Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers More Ques

I. Multiple choice question

Question 1. The decimal form of \( -\frac{3}{4} \) is .........
(a) – 0.75
(b) – 0.50
(c) – 0.25
(d) – 0.125
Answer: (a) – 0.75
In simple words: To find the decimal form of a fraction, divide the numerator by the denominator. Here, 3 divided by 4 is 0.75, and since the fraction is negative, the answer is -0.75.

🎯 Exam Tip: Remember that dividing 3 by 4 gives 0.75, and the negative sign simply applies to the result.

Question 2. If a number has a non-terminating and non-recurring decimal expansion, then it is.........
(a) a rational number
(b) a natural number
(c) an irrational number
(d) an integer
Answer: (c) an irrational number
In simple words: An irrational number is a number that cannot be written as a simple fraction and whose decimal form goes on forever without repeating any pattern. It just keeps going with new digits all the time.

🎯 Exam Tip: Rational numbers either terminate (like 0.5) or repeat (like 0.333...). Irrational numbers do neither.

Question 3. Which one of the following has terminating decimal expansion?
(a) \( \frac{7}{9} \)
(b) \( \frac{8}{15} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{5}{32} \)
Answer: (d) \( \frac{5}{32} \)
In simple words: A fraction has a decimal that stops (terminates) if the prime factors of its denominator are only 2s and 5s. For 32, the prime factors are just 2s (`\( 2 \times 2 \times 2 \times 2 \times 2 = 32 \)`), so its decimal will terminate.

🎯 Exam Tip: To check if a fraction has a terminating decimal, simplify it first, then look at the prime factors of the denominator. If they are only 2s and 5s, it terminates.

Question 4. Which of the following are irrational numbers?
(i) \( \sqrt{2+\sqrt{3}} \)
(ii) \( \sqrt{4+\sqrt{25}} \)
(iii) \( \sqrt[3]{5+\sqrt{7}} \)
(iv) \( \sqrt{4+\sqrt{7}} \)
Answer: (d) (i), (iii) and (iv)
In simple words: Irrational numbers cannot be written as simple fractions and have decimals that never end or repeat. We look for square roots of numbers that are not perfect squares or sums/differences involving such numbers. Option (ii) is rational because `\( \sqrt{25} = 5 \)`, so `\( \sqrt{4+5} = \sqrt{9} = 3 \)`, which is a whole number.

🎯 Exam Tip: Always simplify expressions fully before determining if they are rational or irrational, especially those involving square roots.

Question 5. Irrational number has a
(a) terminating decimal
(b) no decimal part
(c) non-terminating and recurring decimal
(d) non-terminating and non-recurring decimal
Answer: (d) non-terminating and non-recurring decimal
In simple words: An irrational number, when written as a decimal, will go on forever without any repeating pattern of digits. It never stops and never repeats.

🎯 Exam Tip: This is the key definition of an irrational number. Make sure to distinguish it from rational numbers, which either terminate or have repeating decimal parts.

Question 6. If \( \frac{1}{7} = 0.142857 \), then the value of \( \frac{3}{7} \) is........
(a) 0.285741
(b) 0.428571
(c) 0.285714
(d) 0.574128
Answer: (b) 0.428571
In simple words: Since we know what `\( \frac{1}{7} \)` is as a decimal, to find `\( \frac{3}{7} \)`, we simply multiply the decimal value of `\( \frac{1}{7} \)` by 3. This is like having three parts of one-seventh.

🎯 Exam Tip: When you know the decimal value of a unit fraction (like `\( \frac{1}{n} \)`), you can find the decimal value of its multiples (like `\( \frac{k}{n} \)` ) by multiplying the unit fraction's decimal by `k`.

Question 7. Which of the following are not rational numbers?
(a) \( 7\sqrt{5} \)
(b) \( \sqrt{16}/2 \)
(c) \( \sqrt{36} - 9 \)
(d) \( \pi + 2 \)
Answer: (a) \( 7\sqrt{5} \) and (d) \( \pi + 2 \)
In simple words: Rational numbers can be written as fractions, but irrational numbers cannot. Options (a) and (d) include irrational numbers like `\( \sqrt{5} \)` and `\( \pi \)`, making the whole expression irrational. Options (b) and (c) simplify to rational numbers: `\( \sqrt{16}/2 = 4/2 = 2 \)` and `\( \sqrt{36} - 9 = 6 - 9 = -3 \)` .

