CBSE Class 7 Mathematics Practice Worksheet Set 03

Question 1. Identify the pattern and write the next three numbers in the series:
\( \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 1\frac{1}{4}, 1\frac{1}{2}, \text{-------}, \text{-------}, \text{-------} \)
Answer: Convert all terms to fractions with a common denominator of 4:
\( \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \frac{5}{4}, \frac{6}{4}, \dots \)
Notice that each term increases by adding \( \frac{1}{4} \) to the previous number.
Rule: Add \( \frac{1}{4} \) to find the next term.
- Next term = \( \frac{6}{4} + \frac{1}{4} = \frac{7}{4} = 1\frac{3}{4} \)
- Following term = \( \frac{7}{4} + \frac{1}{4} = \frac{8}{4} = 2 \)
- Next following term = \( \frac{8}{4} + \frac{1}{4} = \frac{9}{4} = 2\frac{1}{4} \)
Thus, the next three numbers in the series are \( 1\frac{3}{4}, 2, 2\frac{1}{4} \) (or \( \frac{7}{4}, 2, \frac{9}{4} \)).
In simple words: Each fraction goes up by one-quarter. The next three numbers are 1 3/4, 2, and 2 1/4.

Exam Tip: Rewrite fractions with a common denominator like 4 to spot the pattern quickly.

Question 2. Solve the following expression:
\( \frac{2}{3} \div \left[ 1\frac{2}{5} - 2\frac{1}{15} \right] \)
Answer: First, convert the mixed fractions inside the bracket into improper fractions:
\( 1\frac{2}{5} = \frac{7}{5} \)
\( 2\frac{1}{15} = \frac{31}{15} \)
Now, find a common denominator of 15 to subtract inside the bracket:
\( \frac{7}{5} = \frac{7 \times 3}{5 \times 3} = \frac{21}{15} \)
\( 1\frac{2}{5} - 2\frac{1}{15} = \frac{21}{15} - \frac{31}{15} = \frac{21 - 31}{15} = -\frac{10}{15} = -\frac{2}{3} \).
Now perform the division:
\( \frac{2}{3} \div \left(-\frac{2}{3}\right) = \frac{2}{3} \times \left(-\frac{3}{2}\right) = -1 \).
Hence, the value of the expression is \( -1 \).
In simple words: Solve the inside of the brackets first to get -2/3. Dividing 2/3 by -2/3 gives -1.

Exam Tip: Remember BODMAS: always solve operations inside the brackets before carrying out division.

Question 3. When a fraction is divided by the difference of 1/2 and 1/6 , we get 2/3 . What Is the fraction?
Answer: Let the required fraction be \( x \).
First, find the difference between \( \frac{1}{2} \) and \( \frac{1}{6} \):
\( \text{Difference} = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \).
According to the given condition:
\( x \div \frac{1}{3} = \frac{2}{3} \)
\( x \times 3 = \frac{2}{3} \)

\( \implies x = \frac{2}{3} \div 3 = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} \).
Therefore, the required fraction is \( \frac{2}{9} \).
In simple words: The difference between 1/2 and 1/6 is 1/3. Dividing 2/9 by 1/3 gives 2/3, so the answer is 2/9.

Exam Tip: To find the original number in a division problem, multiply the quotient by the divisor.

Question 4. Find the mean , median, mode and range of the following data:
(a) 3,4,3,5,3,6,3,8,4
(b) 6,14,5,13,11,7,8,8
Answer:
(a) For the data: 3, 4, 3, 5, 3, 6, 3, 8, 4
- Number of observations (\( n \)) = 9.
- Sum of observations = \( 3 + 4 + 3 + 5 + 3 + 6 + 3 + 8 + 4 = 39 \).
- Mean = \( \frac{\text{Sum}}{n} = \frac{39}{9} = \frac{13}{3} \approx 4.33 \) (or \( 4\frac{1}{3} \)).
- Sorted data in ascending order: 3, 3, 3, 3, 4, 4, 5, 6, 8.
- Median = \( \left(\frac{9 + 1}{2}\right)\text{th term} = 5\text{th term} = 4 \).
- Mode = 3 (since 3 appears most frequently, 4 times).
- Range = Highest value - Lowest value = \( 8 - 3 = 5 \).

(b) For the data: 6, 14, 5, 13, 11, 7, 8, 8
- Number of observations (\( n \)) = 8.
- Sum of observations = \( 6 + 14 + 5 + 13 + 11 + 7 + 8 + 8 = 72 \).
- Mean = \( \frac{\text{Sum}}{n} = \frac{72}{8} = 9 \).
- Sorted data in ascending order: 5, 6, 7, 8, 8, 11, 13, 14.
- Median = Average of 4th and 5th terms = \( \frac{8 + 8}{2} = 8 \).
- Mode = 8 (since 8 appears most frequently, 2 times).
- Range = Highest value - Lowest value = \( 14 - 5 = 9 \).
In simple words: For (a), the mean is 4.33, median is 4, mode is 3, and range is 5. For (b), the mean is 9, median is 8, mode is 8, and range is 9.

Exam Tip: Always arrange the numbers in ascending order first to find both the median and the range accurately.

Question 5. Find the value of the following by using the law of exponents:
\[ \frac{81 \times 7^3 \times 100}{10^2 \times 3^4 \times 7} \]
Answer: Express composite numbers in base form with exponents:
\( 81 = 3^4 \)
\( 100 = 10^2 \)
Substitute these into the expression:
\[ = \frac{3^4 \times 7^3 \times 10^2}{10^2 \times 3^4 \times 7^1} \]
Apply the quotient rule of exponents, \( \frac{a^m}{a^n} = a^{m-n} \):
\[ = 3^{4-4} \times 7^{3-1} \times 10^{2-2} \]
\[ = 3^0 \times 7^2 \times 10^0 \]
Since any non-zero number raised to power 0 equals 1 (\( a^0 = 1 \)):
\[ = 1 \times 49 \times 1 = 49 \].
Therefore, the value is 49.
In simple words: Rewrite 81 as 3⁴ and 100 as 10². Subtract the powers of matching bases to get 7², which equals 49.

Exam Tip: Remember that any base with an exponent of zero equals 1, so \( 3^0 = 1 \) and \( 10^0 = 1 \).

Question 6. If 81÷ 3^x = 9, find the value of x.
Answer: Write the given equation in fraction form:
\( \frac{81}{3^x} = 9 \)
Rearrange the terms:
\( 3^x = \frac{81}{9} \)
\( 3^x = 9 \)
Express 9 as a power of 3:
\( 3^x = 3^2 \)
Since the bases are identical on both sides, equate the exponents:
\( x = 2 \).
Hence, the value of \( x \) is 2.
In simple words: 81 divided by 9 is 9. Since 3 to the power 2 is 9, x must be 2.

Exam Tip: Convert both sides of the equation to the same base so you can compare the powers directly.