by all of them HCF (240, 90, 120) = 2 x 3 x 5 =30

Q3: Read the source below and answer the questions that follow:

Teaching Mathematics through activities is a powerful approach that enhances students' understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announced the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to the second student. The second student also multiplied it by a prime number and passed it to the third student. In this way by multiplying by a prime number, the last student got 173250.
Now, Mukta asked some questions as given below to the students:

i. What is the least prime number used by students?
ii. How many students are in the class?
iii. What is the highest prime number used by students?
iv. Which prime number has been used maximum times?

Ans:
i. 

So least prime no. used by students = 3(because 2 is announced by the teacher, so the least number used by the students is 3)
ii. As the last student got  173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11
there are 7 factors other than 2, which is announced by teacher. So, Number of student = 7
iii. Highest prime number used by student = 11
iv. Prime number 5 is used maximum times i.e., 3 times.

Q4: Read the source below and answer the questions that follow:

Ms. Asha planned a prime factorization activity for her Class 10 students. She announced the number 3 in her class and asked the first student to multiply it by a prime number and pass it to the next student. Each student multiplied the result by a prime number and passed it further. The last student ended up with the number 231000.
Now, Ms. Asha asked the following questions:

i. What is the least prime number used by students?
ii. How many students are in the class?
iii. What is the highest prime number used by students?
iv. Which prime number has been used maximum times?

Ans: 
i. The least prime number used is 2.
ii. The total number of students is 9 (since there are 9 prime factors).
iii. The highest prime number used is 11.
iv. The prime number used the maximum times is 2 and 5 (each used 3 times).