🎯 Exam Tip: A sum or product of a rational number and an irrational number (where the rational number is not zero) is always irrational. Simplify all terms to their most basic form to identify them correctly.

Question 8. The product of \( 2\sqrt{5} \) and \( 6\sqrt{5} \) is..........
(a) \( 12\sqrt{5} \)
(b) 60
(c) 40
(d) \( 8\sqrt{5} \)
Answer: (b) 60
In simple words: To multiply numbers with square roots, multiply the numbers outside the root together and the numbers inside the root together. Here, `\( (2 \times 6) \times (\sqrt{5} \times \sqrt{5}) = 12 \times 5 = 60 \)` . The product of a square root with itself is the number inside the root.

🎯 Exam Tip: Remember that `\( \sqrt{a} \times \sqrt{a} = a \)` for any positive number `a`. This property simplifies calculations greatly.

Question 9. The rational number lying between \( \frac{1}{5} \) and \( \frac{1}{2} \)
(a) \( \frac{7}{20} \)
(b) \( \frac{2}{10} \)
(c) \( \frac{2}{7} \)
(d) \( \frac{3}{10} \)
Answer: (a) \( \frac{7}{20} \)
In simple words: To find a rational number between two fractions, you can average them. Adding `\( \frac{1}{5} \)` and `\( \frac{1}{2} \)` gives `\( \frac{7}{10} \)`, and then dividing by 2 (multiplying by `\( \frac{1}{2} \)` ) gives `\( \frac{7}{20} \)` . You can also convert them to decimals or common denominators to compare.

🎯 Exam Tip: To find a number between two given numbers, you can often add them and divide by two. Converting to decimals (e.g., `\( 0.2 \)` and `\( 0.5 \)` ) can also help visually.

Question 10. The value of \( 0.\overline{03} + 0.\overline{03} \) is ..........
(a) \( 0.\overline{09} \)
(b) \( 0.\overline{0303} \)
(c) \( 0.\overline{06} \)
(d) \( 0.\overline{006} \)
Answer: (c) \( 0.\overline{06} \)
In simple words: When you add two repeating decimals that have the same repeating block and start at the same position, you can simply add the numbers in the repeating block. Here, `\( 0.\overline{03} \)` is `\( \frac{3}{99} \)`, so `\( \frac{3}{99} + \frac{3}{99} = \frac{6}{99} \)`, which is `\( 0.\overline{06} \)` .

🎯 Exam Tip: Convert repeating decimals to fractions (e.g., `\( 0.\overline{xy} = \frac{xy}{99} \)` ) to perform arithmetic operations accurately before converting back to decimal form if needed.

Question 11. The sum of \( \sqrt{343} + \sqrt{567} \) is
(a) \( 18\sqrt{3} \)
(b) \( 16\sqrt{7} \)
(c) \( 15\sqrt{3} \)
(d) \( 14\sqrt{7} \)
Answer: (b) \( 16\sqrt{7} \)
In simple words: To add square roots, first simplify each square root to its simplest form by taking out any perfect square factors. After simplifying, if the numbers under the square root are the same, you can add their coefficients. Both `\( \sqrt{343} \)` and `\( \sqrt{567} \)` simplify to terms with `\( \sqrt{7} \)` .

🎯 Exam Tip: Always simplify square roots as much as possible by factoring out perfect squares (e.g., `\( \sqrt{343} = \sqrt{49 \times 7} = 7\sqrt{7} \)` and `\( \sqrt{567} = \sqrt{81 \times 7} = 9\sqrt{7} \)` ) before attempting to add or subtract them.

Question 12. If \( \sqrt{363} = x\sqrt{3} \) then x = ..........
(a) 8
(b) 9
(c) 10
(d) 11
Answer: (d) 11
In simple words: To find `x`, we need to simplify `\( \sqrt{363} \)` by finding any perfect square factors. Since `\( 363 = 121 \times 3 \)` and `\( 121 \)` is `\( 11^2 \)`, we can write `\( \sqrt{363} \)` as `\( \sqrt{121 \times 3} \)` , which simplifies to `\( 11\sqrt{3} \)` . So, `x` is 11.

