Rational Numbers worksheet with multiple choice questions on mathematical operations and concepts.

Let's solve each of the multiple-choice questions step by step and explain the solutions clearly.

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Question (45):

If
$$
\frac{3}{7} + x + \left(\frac{-8}{21}\right) + \frac{5}{22} = \frac{2}{462},
$$
find $x$.

#### Step 1: Simplify the known terms.
We want to isolate $x$, so:
$$
x = \frac{2}{462} - \frac{3}{7} + \frac{8}{21} - \frac{5}{22}
$$

Let’s compute this carefully. First, find a common denominator.

Note: $462 = 2 \times 3 \times 7 \times 11$, which is divisible by 7, 21, and 22.

So, let’s convert all fractions to have denominator 462.

- $\frac{3}{7} = \frac{3 \times 66}{7 \times 66} = \frac{198}{462}$
- $\frac{8}{21} = \frac{8 \times 22}{21 \times 22} = \frac{176}{462}$
- $\frac{5}{22} = \frac{5 \times 21}{22 \times 21} = \frac{105}{462}$

Now plug into the equation:
$$
x = \frac{2}{462} - \frac{198}{462} + \frac{176}{462} - \frac{105}{462}
$$
$$
x = \frac{2 - 198 + 176 - 105}{462} = \frac{(2 + 176) - (198 + 105)}{462} = \frac{178 - 303}{462} = \frac{-125}{462}
$$

✔ So, $x = -\frac{125}{462}$

Answer: (C) $-\frac{125}{462}$

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Question (46):

The product of two rational numbers is $\frac{-28}{81}$. One number is $\frac{14}{27}$. Find the other.

Let the unknown number be $x$. Then:
$$
\frac{14}{27} \cdot x = \frac{-28}{81}
$$
Solve for $x$:
$$
x = \frac{-28}{81} \div \frac{14}{27} = \frac{-28}{81} \times \frac{27}{14}
$$

Simplify:
- $28$ and $14$: $28/14 = 2$
- $27$ and $81$: $27/81 = 1/3$

So:
$$
x = \frac{-2}{3} \times \frac{1}{1} = -\frac{2}{3}
$$

Wait — let's do it step by step:

$$
x = \frac{-28 \times 27}{81 \times 14} = \frac{-756}{1134}
$$

Simplify:
Divide numerator and denominator by 126:
- $756 ÷ 126 = 6$
- $1134 ÷ 126 = 9$

Wait, better to simplify earlier.

From:
$$
\frac{-28}{81} \times \frac{27}{14} = \frac{-28 \times 27}{81 \times 14}
$$

Break down:
- $28 = 4 \times 7$
- $14 = 2 \times 7$
- $27 = 3^3$
- $81 = 3^4$

So:
$$
= \frac{- (4 \times 7) \times 27}{81 \times (2 \times 7)} = \frac{-4 \times 27}{81 \times 2} = \frac{-108}{162} = -\frac{2}{3}
$$

✔ So, $x = -\frac{2}{3}$

Answer: (B) $-\frac{2}{3}$

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Question (47):

Consider:
- Statement 1: $\sqrt{(a-b)^2} = a - b$
- Statement 2: $\sqrt{(a-b)^2} = b - a$

Which is correct?

Recall: $\sqrt{x^2} = |x|$, so:
$$
\sqrt{(a-b)^2} = |a - b|
$$

This equals:
- $a - b$ if $a \geq b$
- $b - a$ if $b > a$

So neither statement is always true.

For example:
- If $a = 5, b = 3$: $|5 - 3| = 2 = 5 - 3$ → Statement 1 is true
- But if $a = 3, b = 5$: $|3 - 5| = 2 = 5 - 3 = b - a$ → Statement 2 is true in this case

But since both are not always true, and the question asks which is/are correct, meaning always true, then neither is universally correct.

So the correct answer is:

Answer: (D) None of these

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Question (48):

Cost of $7\frac{2}{3}$ metres of rope is Rs. $12\frac{3}{4}$. Find cost per metre.

Convert mixed numbers:
- $7\frac{2}{3} = \frac{23}{3}$
- $12\frac{3}{4} = \frac{51}{4}$

Cost per metre = Total cost ÷ Length
$$
= \frac{51}{4} \div \frac{23}{3} = \frac{51}{4} \times \frac{3}{23} = \frac{153}{92}
$$

So, cost per metre = Rs. $\frac{153}{92}$

Answer: (B) Rs. $\frac{153}{92}$

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Question (49):

Find $x$ such that $\frac{3}{7} = \frac{x}{42}$

Cross-multiply:
$$
3 \times 42 = 7 \times x \Rightarrow 126 = 7x \Rightarrow x = \frac{126}{7} = 18
$$

Answer: (A) 18

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Question (50):

Find the value of $\frac{7}{12} + \frac{19}{10}$

Find LCM of 12 and 10 = 60

Convert:
- $\frac{7}{12} = \frac{7 \times 5}{60} = \frac{35}{60}$
- $\frac{19}{10} = \frac{19 \times 6}{60} = \frac{114}{60}$

Add:
$$
\frac{35 + 114}{60} = \frac{149}{60}
$$

Answer: (C) $\frac{149}{60}$

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✔ Final Answers:

| Q.No | Answer |
|------|--------|
| 45 | (C) $-\frac{125}{462}$ |
| 46 | (B) $-\frac{2}{3}$ |
| 47 | (D) None of these |
| 48 | (B) Rs. $\frac{153}{92}$ |
| 49 | (A) 18 |
| 50 | (C) $\frac{149}{60}$ |

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