Full Solution - Test No 1
Test No. 1 - Detailed Solution
Q1: Find two rational numbers between 2 and 3.
Method: Mid-Point Formula
To find a rational number exactly in the middle of two numbers \(a\) and \(b\), we use the formula: \( \frac{a+b}{2} \)
Step 1: Find the first number.
Add 2 and 3, then divide by 2:
\( \text{First Number} = \frac{2 + 3}{2} = \frac{5}{2} = \mathbf{2.5} \)
Step 2: Find the second number.
Now, find the number between 2 and our new number (2.5):
\( \text{Second Number} = \frac{2 + 2.5}{2} = \frac{4.5}{2} = \mathbf{2.25} \)
(Alternatively, in fraction form: \( \frac{2 + \frac{5}{2}}{2} = \frac{9}{4} \))
Answer: The two rational numbers are 2.5 and 2.25.
Q2: Write the following in scientific notation.
Rule: Scientific Notation is \( a \times 10^n \), where \( 1 \le a < 10 \).
• Move decimal Left = Positive Power (+)
• Move decimal Right = Negative Power (-)
(i) \( 2000000 \)
The decimal point is currently at the end: \( 2000000.0 \)
We need to move it to get a number between 1 and 10 (which is 2).
Move decimal 6 places to the Left.
\( 2.0 \times 10^{+6} \)
Answer: \( 2 \times 10^6 \)
(ii) \( 0.0000009 \)
We need to move the decimal point so that we have a non-zero digit before the decimal (which is 9).
Move decimal 7 places to the Right.
Since we moved Right, the power is negative.
\( 9.0 \times 10^{-7} \)
Answer: \( 9 \times 10^{-7} \)
(iii) \( 73 \times 10^{-3} \)
This is not in standard form because 73 is greater than 10.
Step 1: Convert 73 into scientific notation first.
\( 73 = 7.3 \times 10^1 \)
Step 2: Substitute this back into the original expression.
\( (7.3 \times 10^1) \times 10^{-3} \)
Step 3: Combine the powers of 10 (Add exponents: \( 1 + (-3) \)).
\( 7.3 \times 10^{1 - 3} \)
\( 7.3 \times 10^{-2} \)
Answer: \( 7.3 \times 10^{-2} \)
Q3: Write two proper subsets of the following sets.
Definition: A Proper Subset of a set A is any subset of A that is not equal to A itself.
(i) \( \{a, b, c\} \)
First, let's list some possible subsets:
Possible Subsets: \( \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \phi \)
Now, select any two from the list above:
Selected Answer: \( \{a\}, \{a, b\} \)
(ii) \( \mathbb{Z} \) (Set of Integers)
The set of integers contains infinite elements: \( \{ \dots, -2, -1, 0, 1, 2, \dots \} \).
Let's list a few finite subsets:
Possible Subsets: \( \{0\}, \{1, 2\}, \{-5, 5\}, \{0, 1, 2, 3\} \)
Select any two:
Selected Answer: \( \{0, 1\}, \{1, 2, 3\} \)
(iii) \( \mathbb{N} \) (Set of Natural Numbers)
Natural numbers are \( \{1, 2, 3, \dots \} \).
Let's list a few finite subsets:
Possible Subsets: \( \{1\}, \{2, 4\}, \{1, 3, 5\}, \{100\} \)
Select any two:
Selected Answer: \( \{1, 2\}, \{5, 10\} \)
Q4: Factorize \( x^2 + 9x + 14 \).
Expression: \( x^2 + 9x + 14 \)
Logic: We need to find two numbers that:
1. Multiply to give 14 (Last term).
2. Add to give 9 (Middle term coefficient).
The factors of 14 are:
• \( 1 \times 14 \) (Sum = 15) - Incorrect
• \( 2 \times 7 \) (Sum = 9) - Correct
Step 1: Split the middle term \( 9x \) into \( 2x \) and \( 7x \).
\( = x^2 + 2x + 7x + 14 \)
Step 2: Group terms and take out common factors.
Take \( x \) common from the first two terms and \( 7 \) from the last two.
\( = x(x + 2) + 7(x + 2) \)
Step 3: Take \( (x + 2) \) as the common factor.
\( = (x + 2)(x + 7) \)
Final Answer: \( (x + 2)(x + 7) \)