🎯 Exam Tip: Look for perfect square factors within the number under the square root. Dividing by small prime numbers like 3 can help identify these factors quickly.

Question 13. The rationalising factor of \( \frac{1}{\sqrt{7}} \) is ...........
(a) 7
(b) \( \sqrt{7} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{1}{\sqrt{7}} \)
Answer: (b) \( \sqrt{7} \)
In simple words: A rationalizing factor is what you multiply a number by to remove any irrational parts, usually square roots, from the denominator. For `\( \frac{1}{\sqrt{7}} \)`, multiplying by `\( \sqrt{7} \)` will turn the denominator into a whole number, 7. This is done to make calculations easier and to have a rational denominator.

🎯 Exam Tip: To rationalize a fraction with `\( \sqrt{a} \)` in the denominator, multiply both the numerator and the denominator by `\( \sqrt{a} \)` .

Question 14. The value of \( \left(\frac{1}{3^5}\right)^4 \) is ........
(a) \( 3^{20} \)
(b) \( 3^{-20} \)
(c) \( \frac{1}{3^{-20}} \)
(d) \( \frac{1}{3^{9}} \)
Answer: (b) \( 3^{-20} \)
In simple words: When you have a power raised to another power, you multiply the exponents. Also, `\( \frac{1}{a^m} \)` can be written as `\( a^{-m} \)` . So, `\( \left(\frac{1}{3^5}\right)^4 = (3^{-5})^4 = 3^{-5 \times 4} = 3^{-20} \)` .

🎯 Exam Tip: Remember the exponent rules: `\( (a^m)^n = a^{mn} \)` and `\( \frac{1}{a^m} = a^{-m} \)` . These rules are fundamental for simplifying expressions with powers.

Question 15. What is \( 3.976 \times 10^{-4} \) written in decimal form?
(a) 0.003976
(b) 0.0003976
(c) 39760
(d) 0.03976
Answer: (b) 0.0003976
In simple words: When you multiply a number by `\( 10^{-4} \)` , it means you move the decimal point 4 places to the left. Starting from 3.976, moving the decimal 4 places to the left gives 0.0003976. This makes the number smaller, which is what negative exponents do.

🎯 Exam Tip: For negative powers of 10, move the decimal point to the left. The exponent tells you how many places to move. For positive powers, move it to the right.

II. Answer the following Questions.

Question 1. Find any seven rational numbers between \( \frac{5}{8} \) and \( -\frac{5}{6} \)
Answer: First, we convert the given rational numbers to have the same denominator.
The L.C.M. of 8 and 6 is 24.
So, \( \frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24} \)
And \( -\frac{5}{6} = \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24} \)
Now we need to find seven rational numbers between \( -\frac{20}{24} \) and \( \frac{15}{24} \) .
Some of the rational numbers between \( -\frac{20}{24} \) and \( \frac{15}{24} \) are:
\( -\frac{19}{24}, -\frac{18}{24}, -\frac{17}{24}, ..., 0, \frac{1}{24}, \frac{2}{24}, ..., \frac{14}{24} \).
We can pick any seven from this range. For example:
\( \frac{1}{24}, \frac{2}{24}, \frac{3}{24}, \frac{4}{24}, \frac{5}{24}, \frac{6}{24}, \frac{7}{24} \)
In simple words: To find numbers between two fractions, change them to have the same bottom number (denominator). Then, pick any numbers between their top numbers (numerators). There are many numbers between `\( -\frac{20}{24} \)` and `\( \frac{15}{24} \)` like `\( \frac{1}{24} \)` or `\( -\frac{5}{24} \)` .

🎯 Exam Tip: Always find the L.C.M. of the denominators to convert fractions to equivalent ones with a common denominator. This makes it easier to compare and find numbers in between.

