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Worksheet: Real Numbers

#### Multiple Choice Questions:

1. Question: $ 1.234\overline{8} $ is:
- (a) An integer
- (b) A rational number
- (c) An irrational number
- (d) A natural number

Solution:
- The number $ 1.234\overline{8} $ has a repeating decimal ($ \overline{8} $), which means it can be expressed as a fraction. Any number with a repeating decimal is a rational number.
- Therefore, the correct answer is: (b) A rational number.

2. Question: $ 3.24636363\ldots $ is:
- (a) an integer
- (b) An irrational number
- (c) A rational number
- (d) Not a real number

Solution:
- The number $ 3.24636363\ldots $ has a repeating decimal pattern ($ 63 $), which means it can be expressed as a fraction. Any number with a repeating decimal is a rational number.
- Therefore, the correct answer is: (c) A rational number.

3. Question: Which of the following numbers has the terminating decimal representation?
- (a) $ \frac{1}{7} $
- (b) $ \frac{1}{3} $
- (c) $ \frac{3}{5} $
- (d) $ \frac{17}{3} $

Solution:
- A fraction has a terminating decimal if and only if its denominator (in lowest terms) has no prime factors other than 2 or 5.
- (a) $ \frac{1}{7} $: The denominator is 7, which is not a power of 2 or 5. Hence, it does not terminate.
- (b) $ \frac{1}{3} $: The denominator is 3, which is not a power of 2 or 5. Hence, it does not terminate.
- (c) $ \frac{3}{5} $: The denominator is 5, which is a power of 5. Hence, it terminates.
- (d) $ \frac{17}{3} $: The denominator is 3, which is not a power of 2 or 5. Hence, it does not terminate.
- Therefore, the correct answer is: (c) $ \frac{3}{5} $.

4. Question: If the H.C.F. of 408 and 1032 is expressible in the form $ 1032m - 408 \times 5 $, then $ m $ will be:
- (a) $-2$
- (b) $2$
- (c) $4$
- (d) $-4$

Solution:
- First, find the H.C.F. of 408 and 1032 using the Euclidean algorithm:
- $ 1032 = 408 \times 2 + 216 $
- $ 408 = 216 \times 1 + 192 $
- $ 216 = 192 \times 1 + 24 $
- $ 192 = 24 \times 8 + 0 $
- The H.C.F. is 24.
- According to the problem, the H.C.F. (24) is expressible in the form $ 1032m - 408 \times 5 $.
- Substitute the values: $ 24 = 1032m - 408 \times 5 $.
- Simplify: $ 24 = 1032m - 2040 $.
- Add 2040 to both sides: $ 2064 = 1032m $.
- Divide by 1032: $ m = 2 $.
- Therefore, the correct answer is: (b) $2$.

5. Question: A rectangular field is $ 150 \, \text{m} \times 60 \, \text{m} $. Two cyclists Anuj and Aman start together and can cycle at speeds of $ 21 \, \text{m/min} $ and $ 28 \, \text{m/min} $, respectively. They cycle along the rectangular track, around the field from the same point and at the same moment. After how many minutes will they meet again at the starting point?
- (a) $ 4 \, \text{mins} $
- (b) $ 5 \, \text{mins} $
- (c) $ 120 \, \text{mins} $
- (d) $ 60 \, \text{mins} $

Solution:
- First, calculate the perimeter of the rectangular field:
- Perimeter = $ 2 \times (150 + 60) = 2 \times 210 = 420 \, \text{m} $.
- The time taken by each cyclist to complete one round of the field:
- Time for Anuj = $ \frac{420}{21} = 20 \, \text{mins} $.
- Time for Aman = $ \frac{420}{28} = 15 \, \text{mins} $.
- To find when they will meet again at the starting point, we need the least common multiple (LCM) of their times:
- LCM of 20 and 15:
- Prime factorization: $ 20 = 2^2 \times 5 $, $ 15 = 3 \times 5 $.
- LCM = $ 2^2 \times 3 \times 5 = 60 $.
- Therefore, they will meet again after $ 60 \, \text{mins} $.
- The correct answer is: (d) $ 60 \, \text{mins} $.

#### Fill in the blanks:

6. Question: The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, the other number is _______.

Solution:
- Use the relationship between HCF and LCM of two numbers:
- $ \text{HCF} \times \text{LCM} = \text{Product of the two numbers} $.
- Let the two numbers be $ a $ and $ b $. Given: HCF = 145, LCM = 2175, and $ a = 725 $.
- Substitute the values: $ 145 \times 2175 = 725 \times b $.
- Solve for $ b $: $ b = \frac{145 \times 2175}{725} $.
- Simplify: $ b = \frac{145 \times 2175}{725} = \frac{145 \times 2175}{725} = 435 $.
- Therefore, the other number is: 435.

7. Question: The L.C.M and H.C.F. of 6 and 20 are ________ and ________ respectively.

Solution:
- Find the HCF of 6 and 20:
- Prime factorization: $ 6 = 2 \times 3 $, $ 20 = 2^2 \times 5 $.
- HCF = $ 2 $.
- Find the LCM of 6 and 20:
- LCM = $ 2^2 \times 3 \times 5 = 60 $.
- Therefore, the L.C.M. and H.C.F. are: 60 and 2, respectively.

Final Answers:

1. (b) A rational number
2. (c) A rational number
3. (c) $ \frac{3}{5} $
4. (b) $2$
5. (d) $ 60 \, \text{mins} $
6. 435
7. 60 and 2

$$
\boxed{(b), (c), (c), (b), (d), 435, 60 \text{ and } 2}
$$