Question 2. Find any three rational numbers between \( \frac{1}{2} \) and \( \frac{1}{5} \)
Answer: We can find rational numbers between two given numbers by averaging them.
1. Rational number between \( \frac{1}{2} \) and \( \frac{1}{5} \):
\( = \frac{1}{2} \left(\frac{1}{2} + \frac{1}{5}\right) \)
\( = \frac{1}{2} \left(\frac{5+2}{10}\right) \)
\( = \frac{1}{2} \times \frac{7}{10} = \frac{7}{20} \)
2. Rational number between \( \frac{1}{2} \) and \( \frac{7}{20} \):
\( = \frac{1}{2} \left(\frac{1}{2} + \frac{7}{20}\right) \)
\( = \frac{1}{2} \left(\frac{10+7}{20}\right) \)
\( = \frac{1}{2} \times \frac{17}{20} = \frac{17}{40} \)
3. Rational number between \( \frac{1}{2} \) and \( \frac{17}{40} \):
\( = \frac{1}{2} \left(\frac{1}{2} + \frac{17}{40}\right) \)
\( = \frac{1}{2} \left(\frac{20+17}{40}\right) \)
\( = \frac{1}{2} \times \frac{37}{40} = \frac{37}{80} \)
Thus, three rational numbers are \( \frac{7}{20}, \frac{17}{40} \) and \( \frac{37}{80} \). These numbers lie in increasing order, demonstrating the density property of rational numbers.
In simple words: One way to find a number in the middle of two others is to add them up and divide by two. We can do this three times: first find a number between `\( \frac{1}{2} \)` and `\( \frac{1}{5} \)`, then between `\( \frac{1}{2} \)` and that new number, and so on.

🎯 Exam Tip: The "mean method" (averaging two numbers) is a reliable way to find a rational number between any two given rational numbers. You can apply it repeatedly to find as many as you need.

Question 3. Represent \( -\frac{2}{11}, -\frac{5}{11} \) and \( -\frac{9}{11} \) on the number lines.
Answer:

To represent \( -\frac{2}{11}, -\frac{5}{11} \) and \( -\frac{9}{11} \) on the number line, we first divide the section between 0 and -1 into 11 equal parts. Each part will represent a distance of \( \frac{1}{11} \) to the left of 0.
The point A represents \( -\frac{2}{11} \) (2 divisions to the left of 0).
The point B represents \( -\frac{5}{11} \) (5 divisions to the left of 0).
The point C represents \( -\frac{9}{11} \) (9 divisions to the left of 0).
In simple words: Draw a line and mark 0 and -1. Divide the space between 0 and -1 into 11 equal small parts. Then, count 2 parts to the left for `\( -\frac{2}{11} \)` (label it A), 5 parts for `\( -\frac{5}{11} \)` (label it B), and 9 parts for `\( -\frac{9}{11} \)` (label it C).

🎯 Exam Tip: For fractions with a common denominator, always divide the relevant segment (e.g., 0 to 1, or 0 to -1) into that many equal parts. Then, count the numerator's value from 0 in the correct direction.

Question 4. Express the following in the form \( \frac{p}{q} \), where p and q are integers and q \( \neq \) 0.
(i) \( 0.\overline{47} \)
Answer:
Let \( x = 0.474747... \)(1)
Since two digits are repeating, multiply by 100:
\( 100x = 47.4747... \)(2)
Subtract (1) from (2):
(2) – (1) \( \implies \) \( 100x - x = 47.4747... - 0.4747... \)
\( 99x = 47 \)
\( x = \frac{47}{99} \)
Therefore, \( 0.\overline{47} = \frac{47}{99} \).
In simple words: To change a repeating decimal like `\( 0.\overline{47} \)` into a fraction, we can set it equal to `x`. Then multiply `x` by a power of 10 that moves the repeating part to the left of the decimal. Subtract the original `x`, and solve for `x` as a fraction.

🎯 Exam Tip: When setting up equations for repeating decimals, make sure the repeating part aligns after the decimal point so it cancels out during subtraction.

(ii) \( 0.5\overline{7} \)
Answer:
Let \( x = 0.57777... \)(1)
Multiply by 10 to move the non-repeating part to the left of the decimal:
\( 10x = 5.77777... \)(2)
Multiply by 100 to move one repeating block to the left of the decimal:
\( 100x = 57.7777... \)(3)
Subtract (2) from (3):
(3) – (2) \( \implies \) \( 100x - 10x = 57.7777... - 5.7777... \)
\( 90x = 52 \)
\( x = \frac{52}{90} \)
Simplify the fraction by dividing both numerator and denominator by 2:
\( x = \frac{26}{45} \)
Therefore, \( 0.5\overline{7} = \frac{26}{45} \).
In simple words: For a decimal with a non-repeating part followed by a repeating part, we first move the non-repeating part past the decimal point, then move one full repeating block past the decimal. Subtracting these two equations helps remove the repeating part, leaving a simple equation to solve for `x`.

🎯 Exam Tip: When a decimal has both non-repeating and repeating parts, you need two multiplications by powers of 10: one to place the non-repeating part to the left of the decimal, and another to place the first full repeating block to the left.

(iii) \( 0.2\overline{45} \)
Answer:
Let \( x = 0.2454545... \)(1)
Multiply by 10 to move the non-repeating digit to the left of the decimal:
\( 10x = 2.454545... \)(2)
Multiply by 1000 to move one repeating block after the non-repeating digit to the left of the decimal:
\( 1000x = 245.4545... \)(3)
Subtract (2) from (3):
(3) – (2) \( \implies \) \( 1000x - 10x = 245.4545... - 2.4545... \)
\( 990x = 243 \)
\( x = \frac{243}{990} \)
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor (which is 9):
\( x = \frac{27}{110} \)
Therefore, \( 0.2\overline{45} = \frac{27}{110} \).
In simple words: This is similar to the previous problem. We first use `\( 10x \)` to shift the non-repeating digit (2) before the decimal. Then, we use `\( 1000x \)` to shift the next full repeating block (`45`) before the decimal. Subtracting these gives us the fraction.

🎯 Exam Tip: Always make sure to simplify the resulting fraction to its lowest terms. This shows a complete understanding of the conversion process.

Question 5. Without actual division classify the decimal expansion of the following numbers as terminating or non-terminating and recurring.
(i) \( \frac{7}{16} \)
Answer:
(i) \( \frac{7}{16} \)
First, find the prime factors of the denominator:
\( 16 = 2 \times 2 \times 2 \times 2 = 2^4 \)
We can write this in the form \( \frac{P}{2^m \times 5^n} \):
\( \frac{7}{16} = \frac{7}{2^4} = \frac{7}{2^4 \times 5^0} \)
Since the denominator has only 2 as a prime factor (or 5 with exponent 0), \( \frac{7}{16} \) has a terminating decimal expansion. This is because we can easily multiply by a power of 5 to get a denominator of 10, 100, etc.
In simple words: To see if a fraction's decimal stops (terminates) without actually dividing, look at the prime numbers that make up its bottom number. If these prime numbers are only 2s or only 5s (or both), then the decimal will stop. For 16, it's just 2s, so the decimal ends.

🎯 Exam Tip: A rational number `\( \frac{p}{q} \)` (in simplest form) will have a terminating decimal expansion if and only if the prime factorization of its denominator `q` contains only powers of 2 and/or 5.

(ii) \( \frac{13}{150} \)
Answer:
(ii) \( \frac{13}{150} \)
First, find the prime factors of the denominator:
\( 150 = 2 \times 75 = 2 \times 3 \times 25 = 2 \times 3 \times 5^2 \)
The denominator contains a prime factor of 3, which is not 2 or 5. Therefore, it is not in the form of \( \frac{P}{2^m \times 5^n} \).
So, \( \frac{13}{150} \) has a non-terminating and recurring decimal expansion.
In simple words: For `\( \frac{13}{150} \)`, the bottom number 150 has prime factors 2, 3, and 5. Because there is a '3' in the factors, the decimal will not stop. Instead, it will go on forever and repeat a pattern.

🎯 Exam Tip: If the denominator of a simplified fraction contains any prime factor other than 2 or 5, its decimal expansion will be non-terminating and repeating.

(iii) \( -\frac{11}{75} \)
Answer:
(iii) \( -\frac{11}{75} \)
First, find the prime factors of the denominator:
\( 75 = 3 \times 25 = 3 \times 5^2 \)
The denominator contains a prime factor of 3, which is not 2 or 5. Therefore, it is not in the form of \( \frac{P}{2^m \times 5^n} \).
So, \( -\frac{11}{75} \) has a non-terminating and recurring decimal expansion. The negative sign does not change this classification.
In simple words: When we break down 75 (the bottom number) into its prime factors, we get 3 and 5. Since there's a 3, the decimal for `\( -\frac{11}{75} \)` will not end but will keep repeating. The negative sign only affects the value, not whether it terminates or repeats.

🎯 Exam Tip: The sign of a rational number does not affect whether its decimal expansion is terminating or non-terminating recurring. Only the prime factors of the denominator matter.

(iv) \( \frac{17}{200} \)
Answer:
(iv) \( \frac{17}{200} \)
First, find the prime factors of the denominator:
\( 200 = 2 \times 100 = 2 \times 10^2 = 2 \times (2 \times 5)^2 = 2 \times 2^2 \times 5^2 = 2^3 \times 5^2 \)
The prime factors of the denominator are only 2 and 5. Therefore, it is in the form of \( \frac{P}{2^m \times 5^n} \).
So, \( \frac{17}{200} \) has a terminating decimal expansion.
In simple words: For `\( \frac{17}{200} \)`, the bottom number 200 can be made by only multiplying 2s and 5s together. Because of this, when you divide 17 by 200, the decimal will eventually stop.

🎯 Exam Tip: Even if there are many 2s and 5s in the denominator's prime factorization, as long as there are no other prime numbers, the decimal will terminate.

Question 6. Find the value of \( \sqrt{27} + \sqrt{75} - \sqrt{108} + \sqrt{48} \)
Answer:
We simplify each square root term:
\( \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \)
\( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \)
\( \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \)
\( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \)
Now, substitute these simplified terms back into the expression:
\( \sqrt{27} + \sqrt{75} - \sqrt{108} + \sqrt{48} \)
\( = 3\sqrt{3} + 5\sqrt{3} - 6\sqrt{3} + 4\sqrt{3} \)
Combine the coefficients of `\( \sqrt{3} \)`:
\( = (3 + 5 - 6 + 4)\sqrt{3} \)
\( = (8 - 6 + 4)\sqrt{3} \)
\( = (2 + 4)\sqrt{3} \)
\( = 6\sqrt{3} \)
To get a numerical value, we use `\( \sqrt{3} \approx 1.732 \)`:
\( = 6 \times 1.732 \)
\( = 10.392 \)
In simple words: To add or subtract square roots, we first need to simplify each root by finding any perfect square numbers hidden inside. For example, `\( \sqrt{27} \)` becomes `\( 3\sqrt{3} \)` . Once all roots are simplified to the same basic square root (like `\( \sqrt{3} \)` in this case), we can add and subtract the numbers in front of them.

🎯 Exam Tip: Always look for the largest perfect square factor for each number under the square root. This makes the simplification efficient and reduces chances of error. Remember that you can only combine square roots that have the same number under the root sign.

Question 7. Evaluate \( \frac{\sqrt{2}+1}{\sqrt{2}-1} \)
Answer:
To evaluate this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is `\( \sqrt{2}+1 \)`.
\( \frac{\sqrt{2}+1}{\sqrt{2}-1} = \frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} \)
Apply the difference of squares formula `\( (a-b)(a+b) = a^2 - b^2 \)` in the denominator and `\( (a+b)^2 = a^2 + 2ab + b^2 \)` in the numerator:
\( = \frac{(\sqrt{2}+1)^2}{(\sqrt{2})^2 - (1)^2} \)
\( = \frac{(\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2}{2 - 1} \)
\( = \frac{2 + 2\sqrt{2} + 1}{1} \)
\( = 3 + 2\sqrt{2} \)
To get a numerical value, we use `\( \sqrt{2} \approx 1.414 \)`:
\( = 3 + 2 \times 1.414 \)
\( = 3 + 2.828 \)
\( = 5.828 \)
In simple words: When you have a square root in the bottom part of a fraction, we can get rid of it by multiplying the top and bottom by a special number called the "conjugate". For `\( \sqrt{2}-1 \)`, its conjugate is `\( \sqrt{2}+1 \)` . This process is called rationalization, and it helps to simplify the expression and calculate its value.

🎯 Exam Tip: Always remember to multiply both the numerator and the denominator by the conjugate to keep the value of the expression unchanged. The conjugate of `\( a-\sqrt{b} \)` is `\( a+\sqrt{b} \)` and vice-versa.

Question 8. \( \frac{(0.003)^7 \times (0.0002)^5}{(0.001)^3} \) in scientific notation.
Answer:
First, convert each number into scientific notation:
\( 0.003 = 3 \times 10^{-3} \)
\( 0.0002 = 2 \times 10^{-4} \)
\( 0.001 = 1 \times 10^{-3} \)
Now substitute these into the expression:
\( \frac{(3 \times 10^{-3})^7 \times (2 \times 10^{-4})^5}{(1 \times 10^{-3})^3} \)
Apply the power rules `\( (ab)^n = a^n b^n \)` and `\( (a^m)^n = a^{mn} \)`:
\( = \frac{3^7 \times (10^{-3})^7 \times 2^5 \times (10^{-4})^5}{1^3 \times (10^{-3})^3} \)
\( = \frac{3^7 \times 10^{-21} \times 2^5 \times 10^{-20}}{1 \times 10^{-9}} \)
Calculate `\( 3^7 \)` and `\( 2^5 \)`:
\( 3^7 = 2187 \)
\( 2^5 = 32 \)
Substitute these values:
\( = \frac{2187 \times 10^{-21} \times 32 \times 10^{-20}}{10^{-9}} \)
Combine the numerical parts and the powers of 10:
\( = (2187 \times 32) \times \frac{10^{-21} \times 10^{-20}}{10^{-9}} \)
\( = 69984 \times 10^{(-21) + (-20) - (-9)} \)
\( = 69984 \times 10^{-41 + 9} \)
\( = 69984 \times 10^{-32} \)
Finally, express 69984 in scientific notation and combine the powers of 10:
\( 69984 = 6.9984 \times 10^4 \)
\( = 6.9984 \times 10^4 \times 10^{-32} \)
\( = 6.9984 \times 10^{4-32} \)
\( = 6.9984 \times 10^{-28} \)
In simple words: To solve this, first change all numbers to scientific notation (like `\( 0.003 \)` becomes `\( 3 \times 10^{-3} \)` ). Then, use rules for exponents: when raising a power to a power, multiply; when multiplying powers, add; when dividing powers, subtract. Finally, make sure your answer is in standard scientific notation with one non-zero digit before the decimal point.

🎯 Exam Tip: Always convert all numbers to scientific notation before performing operations. Pay close attention to the rules for adding, subtracting, and multiplying exponents, especially with negative exponents.

Question 9. Write
(a) \( 9.87 \times 10^9 \)
(b) \( 4.134 \times 10^{-4} \)
(c) \( 1.432 \times 10^{-9} \)
in decimal form.
Answer:
(a) For \( 9.87 \times 10^9 \): Since the exponent is positive 9, move the decimal point 9 places to the right.
\( 9.87 \times 10^9 = 9,870,000,000 \)
(b) For \( 4.134 \times 10^{-4} \): Since the exponent is negative 4, move the decimal point 4 places to the left.
\( 4.134 \times 10^{-4} = 0.0004134 \)
(c) For \( 1.432 \times 10^{-9} \): Since the exponent is negative 9, move the decimal point 9 places to the left.
\( 1.432 \times 10^{-9} = 0.000000001432 \)
In simple words: When you see a number like `\( 10^9 \)` , it means move the decimal point 9 steps to the right, adding zeros as needed. When you see `\( 10^{-4} \)` , it means move the decimal point 4 steps to the left, adding zeros after the decimal point. Positive exponents make the number bigger, negative exponents make it smaller.

🎯 Exam Tip: The sign of the exponent on 10 tells you the direction to move the decimal point: positive means right (making the number larger), and negative means left (making the number smaller). The number in the exponent tells you how many places to move.